1.1. $v=l / 2 \tau=3.0 \mathrm{~km}$ per hour.APPENDICES
1. Basic Trigonometrical formulas
1.2. $\langle v\rangle=2 v_{0}\left(v_{1}+v_{2}\right) /\left(2 v_{0}+v_{1}+v_{2}\right)$.
1.3. $\Delta t=\tau \sqrt{1-4\langle v\rangle / w \tau}=15 \mathrm{~s}$.
1.4. (a) $10 \mathrm{~cm} / \mathrm{s}$; (b) $25 \mathrm{~cm} / \mathrm{s}$; (c) $t_{0}=16 \mathrm{~s}$.
1.5. $\left(r_{1}-r_{2}\right) /\left|r_{1}-r_{2}\right|=\left(v_{2}-v_{1}\right) /\left|v_{2}-v_{1}\right|$.
1.6. $v^{\prime}=\sqrt{v_{0}^{2}+v^{2}+2 v_{0} v \cos \varphi} \approx 40 \mathrm{~km}$ per hour, $\varphi^{\prime}=19^{\circ}$.
1.7. $u=\frac{v_{0}}{\left(1-v_{0}^{2} / v^{\prime 2}\right)^{-1 / 2}-1}=3.0 \mathrm{~km}$ per hour.
1.8. $\tau_{A} / \tau_{B}=\eta / V \overline{\eta^{2}-1}=1.8$.
1.9. $\theta=\arcsin (1 / n)+\pi / 2=120^{\circ}$.
1.10. $l=v_{0} t V \overline{2(1-\sin \theta)}=22 \mathrm{~m}$.
1.11. $l=\left(v_{1}+v_{2}\right) \sqrt{v_{1} v_{2}} / g=2.5 \mathrm{~m}$.
1.12. $t=2 a / 3 v$.
1.13. It is seen from Fig. $1 a$ that the points $A$ and $B$ converge with velocity $v-u \cos \alpha$, where the angle $\alpha$ varies with time. The
Fig. 1.
points merge provided the following two conditions are met:
\[
\int_{0}^{\tau}(v-u \cos \alpha) d t=l, \quad \int_{0}^{\tau} v \cos \alpha d t=u \tau,
\]
where $\tau$ is the sought time. It follows from these two equations that.
\[
\tau=v l /\left(v^{2}-u^{2}\right) \text {. }
\]
1.14. $x_{1}-x_{2}=l-w \tau(t+\tau / 2)=0.24 \mathrm{~km}$. Toward the train with velocity $V=4.0 \mathrm{~m} / \mathrm{s}$.
1.15. (a) $0.7 \mathrm{~s}$; (b) 0.7 and $1.3 \mathrm{~m}$ respectively.
1.16. $t_{m}=\frac{v_{1} l_{1}+v_{2} l_{2}}{v_{1}^{2}+v_{2}^{2}}, \quad l_{\min }=\frac{\left|l_{1} v_{2}-l_{2} v_{1}\right|}{\sqrt{v_{1}^{2}+v_{2}^{2}}}$.
1.17. $C D=l / \sqrt{\eta^{2}-1}$.
1.18. See Fig. $1 b$.
1.19. (a) $\langle v\rangle=\pi R / \tau=50 \mathrm{~cm} / \mathrm{s}$;
(c) $|\langle\mathbf{w}\rangle|=2 \pi R / \tau^{2}=10 \mathrm{~cm} / \mathrm{s}^{2}$.
1.20. (a) $\mathbf{v}=\mathrm{a}(1-2 \alpha t), \quad \mathbf{w}=-2 \alpha \mathrm{a}=\mathrm{const}$; (b) $\Delta t=1 / \alpha$, $s=a / 2 \alpha$.
1.21. (a) $x=v_{0} t(1-t / 2 \tau), x=0.24,0$ and $-4.0 \mathrm{~m}$;
(b) $1.1,9$ and $11 \mathrm{~s}$; 24 and $34 \mathrm{~cm}$ respectively.
(c) $s=\left\{\begin{array}{l}(1-t / 2 \tau) v_{0} t \text { for } t \leqslant \tau, \\ {\left[1+(1-t / \tau)^{2}\right] v_{0} t / 2 \text { for } t \geqslant \tau .}\end{array}\right.$
1.22. (a) $v=\alpha^{2} t / 2, w=\alpha^{2} / 2$;
(b) $\langle v\rangle=\alpha \sqrt{s} / 2$.
1.23. (a) $s=(2 / 3 a) v_{0}^{3 / 2}$;
(b) $t=2 V \overline{v_{0}} / a$.
1.24. (a) $y=-x^{2} b / a^{2} ; \quad$ (b) $\mathbf{v}=a \mathbf{i}-2 b t \mathbf{j}, \quad \mathbf{w}=-2 b \mathbf{j}, \quad v=$ $=\sqrt{a^{2}+4 b^{2} t^{2}}, w=2 b$; (c) $\tan \alpha=a / 2 b t$; (d) $\langle\mathbf{v}\rangle=a \mathbf{i}-b t \mathbf{j},|\langle\mathbf{v}\rangle|=$ $=\sqrt{a^{2}+b^{2} t^{2}}$.
1.25. (a) $y=x-x^{2} \alpha / a$;
(b) $v=a \sqrt{1+(1-2 \alpha t)^{2}}, w=2 \alpha a=$ $=$ const; (c) $t_{0}=1 / \alpha$.
1.26. (a) $s=a \omega \tau$; (b) $\pi / 2$.
1.27. $v_{0}=V \overline{\left(1+a^{2}\right) w / 2 b}$.
1.28. (a) $\mathbf{r}=\mathbf{v}_{0} t+\mathbf{g} t^{2} / 2$; (b) $\langle\mathbf{v}\rangle_{t}=$ $=\mathbf{v}_{0}+\mathrm{g} t / 2, \quad\langle\mathbf{v}\rangle=\mathbf{v}_{0}-\mathbf{g}\left(\mathbf{v}_{0} \mathrm{~g}\right) / g^{2}$.
1.29. (a) $\tau=2\left(v_{0} / g\right) \sin \alpha$;
(b) $h=\left(v_{0}^{2} / 2 g\right) \sin ^{2} \alpha, l=\left(v_{0}^{2} / g\right) \sin 2 \alpha$, $\alpha=76^{\circ}$;
(c) $y=x \tan \alpha-\left(g / 2 v_{0}^{2} \cos ^{2} \alpha\right) x^{2}$;
(d) $R_{1}=v_{0}^{2} / g \cos \alpha, R_{2}=\left(v_{0}^{2} / g\right) \cos ^{2} \alpha$.
1.30. See Fig. 2.
1.31. $l=8 h \sin \alpha$.
Fig. 2.
1.32. 0.41 or $0.71 \mathrm{~min}$ later, depending on the initial angle.
1.33. $\Delta t=\frac{2 v_{0}}{g} \frac{\sin \left(\theta_{1}-\theta_{2}\right)}{\cos \theta_{1}+\cos \theta_{2}}=11 \mathrm{~s}$.
1.34. (a) $x=\left(a / 2 v_{0}\right) y^{2}$;
(b) $\quad w=a v_{0}, \quad w_{\tau}=a^{2} y / \sqrt{1+\left(a y / v_{0}\right)^{2}}$, $w_{n}=a v_{0} / \sqrt{1+\left(a y / v_{0}\right)^{2}}$.
1.35. (a) $y=(b / 2 a) x^{2}$;
(b) $\quad R=v^{2} / w_{n}=v^{2} / \sqrt{w^{2}-w_{\tau}^{2}}=$ $=(a / b)\left[1+(x b / a)^{2}\right]^{3 / 2}$.
1.36. $v=V \overline{2 a x}$.
1.37. $w=a \sqrt{1+(4 \pi n)^{2}}=0.8 \mathrm{~m} / \mathrm{s}^{2}$.
1.38. (a) $v=v_{0} /\left(1+v_{0} t / R\right)=v_{0} \mathrm{e}^{-8 / R}$;
(b) $w=V \overline{2} v_{0}^{2} / R \mathrm{e}^{2 s / R}=$ $=V \overline{2} v^{2} / R$.
1.39. $\tan \alpha=2 s / R$.
1.40. (a) $w_{0}=a^{2} \omega^{2} / R=2.6 \mathrm{~m} / \mathrm{s}^{2}, w_{a}=a \omega^{2}=3.2 \mathrm{~m} / \mathrm{s}^{2}$; (b) $w_{\text {min }}=$ $=a \omega^{2} V \overline{1-(R / 2 a)^{2}}=2.5 \mathrm{~m} / \mathrm{s}^{2}, l_{m}= \pm a \sqrt{1-R^{2} / 2 a^{2}}= \pm 0.37 \mathrm{~m}$.
282
1.41. $R=a^{3} / 2 b s, w=a \sqrt{1+\left(4 b s^{\left.2 / a^{3}\right)^{2}}\right.}$.
1.42. (a) $w=2 a v^{2}, R=1 / 2 a$; (b) $w=b v^{2} / a^{2}, R=a^{2} / b$.
1.43. $v=2 R \omega=0.40 \mathrm{~m} / \mathrm{s}, w=4 R \omega^{2}=0.32 \mathrm{~m} / \mathrm{s}^{2}$.
1.44. $w=(v / t) \sqrt{1+4 a^{2} t^{4}}=0.7 \mathrm{~m} / \mathrm{s}^{2}$.
1.45. $\omega=2 \pi n v / l=2.0 \cdot 10^{3} \mathrm{rad} / \mathrm{s}$.
1.46. (a) $\langle\omega\rangle=2 a / 3=4 \mathrm{rad} / \mathrm{s},\langle\beta\rangle=\sqrt{3 a b}=6 \mathrm{rad} / \mathrm{s}^{2}$;
(b) $\beta=$ $=2 \sqrt{3 a b}=12 \mathrm{rad} / \mathrm{s}^{2}$.
1.47. $t=\sqrt[3]{(4 / a) \tan \alpha}=7 \mathrm{~s}$.
1.48. $\langle\omega\rangle=\omega_{0} / 3$.
1.49. (a) $\varphi=\left(1-\mathrm{e}^{-a t}\right) \omega_{0} / a$; (b) $\omega=\omega_{0} \mathrm{e}^{-a t}$.
1.50. $\omega_{z}= \pm \sqrt{2 \beta_{0} \sin \varphi}$, see Fig. 3 .
1.51. (a) $y=v^{2} / \beta x$ (hyperbola); (b) $y=\sqrt{2 w x} / \omega$ (parabola).
1.52. (a) $w_{A}=v^{2} / R=2.0 \mathrm{~m} / \mathrm{s}^{2}$, the vector $\mathbf{w}_{A}$ is permanently directed to the centre of the wheel; (b) $s=8 R=4.0 \mathrm{~m}$.
1.53. (a) $v_{A}=2 w t=10.0 \mathrm{~cm} / \mathrm{s}, v_{B}=$ $=\sqrt{2} w t=7.1 \mathrm{~cm} / \mathrm{s}, \quad v_{o}=0 ;$ (b) $w_{A}=$ $=2 w \sqrt{1+\left(w t^{2} / 2 R\right)^{2}}=5.6 \mathrm{~cm} / \mathrm{s}^{2}, \quad w_{B}=$ $=w \sqrt{1+\left(1-w t^{2} / R\right)^{2}}=2.5 \mathrm{~cm} / \mathrm{s}^{2}, \quad w_{O}=$ $=w^{2} t^{2} / R=2.5 \mathrm{~cm} / \mathrm{s}^{2}$.
1.54. $R_{A}=4 r, R_{B}=2 \sqrt{2} r$.
Fig. 3.
1.55. $\omega=\sqrt{\omega_{1}^{2}+\omega_{2}^{2}}=5 \mathrm{rad} / \mathrm{s}, \quad \beta=\omega_{1} \omega_{2}=12 \mathrm{rad} / \mathrm{s}^{2}$.
1.56. (a) $\omega=a t \sqrt{1+(b t / a)^{2}}=8 \mathrm{rad} / \mathrm{s}, \quad \beta=a \sqrt{1+(2 b t / a)^{2}}=$ $=1.3 \mathrm{rad} / \mathrm{s}^{2} ;$ (b) $17^{\circ}$.
1.57. (a) $\omega=v / R \cos \alpha=2.3 \mathrm{rad} / \mathrm{s}, 60^{\circ}$; $=2.3 \mathrm{rad} / \mathrm{s}^{2}$.
1.58. $\omega=\omega_{0} \sqrt{1+\left(\beta_{0} t / \omega_{0}\right)^{2}}=0.6 \mathrm{rad} / \mathrm{s}$,
(b) $\beta=(v / R)^{2} \tan \alpha=$ $=0.2 \mathrm{rad} / \mathrm{s}^{2}$.
1.59. $\Delta m=2 m w /(g+w)$.
$\beta=\beta_{0} \sqrt{1+\omega_{0}^{2} t^{2}}=$
1.60. $\mathbf{w}=\frac{m_{0}-k\left(m_{1}+m_{2}\right)}{m_{0}+m_{1}+m_{2}} \mathbf{g}, \quad T=\frac{(1+k) m_{0}}{m_{0}+m_{1}+m_{2}} m_{2} g$.
1.61. (a) $F=\frac{\left(k_{1}-k_{2}\right) m_{1} m_{2} g \cos \alpha}{m_{1}+m_{2}}$;
(b) $\tan \alpha_{\min }=\frac{k_{1} m_{1}+k_{2} m_{2}}{m_{1}+m_{2}}$.
1.62. $k=\left[\left(\eta^{2}-1\right) /\left(\eta^{2}+1\right)\right] \tan \alpha=0.16$.
1.63. (a) $m_{2} / m_{1}>\sin \alpha+k \cos \alpha$; (b) $m_{2} / m_{1}<\sin \alpha-k \cos \alpha$;
(c) $\sin \alpha-k \cos \alpha<m_{2} / m_{1}<\sin \alpha+k \cos \alpha$.
1.64. $\mathbf{w}_{2}=\mathbf{g}(\eta-\sin \alpha-k \cos \alpha) /(\eta+1)=0.05 \mathrm{~g}$.
1.65. When $t \leqslant t_{0}$, the accelerations $w_{1}=w_{2}=a t /\left(m_{1}+m_{2}\right)$; when $t \geqslant t_{0} \quad w_{1}=k g m_{2} / m_{1}, \quad w_{2}=\left(a t-k m_{2} g\right) / m_{2}$. Here $t_{0}=$ $=k g m_{2}\left(m_{1}+m_{2}\right) / a m$. See Fig. 4 .
1.66. $\tan 2 \alpha=-1 / k, \alpha=49^{\circ} ; t_{\min }=1.0 \mathrm{~s}$.
1.67. $\tan \beta=k ; \quad T_{\text {min }}=m g(\sin \alpha+k \cos \alpha) / V \overline{1+k^{2}}$.
1.68. (a) $v=\frac{m g^{2} \cos \alpha}{2 a \sin ^{2} \alpha}$;
(b) $s=\frac{m^{2} g^{3} \cos \alpha}{6 a^{2} \sin ^{3} \alpha}$.
1.69. $v=\sqrt{(2 g / 3 a) \sin \alpha}$.
1.70. $\tau=\sqrt{2 l /(3 w+k g)}$.
1.71. (a) $\mathbf{w}_{1}=\frac{\left(m_{1}-m_{2}\right) \mathrm{g}+2 m_{2} \mathrm{w}_{0}}{m_{1}+m_{2}}, \quad \mathbf{w}_{1}^{\prime}=\frac{m_{1}-m_{2}}{m_{1}+m_{2}}\left(\mathrm{~g}-\mathbf{w}_{0}\right)$ :
(b) $\mathbf{F}=\frac{4 m_{1} m_{2}}{m_{1}+m_{2}}\left(\mathrm{~g}-\mathbf{w}_{0}\right)$.
1.72. $w=2 \mathrm{~g}(2 \eta-\sin \alpha) /(4 \eta+1)$.
1.73. $\mathrm{w}_{1}=\frac{4 m_{1} m_{2}+m_{0}\left(m_{1}-m_{2}\right)}{4 m_{1} m_{2}+m_{0}\left(m_{1}+m_{2}\right)} \mathbf{g}$.
1.74. $F_{j r}=2 l m M /(M-m) t^{2}$.
1.75. $t=\sqrt{2 l(4+\eta) / 3 g(2-\eta)}=1.4 \mathrm{~s}$.
1.76. $H=6 h \eta /(\eta+4)=0.6 \mathrm{~m}$.
Fig. 4.
Fig. 5.
1.77. $w_{A}=g /\left(1+\eta \cot ^{2} \alpha\right), w_{B}=g /(\tan \alpha+\eta \cot \alpha)$.
1.78. $w=g V \overline{2} /(2+k+M / m)$.
1.79. $w_{\min }=g(1-k) /(1+k)$.
1.80. $w_{\max }=g(1+k \cot \alpha) /(\cot \alpha-k)$.
1.81. $w=g \sin \alpha \cos \alpha /\left(\sin ^{2} \alpha+m_{1} / m_{2}\right)$.
1.82. $w=\frac{m g \sin \alpha}{M+2 m(1-\cos \alpha)}$.
1.83. (a) $|\langle\mathbf{F}\rangle|=2 V \overline{2} m v^{2} / \pi R$;
(b) $|\langle\mathbf{F}\rangle|=m w_{\tau}$.
1.84. $2.1,0.7$ and $1.5 \mathrm{kN}$.
1.85. (a) $w=g \sqrt{1+3 \cos ^{2} \theta}, \quad T=3 m g \cos \theta$;
(b) $T=m g V \overline{3}$;
(c) $\cos \theta=1 / \sqrt{3}, \quad \theta=54.7^{\circ}$.
1.86. $\approx 53^{\circ}$.
1.87. $\theta=\arccos (2 / 3) \approx 48^{\circ}, \quad v=V \overline{2 g R / 3}$.
1.88. $\varepsilon=1 /\left(x / m \omega^{2}-1\right)$. Is independent of the rotation direction.
1.89. $r=R / 2, \quad v_{\max }=1 / 2 \sqrt{k g R}$.
1.90. $s=1 / 2 R V \overline{\left(k g / w_{\tau}\right)^{2}-1}=60 \mathrm{~m}$.
1.91. $v \leqslant \alpha \sqrt{\mathrm{kg} / a}$.
1.92. $T=\left(\cot \theta+\omega^{2} R / g\right) m g / 2 \pi$.
1.93. (a) Let us examine a small element of the thread in contact with the pulley (Fig. 5). Since the element is weightless, $d T=$ $=d F_{f r}=k d F_{n}$ and $d F_{n}=T d \alpha$. Hence, $d T / T=k d \alpha$. Integrat-
ing this equation, we obtain $k=\left(\ln \eta_{0}\right) / \pi$; (b) $w=g\left(\eta-\eta_{0}\right) /\left(\eta+\eta_{0}\right)$. 1.94. $F=\left(m v_{0}^{2} / R\right) \cos ^{2} \alpha$.
1.95. $\mathbf{F}=-m \omega^{2} \mathbf{r}$, where $\mathbf{r}$ is the radius vector of the particle relative to the origin of coordinates; $F=m \omega^{2} V \overline{x^{2}+y^{2}}$.
1.96. (a) $\Delta \mathbf{p}=m \mathbf{g} t$; (b) $|\Delta \mathbf{p}|=$ $=-2 m\left(\mathrm{v}_{0} \mathrm{~g}\right) / \mathrm{g}$.
1.97.
(a)
$\mathbf{p}=\mathbf{a} \tau^{3} / 6$
(b) $s=$ $=a \tau^{4} / 12 \mathrm{~m}$.
1.98. $s=(\omega t-\sin \omega t) F_{0} / m \omega^{2}$.
see Fig. 6.
1.99. $t=\pi / \omega ; \quad s=2 F_{0} / m \omega^{2}$; $v_{\max }=F_{0} / m \omega$.
1.100. (a) $v=v_{0} \mathrm{e}^{-t r / m}, \quad t \rightarrow \infty$;
(b) $v=v_{0}-s r / m, \quad s_{\text {total }}=$
$=\frac{m v_{0}}{r}$;
(c) $\langle v\rangle=v_{0} \frac{\eta-1}{\eta \ln \eta}$.
1.101. $t=\frac{h\left(v_{0}-v\right)}{v_{0} v \ln \left(v_{0} / v\right)}$.
1.102. $s=\frac{2}{a} \tan \alpha, \quad v_{\max }=\sqrt{\frac{g}{a} \sin \alpha \tan \alpha}$.
Instruction. To reduce the equation to the form which is convenient to integrate, the acceleration must be represented as $d v / d t$ and then a change of variables made according to the formula $d t=d x / v$.
1.103. $s=1 / 6 a\left(t-t_{0}\right)^{3} / \mathrm{m}$, where $t_{0}=k m g / a$ is the moment of time at which the motion starts. At $t \leqslant t_{0}$ the distance is $s=0$.
1.104. $v^{\prime}=v_{0} / \sqrt{1+k v_{0}^{2} / m g}$.
1.105. (a) $v=(2 F / m \omega)|\sin (\omega t / 2)|$;
(b) $\Delta s=8 F / m \omega^{2},\langle v\rangle=$ $=4 F / \pi m \omega$.
1.106. $v=v_{0} /(1+\cos \varphi)$. Instruction. Here $w_{\tau}=-w_{x}$, and therefore $v=-v_{x}+$ const. From the initial condition it follows that const $=v_{0}$. Besides, $v_{x}=v \cos \varphi$.
1.107. $w=[1-\cos (l / R)] R g / l$.
1.108. (a) $v=\sqrt{2 g R / 3}$;
(b) $\cos \theta_{0}=\frac{2+\eta \sqrt{5+9 \eta^{2}}}{3\left(1+\eta^{2}\right)}$, where $\eta=$ $=w_{0} / g, \theta_{0} \approx 17^{\circ}$.
1.109. For $n<1$, including negative values.
1.110. When $\omega^{2} R>g$, there are two steady equilibrium positions: $\theta_{1}=0$ and $\theta_{2}=\arccos \left(g / \omega^{2} R\right)$. When $\omega^{2} R<g$, there is only one equilibrium position: $\theta_{1}=0$. As long as there is only one lower equilibrium position, it is steady. Whenever the second equilibrium position appears (which is permanently steady) the lower one becomes unsteady.
1.111. $h \approx\left(\omega s^{2} / v\right) \sin \varphi=7 \mathrm{~cm}$, where $\omega$ is the angular velocity of the Earth’s rotation.
1.112. $F=m \sqrt{g^{2}+\omega^{4} r^{2}+\left(2 v^{\prime} \omega\right)^{2}}=8 \mathrm{~N}$.
1.113. $F_{\text {cor }}=2 m \omega^{2} r \sqrt{1+\left(v_{0} / \omega r\right)^{2}}=2.8 \mathrm{~N}$.
1.114. (a) $\omega^{\prime}=\omega^{2} R:$ (b) $F_{\text {in }}=m \omega^{2} r \sqrt{(2 R / r)^{2}-1}$.
1.115. $F_{e f}=m \omega^{2} R V \overline{5 / 9}-8 \mathrm{~N}, \quad F_{i e m}=2 / \omega^{2} R V \overline{5+8 g / 3 \omega^{2} R}=$ $=17 \mathrm{~N}$.
1.116. (a) $F=2 m v \omega$ sin $\varphi=3.8 \mathrm{kN}$, on the right rail; (b) along the parallel from the east to the west with the velocity $v=$ $=\frac{1}{2} \omega R \cos \varphi \approx 420 \mathrm{~km}$ per hour. Here $\theta$ is the angular rotation velocity of the Earth about its axis, $\boldsymbol{R}$ is its radius.
1.117. Will deviate to the east by the distance $x \approx$ $\approx \frac{2}{3} \omega h \sqrt{2 \mathrm{~h} / \mathrm{g}}=24 \mathrm{~cm}$. Hcre $\omega$ is the angular velocity of the Earth’s rotation about its axis.
1.118. $A=F\left(r_{3}-r_{2}\right)=-17 \mathrm{~J}$.
1.119. $A=\operatorname{mat}^{4} / 8$.
1.120. $F=2 a s \sqrt{1+(s / F)^{2}}$.
1.121. $A=m g(h+k h)$.
1.122. $A=-k m g l /(1-k \cot \alpha)=-0.05 \mathrm{~J}$
1.123. $F_{\mathrm{min}}=\left(m_{1}+m_{1} / 2\right) \mathrm{kg}$.
1.124. $A=-(1-\eta) \eta m g l / 2=-1.3 \mathrm{~J}$.
1.125. $(P)=0, P=m g\left(g t-v_{0} \sin \alpha\right)$.
1.126. $P=m$ Rat, $(P)=m$ Rat $/ 2$.
1.127. (a) $(P)=-k m g v_{g} / 2=-2 \mathrm{~W}$; (b) $P_{\max }=-1 / g m t, V \overline{\alpha_{g}}$.
1.128. $A=1 / 2 m \omega^{2}\left(r_{t}^{2}-r\right)^{2}=$ $=0.20 \mathrm{~J}$.
1.129. $A$ mos $-1 / k(\Delta l)^{2}$, where $k=k_{1} k_{2} /\left(k_{1}+k_{2}\right)$.
1.130. $A=3 m g / 4 a, \Delta U=$ $=m g / 2 a$.
1.131. (a) $r_{0}=2 a / b$, steady:
(b) $F_{\text {max }}=b^{0} / 27 a^{4}$, see Fig. ?:
1.132. (a) No; (b) ellipses whose ratio of semiaxes is $a / b=$ $=V \overline{\beta / \alpha}$; also ellipses, but with $a / b=\beta / a$.
1.133. The latter field is potential.
1.134. $t=v_{2}^{2} / 2 g(\sin a+k \cos \alpha), A=-m v_{0}^{\prime} k / 2(k+\tan a)$.
1.135. $\boldsymbol{h}=\mathrm{H} / 2 ; s_{\operatorname{mex}}=\boldsymbol{H}$.
1.136. $v=\pi / \sqrt{\mathrm{s}^{h / 3}}$.
1.137. $v_{\text {mis }}=V{ }_{5 g l} ; \quad T=3 m_{g}$.
1.138. $t=1,2 v_{0} R$.
1.139. $\Delta l=(1+V \overline{1+2 k l / m g}) m g / k$.
1.140. $v=V \overline{19 g} \sqrt{32}=1.7 \mathrm{~m} / \mathrm{s}$.
1.141. $A=\frac{k \sin 6}{2} \frac{1-\operatorname{ces} \theta}{(\sin 6+k \operatorname{ces} \theta \operatorname{ces} b}=0.09 \mathrm{~J}$.
1.142. $A-x / \eta(1+\eta) / 2(1-\eta)^{2}$, where $\eta=m \omega^{2} / \mathrm{x}$.
1.143. $w_{c}=g\left(m_{1}-m_{n}\right)^{2} /\left(m_{1}+m_{2}\right)^{2}$.
1.145. $r=\left(g^{2}\right) \tan \theta=0.8 \mathrm{~cm}, T=m g / \cos \theta=5 \mathrm{~N}$.
1.146. (a) $F_{t r}-m g\left|\sin a+\left(\omega^{2} / / g\right) \cos \alpha\right|=6 \mathrm{~N}$.
(b) $\bullet<$ $<V \frac{\mathrm{ran}}{g(k-\tan \alpha) / l(1+k \tan \alpha)}=2 \mathrm{rad} / \mathrm{s}$.
1.147. (a) $\mathbf{v}=\left(m_{1} \mathrm{v}_{1}+m_{2} \mathrm{v}_{2}\right) /\left(m_{1}+m_{2}\right)$; (b) $T=\mu\left(\mathrm{v}_{1}-\right.$ $\left.-v_{2}\right)^{1 / 2}$, where $\mu=m_{1} m_{2}\left(m_{1}+m_{2}\right)$.
1.148. $E=\vec{E}+m V^{2} / 2$.
1.149. $\tilde{E}=\mu\left(v_{1}^{l}+v_{0}\right) / 2$, where $\mu=m_{1} m_{v} /\left(m_{1}+m_{1}\right)$.
1.150. $\mathbf{p}=\mathrm{p}_{2}+m \mathrm{~g} t$, where $\mathrm{p}_{0}=m \mathrm{v}_{1}+m_{2} \mathrm{v}_{2}, m=m_{1}+m_{2}$ $\mathbf{r}_{c}=\mathbf{v}_{4} t+g t^{4} / 2$, where $\mathbf{v}_{6}=\left(m_{1} \mathbf{v}_{1}+m_{2} \mathbf{v}_{2}\right) /\left(m_{1}+m_{2}\right)$.
1.151. $v_{c}=x V \overline{x m_{2}}\left(m_{1}+m_{2}\right)$.
1.152. (a) $l_{\text {mex }}=l_{0}+F / x_{i}, \quad l_{\text {min }}=l_{4}$
(b) $I_{\text {max }}=l_{6}+$
$+2 m_{1} F / x\left(m_{1}+m_{2}\right), i_{\text {min }}=i_{5}$
1.153. (a) $\Delta l>3 m g / x$; (b) $h=(1+x \Delta l / m g)^{2} m g / 8 x=8 m g / x$. 1.154. $\mathrm{v}_{1}=-m \mathrm{v} /(M-m), \mathrm{v}_{2}=M \mathrm{v} /(M-m)$.
1.155. $\mathrm{v}_{\text {reer }}=\mathrm{v}_{0}-\frac{\mathrm{m}}{\boldsymbol{M}+\mathrm{m}} \mathrm{w} ; \mathrm{v}_{\text {ferm }}=\mathrm{v}_{0}+\frac{\mathrm{m}}{(\boldsymbol{M}+\mathrm{m})} \mathrm{u}$.
1.156. (1) $v_{1}=-\frac{2 m}{M+2 m}$;
(2) $\mathbf{v}_{2}=-\frac{m(2 M+3 m)}{(M+m)(M+2 m)} u$, $v_{2} / v_{1}-1+m / 2(M+m)>1$.
1.158. $\Delta p=m \sqrt{2 g h}(\eta+1) /(\eta-1)=0.2 \mathrm{~kg} \cdot \mathrm{m} / \mathrm{s}$.
1.159. (a) $\mathrm{I}=-\frac{\mathrm{n}}{N+\mathrm{m}} \mathrm{r}^{\prime}$;
(b) $\mathbf{F}=-\frac{\mathrm{m}}{M+\mathrm{M}} \frac{d \mathrm{v}}{d i}$.
1.160. $\mathrm{I}=\mathrm{ml}^{\prime} / 2 \mathrm{M}$.
1.161. $\tau=(p \cos \alpha-M \sqrt{2 g l \sin \alpha}) / M_{g} \sin \alpha$.
1.162. (a) $
u=(2 M / m) V \frac{g}{i} \sin (\theta / 2)$;
(b) $\eta \approx 1-m / M$.
1.163. $h=M v^{2} / 2 g(M+m)$.
1.164. (1) $A=-\mu$ ch, where $\mu=m \cdot M /(m+M)$; (2) Yes.
1.166. $\mathrm{v}=1.01+2.0 \mathrm{j}-4.0 \mathrm{k}, v \approx 4.6 \mathrm{~m} / \mathrm{s}$.
1.167. $\Delta T=-\mu\left(v_{1}-v_{4}\right)^{2 / 2}$, where $\mu=m_{1} m_{2} /\left(m_{1}+m_{2}\right)$.
1.168. (a) $\eta-2 m_{1} /\left(m_{1}+m_{2}\right)$; (b) $\eta=4 m_{2} m_{2} /\left(m_{1}+m_{2}\right)$ ?
1.169. (a) $m_{2} / m_{1}=1 / 3$; (b) $m_{1} / m_{1}=1+2 \cos \theta=2.0$.
1.170. $\eta=1 / 2 \cos ^{2} a=0.25$.
1.171. $v_{\text {mex }}=v(1+\sqrt{2(\eta-1)})=1.0 \mathrm{~km}$ per second.
1.172. Will continve moving in the same direction, although this time with the velocity $v^{\prime}=(1-\sqrt{1-2 \eta}) v / 2$. For $\eta<1$ the velocity $v^{\prime} \approx \pi / 2=5 \mathrm{~cm} / \mathrm{s}$.
1.173. $\Delta T / T=(1+m / M) \tan ^{2} \theta+m / M-1=-40 \%$.
1.174. (a) $p=\mu \sqrt{v_{1}^{3}+v_{1}^{2} ;}$ (b) $T=1 / 2 \mu\left(v_{1}^{4}+v_{0}^{0}\right)$. Here $\mu=$ $=m_{1} m_{2} /\left(m_{1}+m_{2}\right)$.
1.175. $\sin \theta_{\operatorname{man}}=m_{3} / m_{1}$.
1.176. $v^{\prime}=-\mathrm{v}\left(2-\eta^{7}\right) /\left(6-\eta^{7}\right)$. Respectively at smaller $n$ equal, or greater than $y \stackrel{\rightharpoonup}{2}$.
1.i78. Suppose that at a certain moment $t$ the rocket has the mass $m$ and the velocity $v$ relative to the reference frame employed. Consider the inertial relerence frame moving with the same velocity as the rocket has at a given moment. In this reference frame the momentum increment that the system \”rocket-ejected portion of gas\” acquires during the time $d t$ is equal to $d \mathbf{p}=m d \mathbf{v}+\mu d t \cdot \mathbf{u}=\mathbf{F} d t$. What follows is evident.
1.179. $\mathrm{v}=-\mathbf{u} \ln \left(m_{0} / m\right)$.
1.180. $m=m_{0} \mathrm{e}^{-w t / u}$.
1.181. $\alpha=\left(u / v_{0}\right) \ln \left(m_{0} / m\right)$.
1.182. $\mathbf{v}=\frac{F}{\mu} \ln \frac{m_{0}}{m_{0}-\mu t}, \quad \mathbf{w}=\frac{\mathbf{F}}{m_{0}-\mu t}$.
1.183. $\mathbf{v}=\mathbf{F} t / m_{0}\left(1+\mu t / m_{0}\right), \quad \mathbf{w}=\mathbf{F} / m_{0}\left(1+\mu t / m_{0}\right)^{2}$.
1.184. $v=\sqrt{2 g h \ln (l / h)}$.
1.185. $\mathrm{N}=2 \mathrm{~b} \sqrt{\mathrm{a} / b}$.
1.186. $M=1 / 2 m g v_{0} t^{2} \cos \alpha ; \quad M=\left(m v_{0}^{3} / 2 g\right) \sin ^{2} \alpha \cos \alpha=$ $=37 \mathrm{~kg} \cdot \mathrm{m}^{2} / \mathrm{s}$.
1.187. (a) Relative to all points of the straight line drawn at right angles to the wall through the point $O$;
(b) $|\Delta \mathbf{M}|=2 m v l \cos \alpha$.
1.188. Relative to the centre of the circle.
$|\Delta \mathbf{M}|=2 \sqrt{\left.1-(g / \omega)^{2} l\right)^{2}} \mathrm{mgl} / \omega$.
1.189. $|\Delta \mathbf{M}|=h m V$.
1.190. $M=m \omega v_{0}^{2} t^{2}$.
1.191. $m=2 k r_{1}^{2} / v_{2}^{2}$.
1.192. $v_{0}=\sqrt{2 g l / \cos \theta}$.
1.193. $F=m \omega_{0}^{2} r_{0}^{4} / r^{3}$.
1.194. $M_{z}=$ Rmgt.
1.195. $M^{2}=R m g t \sin \alpha$. Will not change.
1.196. $\mathbf{M}^{\prime}=\mathbf{M}-\left[\mathbf{r}_{0} \mathbf{p}\right]$. In the case when $\mathbf{p}=0$, i.e. in the frame of the centre of inertia.
1.198. $\widetilde{M}=1 / 3 l m v_{0}$.
1.199. $\varepsilon_{\max } \approx m v_{0}^{2} / x l_{0}^{2}$. The problem is easier to solve in the frame of the centre of inertia.
1.200. $T=2 \pi \gamma M / v^{3}=225$ days.
1.201. (a) 5.2 times; (b) $13 \mathrm{~km} / \mathrm{s}, 2.2 \cdot 10^{-4} \mathrm{~m} / \mathrm{s}^{2}$.
1.202. $T=\pi \sqrt{(r+R)^{3} / 2 \gamma M}$. It is sufficient to consider the motion along the circle whose radius is equal to the major semi-axis of the given ellipse, i.e. $(r+R) / 2$, since in accordance with Kepler’s laws the period of revolution is the same.
1.203. Falling of the body on the Sun can be considered as the motion along a very elongated (in the limit, degenerated) ellipse whose major semi-axis is practically equal to the radius $R$ of the Earth’s orbit. Then from Kepler’s laws, $(2 \tau / T)^{2}=[(R / 2) / R]^{3}$, where $\tau$ is the falling time (the time needed to complete half a revolution along the elongated ellipse), $T$ is the period of the Earth’s revolution around the Sun. Hence, $\tau=T / 4 \sqrt{2}=65$ days.
1.204. Will not change.
1.205. $l=\sqrt[3]{\gamma M(T / 2 \pi)^{2}}$.
1.206. (a) $U=-\gamma m_{1} m_{2} / r$ $F=\gamma m M / a(a+l)$.
1.207. $M=m \sqrt{2 \gamma m_{S} r_{1} r_{2} /\left(r_{1}+r_{2}\right)}$, where $m_{S}$ is the mass of the Sun.
1.208. $E=T+U=-\gamma m m_{S} / 2 a$, where $m_{S}$ is the mass of the Sun.
1.209. $r_{m}=\frac{r_{0}}{2-\eta}\left[1 \pm \sqrt{1-(2-\eta) \eta \sin ^{2} \alpha}\right]$, where $\quad \eta=$ $=r_{0} v_{0}^{2} / \gamma m_{S}, m_{S}$ being the mass of the Sun.
1.210. $r_{\text {min }}=\left(\gamma m_{S} / v_{0}^{2}\right)\left[\sqrt{1+\left(l v_{0}^{2} / \gamma m_{S}\right)^{2}}-1\right]$, where $m_{S}$ is the mass of the Sun.
1.211. (a) First let us consider a thin spherical layer of radius $\rho$ and mass $\delta M$. The energy of interaction of the particle with an elementary belt $\delta S$ of that layer is equal to (Fig. 8)
\[
d U=-\gamma(m \delta M / 2 l) \sin \theta d \theta .
\]

According to the cosine theorem in the triangle $O A P l^{2}=\rho^{2}+$ $+r^{2}-2 \rho r \cos \theta$. Having determined the differential of this expression, we can reduce Eq. $\left(^{*}\right)$ to the form that is convenient for integration. After integrating over the whole layer we obtain $\delta U=$ $=-\gamma m \delta M / r$. And finally, integrating over all layers of the sphere, we obtain $U=-\gamma m M / r$; (b) $F_{r}=-\partial U / \partial r=-\gamma m M / r^{2}$.
Fig. 8.
Fig. 9.
1.212. First let us consider a thin spherical layer of substance (Fig. 9). Construct a cone with a small angle of taper and the vertex at the point $A$. The ratio of the areas cut out by the cone in the layer is $d S_{1}: d S_{2}=r_{1}^{2}: r_{2}^{2}$. The masses of the cut volumes are proportional to their areas. Therefore these volumes will attract the particle $A$ with forces equal in magnitude and opposite in direction. What follows is obvious.
1.213. $A=-3 / 2 \gamma m M / R$.
1.214. $\mathbf{G}=\left\{\begin{array}{ll}-\left(\gamma M / R^{3}\right) \mathbf{r} & \text { for } r \leqslant R, \\ -\left(\gamma M / r^{3}\right) \mathbf{r} & \text { for } r \geqslant R ;\end{array}\right.$
\[
\varphi=\left\{\begin{array}{ll}
-3 / 2\left(1-r^{2} / 3 R^{2}\right) \gamma M / R & \text { for } r \leqslant R, \\
-\gamma M / r & \text { for } r \geqslant R .
\end{array}\right.
\]
1.215. $\mathrm{G}=-4 / 3 \pi \gamma \rho$. The field inside the cavity . See Fig. 10 . 1.216. $p=3 / 8\left(1-r^{2} / R^{2}\right) \gamma M^{2} / \pi R^{4}$. About $1.8 \cdot 10^{6}$ atmospheres.
1.217. (a) Let us subdivide the spherical layer into small elements, each of mass $\delta \mathrm{m}$. In this case the energy of interaction of each element with all others is $\delta U=-\gamma m \delta m / R$. Summing over all
Fig. 10.
elements and taking into account that each pair of interacting elements appears twice in the result, we obtain $U=-v m^{2} / 2 R$; (b) $U=-3 \mathrm{\gamma m}^{2} / 5 R$.
1.218. $\Delta t \approx \frac{2 \pi}{\sqrt{\gamma M}} \frac{r^{3 / 2}}{3 \Delta r / 2 r+\delta}=\left\{\begin{array}{l}4.5 \text { days }(\delta=0), \\ 0.84 \text { hour }(\delta=2) .\end{array}\right.$
1.219. $w_{1}: w_{2}: w_{3}=1: 0.0034: 0.0006$.
1.220. $32 \mathrm{~km} ; 2650 \mathrm{~km}$.
1.221. $h=R /\left(2 g R / v_{0}^{2}-1\right)$.
1.222. $h=R\left(g R / v^{2}-1\right)$.
1.223. $r=\sqrt[3]{\gamma M(T / 2 \pi)^{2}}=4.2 \cdot 10^{4} \mathrm{~km}$, where $M$ and $T$ are the mass of the Earth and its period of revolution about its own axis respectively; $3.1 \mathrm{~km} / \mathrm{s}, 0.22 \mathrm{~m} / \mathrm{s}^{2}$.
1.224. $M=\left(4 \pi^{2} R^{3} / \gamma T^{2}\right)(1+T / \tau)^{2}=6 \cdot 10^{24} \mathrm{~kg}$, where $T$ is the period of revolution of the Earth about its own axis.
1.225. $v^{\prime}=\frac{2 \pi R}{T}+\sqrt{\frac{\gamma M}{R}}=7.0 \mathrm{~km} / \mathrm{s}, \quad w^{\prime}=\frac{\gamma M}{R^{2}}\left(1+\frac{2 \pi R}{T} \times\right.$ $\left.\times \sqrt{\frac{R}{\gamma M}}\right)=4.9 \mathrm{~m} / \mathrm{s}^{2}$. Here $M$ is the mass of the Earth, $T$ is its period of revolution about its own axis.
1.226. 1.27 times.
1.227. The decrease in the total energy $E$ of the satellite over the time interval $d t$ is equal to $-d E=F v d t$. Representing $E$ and $v$ as functions of the distance $r$ between the satellite and the centre of the Moon, we can reduce this equation to the form convenient for integration. Finally, we get $\tau \approx(\sqrt{\eta}-1) \mathrm{m} / \alpha \sqrt{g R}$
1.228. $v_{1}=1.67 \mathrm{~km} / \mathrm{s}, v_{2}=2.37 \mathrm{~km} / \mathrm{s}$.
1.229. $\Delta v=\sqrt{\gamma M / R}(1-\sqrt{2})=-0.70 \mathrm{~km} / \mathrm{s}$, where $M$ and $R$ are the mass and the radius of the Moon.
1.230. $\Delta v=\sqrt{g R}(\sqrt{2}-1)=3.27 \mathrm{~km} / \mathrm{s}$, where $g$ is the standard free-fall acceleration, $R$ is the radius of the Earth.
1.231. $r=n R /(1+\sqrt{\eta})=3.8 \cdot 10^{4} \mathrm{~km}$.
1.232. $A \approx \gamma m\left(M_{1} / R_{1}+M_{2} / R_{2}\right)=1.3 \cdot 10^{8} \mathrm{~kJ}$, where $M$ and $R$ are the mass and the radius of the Earth and the Moon.
1.233. $v_{3} \approx \sqrt{2 v_{1}^{2}+(V \overline{2}-1)^{2} V_{1}^{2}} \approx 17 \mathrm{~km} / \mathrm{s}$. Here $v_{1}^{2}=$ $=\gamma M_{E} / R, M_{E}$ and $R$ are the mass and the radius of the Earth; $V_{1}^{2}=\gamma M_{S} / r, M_{S}$ is the mass of the Sun, $r$ is the radius of the Earth’s orbit.
1.234. $l=2 a F_{2} / m w=1.0 \mathrm{~m}$.
1.235. $\mathbf{N}=(a B-b A) \mathbf{k}$, where $\mathbf{k}$ is the unit vector of the $\boldsymbol{z}$ axis; $l=|a B-b A| / \sqrt{A^{2}+B^{2}}$.
1.236. $l=|a A-b B| / \sqrt{A^{2}+B^{2}}$.
1.237. $F_{\text {res }}=2 F$. This force is parallel to the diagonal $A C$ and is applied at the midpoint of the side $B C$.
1.238. (a) $I=1 / 3 m l^{2}$; (b) $I=1 / 3 m\left(a^{2}+b^{2}\right)$.
1.239. (a) $I=1 / 2 \pi \rho b R^{4}=2.8 \mathrm{~g} \cdot \mathrm{m}^{2}$; (b) $I=3 / 10 m R^{2}$.
1.240. $I=1 / 4 m R^{2}$.
1.241. $I=(37 / 72) m R^{2}=0.15 \mathrm{~kg} \cdot \mathrm{m}^{2}$.
1.242. $I=2 / 3 m R^{2}$.
1.243. (a) $\omega=g t / R(1+M / 2 m)$; (b) $T=m g^{2} t^{2} / 2(1+M / 2 m)$.
1.244. $T=1 / 2 m g, w_{0}=g m r^{2} / I$.
1.245. $\omega=\sqrt{6 \bar{F} \sin \varphi / m l}$.
1.246. $\beta=\frac{\left|m_{2}-m_{1}\right| g}{\left(m_{1}+m_{2}+m / 2\right) R}, \quad \frac{T_{1}}{T_{2}}=\frac{m_{1}\left(m+4 m_{2}\right)}{m_{2}\left(m+4 m_{1}\right)}$.
1.247. $A=-\frac{\left(m_{2}-k m_{1}\right) k m_{1} g^{2} t^{2}}{m+2\left(m_{1}+m_{2}\right)}$.
1.248. $n=\left(1+k^{2}\right) \omega_{0}^{2} R / 8 \pi k(k+1) g$.
1.249. $t=3 / 4 \omega R / k g$.
1.250. $\langle\omega\rangle=1 / 3 \omega_{0}$.
1.251. $\beta=2 m g x / R l(M+2 m)$.
1.252. (a) $k \geqslant 2 / 7 \tan \alpha$; (b) $T=5 / 14 g^{2} t^{2} \sin ^{2} \alpha$.
1.253. (a) $T=1 / 6 m g=13 \mathrm{~N}, \beta=2 / 3 \quad \mathrm{~g} / R=5 \cdot 10^{2} \mathrm{rad} / \mathrm{s}^{2}$;
(b) $P=2 / 3 m g^{2} t$.
1.254. $\mathbf{w}^{\prime}=2 / 3\left(g-w_{0}\right), F=1 / 3 m\left(g-w_{0}\right)$.
1.255. $w=g \sin \alpha /\left(1+I / m r^{2}\right)=1.6 \mathrm{~m} / \mathrm{s}^{2}$.
1.256. $F_{\max }=3 \mathrm{kmg} /(2-3 k) ; w_{\max }=2 \mathrm{~kg} /(2-3 k)$.
1.257. (a) $w_{x}=\frac{F(\cos \alpha-r / R)}{m(1+\gamma)}$;
(b) $A=\frac{F^{2} t^{2}(\cos \alpha-r / R)^{2}}{2 m(1+\gamma)}$.
1.258. $T=1 / 10 \mathrm{mg}$.
1.259. $\mathrm{w}=3 \mathrm{~g}(M+3 m) /\left(M+9 m+I / R^{2}\right)$.
1.260. (a) $w=\frac{F\left(3 m_{1}+2 m_{2}\right)}{m_{1}\left(m_{1}+m_{2}\right)}$;
(b) $T=\frac{F^{2} t^{2}\left(3 m_{1}+2 m_{2}\right)}{2 m_{1}\left(m_{1}+m_{2}\right)}$.
1.261. $w_{1}=F /\left(m_{1}+2 / 7 m_{2}\right) ; w_{2}=2 / 7 w_{1}$.
1.262. (a) $t=1 / 3 \omega_{0} R / \mathrm{kg}$; (b) $A=-1 / 6 m \omega_{0}^{2} R^{2}$.
1.263. $\omega=\sqrt{10 g(R+r) / 17 r^{2}}$.
1.264. $v_{0}=\sqrt{1 / 3 g R(7 \cos \alpha-4)}=1.0 \mathrm{~m} / \mathrm{s}$.
1.265. $v_{0}=\sqrt{8 g R}$.
1.266. $T=m v^{2}$.
1.267. $T=7 / 10 m v^{2}\left(1+2 / r^{2} / R^{2}\right)$.
1.269. $N=1 / 2 \omega^{1 / 0} \sin 2 \theta$.
1.270. $\cos \theta=3 / 2 / \omega^{2} \mathrm{~L}$
1.271. $\Delta x=1 / 2, k n$.
1.272. $v^{\prime}=\theta_{0} L / \sqrt{1+3 m / M}$.
1.273. $F=\%, n / m l=9 \mathrm{~N}$.
1.274. (a) $v^{\prime}=\frac{3 m-4 M}{3+6 M} v$
1.275. (a) $v=(M / m) V$ i/al $\sin (a / 2)$;
(b) $\Delta p=M \sqrt{T / \Omega}$ sin $(a / 2) ;$ (c) $x \approx 2 / d$.
1.276. (a) $\theta=(1+2 m / M) \omega_{0}:(\mathrm{b}) A=1 / 2 m \omega_{0}^{3} R^{2}(1+2 m / M)$.
1.277. (a) $\varphi=-\frac{2 m_{1}}{2 m_{1}+m_{4}} \varphi^{\prime}$;
1.278. (a) $\bullet=\frac{I_{1} e_{2}+T_{r} e_{i}}{T_{1}+I_{1}}$;
(b) $N_{t}=-\frac{m_{1} m_{1} R}{2 m_{1}+m_{1}} \frac{d r}{d i}$.
(b) $A=-\frac{I_{1} J_{1}+m_{1}}{2\left(L_{1}+T_{2}\right.}\left(\omega_{1}-\omega_{2}\right)^{2}$.
1.279. $\left.v^{\prime}=v(4-\eta) / 4+\eta\right), \omega-12 w / I^{2}(4+\eta)$. For $\eta=4$ and $\eta \geqslant 4$.
(b) $N=$ – $\Gamma_{0} \cdot\left(I+I_{0}\right)$.
1.281. $\omega=\sqrt{2 g \pi}-6.0 \mathrm{rad} / \mathrm{s} ; F=m g l_{6} / l=25 \mathrm{~N}$.
1.282. (a) $M=1 / 1 a m \omega l^{2} \sin \theta, M, M$ sin $\theta$. (b) $|\Delta M|=$ $\mathrm{x} \sin 28$.
1.283. (a) $\omega^{\prime}=m q l / I \omega=0.7 \mathrm{rad} / \mathrm{s}$; (b) $F=m \theta^{\prime} l \sin \theta=10 \mathrm{mN}$. See Fig. 11 . 1.284. $\quad=-(g+u) U / \pi n h^{2}=3 \times$ $\times 10^{4} \mathrm{rad} / \mathrm{s}$.
1.285. $\quad \omega^{\prime}=m l \sqrt{\varepsilon^{2}+\omega^{2}} / I \omega=$ $=0.8 \mathrm{rad} / \mathrm{s}$. The vector $\omega$ forms the angle $\theta=\arctan (w / g)=6^{\circ}$ with the vertical.
1.286. $F^{\prime}=2 / 3 R^{2} \omega \omega^{\prime} / l=0.30 \mathrm{kN}$.
1.287. $F_{\text {zax }}=\pi m r^{2} q_{m}=0 / I T=0.09 \mathrm{kN}$.
1.288. $\hat{N}^{20}=2 \pi \mathrm{m} / \mathrm{h}=6 \mathrm{kN} \cdot \mathrm{m}$.
1.289. $F_{\text {edu }}=2 \pi n J v / R i=1.4 \mathrm{kN}$. The force exerted on the outside rail increases by this value while that exerted on the inside one decreases by the same value.
1.290. $P=a E \Delta T=2.2 \cdot 10^{\circ} \mathrm{atm}$, where $a$ is the thermal expansion coefficient.
1.291. (a) $P \approx \sigma_{m} \Delta r / r=20$ atm; (b) $p \approx 2 \sigma_{m} \Delta r / r=40$ atm. Here $\sigma_{m}$ is the glass strength.
1.292. $n=\sqrt{2 \sigma_{m} / \rho / a l}=0.8 \cdot 10^{a}$ rps, where $\sigma_{m}$ is the tensile strength, and $p$ is the density of copper.
1.293. $n=\sqrt{\sigma_{-} / \rho} / 2 \pi R=23 \mathrm{rps}$, where $\sigma_{m}$ is the tensile strength, and $\rho$ is the density of lead.
1.294. $x \approx 1 \sqrt[3]{m / 2 \pi d^{2} E}=2.5 \mathrm{~cm}$
1.295. $\varepsilon=1 / 2 F J B S$.
1.296. $T=1 / 2 m \omega^{2 l}\left(1-r^{2} / l^{2}\right), \Delta l=1 / 3 \omega^{2} / B / E$, where $p$ is the density of copper.
1.297. $\Delta V^{\prime}=(1-2 \mu) F L / E=1.6 \mathrm{~mm}^{2}$, where $\mu$ is Poisson’s ratio for copper.
1.298. (a) $\Delta l=1 / 2 p l^{2} / E$; (b) $\Delta V / V=(1-2 \mu) \Delta L / l$, where $p$ is the density, and $\mu$ is Poisson’s ratio for copper.
1.299. (a) $\Delta V / V=-3(1-2 \mu) p / E$; (b) $\beta=3(1-2 \mu) / E$.
1.300. $R=1 / 6 E h^{3} / p g^{n}=0.12 \mathrm{~km}$, where $p$ is the density of steel.
1.301. (a) Here $N$ is independent of $x$ and equal to $N_{6}$. Integrating twice the initial equation with regard to the boundory conditions $d y / d x(0)=0$ and $y(0)=0$, we obtain $y=\left(N_{2} / 2 E I\right) x^{4}$. This is the equation of a parabola. The bending deflection is $\lambda$ $=N_{0} I^{2} / 2 E I$, where $I=a^{4} / 12$.
(b) In this case $N(x)=F(l-x)$ and $y=(F / 2 E I)(l-x / 3) x^{2}$; $\lambda=F P / 3 E I$, where $I$ is of the same magnitude as in (a).
1.302. $\lambda=F P / 48 E I$.
1.303. (a) $\lambda=3 / 2 g^{4} / E h^{2}$;
(b) $\lambda=4 / 2 \rho l^{4} / E h^{2}$. Here $\rho$ is the density of steel.
1.304. $\lambda=0 / \mathrm{Ap} / \mathrm{F}^{2} / E h^{2}$, where $\rho$ is the deasity of steel.
1.305. (a) $\varphi=\left(L / 2 \pi v^{3} \Delta r G\right) \cdot N$; (b) $\varphi=\left(2 L / \pi r^{+} G\right) \cdot N$.
1.306. $N=\pi(d-d i) G e / 32 i=0.5 \mathrm{kN} \cdot \mathrm{m}$.
1.307. $P=1 / 2$ areGes $-17 \mathrm{~kW}$.
1.308. $N=1 / 2 \beta m\left(r^{4}-r^{2}\right) /\left(r^{2}-r^{2}\right)$.
1.309. $U=1 / 2 m E e^{1} / p_{0}=0.04 \mathrm{~kJ}$, Where $p$ is the density of steel.
1.316. (a) $U=1 / \mathrm{sr}^{2} \mathrm{P}^{3} \mathrm{~g}^{2} / E$;
(b) $U=2 /, u^{2} I E(\Delta L / L)^{2}$. Here $p$ is the density of steel.
1.311. $A \approx 1 /, \pi^{2} h \delta^{3} E / l=0.08 \mathrm{~kJ}$.
1.312. $U=1 \% 4 r^{2} G e^{2} / l=7 \mathrm{~J}$.
1.313. $u=1 / 2 G \mathrm{y}^{2} \mathrm{r}^{2} / \mathrm{r}$.
1.314. $u=1 / 2 \beta(\rho g h)^{2}=23.5 \mathrm{~kJ} / \mathrm{m}^{2}$, where $\beta$ is the compressibility.
1.315. $p_{1}>p_{2}, v_{1}<v_{2}$. The density of streamlines grows on transition from point 1 to point 2.
1.316. $Q=s_{1} s_{2} V \overline{2 q \Delta h}\left(s_{i}-S_{1}\right)$.
1.317. $Q=S \sqrt{2 g \Delta h_{0} / p}$.
1.318. $y=\sqrt{2 g\left(h_{1}+h_{3} / \rho_{2}\right)}=3 \mathrm{~m} / \mathrm{s}$, where $\rho_{1}$ and $\rho_{2}$ are the densities of water and kerosene.
1.319. $h=25 \mathrm{~cm} ; l_{m p z}=50 \mathrm{~cm}$.
1.320. $\boldsymbol{h}=1 / 2 v^{2 / g}-\hat{h}_{0}=20 \mathrm{~cm}$.
1.321. $P=P_{0}+p g h\left(1-R_{1} / r^{m}\right)$, where $R_{1}<r<R_{4}, P_{0}$ is the atmospheric pressure.
1.322. $A=1 / s V^{2} / s^{2} 4$, where $p$ is the density of water.
1.323. $\tau=\sqrt{2 h / g} S / \mathrm{s}$.
1.324. $v=$ ah $\sqrt{ } \frac{5}{21 / h-1}$.
1.326. $F=2 p g \Delta h=0.50 \mathrm{~N}$.
1.327. $F=p g b l(2 h-D)=5 \mathrm{~N}$.
1.328. $N=\rho l Q^{2} / \pi r^{2}=0.7 \mathrm{~N} \cdot \mathrm{m}$.
1.329. $F=\rho g h(S-s)^{2} / S=6 \mathrm{~N}$
1.330. (a) The paraboloid of revolution: $z=\left(\omega^{2} / 2 g\right) r^{2}$, where $z$ is the height measured from the surface of the liquid along the axis of the vessel, $r$ is the distance from the rotation axis; (b) $p=p_{0}+$ $+1 / 2 \rho \omega^{2} r^{2}$.
1.331. $P=\pi \eta \omega^{2} R^{4} / h=9 \mathrm{~W}$.
1.332. $v=v_{0} \frac{\ln \left(r / R_{2}\right)}{\ln \left(R_{1} / R_{2}\right)}$.
1.333. (a) $\omega=\omega_{2} \frac{R_{1}^{2} R_{2}^{2}}{R_{2}^{2}-R_{1}^{2}}\left(\frac{1}{R_{1}^{2}}-\frac{1}{r^{2}}\right)$;
(b) $N=4 \pi \eta \omega_{2} \frac{R_{1}^{2} R_{2}^{2}}{R_{2}^{2}-R_{1}^{2}}$
1.334. (a) $Q=1 / 2 \pi v_{0} R^{2}$;
(b) $T=1 / 6 \pi l R^{2} \rho v_{0}^{2}$;
(c) $F_{f r}=$ $=4 \pi \eta l v_{0} ;$ (d) $\Delta p=4 \eta l v_{0} / R^{2}$.
1.335. The additional head $\Delta h=5 \mathrm{~cm}$ at the left-hand end of the tube imparts kinetic energy to the liquid flowing into the tube. From the condition $\rho v^{2} / 2=\rho g \Delta h$ we get $v=\sqrt{2 g \Delta h}=1.0 \mathrm{~m} / \mathrm{s}$. 1.336. $\mathrm{e}^{\alpha \Delta x}=5$.
1.337. $v_{2}=v_{1} \frac{r_{1} \rho_{1} \eta_{2}}{r_{2} \rho_{2} \eta_{1}}=5 \mu \mathrm{m} / \mathrm{s}$.
1.338. $d=\sqrt[3]{\frac{18 \mathrm{Re} \eta^{2}}{\left(\rho-\rho_{0}\right) \rho_{0} g}}=5 \mathrm{~mm}$, where $\rho_{0}$ and $\rho$ are the densities of glycerin and lead.
1.339. $t=-\frac{\rho d^{2}}{18 \eta} \ln n=0.20 \mathrm{~s}$.
1.340. $v=c \sqrt{\eta(2-\eta)}=0.1 c$, where $c$ is the velocity of light.
1.341. (a) $P=a\left(1+\sqrt{4-3 \beta^{2}}\right)$;
(b) $P=a\left(\sqrt{1-\beta^{2}}+\sqrt{4-\beta^{2}}\right)$. Here $\beta=V / c$.
1.342. $l_{0}=l \sqrt{\left(1-\beta^{2} \sin ^{2} \theta\right) /\left(1-\beta^{2}\right)}=1.08 \mathrm{~m}$, where $\beta=v / c$.
1.343. (a) $\tan \theta^{\prime}=\frac{\tan \theta}{\sqrt{1-\beta^{2}}}$. Hence $\theta^{\prime}=59^{\circ}$;
(b) $S=$ $=S_{0} \sqrt{1-\beta^{2} \cos ^{2} \theta}=3.3 \mathrm{~m}^{2}$. Here $\beta=v / c$.
1.344. $v=c \sqrt{\left(2-\frac{\Delta t}{t}\right) \frac{\Delta t}{t}}=0.6 \cdot 10^{8} \mathrm{~m} / \mathrm{s}$.
1.345. $l_{0}=c \Delta t^{\prime} \sqrt{1-\left(\Delta t / \Delta t^{\prime}\right)^{2}}=4.5 \mathrm{~m}$.
1.346. $s=c \Delta t \sqrt{1-\left(\Delta t_{0} / \Delta t\right)^{2}}=5 \mathrm{~m}$.
1.347. (a) $\Delta t_{0}=(l / v) \sqrt{1-(v / c)^{2}}=1.4 \mu \mathrm{s}$;
(b) $l^{\prime}=l \sqrt{1-(v / c)^{2}}=0.42 \mathrm{~km}$.
1.348. $l_{0}=v \Delta t / \sqrt{1-(v / c)^{2}}=17 \mathrm{~m}$.
1.349. $l_{0}=\sqrt{\Delta x_{1} \Delta x_{2}}=6.0 \mathrm{~m}, v=c \sqrt{1-\Delta x_{1} / \Delta x_{2}}=2.2 \cdot 10^{8} \mathrm{~m} / \mathrm{s}$.
1.350. $v=\frac{2 l_{0} / \Delta t}{1+\left(l_{0} / c \Delta t\right)^{2}}$.
1.351. The forward particle decayed $\Delta t=l \beta / c\left(1-\beta^{2}\right)=20 \mu \mathrm{s}$ later, where $\beta=v / c$.
1.352. (a) $l_{0}=\frac{x_{A}-x_{B}-v\left(t_{A}-t_{B}\right)}{\sqrt{1-(v / c)^{2}}}$;
(b) $t_{A}-t_{B}=\left(1-\sqrt{1-(v / c)^{2}}\right) l_{0} / v$ or $t_{B}-t_{A}=\left(1+\sqrt{1-(v / c)^{2}}\right) l_{0} / v$. 1.353. (a) $t(B)=l_{0} / v, \quad t\left(B^{\prime}\right)=\left(l_{0} / v\right) \sqrt{1-(v / c)^{2}}$;
(b) $t(A)=$ $=\left(l_{0} / v\right) \sqrt{1-(v / c)^{2}}, t\left(A^{\prime}\right)=l_{0} / v$.
1.354. See Fig. 12 showing the positions of hands \”in terms of $K$ clocks\”.
Fig. 12.
1.355. $\dot{x}=\left(1-\sqrt{1-\beta^{2}}\right) c / \beta$, where $\beta=V / c$.
1.356. It should be shown first that if $\Delta t=t_{2}-t_{1}>0$, then $\Delta t^{\prime}=t_{2}^{\prime}-t_{1}^{\prime}>0$.
1.357. (a) $13 \mathrm{~ns}$; (b) $4.0 \mathrm{~m}$. Instruction. Employ the invariance of the interval.
1.358. $v^{\prime}=\frac{\sqrt{\left(v_{x}-V\right)^{2}+v_{y}^{2}\left(1-V^{2} / c^{2}\right)}}{1-v_{x} V / c^{2}}$.
1.359. (a) $v=v_{1}+v_{2}=1.25 c$; (b) $v=\left(v_{1}+v_{2}\right) /\left(1+v_{1} v_{2} / c^{2}\right)=$ $=0.91 c$.
1.360. $l=l_{0}\left(1-\beta^{2}\right) /\left(1+\beta^{2}\right)$, where $\beta=v / c$.
1.361. $v=\sqrt{v_{1}^{2}+v_{2}^{2}-\left(v_{1} v_{2} / c^{2}\right)}$.
1.362. $s=\Delta t_{0} \sqrt{\frac{V^{2}+\left(1-\beta^{2}\right) v^{\prime 2}}{\left(1-\beta^{2}\right)\left(1-v^{\prime 2} / c^{2}\right)}}$, where $\beta=V / c$.
1.363. $\tan \theta^{\prime}=\frac{\sqrt{1-\beta^{2}} \sin \theta}{\cos \theta-V / c}$, where $\beta=V / c$.
1.364. $\tan \theta=v^{\prime} V / c^{2} \sqrt{1-(V / c)^{2}}$.
1.365. (a) $w^{\prime}=w\left(1-\beta^{2}\right)^{3 / 2} /(1-\beta v / c)^{3}$; (b) $w^{\prime}=w\left(1-\beta^{2}\right)$. Here $\beta=V / c$.
1.366. Let us make use of the relation between the acceleration $w^{\prime}$ and the acceleration $w$ in the reference frame fixed to the Earth:
\[
w^{\prime}=\left(1-v^{2} / c^{2}\right)^{-3 / 2} \frac{d v}{d t} .
\]

This formula is given in the solution of the foregoing problem (item (a)) where it is necessary to assume $V=v$. Integrating the given equation (for $w^{\prime}=$ const), we obtain $v=w^{\prime} t / \sqrt{1+\left(w^{\prime} t / c\right)^{2}}$. The sought distance is $l=\left(\sqrt{1+\left(w^{\prime} t / c\right)^{2}}-1\right) c^{2} / w^{\prime}=0.91$ lightyear; $(c-v) / c=1 / 2\left(c / w^{\prime} t\right)^{2}=0.47 \%$.
1.367. Taking into account that $v=w^{\prime} t / \sqrt{1+\left(w^{\prime} t / c\right)^{2}}$, we get $\tau_{0}=\int_{0}^{\tau} \frac{d t}{\sqrt{1+\left(w^{\prime} t / c\right)^{2}}}=\frac{c}{w^{\prime}} \ln \left[\frac{w^{\prime} \tau}{c}+\sqrt{1+\left(\frac{w^{\prime} \tau}{c}\right)^{2}}\right]=3.5$ months.
1.368. $m / m_{0} \approx 1 / \sqrt{2(1-\beta)} \approx 70$, where $\beta=v / c$.
1.369. $v=c \sqrt{\eta(2+\eta)} /(1+\eta)=0.6 c$, where $c$ is the velocity of light. The definition of density as the ratio of the rest mass of a body to its volume is employed here.
1.370. $(c-v) / c=1-\left[1+\left(m_{0} c / p\right)^{2}\right]^{-1 / 2}=0.44 \%$.
1.371. $v=(c / \eta) \sqrt{\eta^{2}-1}=1 / 2 c \sqrt{3}$.
1.372. $A=0.42 \mathrm{~m}_{0} c^{2}$ instead of $0.14 \mathrm{~m}_{0} c^{2}$.
1.373. $v=1 / 2 \mathrm{c} V \overline{3}=2.6 \cdot 10^{8} \mathrm{~m} / \mathrm{s}$.
1.374. For $\varepsilon \ll 1$ the ratio is $T / m_{0} c^{2} \leqslant 4 / 3 \varepsilon \approx 0.013$.
1.375. $p=\sqrt{T\left(T+2 m_{0} c^{2}\right) / c}=1.09 \mathrm{GeV} / c$, where $c$ is the velocity of light.
1.376. $F=(I / e c) \sqrt{T\left(T+2 m_{0} c^{2}\right)}, P=T I / e$.
1.377. $p=2 n m v^{2} /\left(1-v^{2} / c^{2}\right)$.
1.378. $v=F c t / \sqrt{m_{0}^{2} c^{2}+F^{2} t^{2}}, \quad l=\sqrt{\left(m_{0} c^{2} / F\right)^{2}+c^{2} t^{2}}-m_{0} c^{2} / F$.
1.379. $F=m_{0} c^{2} / a$.
1.380. (a) In two cases: $\mathbf{F} \| \mathbf{v}$ and $\mathbf{F} \perp \mathbf{v}$; (b) $\mathbf{F}_{\perp}=m_{0} \mathbf{w} \sqrt{1-\beta^{2}}$, $\mathbf{F}_{\text {II }}=m_{0} \mathbf{w} /\left(1-\beta^{2}\right)^{3 / 2}$, where $\beta=v / c$.
1.382. $\varepsilon^{\prime}=\varepsilon \sqrt{(1-\beta) /(1+\beta)}$, where $\beta=V / c, V={ }^{3} / 5$.
1.383. $E^{2}-p^{2} c^{2}=m_{0}^{2} c^{4}$, where $m_{0}$ is the rest mass of the particle.
1.384. (a) $\widetilde{T}=2 m_{0} c^{2}\left(\sqrt{1+T / 2 m_{0} c^{2}}-1\right)=777 \mathrm{MeV}$, $\tilde{p}=\sqrt{\frac{1 / 2}{} m_{0} T}=940 \mathrm{MeV} / c ;$ (b) $V=c \sqrt{T /\left(T+2 m_{0} c^{2}\right)}=2.12 \cdot 10^{8} \mathrm{~m} / \mathrm{s}$.
1.385. $M_{0}=\sqrt{2 m_{0}\left(T+2 m_{0} c^{2}\right)} / c, V=c V \overline{T /\left(T+2 m_{0} c^{2}\right)}$.
1.386. $T^{\prime}=2 T\left(T+2 m_{0} c^{2}\right) / m_{0} c^{2}=1.43 \cdot 10^{3} \mathrm{GeV}$.
1.387. $E_{1 \text { max }}=\frac{m_{0}^{2}+m_{1}^{2}-\left(m_{2}+m_{3}\right)^{2}}{2 m_{0}} c^{2}$. The particle $m_{1}$ has the highest energy when the energy of the system of the remaining two particles $m_{2}$ and $m_{3}$ is the lowest, i.e. when they move as a single whole.
1.388. $v / c=\frac{1-\left(m / m_{0}\right)^{2 u / c}}{1+\left(m / m_{0}\right)^{2 u / c}}$. Use the momentum conservation law (as in solving Problem 1.178) and the relativistic formula for velocity transformation.
2.1. $m=\rho V \Delta p / p_{0}=30 \mathrm{~g}$, where $p_{0}$ is the standard atmospheric pressure.
2.2. $p=1 / 2\left(p_{1} T_{2} / T_{1}-\Delta p\right)=0.10$ atm.
2.3. $m_{1} / m_{2}=\left(1-a / M_{2}\right) /\left(a / M_{1}-1\right)=0.50$, where $a=$ $=m R T / p V$.
2.4. $\rho=\frac{p_{0}\left(m_{1}+m_{2}\right)}{R T\left(m_{1} / M_{1}+m_{2} / M_{2}\right)}=1.5 \mathrm{~g} / 1$.
2.5. (a) $p=\left(v_{1}+v_{2}+v_{3}\right) R T / V=2.0$ atm; (b) $M=$ $=\left(v_{1} M_{1}+v_{2} M_{2}+v_{3} M_{3}\right) /\left(v_{1}+v_{2}+v_{3}\right)=36.7 \mathrm{~g} / \mathrm{mol}$.
2.6. $T=T_{0} \eta^{\prime}\left(\eta^{2}-1\right) / \eta\left(\eta^{\prime 2}-1\right)=0.42 \mathrm{kK}$.
2.7. $n=\frac{\ln \eta}{\ln (1+\Delta V / V)}$.
2.8. $p=p_{0} \mathrm{e}^{-C t / V}$.
2.9. $t=(V / C) \ln \eta=1.0 \mathrm{~min}$.
2.10. $\Delta T=\left(m g+p_{0} \Delta S\right) l / R=0.9 \mathrm{~K}$.
2.11. (a) $T_{\max }=2 / 3\left(p_{0} / R\right) \sqrt{p_{0} / 3 \alpha}$;
(b) $T_{\max }=p_{0} / \mathrm{e} \beta R$.
2.12. $p_{\min }=2 R \sqrt{\alpha T_{0}}$.
2.13. $d T / d h=-M g / R=-33 \mathrm{mK} / \mathrm{m}$.
2.14. $d T / d h=-M g(n-1) / n R$.
2.15. 0.5 and $2 \mathrm{~atm}$.
2.16. (a) $h=R T / M g=8.0 \mathrm{~km}$; (b) $h \approx \eta R T / M g=0.08 \mathrm{~km}$.
2.17. $m=\left(1-\mathrm{e}^{-M g h / R T}\right) p_{0} S / g$.
2.18. $h_{C}=\int_{0}^{\infty} h \rho d h / \int_{0}^{\infty} \rho d h=R T / M g$.
2.19. (a) $p=p_{0}(1-a h)^{n}, h<1 / a$; (b) $p=p_{0} /(1+a h)^{n}$. Here $n=M g / a R T_{0}$.
2.20. $p=p_{0} \mathrm{e}^{M \omega 2 r 2 / 2 R T}$.
2.21. $p_{i d}=\rho R T / M=280 \mathrm{~atm} ; p=\rho R T /(M-\rho b)-a \rho^{2} / M^{2}=$ $=80 \mathrm{~atm}$.
2.22. (a) $T=a(V-b)(1+\eta) / R V(\eta V+b)=133 \mathrm{~K}$; (b) $p=R T /(V-b)-a / V^{2}=9.9 \mathrm{~atm}$.
2.23. $a=V^{2}\left(T_{1} p_{2}-T_{2} p_{1}\right) /\left(T_{2}-T_{1}\right)=185 \mathrm{~atm} \cdot 1^{2} / \mathrm{mol}^{2}, b=$ $=V-R\left(T_{2}-T_{1}\right) /\left(p_{2}-p_{1}\right)=0.042 \mathrm{l} / \mathrm{mol}$.
2.24. $x=V^{2}(V-b)^{2} /\left[R T V^{3}-2 a(V-b)^{2}\right]$.
2.25. $T>a / b R$.
2.26. $U=p V /(\gamma-1)=10 \mathrm{MJ}$.
2.27. $\Delta T=1 / 2 M v^{2}(\gamma-1) / R$.
2.28. $T=T_{1} T_{2}\left(p_{1} V_{1}+p_{2} V_{2}\right) /\left(p_{1} V_{1} T_{2}+p_{2} V_{2} T_{1}\right) ; \quad p=$ $=\left(p_{1} V_{1}+p_{2} V_{2}\right) /\left(V_{1}+V_{2}\right)$.
2.29. $\Delta U=-p_{0} V \Delta T / T_{0}(\gamma-1)=-0.25 \mathrm{~kJ}, Q^{\prime}=-\Delta U$.
2.30. $Q=A \gamma /(\gamma-1)=7 \mathrm{~J}$.
2.31. $A=R \Delta T=0.60 \mathrm{~kJ}, \Delta U=Q-R \Delta T=1.00 \mathrm{~kJ}$, $\gamma=Q /(Q-R \Delta T)=1.6$.
2.32. $Q=
u R T_{0}(1-1 / n)=2.5 \mathrm{~kJ}$.
2.33. $\gamma=\frac{v_{1} \gamma_{1}\left(\gamma_{2}-1\right)+v_{2} \gamma_{2}\left(\gamma_{1}-1\right)}{v_{1}\left(\gamma_{2}-1\right)+v_{2}\left(\gamma_{1}-1\right)}=1.33$.
2.34. $c_{V}=0.42 \mathrm{~J} /(\mathrm{g} \cdot \mathrm{K}), c_{p}=0.65 \mathrm{~J} /(\mathrm{g} \cdot \mathrm{K})$.
2.35. $A=R T(n-1-\ln n)$.
2.36. $A^{\prime}=p_{0} V_{0} \ln \left[(\eta+1)^{2} / 4 \eta\right]$.
2.37. $\gamma=1+(n-1) /\left(Q /
u R T_{0}-\ln n\right)=1.4$.
2.38. See Fig. 13 where $V$ is an isochore, $p$ is an isobaric line, $T$ is an isothermal line, and $S$ is an adiabatic line.
2.39. (a) $T=T_{0} \eta^{(\gamma-1) / \gamma}=0.56 \mathrm{kK}$;
(b) $A^{\prime}=R T_{0}\left(\eta^{(\gamma-1) / \gamma}\right.$
2.40. The work in the adiabatic process is $n=\left(\eta^{\gamma-1}-\right.$ $-1) /(\gamma-1) \ln \eta=1.4$ times greater.
2.41. $T=T_{0}\left[(\eta+1)^{2} / 4 \eta\right](\gamma-1) / 2$.
2.42. $v=\sqrt{2 \gamma R T /(\gamma-1) M}=3.3 \mathrm{~km} / \mathrm{s}$.
2.43. $Q=R \Delta T(2-\gamma) /(\gamma-1)$.
2.45. $C_{n}=R(n-\gamma) /(n-1)(\gamma-1) ; C_{n}<0$ for $1<n<\gamma$.
2.46. $C=R(n-\gamma) /(n-1)(\gamma-1)=-4.2 \mathrm{~J} /(\mathrm{K} \cdot \mathrm{mol})$, where $n=\ln \beta / \ln \alpha$.
2.47. (a) $Q=R(n-\gamma) \Delta T /(n-1)(\gamma-1)=0.11 \mathrm{~kJ}$; (b) $A=-R \Delta T /(n-1)=0.43 \mathrm{~kJ}$.
Fig. 13.
2.48. (a) $\Delta U=\alpha V_{0}^{2}\left(\eta^{2}-1\right) /(\gamma-1)$;
(b) $A=1 / 2 \alpha V_{0}^{2}\left(\eta^{2}-1\right)$; (c) $C=1 / 2 R(\gamma+1) /(\gamma-1)$.
2.49. (a) $C=-R /(\gamma-1)$;
(b) $T V^{(\gamma-1) / 2}=$ const;
(c) $A=$ $=2 R T_{0}\left(1-\eta^{(1-\gamma) / 2}\right) /(\gamma-1)$.
2.50. (a) $A=(1-\alpha) R \Delta T$
(b) $C=R /(\gamma-1)+R(1-\alpha)$; $C<0$ for $\alpha>\gamma /(\gamma-1)$.
2.51. (a) $A=\Delta U(\gamma-1) / \alpha ; Q=\Delta U[1+(\gamma-1) / \alpha]$; (b) $C=$ $=R /(\gamma-1)+R / \alpha$.
2.52. (a) $C=C_{V}+R / \alpha V$; (b) $C=C_{V}+R /(1+\alpha V)$.
2.53. (a) $C=\gamma R /(\gamma-1)+\alpha R / p_{0} V$;
(b) $\Delta U=p_{0}\left(V_{2}-\right.$ $\left.-V_{1}\right) /(\gamma-1) ; A=p_{0}\left(V_{2}-V_{1}\right)+\alpha \ln \left(V_{2} / V_{1}\right) ; Q=\gamma p_{0}\left(V_{2}-\right.$ $\left.-V_{1}\right) /(\gamma-1)+\alpha \ln \left(V_{2} / V_{1}\right)$.
2.54. (a) $C=C_{p}+R T_{0} / \alpha V$;
(b) $Q=\alpha C_{p}\left(V_{2}-V_{1}\right)+$ $+R T_{0} \ln \left(V_{2} / V_{1}\right)$.
2.55. (a) $V \mathrm{e}^{-\alpha T / R}=\mathrm{const}$;
(b) $T \mathrm{e}^{R / \beta V}=$ const;
(c) $V-a T=$ $=$ const.
2.56. (a) $A=\alpha \ln \eta-R T_{0}(\eta-1) /(\gamma-1)$; (b) $p V^{\gamma} \mathrm{e}^{\alpha(\gamma-1) / p V}=$ $=$ const.
2.57. $A=R T \ln \frac{V_{2}-b}{V_{1}-b}+a\left(\frac{1}{V_{2}}-\frac{1}{V_{1}}\right)$, where $a$ and $b$ are Van der Waals constants.
2.58. (a) $\Delta U=a / V_{1}-a / V_{2}=0.11 \mathrm{~kJ}$;
(b) $Q=R T \ln \frac{V_{2}-b}{V_{1}-b}=$ $=3.8 \mathrm{~kJ}$.
298
2.59. (a) $T(V-b)^{R / c_{V}}=$ const;
(b)
$C_{p}-C_{V}=\frac{R}{1-2 a(V-b)^{2} / R T V^{3}}$.
2.60. $\Delta T=-\frac{v a V_{2}(\gamma-1)}{R V_{1}\left(V_{1}+V_{2}\right)}=-3.0 \mathrm{~K}$.
2.61. $Q=v^{2} a\left(V_{2}-V_{1}\right) / V_{1} V_{2}=0.33 \mathrm{~kJ}$.
2.62. $n=p / k T=1 \cdot 10^{5} \mathrm{~cm}^{-3} ;\langle l\rangle=0.2 \mathrm{~mm}$.
2.63. $p=(1+\eta) m R T / M V=1.9 \mathrm{~atm}$, where $M$ is the mass of an $\mathrm{N}_{2}$ mole.
2.64. $n=\left(p / k T-\rho / m_{2}\right) /\left(1-m_{1} / m_{2}\right)=1.6 \cdot 10^{19} \mathrm{~cm}^{-3}$, where $m_{1}$ and $m_{2}$ are the masses of helium and nitrogen molecules.
2.65. $p=2 n m v^{2} \cos ^{2} \theta=1.0 \mathrm{~atm}$, where $m$ is the mass of a nitrogen molecule.
2.66. $i=2 /\left(\rho v^{2} / p-1\right)=5$.
2.67. $v / v_{s q}=\sqrt{(i+2) / 3 i}$; (a) 0.75 ; (b) 0.68 .
2.68. $\langle\varepsilon\rangle=\left\{\begin{array}{l}(3 N-3) k T \text { for volume molecules. } \\ (3 N-5 / 2) k T \text { for linear molecules. }\end{array}\right.$
$1 / 2(N-1)$ and $1 /(2 N-5 / 3)$ respectively.
2.69. (a) $C_{V}=7 / 2 R, \quad \gamma=9 / 7$; (b) $C_{V}=(3 N-5 / 2) R, \gamma=$ $=(6 N-3) /(6 N-5)$; (c) $C_{v}=3(N-1) R, \gamma=$ $=(N-2 / 3) /(N-1)$.
2.70. $A / Q=\left\{\begin{array}{l}1 /(3 N-2) \text { for volume molecules, } \\ 1 /(3 N-3 / 2) \text { for linear molecules. }\end{array}\right.$

For monoatomic molecules $A / Q=2 / 5$.
2.71. $M=R /\left(c_{p}-c_{\mathrm{v}}\right)=32 \mathrm{~g} / \mathrm{mol}: i=2 /\left(c_{p} / c_{V}-1\right)=5$.
2.72. (a) $i=2\left(C_{p} / R-1\right)=5$; (b) $i=2[C / R+1 /(n-1)]=$ $=3$, where $n=1 / 2$ is the polytropic index.
2.73. $\gamma=\left(5 v_{1}+7 v_{2}\right) /\left(3 v_{1}+5 v_{2}\right)$.
2.74. Increases by $\Delta p / p=M v^{2} / i R T=2.2 \%$, where $i=5$.
2.75. (a) $v_{s q}=\sqrt{3 R T / M}=0.47 \mathrm{~km} / \mathrm{s},\langle\varepsilon\rangle={ }^{3} / 2 k T=6.0 \cdot 10^{-21} \mathrm{~J}$;
(b) $v_{s q}=3 \sqrt{2 k T / \pi \rho d^{3}}=0.15 \mathrm{~m} / \mathrm{s}$.
2.76. $\eta^{i}=7.6$ times.
2.77. $Q=1 / 2\left(\eta^{2}-1\right) \mathrm{imRT} / M=10 \mathrm{~kJ}$.
2.78. $\omega_{s q}=\sqrt{2 k T / I}=6.3 \cdot 10^{12} \mathrm{rad} / \mathrm{sec}$.
2.79. $\langle\varepsilon\rangle_{\text {rot }}=k T_{0} \eta^{2 / i}=0.7 \cdot 10^{-20} \mathrm{~J}$.
2.80. Decreases $\eta^{(i+1) / i}$ times, where $i=5$.
2.81. Decreases $\eta^{(i-1) /(i-2)}=2.5$ times.
2.82. $C=1 / 2 R(i+1)=3 R$.
2.83. $v_{p r}=\sqrt{2 p / \rho}=0.45 \mathrm{~km} / \mathrm{s},\langle v\rangle=0.51 \mathrm{~km} / \mathrm{s}, \quad v_{s q}=$ $=0.55 \mathrm{~km} / \mathrm{s}$.
2.84. (a) $\delta N / N=(8 / \sqrt{\pi}) \mathrm{e}^{-1} \delta \eta=1.66 \%$;
(b) $\delta N / N=12 \sqrt{3 / 2 \pi} \mathrm{e}^{-3 / 2} \delta \eta=1.85 \%$.
2.85. (a) $T=\frac{m(\Delta v)^{2}}{k(\sqrt{3}-\sqrt{2})^{2}}=380 \mathrm{~K}$;
(b) $T=\frac{m v^{2}}{2 k}=340 \mathrm{~K}$.
2.86. (a) $T=\frac{m\left(v_{2}^{2}-v_{1}^{2}\right)}{4 k \ln \left(v_{2} / v_{1}\right)}=330 \mathrm{~K}$;
(b) $v=\sqrt{\frac{3 k T_{0}}{m} \frac{\eta \ln \eta}{\eta-1}}$.
2.87. $T=\frac{m_{\mathrm{N}}(\Delta v)^{2}}{2 k\left(1-\sqrt{m_{\mathrm{N}} / m_{0}}\right)^{2}}=0.37 \mathrm{kK}$.
2.88. $v=\sqrt{\frac{3 k T \ln \left(m_{2} / m_{1}\right)}{m_{2}-m_{1}}}=1.61 \mathrm{~km} / \mathrm{s}$.
2.89. $T=1 / 3 m v^{2} / k$.
2.90. $d N / N=\left(\frac{m}{2 \pi k T}\right)^{3 / 2} \mathrm{e}^{-m v 2 / 2 k T} 2 \pi v_{\perp} d v_{\perp} d v_{x}$.
2.91. $\left\langle v_{x}\right\rangle=0$, $\left\langle\left|v_{x}\right|\right\rangle=\sqrt{2 k T / \pi m}$.
2.92. $\left\langle v_{x}^{2}\right\rangle=k T / m$.
2.93. $v=1 / 4\langle v\rangle$, where $\langle v\rangle=\sqrt{8 k T / \pi m}$.
2.94. $p=\int_{0}^{\infty} 2 m v_{x} \cdot v_{x} d n\left(v_{x}\right)=n k T, \quad$ where $\quad d n\left(v_{x}\right)=$ $=(m / 2 \pi k T)^{1 / 2} n \cdot \mathrm{e}^{-m v_{x}^{2} / 2 k T} d v_{x}$.
2.95. $(1 / v\rangle=\sqrt{2 m / \pi k T}=4 \pi\langle v\rangle$.
2.96. $d N / N=2 \pi(\pi k T)^{-3 / 2} e^{-\varepsilon / k T} \sqrt{\varepsilon} d \varepsilon ; \varepsilon_{p r}=1 / 2 k T$; no.
2.97. $\delta N / N=3 \sqrt{6 \pi} \mathrm{e}^{-3 / 2} \delta \eta=0.9 \%$.
2.98. $\frac{\Delta N}{N}=\frac{2 \pi}{(\pi k T)^{3 / 2}} \int_{\varepsilon_{0}}^{\infty} \sqrt{\varepsilon} e^{-\varepsilon / k T} d \varepsilon$.

The principal contribution to the value of the integral is provided by the smallest values of $\varepsilon$, namely $\varepsilon \approx \varepsilon_{0}$. The slowly varying factor $\sqrt{\varepsilon}$ can be taken from under the radical sign if ascribed the constant value $\sqrt{\varepsilon_{0}}$. Then
\[
\Delta N / N=2 \sqrt{\varepsilon_{0} / \pi k T} \mathrm{e}^{-\varepsilon_{0} / k T} .
\]
2.99. (a) $v_{p r}=\sqrt{3 k T / m}$;
(b) $\varepsilon_{p r}=k T$.
2.100. $d v=\int_{v=0}^{\infty} d n(d \Omega / 4 \pi) v \cos \theta=n(2 k T / \pi m)^{1 / 2} \sin \theta \cos \theta d \theta$.
2.101. $d v=\int_{\theta=0}^{\pi / 2} d n(d \Omega / 4 \pi) v \cos \theta=\pi(m / 2 \pi k T)^{3 / 2} \mathrm{e}^{-m v^{2} / 2 k T} v^{3} d v$.
2.102. $F=(k T / \Delta h) \ln \eta=0.9 \cdot 10^{-10} \mathrm{~N}$.
2.103. $N_{A}=\left(6 R T / \pi d^{3} \Delta \rho g h\right) \ln \eta \approx 6.4 \cdot 10^{23} \cdot \mathrm{mol}^{-1}$.
2.104. $\eta / \eta_{0}=\mathrm{e}^{\left(M_{2}-M_{1}\right) g h / R T}=1.39$.
2.105. $h=\frac{k T \ln \left(n_{2} / n_{1}\right)}{\left(m_{2}-m_{1}\right) g}$.
2.106. Will not change.
2.107. $\langle U\rangle=k T$. Does not depend.
2.108. $w \approx \eta R T / M l \approx 70 \mathrm{~g}$.
2.109. $M=\frac{2 R T \rho \ln \eta}{\left(\rho-\rho_{0}\right)\left(r_{2}^{2}-r_{1}^{2}\right) \omega^{2}}$.
2.111. (a) $d N=n_{0} \mathrm{e}^{-a r^{2} / k T} 4 \pi r^{2} d r$; (b) $r_{p r}=\sqrt{k T / a}$; (c) $d N / N=$ $=(a / \pi k T)^{3 / 2} \mathrm{e}^{-a r 2 / k T} 4 \pi r^{2} d r$; (d) Will increase $\eta^{3 / 2}$-fold.
2.112. (a) $d N=\left(2 \pi n_{0} / a^{3 / 2}\right) \mathrm{e}^{-U / k T} \sqrt{U} d U$;
(b) $U_{p r}=1 / 2 k T$.
2.113. In the latter case.
2.114. (a) $\eta=1-n^{1-\gamma}=0.25$;
(b) $\eta=1-n^{1 / \gamma-1}=0.18$.
2.115. $\varepsilon=(1-\eta) / \eta=9$.
2.116. $\eta=1-2 T_{3} /\left(T_{1}+T_{2}\right)$.
2.117. $\eta=1-n^{1-\gamma}=60 \%$.
2.118. $\eta=1-n^{-(1-1 / \gamma)}$.
2.119. $\eta=1-(n+\gamma) /(1+\gamma n)$
2.120. In both cases $\eta=1-\frac{\ln n}{n-1}$.
2.121. In both cases $\eta=1-\frac{n-1}{n \ln n}$.
2.122. $\eta=1-\frac{n-1}{n \ln n}$.
2.123. (a) $\eta=1-\gamma \frac{n-1}{n^{\gamma}-1}$; (b) $\eta=1-\frac{n^{\gamma}-1}{\gamma(n-1) n^{\gamma-1}}$
2.124. (a) $\eta=1-\frac{\gamma(n-1)}{n-1+(\gamma-1) n \ln n}$;
(b) $\eta=1-\frac{n-1+(\gamma-1) \ln n}{\gamma(n-1)}$.
2.125. $\eta=\frac{(\tau-1) \ln v}{\tau \ln v+(\tau-1) /(\gamma-1)}$.
2.126. $\eta=\frac{(\tau-1) \ln n}{\tau \ln n+(\tau-1) \underline{\gamma} /(\gamma-1)}$.
2.127. $\eta=1-2 \frac{\gamma+\sqrt{\tau}}{(1+\gamma)(1+\sqrt{\tau})}$.
2.128. The inequality $\int \frac{\delta Q_{1}}{T_{1}}-\int \frac{\delta Q_{2}^{\prime}}{T_{2}} \leqslant 0$ becomes even stronger when $T_{1}$ is replaced by $T_{\max }$ and $T_{2}^{2}$ by $T_{\min }$. Then $Q_{1} / T_{\max }-$ $-Q_{2}^{\prime} / T_{\min }<0$. Hence
\[
\frac{Q_{1}-Q_{2}^{\prime}}{Q_{1}}<\frac{T_{\max }-T_{\min }}{T_{\max }} \text {, or } \eta<\eta_{\text {Carnoi }} \text {. }
\]
2.129. According to the Carnot theorem $\delta A / \delta Q_{1}=d T / T$. Let us find the expressions for $\delta A$ and $\delta Q_{1}$. For an infinitesimal Carnot cycle (e.g. parallelogram 1234 shown in Fig. 14)
\[
\begin{array}{c}
\delta A=d p \cdot d V=(\partial p / \partial T)_{V} d T \cdot d V, \\
\delta Q_{1}=d U_{12}+p d V=\left[(\partial U / \partial V)_{T}+p\right] d V .
\end{array}
\]

Fig. 14.
It remains to substitute the two latter expressions into the former one.
2.130. (a) $\Delta S=\frac{R \ln n}{\gamma-1}=19 \mathrm{~J} /(\mathrm{K} \cdot \mathrm{mol})$;
(b) $\Delta S=\frac{\gamma R \ln n}{\gamma-1}=$ $=25 \mathrm{~J} /(\mathrm{K} \cdot \mathrm{mol})$.
2.131. $n=e^{\Delta S / v K}=2.0$.
2.132. $\Delta S=v R \ln n=20 \mathrm{~J} / \mathrm{K}$.
2.133. $\Delta S=-\frac{m}{M} \frac{\gamma R}{\gamma-1} \ln n=-10 \mathrm{~J} / \mathrm{K}$.
2.134. $\Delta S=(\gamma \ln \alpha-\ln \beta) v R /(\gamma-1)=-11 \mathrm{~J} / \mathrm{K}$.
2.135. $S_{2}-S_{1}=v R\left(\ln \alpha-\frac{\ln \beta}{\gamma-1}\right)=1.0 \mathrm{~J} / \mathrm{K}$.
2.136. $\Delta S=\frac{(n-\gamma) R}{(n-1)(\gamma-1)}$ ln $\mathrm{v}$.
2.137. $\Delta S=\frac{\gamma(\gamma+1)}{\gamma-1} \ln \alpha=46 \mathrm{~J} / \mathrm{K}$.
2.138. $V_{m}=\gamma P_{0} / a(1+\gamma)$.
2.139. $T=T_{0}+(R / a)$ ln $\left(V / V_{0}\right)$.
2.140. $\Delta s=R \ln \left(\left(V_{1}-b\right) /\left(V_{1}-b\right)\right\}$.
2.141. $\Delta S=C_{V} \ln \left(T_{2} / T_{1}\right)+K \ln \left(V_{1}-b\right) /\left(V_{1}-b\right) \mathrm{L}$
2.142. $S=a T^{*} / 3$.
2.143. $\Delta S=m$ la $\ln \left(T_{2} / T_{1}\right)+b\left(T_{1}-\right.$ $\left.\left.-T_{1}\right)\right\}=2.0 \mathrm{~kJ} / \mathrm{K}$.
2.144. $C=S / n ; c<0$ for $n<0$.
2.145. $T=T$, eis-suc, See Fig. 15.
2.146. (a) $C=-a / T$; (b) $Q=a$ in $\left(T_{1} / T_{2}\right)$; (c) $A=a \ln \left(T_{1} / T_{2}\right)+C_{V}\left(T_{1}-T_{2}\right)$.
2.147. (a) $\eta=(n-1) / 2 n$; (b) $n=(n-$ $-1)(n+1)$.
2.148. $\Delta S=v R \ln n=20 \mathrm{~J} / \mathrm{K}$.
2.149. $\Delta U=(2 \gamma-1-1) R T_{0} /(\gamma-1), \Delta S=$ $=R$ ln 2 .
2.150. The pressure will be higher after the fast expansion.
2.151. $\Delta S=\mathrm{v}_{2} R$ ln $(1+n)+\mathrm{v}_{8} R$ ln $(1+1 / n)-5.1 \mathrm{~J} / \mathrm{K}$.
2.152. $\Delta S=m_{1} \varepsilon_{1} \ln \left(T / T_{1}\right)+m_{4} c_{1} \ln \left(T / T_{1}\right)=4.4 \mathrm{~J} / \mathrm{K}$, where $T=\left(m_{1} c_{1} T_{1}+m_{4} c_{1} T_{2}\right) /\left(m_{1} c_{1}+m_{4} c_{2}\right), \quad c_{1}$ and $c_{2}$ are the specific heat capacities of copper and water.
2.153. $\Delta S=C_{v} \ln \frac{\left(r_{1}+r_{v}\right)^{2}}{4 r_{1} T_{r}}>0$.
2.154. (a) $P=1 / 2^{N} ;$ (b) $N=\frac{\log (u / v)}{\log 2^{2}} \approx 80$, where $\tau \approx 10^{-4}$ s is the mean time which takes a helium atom to cover distances of the order of the vessel’s dimensions.
2.155. $\Omega_{p},-N 1 /(N / 2) \Pi^{2}=252 . P_{s h}=\Omega_{p} N^{2 N}=24.6 \%$.
2.156. $P_{n}=\frac{N !}{n !(N-n) 12^{N}}: 1 / 32,5 / 32,10 / 32,10 / 32,5 / 32,1 / 32$ respectively.
2.157. $P_{n}=\frac{N !}{n(N-n) !} p^{n}(1-p)^{N-s}$, where $p=V / V_{0}$.
2.158. $d=\sqrt[3]{6 / \pi n_{9} \eta^{2}}=0.4 \mu \mathrm{m}$, where $n_{0}$ is Loschmidt’s number; $(n)=1 / \eta^{2}=1.0 \cdot 10^{\circ}$.
2.159. Will increase $\Omega / \Omega_{0}=\left(1+\Delta T / T_{6}\right)^{a N} A^{2 / 2}=10^{1.31 \cdot 10^{n}}$ times.
2.160. (a) $\Delta p=4 a / d=13 \mathrm{~atm} ;$ (b) $\Delta p=8 a / d=1.2 \cdot 10^{-1} \mathrm{~atm}$.
2.161. $h=4 a / p g d=21 \mathrm{~cm}$.
2.162. $\alpha=1 / s p_{0} d\left(1-\eta^{2} / n\right) /\left(\eta^{2}-1\right)$.
2.163. $p=p_{0}+p s^{c h}+4 a / d=2.2$ atm.
2.164. $h=p_{p}\left(n^{2}-1\right)+4 a\left(n^{2}-1\right) / d{ }^{2} / \rho s=5 \mathrm{~m}$.
2.165. $\Delta h=4 a|\cos \theta|\left(d_{2}-d_{1}\right) / d_{1} d_{\Perp} \rho=11 \mathrm{~mm}$.
2.166. $\boldsymbol{R}=2 a / \mathrm{pgh}=0.6 \mathrm{~mm}$.
2.167. $x=L /(1+P, d / 4 \alpha)=1.4 \mathrm{~cm}$.
2.168. $a=\left|p g h+p_{0} l(l-h)\right| d / 4 \cos \theta$.
2.169. $h=4 a / p g\left(d_{1}-d_{1}\right)=6 \mathrm{~cm}$.
2.170. $h=2 a \cos \theta / p g a \delta$ e.
2.171. $V_{1}=1 / 4 d^{2} \sqrt{\frac{2 \pi-62(n-1) / \mu^{2}}{n^{2}-1}}=0.9 \mathrm{~cm}^{3} / \mathrm{s}$.
2.172. $R_{1}-R_{1} \approx 1 /$ sph $^{3} / \alpha=0.20 \mathrm{~mm}$.
2.173. $\left.m \approx 2 \pi R^{2} a \mid \cos \theta\right)\left(n^{2}-1\right) / g h=0.7 \mathrm{~kg}$.
2.174. $F \approx 2 a m / h^{\prime}=1.0 \mathrm{~N}$.
2.175. $F=2 \pi R^{2} a / h=0.6 \mathrm{kN}$.
2.176. $F=2 \alpha^{4} / \operatorname{pg}^{4}-13 \mathrm{~N}$.
2.177. $t=2 \ln R^{4} / a r^{4}$.
2.178. $Q=2 \pi a^{2} / p g$.
2.179. (a) $F=$ rad $^{2}-3 \mu \mathrm{J}$; (b) $F=2 \pi a d^{2}=10 \mu \mathrm{J}$.
2.180. $\Delta F=2$ rad $(2-40-1)=-1.5 \mu \mathrm{J}$.
2.181. $A^{\prime}=F+p V$ in $\left(p^{\prime} / p_{0}\right)$, where $F^{2}=8 \pi R^{2} a, p=p_{0}+$ $+4 a / R, V=4 / 2 \pi H^{2}$.
2.182. $C-C_{p}=1 / 2 R /(1+1 / s P r / a)$.
2.184. (a) $\Delta S^{\prime}=-2$ (da/dT) $\Delta \sigma$; (b) $\Delta U=2$ (a $\left.a T d a / d T\right) \times$ $\times \Delta \boldsymbol{\Delta}$.
2.185. $A=\Delta m R T / M=1.2 \mathrm{~J}$.
2.186. $m_{q}=(V-m V) /(V ;-V)-20$ g. $V_{q}-1.0$ l. Here $V_{i}$ is the specific volume of water.
2.187. $m_{t} \approx M_{p_{0}}\left(V_{0}-V\right) / R T=2.0 \mathrm{~g}$, where $p_{0}$ is the standard atmospheric pressure.
2.188. $\eta=(n-1) /(N-1) ; \eta-1 /(N+1)$.
2.189. is $=m q / T=6.0 \mathrm{~kJ} / \mathrm{K} ; \quad \Delta U=m(q-R T / M=2.1 \mathrm{MJ}$, where $T=373 \mathrm{~K}$.
2.190. $\mathrm{h} \approx \frac{(0-\operatorname{mas} n}{\mathrm{p}^{5(1+q / \mathrm{M} T)}}=20 \mathrm{~cm}$, where $c$ is the specific heat capacity of water, $\Delta T=100 \mathrm{~K}, q$ is the specific heat of vaporization of water, $T$ is its boiling temperature.
2.191. $A=m e\left(T-T_{t}\right) R T / q M=25 \mathrm{~J}$, where $e$ is the specific heat capacity of water, $T$ is the initial vapour temperature equal to the water boiling temperature, as is seen from the hypothesis, $q$ is the specific heat of vapour condensation.
2.192. $d \approx 4 a M / \eta p R T=0.2 \mu \mathrm{m}$, where $\rho$ is the density of water.
2.193. $\mu=\eta p_{0} V \overline{M / 2 \pi A T}=0.35 \mathrm{~g} /\left(\mathrm{s} \cdot \mathrm{cm}^{2}\right)$, where $P_{0}$ is the standard atmospheric pressure.
2.194. $p=\mu V 2 \pi F T / M=0.9 \mathrm{nPa}$.
2.195. $\Delta p=a / V^{2} M=1.7 \cdot 10^{4} \mathrm{~atm}$.
2.196. $P_{i} \approx p$. About $2 \cdot 10^{4}$ atm. $-0.043 \mathrm{l} / \mathrm{mol}$.
2.199. $V_{e v}=3 / g R T_{e r} / M p_{e r}=4.7 \mathrm{~cm}^{2} / \mathrm{g}$.
2.200. $\left(\pi+3 / v^{2}\right)(3 v-1)=8 \tau, \tau=1.5$.
2.201. (a) $V_{\max }=3 \mathrm{bm} / M=5.0 \mathrm{l}$;
(b) $p_{\max }=a / 27 b^{2}=230$ atm.
2.202. $T_{c r}=8 / 27 a / b R=0.30 \mathrm{kK}, \quad \rho_{c r}=1 / 3 M / b=0.34 \mathrm{~g} / \mathrm{cm}^{3}$.
2.203. $\eta=8 / 3 M p_{c r} / \rho R T_{c r}=0.25$, where $\rho$ is the density of ether at room temperature.
2.204. Let us apply Eq. (2.4e) to the reversible isothermic cycle 1-2-3-4-5-3-1:
\[
T \oint d S=\oint d U+\oint p d V
\]

Since the first two integrals are equal to zero, $\oint p d V=0$ as well. The latter equality is possible only when areas $I$ and $I I$ are equal.

Note that this reasoning is inapplicable to the cycle 1-2-3-1, for example. It is irreversible since it involves the irreversible transition at point 3 from a single-phase to a diphase state.
2.205. $\eta=c|t| / q=0.25$, where $q$ is the specific heat of melting of ice; at $t=-80^{\circ} \mathrm{C}$.
2.206. $\Delta T=-\left(T \Delta V^{\prime} / q\right) \Delta p=-7.5 \mathrm{mK}$, where $q$ is the specific heat of melting of ice.
2.207. $V_{s v}^{\prime} \approx q \Delta T / T \Delta p=1.7 \mathrm{~m}^{3} / \mathrm{kg}$, where $q$ is the specific heat of vaporization, $T=373 \mathrm{~K}$.
2.208. $p_{s v} \approx p_{0}\left(1+q M \Delta T / R T^{2}\right)=1.04$ atm where $q$ is the specific heat of vaporization, $p_{0}$ is the standard atmospheric pressure, $\Delta T=1.1 \mathrm{~K}$.
2.209. $\Delta m / m=(q M / R T-1) \Delta T / T=5 \%$.
2.210. $p=p_{0} \exp \left[\frac{q M}{R}\left(\frac{1}{T_{0}}-\frac{1}{T}\right)\right]$. These assumptions are admissible in the case of a vapour narrow temperature interval, far below the critical temperature.
2.211. $\eta \approx c p T \Delta V^{\prime} / q^{2}=0.03$, where $c$ is the specific heat capacity of ice, $T \approx 273 \mathrm{~K}, q$ is the specific heat of melting.
2.212. (a) $216 \mathrm{~K}, 5.1 \mathrm{~atm}$; (b) $0.78,0.57$, and $0.21 \mathrm{~kJ} / \mathrm{g}$.
2.213. $\Delta S \approx m\left[c \ln \left(T_{2} / T_{1}\right)+q / T_{2}\right]=7.2 \mathrm{~kJ} / \mathrm{K}$.
2.214. $\Delta s \approx q_{m} / T_{1}+c \ln \left(T_{2} / T_{1}\right)+q_{v} / T_{2}=8.6 \mathrm{~J} /(\mathrm{g} \cdot \mathrm{K})$.
2.215. $\Delta S=m c \ln \left(T / T_{1}\right)=-10 \mathrm{~J} / \mathrm{K}$, where $c$ is the specific heat capacity of copper, $T=273 \mathrm{~K}$ (under these conditions only a part of the ice will melt).
2.216. (a) When $m_{2} c_{2} t_{2}<m_{1} q$, not all the ice will melt and
\[
\Delta S=m_{2} c_{2}\left(\frac{T_{2}}{T_{1}}-1-\ln \frac{T_{2}}{T_{1}}\right)=9.2 \mathrm{~J} / \mathrm{K} ;
\]
(b) When $m_{2} c_{2} t_{2}>m_{1} q$, the ice will melt completely and
\[
\Delta S=\frac{m_{1} q}{T_{1}}+c_{2}\left(m_{1} \ln \frac{T}{T_{1}}-m_{2} \ln \frac{T_{2}}{T}\right)=18 \mathrm{~J} / \mathrm{K},
\]
where $T=\frac{m_{1} T_{1}+m_{2} T_{2}-m_{1} q / c_{2}}{m_{1}+m_{2}}$.
2.217. $\Delta S=m q\left(\frac{1}{T_{1}}-\frac{1}{T_{2}}\right)+m c\left(\frac{T_{2}}{T_{1}}-1-\ln \frac{T_{2}}{T_{1}}\right)=0.48 \mathrm{~J} / \mathrm{K}$.
2.218. $C=C_{p}-q M / T=-74 \mathrm{~J} /(\mathrm{K} \cdot \mathrm{mol})$, where $C_{p}=$ $=R \gamma /(\gamma-1)$.
314
2.219. $\Delta S=q M / T_{2}+C_{p} \ln \left(T_{2} / T_{1}\right)$.
2.220. (a) $\eta \approx 0.37$; (b) $\eta \approx 0.23$.
2.221. $\lambda=\Delta l / \ln \eta$.
2.222. (a) $P=\mathrm{e}^{-\alpha t}$; (b) $\langle t\rangle=1 / \alpha$.
2.223. (a) $\lambda=0.06 \mu \mathrm{m}, \quad \tau=0.13 \mathrm{~ns}$; hours.
(b) $\lambda=6 \mathrm{Mm}, \tau=3.8$
2.224. 18 times.
2.225. $\lambda=\left(2 \pi N_{A} / 3 b\right)^{2 / 3}\left(k T_{0} / \sqrt{2} \pi p_{0}\right)=84 \mathrm{~nm}$.
2.226. $v=\pi d^{2} p_{0} N_{A} \sqrt{2 \gamma / M R T_{0}}=5.5 \mathrm{GHz}$.
2.227. (a) $0.7 \mathrm{~Pa}$; (b) $2 \cdot 10^{14} \mathrm{~cm}^{-3}, 0.2 \mu \mathrm{m}$.
2.228. (a) $v=\sqrt{2} \pi d^{2} n\langle v\rangle=0.74 \cdot 10^{10} \mathrm{~s}^{-1}$;
(b) $v=1 / 2 \sqrt{2} \pi d^{2} n^{2}\langle v\rangle=1.0 \times 10^{29} \quad \mathrm{~s}^{-1} \cdot \mathrm{cm}^{-3}$, where $n=p_{0} / k T_{0}$, $\langle v\rangle=\sqrt{8 R T / \pi M}$.
2.229. (a) $\lambda=\mathrm{const}, v \propto \sqrt{T}$; (b) $\lambda \propto T, v \propto 1 / \sqrt{T}$.
2.230. (a) $\lambda=\mathrm{const}, v$ increases $\sqrt{n}$ times; (b) $\lambda$ decreases $n$ times, $v$ increases $n$ times.
2.231. (a) $\lambda \propto V, v \propto V^{-6 / 5}$;
(b) $\lambda \propto p^{-5 / 7}, v \propto p^{6 / 7}$
(c) $\lambda \propto$ $\propto T^{-5 / 2}, v \propto T^{3}$.
2.232. (a) $\lambda \propto V, \quad v \propto V^{-(n+1) / 2}$;
(c) $\lambda \propto T^{1 /(1-n)}, v \propto T^{(n+1) / 2(n-1)}$.
(b) $\lambda \propto p^{-1 / n}, \quad v \propto p^{(n+1) / 2 n} ;$
2.233. (a) $C=1 / 4 R(1+2 i)=23 \mathrm{~J} /(\mathrm{K} \cdot \mathrm{mol})$;
(b) $C=$ $=1 / 2 R(i+2)=29 \mathrm{~J} /(\mathrm{K} \cdot \mathrm{mol})$.
2.234. $n=n_{0} \mathrm{e}^{-t / \tau}$, where $\tau=4 V / S\langle v\rangle,\langle v\rangle=\sqrt{8 R T / \pi M}$.
2.235. Increases $(1+\eta) /(1+\sqrt{\eta})$ times.
2.236. Increases $\alpha^{3} / \beta=2$ times.
2.237. (a) $D$ increases $n$ times, $\eta=$ const; (b) $D$ increases $n^{3 / 2}$ times, $\eta$ increases $\sqrt{n}$ times.
2.238. $D$ decreases $n^{4 / 5} \approx 6.3$ times, $\eta$ increases $n^{1 / 5} \approx 1.6$ times.
2.239. (a) $n=3$; (b) $n=1$; (c) $n=1$.
2.240. $0.18 \mathrm{~nm}$.
2.241. $d_{\mathrm{Ar}} / d_{\mathrm{He}}=1.7$.
2.242. $N_{1} \approx 2 \pi \eta \omega R^{3} / \Delta R ; p=\sqrt{2} k T / \pi d^{2} n \Delta R=0.7 \mathrm{~Pa}$.
2.243. $\eta=\left(1 / R_{1}^{2}-1 / R_{2}^{2}\right) N_{1} / 4 \pi \omega$.
2.244. $N=1 / 2 \pi \eta \omega a^{4} / h$.
2.245. $N=1 / 3 \omega a^{4} p \sqrt{\pi M / 2 R T}$.
2.246. $\mu=\frac{\pi a^{4} M}{16 \eta R T} \frac{\left|p_{2}^{2}-p_{1}^{2}\right|}{l}$.
2.247. $T=\left(x_{1} T_{1} / l_{1}+\varkappa_{2} T_{2} / l_{2}\right) /\left(\varkappa_{1} / l_{1}+x_{2} / l_{2}\right)$.
2.248. $x=\left(l_{1}+l_{2}\right) /\left(l_{1} / x_{1}+l_{2} / x_{2}\right)$.
2.249. $T(x)=T_{1}\left(T_{2} / T_{1}\right)^{x / l} ; q=(\alpha / l) \ln \left(T_{2} / T_{1}\right)$.
2.250. $\Delta T=(\Delta T)_{0} \mathrm{e}^{-\alpha t}$, where $\alpha=\left(1 / C_{1}+1 / C_{2}\right) S x / l$.
2.251. $T=T_{1}\left\{1+(x / l)\left[\left(T_{2} / T_{1}\right)^{3 / 2}-1\right]\right\}^{2 / 3}$, where $x$ is the distance from the plate maintained at the temperature $T_{1}$.
2.252. $q=\frac{2 i R^{3 / 2}\left(T_{2}^{3 / 2}-T_{1}^{3 / 2}\right)}{9 \pi^{3 / 2} l d^{2} N_{A} \sqrt{M}}=40 \mathrm{~W} / \mathrm{m}^{2}$, where $i=3, d$ is the effective diameter of helium atom.
2.253. $\lambda=23 \mathrm{~mm}>l$, consequently, the gas is ultra-thin; $q=p\langle v\rangle\left(t_{2}-t_{1}\right) / 6 T(\gamma-1)=22 \mathrm{~W} / \mathrm{m}^{2}$, where $\langle v\rangle=$ $=\sqrt{8 R T / \pi M}, T=1 / 2\left(T_{1}+T_{2}\right)$.
2.254. $T=T_{1}+\frac{T_{2}-T_{1}}{\ln \left(R_{2} / R_{1}\right)} \ln \frac{r}{R_{1}}$.
2.255. $T=T_{1}+\frac{T_{2}-T_{1}}{1 / R_{1}-1 / R_{2}}\left(\frac{1}{R_{1}}-\frac{1}{r}\right)$.
2.256. $T=T_{0}+\left(R^{2}-r^{2}\right) w / 4 x$.
2.257. $T=T_{0}+\left(R^{2}-r^{2}\right) w / 6 x$.
3.1. The ratio $F_{e l} / F_{g r}$ is equal to $4 \cdot 10^{42}$ and $1 \cdot 10^{36}$ respectively; $q / m=0.86 \cdot 10^{-10} \mathrm{C} / \mathrm{kg}$.
3.2. About $2 \cdot 10^{15} \mathrm{~N}$.
3.3. $d q / d t=3 / 2 a \sqrt{2 \pi \varepsilon_{0} m g / l}$.
3.4. $q_{3}=-\frac{q_{1} q_{2}}{\left(\sqrt{q_{1}}+\sqrt{q_{2}}\right)^{2}}, \quad \mathbf{r}_{3}=\frac{\mathbf{r}_{1} \sqrt{q_{2}}+\mathbf{r}_{2} \sqrt{q_{1}}}{\sqrt{q_{1}}+\sqrt{q_{2}}}$.
3.5. $\Delta T=\frac{q q_{0}}{8 \pi^{2} \varepsilon_{0} r^{2}}$.
3.6. $\mathbf{E}=2.7 \mathbf{i}-3.6 \mathbf{j}, E=4.5 \mathrm{kV} / \mathrm{m}$.
3.7. $E=\frac{q l}{\sqrt{2} \pi \varepsilon_{0}\left(l^{2}+x^{2}\right)^{3 / 2}}$.
3.8. $E=\frac{q}{2 \pi^{2} \varepsilon_{0} R^{2}}=0.10 \mathrm{kV} / \mathrm{m}$.
3.9. $E=\frac{q l}{4 \pi \varepsilon_{0}\left(r^{2}+l^{2}\right)^{3 / 2}}$. For $l \gg r$ the strength $E \approx \frac{q}{4 \pi \varepsilon_{0} l^{2}}$, as in the case of a point charge. $E_{m a x}=\frac{q}{6 \sqrt{3} \pi \varepsilon_{0} r^{2}}$ for $l=r / \sqrt{2}$.
3.10. $E=\frac{3 q R^{2}}{4 \pi \varepsilon_{0} x^{4}}$.
3.11. $F=\frac{q \lambda}{4 \pi \varepsilon_{0} R}$.
3.12. (a) $E=\frac{\lambda_{0}}{4 \varepsilon_{0} R}$;
(b) $E=\frac{\lambda_{0} R^{2}}{4 \varepsilon_{0}\left(x^{2}+R^{2}\right)^{3 / 2}}$. For $x \gg R$ the strength $E \approx \frac{p}{4 \pi \varepsilon_{0} x^{3}}$, where $p=\pi R^{2} \lambda_{0}$.
3.13. (a) $E=\frac{q}{4 \pi \varepsilon_{0} r \sqrt{a^{2}+r^{2}}}$;
(b) $E=\frac{q}{4 \pi \varepsilon_{0}\left(r^{2}-a^{2}\right)}$. In both cases $E \approx \frac{q}{4 \pi \varepsilon_{0} r^{2}}$ for $r \gg a$.
3.14. $E=\frac{\lambda \sqrt{2}}{4 \pi \varepsilon_{0} y}$. The vector $\mathbf{E}$ is directed at the angle $45^{\circ}$ to the thread.
3.15. (a) $E=\frac{\lambda \sqrt{2}}{4 \pi \varepsilon_{0} R}$
(b) $E=0$.
3.16. $\mathbf{E}=-1 / 3 \mathrm{ar} / \varepsilon_{0}$.
3.17. $\mathbf{E}=-1 / 3 \mathbf{k} \sigma_{0} / \varepsilon_{0}$, where $\mathbf{k}$ is the unit vector of the $z$ axis with respect to which the angle $\theta$ is read off. Clearly, the field inside the given sphere is uniform.
3.18. $\mathbf{E}=-1 /{ }_{6} \mathrm{a} R^{2} / \varepsilon_{0}$.
3.19. $|\Phi|=1 / 2 \lambda R / \varepsilon_{0}$. The sign of $\Phi$ depends on how the direction of the normal to the circle is chosen.
3.20. $|\Phi|=\frac{q}{\varepsilon_{0}}\left(1-\frac{1}{\sqrt{1+(R / l)^{2}}}\right)$. The sign of $\Phi$ depends on how the direction of the normal to the circle is chosen.
3.21. $|\Phi|=1 / 3 \pi \rho r_{0}\left(R^{2}-r_{0}^{2}\right) / \varepsilon_{0}$.
3.22. $E_{\max }=\lambda / \pi \varepsilon_{0} l$.
3.23. $E=1 / 2 \quad \sigma_{0} / \varepsilon_{0}$, with the direction of the vector $\mathbf{E}$ corresponding to the angle $\varphi=\pi$.
3.24. $\Phi=4 \pi R a$.
3.25. (a) $E=\frac{\rho_{0} r}{3 \varepsilon_{0}}\left(1-\frac{3 r}{4 R}\right)$ for $r \leqslant R, E=\frac{\rho_{0} R^{3}}{12 \varepsilon_{0} r^{2}}$ for $r \geqslant R$;
(b) $E_{\max }=1 / 9 \rho_{0} R / \varepsilon_{0}$ for $r_{m}=2 / 3 R$.
3.26. $q=2 \pi R^{2} \alpha, E=1 / 2 \alpha / \varepsilon_{0}$.
3.27. $E=\frac{\rho_{0}}{3 \varepsilon_{0} \alpha r^{2}}\left(1-\mathrm{e}^{-\alpha r^{3}}\right)$. Accordingly, $E \approx \frac{\rho_{0^{\prime}}}{3 \varepsilon_{0}}$ and $E \approx$ $\approx \frac{\rho_{0}}{3 \varepsilon_{0} \alpha r^{2}}$.
3.28. $E=1 / 3 a \rho / \varepsilon_{0}$.
3.29. $\mathbf{E}=1 / 2 \mathrm{a} \rho / \varepsilon_{0}$, where the vector $\mathrm{a}$ is directed toward the axis of the cavity.
3.30. $\Delta \varphi=\frac{q}{2 \pi \varepsilon_{0} R}\left(1-\frac{1}{\sqrt{1+(a / R)^{2}}}\right)$.
3.31. $\varphi_{1}-\varphi_{2}=\frac{\lambda}{2 \pi \varepsilon_{0}} \ln \eta=5 \mathrm{kV}$.
3.32. $\varphi=1 / 2 \sigma R / \varepsilon_{0}, E=1 / 4 \sigma / \varepsilon_{0}$.
3.33. $\varphi=\frac{\sigma l}{2 \varepsilon_{0}}\left(\sqrt{1+(R / l)^{2}}-1\right), E=\frac{\sigma}{2 \varepsilon_{0}}\left(1-\frac{l}{\sqrt{l^{2}+R^{2}}}\right)$. When $l \rightarrow 0$, then $\varphi=\frac{\sigma R}{2 \varepsilon_{0}}, E=\frac{\sigma}{2 \varepsilon_{0}} ;$ when $l \gg R$, then $\varphi \approx \frac{q}{4 \pi \varepsilon_{0} l}$, $E \approx \frac{q}{4 \pi \varepsilon_{0} l^{2}}$, where $q=\sigma \pi R^{2}$.
3.34. $\varphi=\sigma R / \pi \varepsilon_{0}$.
3.35. $\mathbf{E}=-\mathbf{a}$, i.e. the field is uniform.
3.36. (a) $\mathbf{E}=-2 a(x \mathbf{i}-y \mathbf{j})$; (b) $\mathbf{E}=-a(y \mathbf{i}-x \mathbf{j})$. Here $\mathbf{i}, \mathbf{j}$ are the unit vectors of the $x$ and $y$ axes. See Fig. 16 illustrating the case $a>0$.
3.37. $\mathbf{E}=-2(a x \mathbf{i}+a y \mathbf{j}+b z \mathbf{k}), \quad E=2 \sqrt{a^{2}\left(x^{2}+y^{2}\right)+b^{2} z^{2}}$.
(a) An ellipsoid of revolution with semiaxes $\sqrt{\varphi / a}$ and $\sqrt{\varphi / b}$. (b) In the case of $\varphi>0$, a single-cavity hyperboloid of revolution; when $\varphi=0$, a right round cone; when $\varphi<0$, a two-cavity hyperboloid of revolution.
3.38. (a) $\varphi_{0}=\frac{3 q}{8 \pi \varepsilon_{0} R}$;
(b) $\varphi=\varphi_{0}\left(1-\frac{r^{2}}{3 R^{2}}\right), r \leqslant R$.
3.39. $E=\sqrt{E_{r}^{2}+E_{\theta}^{2}}=\frac{p}{4 \pi \varepsilon_{0} r^{3}} \sqrt{1+3 \cos ^{2} \theta}$, where $E_{r}$ is the radial component of the vector $\mathbf{E}$, and $E_{\theta}$ is its component perpendicular to $E_{r}$.
\[
\text { 3.40. } E_{z}=\frac{p}{4 \pi \varepsilon_{0}} \frac{3 \cos ^{2} \theta-1}{r^{3}}, E_{\perp}=\frac{p}{4 \pi \varepsilon_{0}} \frac{3 \sin \theta \cos \theta}{r^{3}} ;
\]
$\mathbf{E} \perp \mathrm{p}$ at the points located on the lateral surface of a cone whose axis is directed along the $z$ axis and whose semi-vertex angle $\theta$ is found
Fig. 16.
from the relation $\cos \theta=1 / V \overline{3}\left(\theta_{1}=54.7^{\circ}, \theta_{2}=123.5^{\circ}\right)$. At these points $E=E_{\perp}=\frac{p \sqrt{2}}{4 \pi \varepsilon_{0} r^{3}}$.
3.41. $R=\sqrt[3]{\frac{p}{4 \pi \varepsilon_{0} E_{0}}}$.
3.42. $\varphi \approx \frac{\lambda l}{2 \pi \varepsilon_{0} r} \cos \theta, E \approx \frac{\lambda l}{2 \pi \varepsilon_{0} r^{2}}$.
3.43. $\varphi=\frac{q l}{4 \pi \varepsilon_{0}} \frac{x}{\left(R^{2}+x^{2}\right)^{3 / 2}}, \quad E_{x}=-\frac{q l}{4 \pi \varepsilon_{0}} \frac{R^{2}-2 x^{2}}{\left(R^{2}+x^{2}\right)^{5 / 2}}$, where
Fig. 17.
Fig. 18.
$E_{x}$ is the projection of the vector $\mathbf{E}$ on the $x$ axis. The functions are plotted in Fig. 17. If $|x| \gg R$, then $\varphi \approx \frac{q l}{4 \pi \varepsilon_{0} x^{2}}$ and $E_{x} \approx \frac{q l}{2 \pi \varepsilon_{0} x^{3}}$.
3.44. $\varphi=\frac{\sigma l}{2 \varepsilon_{0} \sqrt{x^{2}+R^{2}}}, E_{x}=-\frac{\sigma l R^{2}}{2 \varepsilon_{0}\left(x^{2}+R^{2}\right)^{3 / 2}}$. See Fig. 18.
3.45. $\varphi \approx \pm \frac{\sigma l}{2 \varepsilon_{0}}\left(1-\frac{x}{\sqrt{x^{2}+R^{2}}}\right), E \approx \frac{\sigma l R^{2}}{2 \varepsilon_{0}\left(x^{2}+R^{2}\right)^{3 / 2}}$. If $x \gg R$, then $\varphi \approx \pm \frac{p}{4 \sigma \varepsilon_{0} x^{2}}$ and $E \approx \frac{p}{2 \pi \varepsilon_{0} x^{3}}$, where $p=\pi R^{2} \sigma l$. In the formulas for the potential $\varphi$ the plus sign corresponds to the space adjoining the positively charged plate and the minus sign to the space adjoining the negatively charged plate.
3.46. (a) $\mathbf{F}=0$;
(b) $\mathbf{F}=-\frac{\lambda \mathbf{p}}{2 \pi \varepsilon_{0} r^{2}}$
(c) $\mathbf{F}=\frac{\lambda \mathbf{p}}{2 \pi \varepsilon_{0} r^{2}}$.
3.47. $F=\frac{3 p^{2}}{2 \pi \varepsilon_{0} l^{4}}=2.1 \cdot 10^{-16} \mathrm{~N}$.
3.48. $\varphi=-a x y+$ const.
3.49. $\varphi=a y\left(\frac{y^{2}}{3}-x^{2}\right)+$ const.
3.50. $\varphi=-y(a x+b z)+$ const.
3.51. $\rho=6 \varepsilon_{0} a x$.
3.52. $\rho=2 \varepsilon_{0} \Delta \varphi / d^{2} ; E=\rho d / \varepsilon_{0}$.
3.53. $\rho=-6 \varepsilon_{0} a$.
3.54. $q=4 l \sqrt{\pi \varepsilon_{0} k x}$.
3.55. $A=\frac{q^{2}}{16 \pi \varepsilon_{0} l}$.
3.56. (a) $F=\frac{(2 \sqrt{2}-1) q^{2}}{8 \pi \varepsilon_{0} l^{2}}$;
(b) $E=2\left(1-\frac{1}{5 \sqrt{5}}\right) \frac{q}{\pi \varepsilon_{0} l^{2}}$.
3.57. $F=\frac{(2 \sqrt{2}-1) q^{2}}{32 \pi \varepsilon_{0} l^{2}}$.
3.58. $F=\frac{3 p^{2}}{32 \pi \varepsilon_{0} l^{4}}$.
3.59. $\sigma=-\frac{q l}{2 \pi\left(l^{2}+r^{2}\right)^{3 / 2}}, q_{\text {ind }}=-q$.
3.60. (a) $F_{1}=\frac{\lambda^{2}}{4 \pi \varepsilon_{0} l}$;
(b) $\sigma=\frac{l \lambda}{\pi\left(l^{2}+x^{2}\right)}$.
3.61. (a) $\sigma=\frac{\lambda}{2 \pi l}$;
(b) $\sigma(r)=\frac{\lambda}{2 \pi \sqrt{l^{2}+r^{2}}}$.
3.62. (a) $\sigma=\frac{l q}{2 \pi\left(l^{2}+R^{2}\right)^{3 / 2}}$;
(b) $E=\frac{1}{4 \pi \varepsilon_{0}} \frac{q}{4 l^{2}\left[1+{ }^{1 / 4}(R / l)^{2}\right]^{3 / 2}}$, $\varphi=\frac{1}{4 \pi \varepsilon_{0}} \frac{q}{R}\left(1-\frac{1}{\sqrt{1+4(l / R)^{2}}}\right)$.
3.63. $\varphi=\frac{q}{4 \pi \varepsilon_{0} l}$.
3.64. $\varphi=\frac{q}{4 \pi \varepsilon_{0}}\left(\frac{1}{r}-\frac{1}{R_{1}}+\frac{1}{R_{2}}\right)$.
3.65. $q_{2}=-\frac{b}{a} q_{1} ; \varphi=\frac{q_{1}}{4 \pi \varepsilon_{0}} \times\left\{\begin{array}{l}1 / r-1 / a \text { if } a \leqslant r \leqslant b, \\ (1-b / a) r \text { if } r \geqslant b .\end{array}\right.$
3.66. (a) $E_{23}=\Delta \varphi / d, E_{12}=E_{34}=1 / 2 E_{23}$;
(b) $\left|\sigma_{1}\right|=\sigma_{4}=$ $=1 / 2 \varepsilon_{0} \Delta \varphi / d, \sigma_{2}=\left|\sigma_{3}\right|=3 / 2 \varepsilon_{0} \Delta \varphi / d$.
3.67. $q_{1}=-q(l-x) / l, q_{2}=-q x / l$. Instruction. If the charge $q$ is imagined to be uniformly spread over the plane passing through
that charge and parallel to the plates, the charges $q_{1}$ and $q_{2}$ remain, obviously, unchanged. What changes is only their distribution, and the electric field becomes easy to calculate.
3.68. $d F / d S=1 / 2 \sigma^{2} / \varepsilon_{0}$.
3.69. $F=\frac{q^{2}}{32 \pi \varepsilon_{0} R^{2}}=0.5 \mathrm{kN}$.
3.70. $F=1 / 4 \pi R^{2} \sigma_{0}^{2} / \varepsilon_{0}$.
3.71. $N=\frac{n_{0} p}{(\varepsilon-1) \varepsilon_{0} E}=3 \cdot 10^{3}$,
where $n_{0}$ is the concentration of molecules.
3.72. $F=\frac{3 \beta p^{2}}{4 \pi^{2} \varepsilon_{0} l^{2}}$.
3.73. (a) $x=R / \sqrt{2}$;
(b) $x=\left\{\begin{array}{l}1.1 R \text { (attraction) } \\ 0.29 R \text { (repulsion). See Fig. } 19 .\end{array}\right.$
3.74. $\mathbf{P}=\frac{\varepsilon-1}{\varepsilon} \frac{q}{4 \pi r^{3}} \mathbf{r}, q^{\prime}=-\frac{\varepsilon-1}{\varepsilon} q$.
3.76. $q_{\text {inn }}^{\prime}=-q(\varepsilon-1) / \varepsilon, q_{0 u t}^{\prime}=q(\varepsilon-1) / \varepsilon$.
Fig. 20.
3.77. See Fig. 20 .
3.78. $E=\frac{E_{0}}{\varepsilon} \sqrt{\cos ^{2} \alpha_{0}+\varepsilon^{2} \sin ^{2} \alpha_{0}}=5.2 \mathrm{~V} / \mathrm{m}$; $\tan \alpha=\varepsilon \tan \alpha_{0}$, hence, $\alpha=74^{\circ} ; \sigma^{\prime}=\frac{\varepsilon_{0}(\varepsilon-1)}{\varepsilon} E_{0} \cos \alpha_{0}=64 \mathrm{pC} / \mathrm{m}^{2}$.
3.79. (a) $\oint \mathbf{E} d \mathbf{S}=\frac{\varepsilon-1}{\varepsilon} \pi R^{2} E_{0} \cos \theta$;
(b) $\oint \mathbf{D} d \mathbf{r}=-\varepsilon_{0}(\varepsilon-1) \times$ $\times l E_{0} \sin \theta$.
3.80. (a) $E=\left\{\begin{array}{ll}p l / \varepsilon \varepsilon_{0} & \text { for } l<d, \\ \rho d / \varepsilon_{0} & \text { for } l>d,\end{array} \varphi=\left\{\begin{array}{l}-\rho l^{2} / 2 \varepsilon \varepsilon_{0} \text { for } l \leqslant d, \\ -(d / 2 \varepsilon+l-d) \rho d / \varepsilon_{0}\end{array}\right.\right.$ for $l \geqslant d$.

The plots $E_{x}(x)$ and $\varphi(x)$ are shown in Fig. 21. (b) $\sigma^{\prime}=\rho d(\varepsilon-1) / \varepsilon$, $\rho^{\prime}=-\rho(\varepsilon-1) / \varepsilon$.
3.81. (a) $E=\left\{\begin{array}{ll}\rho r / 3 \varepsilon_{0} \varepsilon & \text { for } r<R, \\ \rho R^{3} / 3 \varepsilon_{0} r^{2} & \text { for } r>R ;\end{array}\right.$
(b) $\rho^{\prime}=-\rho(\varepsilon-1) / \varepsilon, \sigma^{\prime}=\rho R(\varepsilon-1) / 3 \varepsilon$. See Fig. 22 .
3.82. $\mathbf{E}=-d \mathbf{P} / 4 \varepsilon_{0} R$.
3.83. $\mathbf{E}=-\mathbf{P}_{0}\left(1-x^{2} / d^{2}\right) / \varepsilon_{0}, \quad U=4 d P_{0} / 3 \varepsilon_{0}$.
\[
\begin{aligned}
& 3.84 . \text { (a) } E_{1}=2 \varepsilon E_{0} /(\varepsilon+1), E_{2}=2 E_{0} /(\varepsilon+1), D_{1}=D_{2}= \\
= & 2 \varepsilon \varepsilon_{0} E_{0} /(\varepsilon+1) ;(\mathrm{b}) E_{1}=E_{0}, E_{2}=E_{0} / \varepsilon, D_{1}=D_{2}=\varepsilon_{0} E_{0} . \\
& 3.85 .(\mathrm{a}) E_{1}=E_{2}=E_{0}, D_{1}=\varepsilon_{0} E_{0}, D_{2}=\varepsilon D_{1} ;(\mathrm{b}) E_{1}=E_{2}= \\
= & 2 E_{0} /(\varepsilon+1), D_{1}=2 \varepsilon_{0} E_{0} /(\varepsilon+1), D_{2}=\varepsilon D_{1} . \\
& 3.86 . E=q / 2 \pi \varepsilon_{0}(\varepsilon+1) r^{2} . \\
& 3.87 . \rho=\rho_{0} \varepsilon /(\varepsilon-1)=1.6 \mathrm{~g} / \mathrm{cm}^{3} \text {, where } \varepsilon \text { and } \rho_{0} \text { are the per- }
\end{aligned}
\]
3.87. $\rho=\rho_{0} \varepsilon /(\varepsilon-1)=1.6 \mathrm{~g} / \mathrm{cm}^{3}$, where $\varepsilon$ and $\rho_{0}$ are the permittivity and density of kerosene.
Fig. 21.
Fig. 22.
3.88. $\sigma_{\text {max }}^{\prime}=(\varepsilon-1) \varepsilon_{0} E=3.5 \mathrm{nC} / \mathrm{m}^{2}, q^{\prime}=\pi R^{2}(\varepsilon-1) \varepsilon_{0} E=$ $=10 \mathrm{pC}$.
3.89. (a) Since the normal component of the vector $\mathbf{D}$ is continuous at the dielectric interface, we obtain
$\sigma^{\prime}=-q l(\varepsilon-1) / 2 \pi r^{3}(\varepsilon+1)$, for $l \rightarrow 0$ and $\sigma^{\prime} \rightarrow 0$;
(b) $q^{\prime}=-q(\varepsilon-1) /(\varepsilon+1)$.
3.90. $F=q^{2}(\varepsilon-1) / 16 \pi \varepsilon_{0} l^{2}(\varepsilon+1)$.
3.91. $D=\left\{\begin{array}{l}q / 2 \pi(1+\varepsilon) r^{2} \text { in vacuum, } \\ \varepsilon q / 2 \pi(1+\varepsilon) r^{2} \text { in dielectric; }\end{array}\right.$
$\left.\begin{array}{l}E=q / 2 \pi \varepsilon_{0}(1+\varepsilon) r^{2} \\ \varphi=q / 2 \pi \varepsilon_{0}(1+\varepsilon) r\end{array}\right\}$ both in vacuum and in dielectric.
3.92. $\sigma^{\prime}=q l(\varepsilon-1) / 2 \pi r^{3} \varepsilon(\varepsilon+1)$; for $l \rightarrow 0$ and $\sigma^{\prime} \rightarrow 0$.
3.93. $\sigma^{\prime}=q l(\varepsilon-1) / 2 \pi r^{3} \varepsilon$.
3.94. $\mathbf{E}_{1}=\mathbf{P} h / \varepsilon_{0} d$ (between the plates), $\mathbf{E}_{2}=-(1-h / d) \mathbf{P} / \varepsilon_{0}$, $\mathbf{D}_{1}=\mathbf{D}_{2}=\mathrm{P} h / d$.
3.95. $\rho^{\prime}=-2 \alpha$, i.e. is independent of $r$.
3.96. (a) $\mathbf{E}=-\mathbf{P} / 3 \varepsilon_{0}$.
3.97. $\mathbf{E}_{0}=\mathbf{E}-\mathbf{P} / 3 \varepsilon_{0}$.
3.98. $\mathbf{E}=3 \mathbf{E}_{0} /(\varepsilon+2), \mathbf{P}=3 \varepsilon_{0} \mathbf{E}_{0}(\varepsilon-1) /(\varepsilon+2)$.
3.99. $\mathbf{E}=-\mathrm{P} / 2 \varepsilon_{0}$.
3.100. $\mathbf{E}=2 \mathbf{E}_{0} /(\varepsilon+1) ; \quad \mathbf{P}=2 \varepsilon_{0} \mathbf{E}_{0}(\varepsilon-1) /(\varepsilon+1)$.
3.101. $C=\frac{4 \pi \varepsilon_{0} \varepsilon R_{1}}{1+(\varepsilon-1) R_{1} / R_{2}}$.
3.102. The strength decreased $1 / 2(\varepsilon+1)$ times; $q=$ $=1 / 2 C \mathscr{E}(\varepsilon-1) /(\varepsilon+1)$.
3.103. (a) $C=\frac{\varepsilon_{0} S}{d_{1} / \varepsilon_{1}+d_{2} / \varepsilon_{2}}$;
(b) $\sigma^{\prime}=\varepsilon_{0} V \frac{\varepsilon_{1}-\varepsilon_{2}}{\varepsilon_{1} d_{2}+\varepsilon_{2} d_{1}}$.
3.104. (a) $C=\varepsilon_{0}\left(\varepsilon_{2}-\varepsilon_{1}\right) S / d \ln \left(\varepsilon_{2} / \varepsilon_{1}\right)$;
(b) $\rho^{\prime}=-q\left(\varepsilon_{2}-\varepsilon_{1}\right) / d S \varepsilon^{2}$.
3.105. $C=4 \pi \varepsilon_{0} a / \ln \left(R_{2} / R_{1}\right)$.
3.106. When $\varepsilon_{1} R_{1} E_{1 m}=\varepsilon_{2} R_{2} E_{2 m}$.
3.107. $V=R_{1} E_{1}\left[\ln \left(R_{2} / R_{1}\right)+\left(\varepsilon_{1} / \varepsilon_{2}\right) \ln \left(R_{3} / R_{2}\right)\right]$.
3.108. $C \approx \pi \varepsilon_{0} \ln (b / a)$.
3.109. $C \approx 2 \pi \varepsilon_{0} / \ln (2 b / a)$.
3.110. $C \approx 2 \pi \varepsilon_{0} \varepsilon a$. Instruction. When $b \gg a$, the charges can be assumed to be distributed practically uniformly over the surfaces of the balls.
3.111. $C \approx 4 \pi \varepsilon_{0} a$.
3.112. (a) $C_{\text {total }}=C_{1}+C_{2}+C_{3}$; (b) $C_{\text {total }}=C$.
3.113. (a) $C=2 \varepsilon_{0} S / 3 d$; (b) $C=3 \varepsilon_{0} S / 2 d$.
3.114. $V \leqslant V_{1}\left(1+C_{1} / C_{2}\right)=9 \mathrm{kV}$.
3.115. $U=\mathscr{E} /\left(1+3 \eta+\eta^{2}\right)=10 \mathrm{~V}$.
3.116. $C_{x}=C(V \overline{5}-1) / 2=0.62 C$. Since the chain is infinite, all the links beginning with the second can be replaced by the capacitance $C_{x}$ equal to the sought one.
3.117. $V_{1}=q / C_{1}=10 \mathrm{~V}, \quad V_{2}=q / C_{2}=5 \mathrm{~V}$, where $q=$ $=\left(\varphi_{\mathrm{A}}-\varphi_{B}+\mathscr{E}\right) C_{1} C_{2} /\left(C_{1}+C_{2}\right)$.
3.118. $V_{1}=\left(\mathscr{E}_{2}-\mathscr{E}_{1}\right) /\left(1+C_{1} / C_{2}\right), V_{2}=\left(\mathscr{E}_{1}-\mathscr{E}_{2}\right) /\left(1+C_{2} / C_{1}\right)$.
3.119. $q=\left|\mathscr{E}_{1}-\mathscr{E}_{2}\right| C_{1} C_{2} /\left(C_{1}+C_{2}\right)$.
3.120. $\varphi_{A}-\varphi_{B}=\mathscr{E} \frac{C_{2} C_{3}-C_{1} C_{4}}{\left(C_{1}+C_{2}\right)\left(C_{3}+C_{4}\right)}$. In the case when $C_{1} / C_{2}=$ $=C_{3} / C_{4}$.
3.121. $q=\frac{V}{1 / C_{1}+1 / C_{2}+1 / C_{3}}=0.06 \mathrm{mC}$.
3.122. $q_{1}=\mathscr{E} C_{2}, q_{2}=-\mathscr{E} C_{1} C_{2} /\left(C_{1}+C_{2}\right)$.
3.123. $q_{1}=\mathscr{E} C_{1}\left(C_{1}-C_{2}\right) /\left(C_{1}+C_{2}\right)=-24 \mu \mathrm{C}$, $q_{2}=\mathscr{E} C_{2}\left(C_{1}-C_{2}\right) /\left(C_{1}+C_{2}\right)=-36 \mu \mathrm{C}, q_{3}=\mathscr{E}\left(C_{2}-C_{1}\right)=+60 \mu \mathrm{C}$.
3.124. $\varphi_{A}-\varphi_{B}=\left(C_{2} \mathscr{E}_{2}-C_{1} \mathscr{E}_{1}\right) /\left(C_{1}+C_{2}+C_{3}\right)$.
3.125. $\varphi_{1}=\frac{\mathscr{E}_{2} C_{2}+\mathscr{E}_{3} C_{3}-\mathscr{E}_{1}\left(C_{2}+C_{3}\right)}{C_{1}+C_{2}+C_{3}}$,
$\varphi_{2}=\frac{\mathscr{E}_{1} C_{1}+\mathscr{E}_{3} C_{3}-\mathscr{E}_{2}\left(C_{1}+C_{3}\right)}{C_{1}+C_{2}+C_{3}}, \quad \varphi_{3}=\frac{\mathscr{E}_{1} C_{1}+\mathscr{E}_{2} C_{2}-\mathscr{E}_{3}\left(C_{1}+C_{2}\right)}{C_{1}+C_{2}+C_{3}}$.
3.126. $C_{\text {total }}=\frac{2 C_{1} C_{2}+C_{3}\left(C_{1}+C_{2}\right)}{C_{1}+C_{2}+2 C_{3}}$.
3.127. (a) $W=(\sqrt{2}+4) q^{2 / 4 \pi \varepsilon_{0} a}$;
(b) $W=(\sqrt{2}-4) q^{2} / 4 \pi \varepsilon_{0} a$;
(c) $W=-V \overline{2} q^{2} / 4 \pi \varepsilon_{0} a$.
3.128. $W=-\frac{2 \ln 2}{4 \pi \varepsilon_{0}} \frac{q^{2}}{a}$.
3.129. $W=-q^{2} / 8 \pi \varepsilon_{0} l$.
3.130. $W=q_{1} q_{2} / 4 \pi \varepsilon_{0} l$.
3.131. $\Delta W=-1 / 2 V^{2} C_{1} C_{2} /\left(C_{1}+C_{2}\right)=-0.03 \mathrm{~mJ}$.
3.132. $Q=\mathscr{E}^{2} C C_{0} /\left(2 C+C_{0}\right)$.
3.133. $Q={ }^{1 / 2} C \mathscr{E}_{2}^{2}$. It is remarkable that the result obtained is independent of $\mathscr{E}_{1}$.
\[
\text { 3.134. } W=W_{1}+W_{2}+W_{12}=\frac{1}{4 \pi \varepsilon_{0}}\left(\frac{q_{1}^{2}}{2 R_{1}}+\frac{q_{2}^{2}}{2 R_{2}}+\frac{q_{1} q_{2}}{R_{2}}\right) .
\]
3.135. (a) $W=3 q^{2} / 20 \pi \varepsilon_{0} R$; (b) $W_{1} / W_{2}=1 / 5$.
3.136. $W=\left(q^{2} / 8 \pi \varepsilon_{0} \varepsilon\right)(1 / a-1 / b)=27 \mathrm{~mJ}$.
3.137. $A=\left(q^{2} / 8 \pi \varepsilon_{0}\right)\left(1 / R_{1}-1 / R_{2}\right)$.
3.138. $A=\frac{q\left(q_{0}+q / 2\right)}{4 \pi \varepsilon_{0}}\left(\frac{1}{R_{1}}-\frac{1}{R_{2}}\right)$.
3.139. $F_{1}=\sigma^{2} / 2 \varepsilon_{0}$.
3.140. $A=\left(q^{2} / 8 \pi \varepsilon_{0}\right)(1 / a-1 / b)$.
3.141. (a) $A=q^{2}\left(x_{2}-x_{1}\right) / 2 \varepsilon_{0} S$;
(b) $A=\varepsilon_{0} S V^{2}\left(x_{2}-x_{1}\right) / 2 x_{1} x_{2}$.
3.142. (a) $A=1 / 2 C V^{2} \eta /(1-\eta)^{2}=1.5 \mathrm{~mJ}$;
(b) $A=1 / 2 C V^{2} \eta \varepsilon(\varepsilon-1) /[\varepsilon-\eta(\varepsilon-1)]^{2}=0.8 \mathrm{~mJ}$.
3.143. $\Delta p=\varepsilon_{0} \varepsilon(\varepsilon-1) V^{2} / 2 d^{2}=7 \mathrm{kPa}=0.07 \mathrm{~atm}$.
3.144. $h=(\varepsilon-1) \sigma^{2} / 2 \varepsilon_{0} \varepsilon \rho \mathrm{g}$.
3.145. $F=\pi R \varepsilon_{0}(\varepsilon-1) V^{2} / d$.
3.146. $N=(\varepsilon-1) \varepsilon_{0} R^{2} V^{2} / 4 d$.
3.147. $I=2 \pi \varepsilon_{0} a E v=0.5 \mu \mathrm{A}$.
3.148. $I \approx 2 \pi \varepsilon_{0}(\varepsilon-1) r v V / d=0.11 \mu \mathrm{A}$.
3.149. (a) $\alpha=\left(\alpha_{1}+\eta \alpha_{2}\right) /(1+\eta)$; (b) $\alpha \approx\left(\alpha_{2}+\eta \alpha_{1}\right) /(1+\eta)$.
3.150. (a) $5 / 6 R$; (b) $7 / 12 R$; (c) $3 / 4 R$.
3.151. $R_{x}=R(V \overline{3}-1)$.
3.152. $R=\left(1+\sqrt{1+4 R_{2} / R_{1}}\right) R_{1} / 2=6 \Omega$. Instruction. Since the chain is infinite, all the links beginning with the second can be replaced by the resistance equal to the sought resistance $R$.
3.153. Imagine the voltage $V$ to be applied across the points $A$ and $B$. Then $V=I R=I_{0} R_{0}$, where $I$ is the current carried by the lead wires, $I_{0}$ is the current carried by the conductor $A B$.

The current $I_{0}$ can be represented as a superposition of two currents. If the current $I$ flowed into point $A$ and spread all over the infinite wire grid, the conductor $A B$ would carry (because of symmetry) the current $I / 4$. Similarly, if the current $I$ flowed into the grid from infinity and left the grid through point $B$, the conductor $A B$ would also carry the current $I / 4$. Superposing both of these solutions, we obtain $I_{0}=I / 2$. Therefore, $R=R_{0} / 2$.
3.154. $R=(\rho / 2 \pi l) \ln (b / a)$.
3.155. $R=\rho(b-a) / 4 \pi a b$. In the case of $b \rightarrow \infty \quad R=\rho / 4 \pi a$.
3.156. $\rho=4 \pi \Delta t a b /(b-a) C \ln \eta$.
3.157. $R=\rho / 2 \pi a$.
3.158. (a) $j=2 a l V / \rho r^{3}$; (b) $R=\rho / 4 \pi a$.
3.159. (a) $j=l V / 2 \rho r^{2} \ln (l / a)$; (b) $R_{1}=(\rho / \pi) \ln (l / a)$.
3.160. $I=V C / \rho \varepsilon \varepsilon_{0}=1.5 \mu \mathrm{A}$.
3.161. $R C=\rho \varepsilon \varepsilon_{0}$.
3.162. $\sigma=D_{n}=D \cos \alpha ; j=D \sin \alpha / \varepsilon \varepsilon_{0} \rho$.
3.163. $I=V S\left(\sigma_{2}-\sigma_{1}\right) / d \ln \left(\sigma_{2} / \sigma_{1}\right)=5$ nA.
3.165. $q=\varepsilon_{0}\left(\rho_{2}-\rho_{1}\right) I$.
3.166. $\sigma=\varepsilon_{0} V\left(\varepsilon_{2} \rho_{2}-\varepsilon_{1} \rho_{1}\right) /\left(\rho_{1} d_{1}+\rho_{2} d_{2}\right), \quad \sigma=0$ if $\varepsilon_{1} \rho_{1}=$ $=\varepsilon_{2} \rho_{2}$.
3.167. $q=\varepsilon_{0} I\left(\varepsilon_{2} \rho_{2}-\varepsilon_{1} \rho_{1}\right)$.
3.168. $\rho=2 \varepsilon_{0} V(\eta-1) / d^{2}(\eta+1)$.
3.169. (a) $R_{1}=2 \pi \alpha / S^{2}$; (b) $E=2 \pi \alpha I / S^{2}$.
3.170. $t=-R C \ln \left(I-V / V_{0}\right)=0.6 \mu \mathrm{s}$.
3.171. $\rho=\tau / \varepsilon_{0} \varepsilon \ln 2=1.4 \cdot 10^{13} \Omega \cdot \mathrm{m}$.
3.172. $I=[(\eta-1) \mathscr{E} / R] \mathrm{e}^{-\eta t / R C}$.
3.173. $V=\mathscr{E} /(\eta+1)=2.0 \mathrm{~V}$.
3.174. $\varphi_{1}-\varphi_{2}=\left(\mathscr{E}_{1}-\mathscr{E}_{2}\right) R_{1} /\left(R_{1}+R_{2}\right)-\mathscr{E}_{1}=-4 \mathrm{~V}$.
3.175. $R=R_{2}-R_{1}, \Delta \varphi=0$ in the source of current with internal resistance $R_{2}$.
3.176. (a) $I=\alpha$; (b) $\varphi_{A}-\varphi_{B}=0$.
3.177. $\varphi_{A}-\varphi_{B}=\left(\mathscr{E}_{1}-\mathscr{E}_{2}\right) R_{1} /\left(R_{1}+R_{2}\right)=-0.5 \mathrm{~V}$.
3.178. $I_{1}=\mathscr{E} R_{2} /\left(R R_{1}+R_{1} R_{2}+R_{2} R\right)=1.2 \mathrm{~A}, I_{2}=I_{1} R_{1} / R_{2}=$ $=0.8 \mathrm{~A}$.
3.179. $V=V_{0} R x /\left[R l+R_{0}(l-x) x / l\right] ;$ for $R \gg R_{0} \quad V \approx$ $\approx V_{0} x / l$.
3.180. $\mathscr{E}=\left(\mathscr{E}_{1} R_{2}+\mathscr{E}_{2} R_{1}\right) /\left(R_{1}+R_{2}\right), \quad R_{i}=R_{1} R_{2} /\left(R_{1}+R_{2}\right)$.
3.181. $I=\left(R_{1} \mathscr{E}_{2}-R_{2} \mathscr{E}_{1}\right) /\left(R R_{1}+R_{1} R_{2}+R_{2} R\right)=0.02 \mathrm{~A}$, the current is directed from the left to the right (see Fig. 3.44).
3.182. (a) $I_{1}=\left[R_{3}\left(\mathscr{E}_{1}-\mathscr{E}_{2}\right)+R_{2}\left(\mathscr{E}_{1}+\mathscr{E}_{3}\right)\right] /\left(R_{1} R_{2}+R_{2} R_{3}+\right.$ $\left.+R_{3} R_{1}\right)=0.06 \mathrm{~A} ;$ (b) $\varphi_{A}-\varphi_{B}=\mathscr{E}_{1}-I_{1} R_{1}=0.9 \mathrm{~V}$.
3.183. $I=\left[\mathscr{E}\left(R_{2}+R_{3}\right)+\mathscr{E}_{0} R_{3}\right] /\left[R\left(R_{2}+R_{3}\right)+R_{2} R_{3}\right]$.
3.184. $\varphi_{\mathrm{A}}-\varphi_{B}=\left[\mathscr{E}_{2} R_{3}\left(R_{1}+R_{2}\right)-\mathscr{E}_{1} R_{1}\left(R_{2}+R_{3}\right)\right] /\left(R_{1} R_{2}+\right.$ $\left.+R_{2} R_{3}+R_{3} R_{1}\right)=-1.0 \mathrm{~V}$.
3.185. $I_{1}=\left[R_{3}\left(\varphi_{1}-\varphi_{2}\right)+R_{2}\left(\varphi_{1}-\varphi_{3}\right)\right] /\left(R_{1} R_{2}+R_{2} R_{3}+\right.$ $\left.+R_{3} R_{1}\right)=0.2 \mathrm{~A}$.
3.186. $I=\frac{V}{R_{2}}\left(\frac{R_{1}+R_{2}}{R_{1}\left[1+R_{2} R_{4}\left(R_{1}+R_{3}\right) / R_{1} R_{3}\left(R_{2}+R_{4}\right)\right]}-1\right)=1.0 \mathrm{~A}$. The current flows from point $C$ to point $D$.
3.187. $R_{A B}=r(r+3 R) /(R+3 r)$.
3.188. $V=1 / 2 \mathscr{E}\left(1-\mathrm{e}^{-2 t / R C}\right)$.
3.189. (a) $Q=4 / 3 q^{2} R / \Delta t$; (b) $Q=1 / 2 \ln 2 \cdot q^{2} R / \Delta t$.
3.190. $R=3 R_{0}$.
3.192. $Q=I(\mathscr{E}-V)=0.6 \mathrm{~W}, P=-I V=-2.0 \mathrm{~W}$.
3.193. $I=V / 2 R ; P_{\max }=V^{2} / 4 R ; \eta=1 / 2$.
3.194. By $2 \eta=2 \%$.
3.195. $T-T_{0}=\left(1-\mathrm{e}^{-k t / C}\right) V^{2} / k R$.
3.196. $R_{x}=R_{1} R_{2} /\left(R_{1}+R_{2}\right)=12 \Omega$.
3.197. $R^{x}=R_{1} R_{2} /\left(R_{1}+R_{2}\right)$;
3.198. $n=\sqrt{N r / R}=3$.
3.199. $Q=1 /{ }_{2} C \mathscr{E}^{2} R_{1} /\left(R_{1}+R_{2}\right)=60 \mathrm{~mJ}$.
3.200. (a) $\Delta W=-1 / 2 C V^{2} \eta /(1-\eta)=-0.15 \mathrm{~mJ}$; (b) $A=$ $=1 / 2 C V^{2} \eta /(1-\eta)=0.15 \mathrm{~mJ}$.
3.201. $\Delta W=-1 / 2(\varepsilon-1) C V^{2}=-0.5 \mathrm{~mJ}$, $=1 / 2(\varepsilon-1) C V^{2}=0.5 \mathrm{~mJ}$.
3.202. $h \approx 1 / 2 \varepsilon_{0}(\varepsilon-1) V^{2} / \rho g d^{2}$, where $\rho$ is the density of water.
3.203. (a) $q=q_{0} \mathrm{e}^{-t / \varepsilon_{0} \varepsilon \rho} ;$ (b) $Q=(1 / a-1 / b) q_{0}^{2} / 8 \pi \varepsilon_{0} \varepsilon$.
3.204. (a) $q=q_{0}\left(1-\mathrm{e}^{-\tau / R C}\right)=0.18 \mathrm{mC}$; $=\left(1-\mathrm{e}^{-2 \tau / R C}\right) q_{0}^{2} / 2 C=82 \mathrm{~mJ}$.
(b) $Q=$
3.205. (a) $I=\left(V_{0} / R\right) \mathrm{e}^{-2 t / R C}$; (b) $Q=1 / 4 C V_{0}^{2}$.
3.206. $e / m=l \omega r / q R=1.8 \cdot 10^{11} \mathrm{C} / \mathrm{kg}$.
3.207. $p=l$ Im $/ e=0.40 \mu \mathrm{N} \cdot \mathrm{s}$.
3.208. $s=$ enl $\langle v\rangle / j \sim 10^{7} \mathrm{~m}$, where $n$ is the concentration of free electrons, $\langle v\rangle$ is the mean velocity of thermal motion of an electron.
3.209. (a) $t=e n l S / I=3 \mathrm{Ms}$;
(b) $F=e n l_{\rho} I=1.0 \mathrm{MN}$, where $\rho$ is the resistivity of copper.
3.210. $E=\left(I / 2 \pi \varepsilon_{0} r\right) V \overline{m / 2 e V}=32 \mathrm{~V} / \mathrm{m}, \Delta \varphi=\left(I / 4 \pi \varepsilon_{0}\right) V \overline{m / 2 e V}=$ $=0.80 \mathrm{~V}$.
3.211. (a) $\rho(x)=-4 / 9 \varepsilon_{0} a x^{-2 / 3}$;
(b) $j=4 / 9 \varepsilon_{0} a^{3 / 2} V \overline{2 e / m}$.
3.212. $n=I d / e\left(u_{0}^{+}+u_{0}^{-}\right) V S=2.3 \cdot 10^{8} \mathrm{~cm}^{-3}$.
3.213. $u_{0}=\omega_{0} l^{2} / 2 V_{0}$.
3.214. (a) $\dot{n}_{i}=I_{s a t} / e V=6 \cdot 10^{9} \mathrm{~cm}^{-3} \cdot \mathrm{s}^{-1}$;
(b) $n=\sqrt{n_{i} / r}=$ $=6 \cdot 10^{7} \mathrm{~cm}^{-3}$.
3.216. $t=\varepsilon_{0} \eta U / \dot{e}_{i} d^{2}=4.6$ days.
3.217. $I=e v_{0} \mathrm{e}^{\alpha d}$.
3.218. $j=\left(e^{\alpha d}-1\right) e n_{i} / \alpha$.
3.219. (a) $B=\mu_{0} I / 2 R=6.3 \mu \mathrm{T}$; (b) $B=\mu_{0} R^{2} I / 2\left(R^{2}+x^{2}\right)^{3 / 2}=$ $=2.3 \mu \mathrm{T}$.
3.220. $B=n \mu_{0} I \tan (\pi / n) / 2 \pi R$, for $n \rightarrow \infty \quad B=\mu_{0} I / 2 R$
3.221. $B=4 \mu_{0} I / \pi d \sin \varphi=0.10 \mathrm{mT}$.
3.222. $B=(\pi-\varphi+\tan \varphi) \mu_{0} I / 2 \pi R=28 \mu \mathrm{T}$.
3.223. (a) $B=\frac{\mu_{0} I}{4 \pi}\left(\frac{2 \pi-\varphi}{a}+\frac{\varphi}{b}\right)$;
(b) $B=\frac{\mu_{0} I}{4 \pi}\left(\frac{3 \pi}{4 a}+\frac{\sqrt{2}}{b}\right)$.
3.224. $B \approx \mu_{0} h I / 4 \pi^{2} R r$, where $r$ is the distance from the cut.
3.225. $B=\mu_{0} I / \pi^{2} R$.
3.226. (a) $B=\left(\mu_{0} / 4 \pi\right)(\pi I / R)$; (b) $B=\left(\mu_{0} / 4 \pi\right)(1+3 \pi / 2) I / R$; (c) $B=\left(\mu_{0} / 4 \pi\right)(2+\pi) I / R$.
3.227. $B=\left(\mu_{0} / 4 \pi\right) I V \overline{2} / l=2.0 \mu \mathrm{T}$.
3.228. (a) $B=\left(\mu_{0} / 4 \pi\right) \sqrt{4+\pi^{2}} I / R=0.30 \mu \mathrm{T}$; (b) $B=\left(\mu_{0} / 4 \pi\right) \times$ $\times \sqrt{2+2 \pi+\pi^{2}} I / R=0.34 \mu \mathrm{T}$; (c) $B=\left(\mu_{0} / 4 \pi\right) V \overline{2} I / R=0.11 \mu \mathrm{T}$.
3.229. (a) $B=\mu_{0} i / 2$; (b) $B=\mu_{0} i$ between the planes and $B=0$ outside the planes.
3.230. $B=\left\{\begin{array}{l}\mu_{\theta} j x \text { inside the plate, } \\ \mu_{0} j d \text { outside the plate. }\end{array}\right.$
3.231. In the half-space with the straight wire, $B=\mu_{0} I / 2 \pi r$, $r$ is the distance from the wire. In the other half-space $B \equiv 0$.
3.232. The given integral is equal to $\mu_{0} I$.
3.233. $\mathbf{B}=\left\{\begin{array}{l}1 / 2 \mu_{0}[\mathrm{jr}] \text { for } r \leqslant R, \\ 1 / 2 \mu_{0}[\mathrm{jr}] R^{2} / r^{2} \text { for } r \geqslant R .\end{array}\right.$
3.234. $\mathbf{B}=1 / 2 \mu_{0}[j l]$, i.e. field inside the cavity is uniform.
3.235. $j(r)=\left(b / \mu_{0}\right)(1+\alpha) r^{\alpha-1}$.
3.236. $B=\mu_{0} n I / \sqrt{1+(2 R / l)^{2}}$.
3.237. (a) $B=1 / 2 \mu_{0} n I\left(1-x / \sqrt{x^{2}+R^{2}}\right)$, where $x>0$ outside the solenoid and $x<0$ inside the solenoid; see Fig. 23 ; (b) $x_{0}=$ $=R(1-2 \eta) / 2 \sqrt{\eta(1-\eta)} \approx 5 R$.
3.238. $B=\left\{\begin{array}{l}\left.\left(\mu_{0} I / h\right) \sqrt{(1-(h / 2 \pi} \vec{R}\right)^{2}=0.3 \mathrm{mT}, r<R, \\ \left(\mu_{0} / 4 \pi\right) 2 I / r, r>R .\end{array}\right.$
3.239. $\eta \approx N / \pi=8 \cdot 10^{2}$.
3.240. $\Phi=\left(\mu_{0} / 4 \pi\right) I=$ $=1.0 \mu \mathrm{Wb} / \mathrm{m}$.
3.241. $\Phi=\Phi_{0} / 2=\mu_{0} n I S / 2$,
where $\Phi_{0}$ is the flux of the vector B through the cross-section of the solenoid far from its ends.
3.242. $\Phi=\left(\mu_{0} / 4 \pi\right) 2 I N h \ln \eta=$ $=8 \mu \mathrm{Wb}$.
3.243. $p_{m}=2 \pi R^{3} B / \mu_{0}=$ $=30 \mathrm{~mA} \cdot \mathrm{m}^{2}$.
3.244. $p_{m}=1 / 2 \mathrm{NI} d^{2}=$ $=0.5 \mathrm{~A} \cdot \mathrm{m}^{2}$.
3.245. (a) $B=\frac{\mu_{0} I N \ln (b / a)}{2(b-a)}=7 \mu \mathrm{T}$;
(b) $p_{m}=1 / 3 \pi I N\left(a^{2}+a b+b^{2}\right)=15 \mathrm{~mA} \cdot \mathrm{m}^{2}$.
3.246. (a) $B=1 / 2 \mu_{0} \sigma \omega R$; (b) $p_{m}=1 / 4 \pi \sigma \omega R^{4}$.
3.247. $B=2 / 3 \mu_{0} \sigma \omega R=29$ pT.
3.248. $p_{m}=1{ }_{5} q R^{2} \omega ; p_{m} / M=q / 2 m$.
3.249. $\mathbf{B}=0$.
3.250. $F_{\mathrm{m}} / F_{\mathrm{e}}=\mu_{0} \varepsilon_{0} v^{2}=(v / c)^{2}=1.00 \cdot 10^{-6}$.
3.251. (a) $F_{1}=\mu_{0} I^{2} / 4 R=0.20 \mathrm{mN} / \mathrm{m}$;
(b) $F_{1}=\mu_{0} I^{2} / \pi l=$ $=0.13 \mathrm{mN} / \mathrm{m}$.
3.252. $B=\pi d^{2} \sigma_{m} / 4 R I=8 \mathrm{kT}$, where $\sigma_{m}$ is the strength of copper.
3.253. $B=(2 \rho g S / I) \tan \theta=10 \mathrm{mT}$, where $\rho$ is the density of copper.
3.254. $B=\Delta m g l / N I S=0.4 \mathrm{~T}$.
3.255. (a) $F=2 \mu_{0} I I_{0} / \pi\left(4 \eta^{2}-1\right)=0.40 \mu \mathrm{N}$;
(b) $A=$ $=\left(\mu_{0} a I I_{0} / \pi\right) \ln [(2 \eta+1) /(2 \eta-1)]=0.10 \mu \mathrm{J}$.
3.256. $R \approx \sqrt{\mu_{0} / \varepsilon_{0}}(\ln \eta) / \pi=0.36 \mathrm{k} \Omega$.
3.257. $F_{1}=\mu_{0} I^{2} / \pi^{2} R$.
3.258. $F_{1}=\frac{\mu_{0}}{4 \pi} \frac{2 I_{1} I_{2}}{b} \ln (1+b / a)$.
3.259. $F_{1}=B^{2} / 2 \mu_{0}$.
3.260. In all three cases $F_{1}=\left(B_{1}^{2}-B_{2}^{2}\right) / 2 \mu_{0}$. The force is directed to the right. The current in the conducting plane is directed beyond the drawing.
3.261. $\Delta p=I B / a=0.5 \mathrm{kPa}$.
3.262. $p=\mu_{0} I^{2} / 8 \pi^{2} R^{2}$.
3.263. $p=1 / 2 \mu_{0} n^{2} I^{2}$.
316
3.264. $I_{l_{i m}}=\sqrt{2 F_{l i m} / \mu_{0} n R}$.
3.265. $P \stackrel{ }{=} v^{2} B^{2} d^{2} R /(R+\rho d / S)^{2}$; when $R=\rho d / S$, the power is $P=P_{\text {max }}=1 /{ }_{4} v^{2} B^{2} d S / \rho$.
3.266. $\stackrel{\max }{U}=1 / 4 \mu_{0} I^{2} / \pi^{2} R^{2} n e=2 \mathrm{pV}$.
3.267. $n=j B / e E=2.5 \cdot 10^{28} \mathrm{~m}^{-1}$; almost $1: 1$.
3.268. $u_{0}=1 / \eta B=3.2 \cdot 10^{-3} \mathrm{~m}^{2} /(\mathrm{V} \cdot \mathrm{s})$.
3.269. (a) $F=0 ;$ (b) $F=\left(\mu_{0} / 4 \pi\right) 2 I p_{m} / r^{2}, \quad \mathbf{F}$ \& ; (c) $F=$ $=\left(\mu_{0} / 4 \pi\right) 2 I p_{m} / r^{2}, \mathbf{F} \uparrow_{t} \mathbf{r}$.
3.270. $F=\left(\mu_{0} / 4 \pi\right) 6 \pi R^{2} I p_{m} x /\left(R^{2}+x^{2}\right)^{5 / 2}$.
3.271. $F=3 / 2 \mu_{0} p_{1 m} p_{2 m} / \pi l^{4}=9 \mathrm{nN}$.
3.272. $I^{\prime} \approx 2 B x^{3} / \mu_{0} R^{2}=0.5 \mathrm{kA}$.
3.273. $B^{\prime}=B \sqrt{\mu^{2} \sin ^{2} \alpha+\cos ^{2} \alpha}$.
3.274. (a) $\oint \mathbf{H} d \mathbf{S}=\pi R^{2} B \cos \theta \cdot(\mu-1) / \mu \mu_{0}$;
(b) $\oint \mathbf{B} d \mathbf{r}=(1-\mu) B l \sin \theta$.
3.275. (a) $I_{\text {sur }}^{\prime}=\chi I$; (b) $I_{\text {vol }}^{\prime}=\chi I$; in opposite directions.
3.276. See Fig. 24.
3.277. $B=\frac{\mu_{0} \mu_{1} \mu_{2}}{\mu_{1}+\mu_{2}} \frac{I}{\pi r}$.
3.278. $\mathbf{B}=2 \mathbf{B}_{0} \mu /(1+\mu)$.
3.279. $B=3 B_{0} \mu /(2+\mu)$.
3.280. $H_{c}=N I / l=6 \mathrm{kA} / \mathrm{m}$.
3.281. $H \approx b B / \mu_{0} \pi d=$ $=0.10 \mathrm{kA} / \mathrm{m}$.
3.282. When $b \ll R$, the permeability is $\mu \approx 2 \pi R B /\left(\mu_{0} N I-\right.$ $-b B)=3.7 \cdot 10^{3}$.
3.283. $H=0.06 \mathrm{kA} / \mathrm{m}, \mu_{\max } \approx 1.0 \cdot 10^{4}$.
3.284. From the theorem on circulation of the vector $\mathbf{H}$ we obtain
\[
B \approx \frac{\mu_{0} N I}{b}-\frac{\mu_{0} \pi d}{b} H=1.51-0.987 H(\mathrm{kA} / \mathrm{m}) .
\]

Besides, $B$ and $H$ are interrelated as shown in Fig. 3.76. The required values of $H$ and $B$ must simultaneously satisfy both relations. Solving this system of equations by means of plotting, we obtain $H \approx 0.26 \mathrm{kA} / \mathrm{m}, B \approx 1.25 \mathrm{~T}$, and $\mu=B / \mu_{0} H \approx 4 \cdot 10^{3}$.
3.285. $F \approx 1 / 2 \chi S B^{2} / \mu_{0}$.
3.286. (a) $x_{m}=1 / V \overline{4 a}$; (b) $\chi=\mu_{0} F_{\text {max }} \sqrt{\mathrm{e} / a} / V B_{0}^{2}=3.6 \cdot 10^{-4}$.
3.287. $A \approx 1 / 2 \chi V B^{2} / \mu_{0}$.
3.288. $\mathscr{E}_{i}=B y \sqrt{8 w / a}$.
3.289. $I=B v l /\left(R+R_{\mu}\right)$, where $R_{\mu}=R_{1} R_{2} /\left(R_{1}+R_{2}\right)$.
3.290. (a) $\Delta \varphi=1 / 2 \omega^{2} a^{2} \mathrm{~m} / \mathrm{e}=3.0 \mathrm{nV}$;
(b) $\Delta \varphi \approx 1 / 2 \omega B a^{2}=$ $=20 \mathrm{mV}$.
3.291. $\int_{A}^{c} \mathbf{E} d \mathbf{r}=-1 /{ }_{2} \omega B d^{2}=-10 \mathrm{mV}$.
Fig. 24.
3.292. $\mathscr{E}_{i}=1 / 2(-1)^{n} B a \beta t$, where $n=1,2, \ldots$ is the number of the half-revolution that the loop performs at the given moment $t$. The plot $\mathscr{E}_{i}(t)$ is shown in Fig. 25 where $t_{n}=\sqrt{2 \pi n / \beta}$.
3.293. $I_{\text {ind }}=\alpha / r$, where $\alpha=1 / 2 \mu_{0} l v I / \pi R$.
3.294. $\mathscr{E}_{i}=\frac{\mu_{0}}{4 \pi} \frac{2 I a^{2} v}{x(x+a)}$.
3.295. $\mathscr{E}_{i}=1 / 2\left(\omega a^{3} B^{3}+2 m g \sin \omega t\right) / a B$.
3.296. $v=\frac{m g R \sin \alpha}{B^{2} l^{2}}$.
3.297. $w=\frac{g \sin \alpha}{1+l^{2} B^{2} C / m}$.
3.298. $\langle P\rangle=1 / 2\left(\pi \omega a^{2} B\right)^{2} / R$.
3.299. $B=1 / 2 q R / N S=0.5 \mathrm{~T}$.
3.300. $q=\frac{\mu_{0} a I}{2 \pi R} \ln \frac{b+a}{b-a}$, i.e. is independent of $L$.
Fig. 25.
3.301. (a) $I=\frac{\mu_{0} I_{0} v}{2 \pi R} \ln \frac{b}{a}$;
(b) $F=\frac{v}{R}\left(\frac{\mu_{0} I_{0}}{2 \pi} \ln \frac{b}{a}\right)^{2}$.
3.302. (a) $s=v_{0} m R / l^{2} B^{2}$,
(b) $Q=1 / 2 m v_{0}^{2}$.
3.303. $v=\frac{F}{\alpha m}\left(1-\mathrm{e}^{-\alpha t}\right)$, where $\alpha=B^{2} l^{2} / m R$.
3.304. (a) In the round conductor the current flows clockwise, there is no current in the connector; (b) in the outside conductor, clockwise; (c) in both round conductors, clockwise; no current in the connector, (d) in the left-hand side of the figure eight, clockwise.
3.305. $I=1 /{ }_{4} \omega B_{0}(a-b) / \rho=0.5$ A.
3.306. $\mathscr{E}_{i m}=1 / 3 \pi a^{2} N \omega B_{0}$.
3.307. $\mathscr{E}_{i}=3 /{ }_{2} w l \dot{B} t^{2}=12 \mathrm{mV}$.
3.308. $E=\left\{\begin{array}{ll}1 / 2 \mu_{0} n \dot{I} r & \text { for } r<a, \\ 1 / 2 \mu_{0} n \dot{I} a^{2} / r & \text { for } r>a .\end{array}\right.$
3.309. $I=1 / 4 \mu_{0} n S d \dot{I} / \rho=2 \mathrm{~mA}$, where $\rho$ is the resistivity of copper.
3.310. $E=1 / 2 a b(\eta-1) /(\eta+1)$.
3.311. $\boldsymbol{\omega}=-\frac{q}{2 m} \mathbf{B}(t)$.
3.312. $F_{1 \max }=\frac{\mu_{0} a^{2} V^{2}}{4 r R l b^{2}}$.
3.313. $Q=1 / a^{2} \tau^{3} / R$.
3.314. $I=1 / 4\left(b^{2}-a^{2}\right) \beta h / \rho$.
3.315. $l=\sqrt{4 \pi l_{0} L / \mu_{0}}=0.10 \mathrm{~km}$.
3.316. $L=\frac{\mu_{0}}{4 \pi} \frac{m R}{l \rho \rho_{0}}$, where $\rho$ and $\rho_{0}$ are the resistivity and the density of copper.
3.317. $t=-\frac{L}{R} \ln (1-\eta)=1.5 \mathrm{~s}$.
3.318. $\tau=\frac{\mu_{0}}{4 \pi} \frac{m}{l \rho \rho_{0}}=0.7 \mathrm{~ms}$, where $\rho$ is the resistivity, $\rho_{0}$ is the density of copper.
3.319. $L_{1}=\frac{\mu \mu_{0}}{2 \pi} \ln \eta=0.26 \mu \mathrm{H} / \mathrm{m}$.
3.320. $L=\frac{\mu_{0}}{2 \pi} \mu N^{2} a \ln \left(1+\frac{a}{b}\right)$.
3.321. $L_{1}=\mu_{0} h / b=25 \mathrm{nH} / \mathrm{m}$.
3.322. $L_{1} \approx \frac{\mu_{0}}{\pi} \ln \eta$.
3.323. (a) $I=\pi a^{2} B / L$; (b) $A=1 / 2 \pi^{2} a^{4} B^{2} / L$.
3.324. $I=I_{0}(1+\eta)=2 \mathrm{~A}$.
3.325. $I=\frac{\pi a B}{\mu_{0}\left(\ln \frac{8 a}{b}-2\right)}=50 \mathrm{~A}$.
3.326. $I=\frac{\mathscr{E}}{R}\left[1+(\eta-1) \mathrm{e}^{-t \eta R / L}\right]$.
3.327. $I=\frac{\mathscr{E}}{R}\left(1-\mathrm{e}^{-t R / 2 L}\right)$.
3.328. $I_{1}=\frac{\mathscr{E} L_{2}}{R\left(L_{1}+L_{2}\right)}, \quad I_{2}=\frac{\mathscr{E} L_{1}}{R\left(L_{1}+L_{2}\right)}$.
3.329. $L_{12}=\frac{\mu_{0} b}{2 \pi} \ln \left(1+\frac{a}{l}\right)$.
3.330. $L_{12}=\frac{\mu_{0} \mu h N}{2 \pi} \ln \frac{b}{a}$.
3.331. (a) $L_{12} \approx 1 / 2 \mu_{0} \pi a^{2} / b$;
(b) $\Phi_{21}=1 / 2 \mu_{0} \pi a^{2} I / b$.
3.332. $p_{m}=2 a R q / \mu_{0} N$.
3.333. $L_{12} \approx 1 / 2 \mu_{0} \pi a^{4} / l^{3}$.
3.334. $I_{2}=\frac{\alpha L_{12}}{R}\left(1-\mathrm{e}^{-t R / L_{2}}\right)$.
3.335. $Q=\frac{L \mathscr{E}^{2}}{2 R^{2}\left(1+R_{0} / R\right)}=3 \mu \mathrm{J}$.
3.336. $W=1 / 2 N \Phi I=0.5 \mathrm{~J}$.
3.337. $W=B H \pi^{2} a^{2} b=2.0 \mathrm{~J}$, where $H=1 / 2 \mathrm{NI} / \pi b$.
3.338. (a) $W_{g a p} / W_{m} \approx \mu b / \pi d=3.0$;
(b) $L \approx \frac{\mu_{0} S N^{2}}{b+\pi d / \mu}=0.15 \mathrm{H}$.
3.339. $W_{1}=\mu_{0} \lambda^{2} \omega^{2} a^{2} / 8 \pi$.
3.340. $E=B / \sqrt{\varepsilon_{0} \mu_{0}}=3 \cdot 10^{8} \mathrm{~V} / \mathrm{m}$.
3.341. $w_{m} / w_{e}=\varepsilon_{0} \mu_{0} \omega^{2} a^{4} / l^{2}=1.1 \cdot 10^{-15}$.
3.343. (a) $L_{\text {total }}=2 L$; (b) $L_{\text {total }}=L / 2$.
3.344. $L_{12}=\sqrt{L_{1} L_{2}}$.
3.346. $W_{12}=\frac{\mu_{0} \pi a^{2}}{2 b} I_{1} I_{2} \cos \theta$.
3.347. (a) $\mathbf{j}_{d}=-\mathbf{j}$; (b) $I_{d}=q / \varepsilon_{0} \varepsilon \rho$.
3.348. The displacement current should be taken into account in addition to the conduction current.
3.349. $E_{m}=I_{m} / \varepsilon_{0} \omega S=7 \mathrm{~V} / \mathrm{cm}$.
3.350. $H=H_{m} \cos (\omega t+\alpha)$, where $H_{m}=\frac{r V_{m}}{2 d} \sqrt{\sigma^{2}+\left(\varepsilon_{0} \varepsilon \omega\right)^{2}}$ and $\alpha$ is determined from the formula $\tan \alpha=\varepsilon_{0} \varepsilon \omega / \sigma$.
3.351. $j_{d}=\left\{\begin{array}{ll}1 / 2 \ddot{B} r & \text { for } r<R, \\ 1 / 2 \ddot{B} R^{2} / r & \text { for } r>R .\end{array}\right.$

Here $\ddot{B}=\mu_{0} n I_{m} \omega^{2} \sin \omega t$.
3.352. (a) $\mathbf{j}_{d}=\frac{2 q \mathbf{v}}{4 \pi r^{3}}$;
(b) $\mathbf{j}_{d}=-\frac{q \mathbf{v}}{4 \pi r^{8}}$.
3.353. $x_{m}=0, j_{d \max }=\frac{q v}{4 \pi a^{3}}$.
3.354. $\mathbf{H}=\frac{q[\mathbf{v r}]}{4 \pi r^{3}}$.
3.355. (a) If $\mathbf{B}(t)$, then $\boldsymbol{
abla} \times \mathbf{E}=-\partial \mathbf{B} / \partial t
eq 0$. The spatial derivatives of the field $\mathbf{E}$, however, may not be equal to zero $(\boldsymbol{
abla} \times \mathbf{E}
eq 0)$ only in the presence of an electric field.
(b) If $\mathbf{B}(t)$, then $\boldsymbol{
abla} \times \mathbf{E}=-\partial \mathbf{B} / \partial t
eq 0$. But in the uniform field $
abla \times \mathbf{E}=0$.
(c) It is assumed that $\mathbf{E}=\mathbf{a} f(t)$, where a is a vector which is independent of the coordinates, $f(t)$ is an arbitrary function of time. Then $-\partial \mathbf{B} / \partial t=\boldsymbol{
abla} \times \mathbf{E}=0$, that is the field $\mathbf{B}$ does not vary with time. Generally speaking, this contradicts the equation $\boldsymbol{
abla} \times \mathbf{H}=$ $=\partial \mathrm{D} / \partial t$ for in this case its left-hand side does not depend on time whereas its right-hand side does. The only exception is the case when $f(t)$ is a linear function. In this case the uniform field $\mathbf{E}$ can be time-dependent.
3.356. Let us find the divergence of the two sides of the equation $\boldsymbol{
abla} \times \mathbf{H}=\mathbf{j}+\partial \mathbf{D} / \partial t$. Since the divergence of a rotor is always equal to zero, we get $0=\boldsymbol{
abla} \cdot \mathbf{j}+\frac{\partial}{\partial t}(\mathbf{V} \cdot \mathbf{D})$. It remains to take into account that $\mathbf{V} \cdot \mathbf{D}=\rho$.
3.357. Let us consider the divergence of the two sides of the first equation. Since the divergence of a rotor is always equal to zero, $\boldsymbol{
abla} \cdot(\partial \mathbf{B} / \partial t)=0$ or $\frac{\partial}{\partial t}(\boldsymbol{
abla} \cdot \mathbf{B})=0$. Hence, $\boldsymbol{
abla} \cdot \mathbf{B}=$ const which does not contradict the second equation.
3.358. $\boldsymbol{
abla} \times \mathbf{E}=-[\omega \mathbf{B}]$.
3.359. $\mathbf{E}^{\prime}=[\mathbf{v B}]$.
3.360. $\sigma=\varepsilon_{0} v B=0.40 \mathrm{pC} / \mathrm{m}^{2}$.
3.361. $\rho=-2 \varepsilon \omega \mathrm{B}=-0.08 \mathrm{nC} / \mathrm{m}^{3}, \quad \sigma=\varepsilon_{0} a \omega \mathrm{B}=2 \mathrm{pC} / \mathrm{m}^{2}$.
3.362. $\mathbf{B}=\frac{\mu_{0}}{4 \pi} \frac{q[\mathbf{v r}]}{r^{3}}$.
3.364. $\mathbf{E}^{\prime}=b r / r^{2}$, where $r$ is the distance from the $z^{\prime}$ axis.
3.365. $\mathbf{B}^{\prime}=\frac{a[\mathbf{r v}]}{c^{2} r^{2}}$, where $r$ is the distance from the $z^{\prime}$ axis.
3.367. (a) $E^{\prime}=E \sqrt{\frac{1-\beta^{2} \cos ^{2} \alpha}{1-\beta^{2}}}=9 \mathrm{kV} / \mathrm{m} ; \quad \tan \alpha^{\prime}=\frac{\tan \alpha}{\sqrt{1-\beta^{2}}}$, whence $\alpha \approx 51^{\circ}$;
(b) $B^{\prime}=\frac{\beta E \sin \alpha}{c \sqrt{1-\beta^{2}}}=14 \mu \mathrm{T}$.
3.368. (a) $E^{\prime}=\frac{\beta B \sin \alpha}{c \sqrt{1-\bar{\beta}^{2}}}=1.4 \mathrm{nV} / \mathrm{m}$;
(b) $B^{\prime}=B \sqrt{\frac{1-\beta^{2} \cos ^{2} \alpha}{1-\beta^{2}}}=0.9 \mathrm{~T}, \alpha^{\prime} \approx 51^{\circ}$.
3.370. $B^{\prime}=B \sqrt{1-(E / c B)^{2}} \approx 0.15 \mathrm{mT}$.
3.371. Suppose the charge $q$ moves in the positive direction of the $x$ axis of the reference frame $K$. Let us pass into the frame $K^{\prime}$ at whose origin of coordinates this charge is at rest (the $x$ and $x^{\prime}$ axes of the two frames coincide and the $y$ and $y^{\prime}$ axes are parallel). In the frame $K^{\prime}$ the field of the charge has the simplest form: $\mathbf{E}^{\prime}=$ $=\frac{1}{4 \pi \varepsilon_{0}} \frac{q}{r^{\prime 3}} \mathbf{r}^{\prime}$, with the following components in the plane $x, y$
\[
E_{x}^{\prime}=\frac{1}{4 \pi \varepsilon_{0}} \frac{q}{r^{\prime 3}} x^{\prime}, \quad E_{y}^{\prime}=\frac{1}{4 \pi \varepsilon_{0}} \frac{q}{r^{\prime 3}} y^{\prime} .
\]

Now let us make the reverse transition to the initial frame $K$. At the moment when the charge $q$ passes through the origin of coordinates of the frame $K$, the $x$ and $y$ projections of the vector $\mathbf{r}$ are related to the $x^{\prime}$ and $y^{\prime}$ projections of the vector $\mathbf{r}^{\prime}$ as
\[
x=r \cos \theta=x^{\prime} \sqrt{1-(v / c)^{2}}, \quad y=r \sin \theta=y^{\prime} .
\]

Besides, in accordance with the formulas that are reciprocal to Eqs. (3.6i),
\[
E_{x}=E_{x}^{\prime}, \quad E_{y}=E_{y}^{\prime} / \sqrt{1-(v / c)^{2}} .
\]

Solving simultaneously all these equations, we obtain
\[
\mathbf{E}=E_{x} \mathbf{i}+E_{y} \mathrm{j}=\frac{1}{4 \pi \varepsilon_{0}} \frac{q \mathbf{r}}{r^{3}} \frac{1-\beta^{2}}{\left(1-\beta^{2} \sin ^{2} \theta\right)^{3 / 2}} .
\]

Note that in this case $(\mathbf{v}=$ const) the vector $\mathbf{E}$ is collinear with the vector $\mathbf{r}$.
3.372. $v=\sqrt[3]{9 / 2 \text { ale } / m}=16 \mathrm{~km} / \mathrm{s}$.
3.373. $\tan \alpha=\frac{a l^{2}}{4} \sqrt{\frac{m}{2 e V^{3}}}$.
3.374. (a) $x=2 E_{0} / a$; (b) $w=q E_{0} / m$.
3.375. $t=\frac{\sqrt{T\left(T+2 m_{0} C^{2}\right)}}{c e E}=3.0 \mathrm{~ns}$.
3.376. $w=\frac{e E}{m_{0}\left(1+T / m_{0} c^{2}\right)^{3}}$.
3.377. (a) $\tan \theta=\frac{e E t}{m_{0} v_{0}} \sqrt{1-\left(v_{0} / c\right)^{2}}$, where $e$ and $m_{0}$ are the charge and the mass of a proton; (b) $v_{x}=v_{0} / \sqrt{1+\left(1-v_{0}^{2} / c^{2}\right)\left(e E t / m_{0} c^{2}\right)^{2}}$.
3.378. $\alpha=\arcsin \left(d B \sqrt{\frac{q}{2 m V}}\right)=30^{\circ}$.
3.379. (a) $v=r e B / m=100 \mathrm{~km} / \mathrm{s}, T=2 \pi \mathrm{m} / e B=6.5 \mu \mathrm{s}$; (b) $v=$
\[
=c / \sqrt{1+\left(m_{0} c / r e B\right)^{2}}=0.51 \mathrm{c}, \quad T=\frac{2 \pi m_{0}}{e B \sqrt{1-\left(v / c^{2}\right)}}=4.1 \mathrm{~ns} .
\]
3.380. (a) $p=q r B ; \quad T=m_{0} c^{2}\left(\sqrt{1+\left(q r \bar{B} / m_{0} c\right)^{2}}-1\right)$;
(c) $w=$ $=\frac{c^{2}}{r\left[1+\left(m_{0} c / q r B\right)^{2}\right]}$.
3.381. $T=\eta m_{0} c^{2}, 5 \mathrm{keV}$ and $9 \mathrm{MeV}$ respectively.
3.382. $\Delta l=2 \pi \sqrt{2 m V / e B^{2}} \cos \alpha=2.0 \mathrm{~cm}$.
3.383. $q / m=\frac{8 \pi^{2} V}{l^{2}\left(B_{2}-B_{1}\right)^{2}}$.
3.384. $r=2 \rho|\sin (\varphi / 2)|$, where $\rho=\frac{m v}{e B} \sin \alpha, \varphi=\frac{l e B}{m v \cos \alpha}$.
3.385. $r_{\max }=a \mathrm{e}^{v_{0} / b}$, where $b=\frac{\mu_{0}}{2 \pi} \frac{e}{m} I$.
3.386. $v=\frac{V}{r B \ln (b / a)}, \quad q / m=\frac{V}{r^{2} B^{2} \ln (b / a)}$.
3.387. (a) $y_{n}=\frac{2 \pi^{2} m E n^{2}}{q B^{2}}$;
(b) $\tan \alpha=\frac{v_{0} B}{2 \pi E n}$.
3.388. $z=l \tan \sqrt{\frac{q B^{2}}{2 m E}} y ;$ for $z \ll 1$ this equation reduces to $y=\left(2 m E / q l^{2} B^{2}\right) z^{2}$.
3.389. $F=m E I / q B=20 \mu \mathrm{N}$.
3.390. $\Delta l=\frac{2 \pi m E}{e B^{2}} \tan \varphi=6 \mathrm{~cm}$.
3.391. $q / m=\frac{a(a+2 b) B^{2}}{2 E \Delta x}$.
3.392. (a) $x=a(\omega t-\sin \omega t) ; y=a(1-\cos \omega t)$, where $a=$ $=m E / q B^{2}, \omega=q B / m$. The trajectory is a cycloid (Fig. 26). The
Fig. 26.
motion of the particle is the motion of a point located at the rim of a circle of radius $a$ rolling without slipping along the $x$ axis so that its centre travels with the velocity $v=E / B$; (b) $s=8 m E / g B^{2}$; (c) $\left\langle v_{x}\right\rangle=E / B$.
3.393. $V=2 \frac{e}{m}\left(\frac{\mu_{0} l}{4 \pi}\right)^{2} \ln \frac{a}{b}$.
3.394. $B \leqslant \frac{2 b}{b^{2}-a^{2}} \sqrt{\frac{2 m}{e} V}$.
3.395. $y=\frac{a}{2 \omega} t \sin \omega t, \quad x=\frac{a}{2 \omega^{2}}(\sin \omega t-\omega t \cos \omega t), \quad$ where $a=q E_{m} / m$. The trajectory has the form of unwinding spiral. 3.396. $V \geqslant 2 \pi^{2} v^{2} m r \Delta r / e=0.10 \mathrm{MV}$.
322
3.397. (a) $T=\frac{(e r B)^{2}}{2 m}=12 \mathrm{MeV}$;
(b) $v_{\min }=\frac{1}{\pi r} \sqrt{\frac{T}{2 m}}=20 \mathrm{MHz}$.
3.398. (a) $t=\frac{\pi^{2} v m r^{2}}{e V}=17 \mu \mathrm{s}$;
(b) $s \approx \frac{4 \pi^{3} v^{2} m r^{2}}{3 \mathrm{eV}}=0.74 \mathrm{~km}$. Instruction. Here $s \sim \sum_{n=1}^{N} v_{n} \sim \sum \sqrt{n}$, where $v_{n}$ is the velocity of the particle after the $n$th passage across the accelerating gap. Since $N$ is large, $\sum_{1}^{N} \sqrt{n} \approx \int_{0}^{N} \sqrt{n} d n$.
3.399. $n=2 \pi v W / e B c^{2}=9$.
3.400. $\omega=\omega_{0} / \sqrt{1+a t}$, where $\omega_{0}=q B / m, a=q B \Delta W / \pi m^{2} c^{2}$.
3.401. $v=1 / 2 \mathrm{rqB} / \mathrm{m}, \rho=r / 2$.
3.402. $N=W / e \dot{\Phi}=5 \cdot 10^{6}$ revolutions, $s=2 \pi r N=8 \cdot 10^{3} \mathrm{~km}$.
3.403. On the one hand,
\[
\frac{d p}{d t}=e E=\frac{e}{2 \pi r} \frac{d \Phi}{d t},
\]
where $p$ is the momentum of the electron, $r$ is the radius of the orbit, $\Phi$ is the magnetic flux acting inside the orbit.

On the other hand, $d p / d t$ can be found after differentiating the relation $p=e r B$ for $r=$ const. It follows from the comparison of the expressions obtained that $d B_{0} / d t=1 / 2 d\langle B\rangle / d t$. In particular, this condition will be satisfied if $B_{0}=1 / 2\langle B\rangle$.
3.404. $r_{0}=\sqrt{2 B_{0} / 3 a}$.
3.405. $d E / d r=\dot{B}\left(r_{0}\right)-1 / 2\langle\dot{B}\rangle=0$.
3.406. $\Delta W=2 \pi r^{2} e B / \Delta t=0.10 \mathrm{keV}$.
3.407. (a) $W=\left(\sqrt{1+\left(r e B / m_{0} c\right)^{2}}-1\right) m_{0} c^{2}$;
(b) $s=W \Delta t / r e B$.
4.1. (a) See Fig. 27; (b) $\left(v_{x} / a \omega\right)^{2}+(x / a)^{2}=1$ and $w_{x}=-\omega^{2} x$.
Fig. 27.
4.2. (a) The amplitude is equal to $a / 2$, and the period is $T=$ $=\pi / \omega$, see Fig. $28 a$; (b) $v_{x}^{2}=4 \omega^{2} x(a-x)$, see Fig. $28 b$.
4.3. $x=a \cos (\omega t+\alpha)=-29 \mathrm{~cm}, \quad v_{x}=-81 \mathrm{~cm} / \mathrm{s}$, where $a=\sqrt{x_{0}^{2}+\left(v_{x 0} / \omega\right)^{2}}, \alpha=\arctan \left(-v_{x 0} / \omega x_{0}\right)$.
4.5. (a) $\langle v\rangle=3 a / T=0.50 \mathrm{~m} / \mathrm{s}$; (b) $\langle v\rangle=6 a / T=1.0 \mathrm{~m} / \mathrm{s}$.
4.6. (a) $\left\langle v_{x}\right\rangle=\frac{2 \sqrt{2}}{3 \pi} a \omega$;
(b) $|\langle\mathrm{v}\rangle|=\frac{2 \sqrt{2}}{3 \pi} a_{0}$;
(c) $\langle v\rangle=$ $=\frac{2(4-\sqrt{2})}{3 \pi} a \omega$.
4.7. $s=\left\{\begin{array}{l}a[n+1-\cos (\omega t-n \pi / 2)], n \text { is even, } \\ a[n+\sin (\omega t-n \pi / 2)], n \text { is odd. }\end{array}\right.$ Here $n$ is a whole number of the ratio $2 \omega t / \pi$.
Fig. 29.
4.8. $s=0.6 \mathrm{~m}$.
4.9. $d P / d x=1 / \pi \sqrt{a^{2}-x^{2}}$
4.10. In both cases $a=7$.
4.11. $v_{\max }=2.73 a \omega$.
4.12. 47.9 and $52.1 \mathrm{~s}^{-1}, 1.5$ ś.
4.13. 18 or $26 \mathrm{~Hz}$.
4.14. (a) $x^{2} / a^{2}+y^{2} / b^{2}=1$, clockwise; (b) $\mathbf{w}=-\omega^{2} \mathbf{r}$.
4.15. (a) $y^{2}=4 x^{2}\left(1-x^{2} / a^{2}\right)$;
(b) $y=a\left(1-2 x^{2} / a^{2}\right)$.
See

Fig. 29 .
4.16. $T=2 \pi \sqrt{m / a^{2} U_{0}}$.
4.17. $T=4 \pi a \sqrt{m a / b^{2}}$.
324
4.18. $T=\pi V \overline{m l / F}=0.2 \mathrm{~s}$.
4.19. $T=2 \pi V \overline{\eta l / g(\eta-1)}=1.1 \mathrm{~s}$.
4.20. $T=2 \sqrt{l / g}[\pi / 2+\arcsin (\alpha / \beta)]$.
4.21. $t=\sqrt{\frac{2 h}{w}} \frac{\sqrt{1+\eta}-\sqrt{1-\eta}}{1-\sqrt{1-\eta}}$ where $\eta=w / g$.
4.22. $T=\sqrt{4 \pi m / \rho g r^{2}}=2.5 \mathrm{~s}$.
4.23. $T=2 \pi \sqrt{\eta(1-\eta) m / x}=0.13 \mathrm{~s}$.
4.24. $T=2 \pi \sqrt{m /\left(x_{1}+x_{2}\right)}$.
4.25. $T=2 \pi V \overline{m / x}$, where $x=x_{1} x_{2} /\left(x_{1}+x_{2}\right)$.
4.26. $\omega=V \overline{2 T_{0} / m l}$.
4.27. $T=2 \pi \sqrt{m / S \rho g(1+\cos \theta)}=0.8 \mathrm{~s}$.
4.28. $T=\pi V \overline{2 l / k g}=1.5 \mathrm{~s}$.
4.29. (a) $\ddot{x}+(g / R) x=0$, where $x$ is the displacement of the body relative to the centre of the Earth, $R$ is its radius, $g$ is the standard free-fall acceleration;
(b) $\tau=\pi \sqrt{R / g}=42 \mathrm{~min}$, (c) $v=\sqrt{g R}=7.9 \mathrm{~km} / \mathrm{s}$.
4.30. $T=2 \pi V \overline{l /|\mathbf{g}-\mathbf{w}|}=0.8 \mathrm{~s}$, where $|g-w|=$ $=\sqrt{g^{2}+w^{2}-2 g w \cos \beta}$.
4.31. $T=2 \pi / \sqrt{x / m-\omega^{2}}=0.7 \mathrm{~s}, \omega \geqslant V \overline{x / m}=10 \mathrm{rad} / \mathrm{s}$.
4.32. $k=4 \pi^{2} a / g T^{2}=0.4$.
4.33. (a) $\theta=3.0^{\circ} \cos 3.5 t$;
(b) $\theta=4.5^{\circ} \sin 3.5 t$; (c) $\theta=$ $=5.4^{\circ} \cos (3.5 t+1.0)$. Here $t$ is expressed in seconds.
4.34. $F=\left(m_{1}+m_{2}\right) g \pm$ $\pm m_{1} a \omega^{2}=60$ and $40 \mathrm{~N}$.
4.35 .
(a) $F=m g(1+$
$\left.+\frac{a \omega^{2}}{g} \cos \omega t\right)$, see Fig. 30 ;
(b) $a_{\text {min }}=g / \omega^{2}=8 \mathrm{~cm}$;
(c) $a=$ $=(\omega \sqrt{2 h / g}-1) g / \omega^{2}=20 \mathrm{~cm}$.
4.36. (a) $y=(1-\cos \omega t) m g / x$, where $\omega=\sqrt{x / m}$; (b) $T_{\text {max }}=$ $=2 m g, \quad T_{\text {min }}=0$.
4.37. $\left(x / r_{0}\right)^{2}+\alpha\left(y / v_{0}\right)^{2}=1$.
4.38. (a) $y=(1-\cos \omega t) w / \omega^{2}$;
(b) $y=(\omega t-\sin \omega t) \alpha / \omega^{3}$.

Here $\omega=\sqrt{\chi / m}$.
4.39. $\Delta h_{\max }=m g / k=10 \mathrm{~cm}, E=m^{2} g^{2} / 2 k=4.8 \mathrm{~mJ}$.
4.40. $a=(m g / x) \sqrt{1+2 h x / m g}, E=m g h+m^{2} g^{2} / 2 x$.
4.41. $a=(m g / x) \sqrt{1+2 h x /(m+M) g}$.
4.42. Let us write the motion equation in projections on the $x$ and $y$ axes:
\[
\ddot{x}=\omega \dot{y}, \ddot{y}=-\omega x \text {, where } \omega=a / m \text {. }
\]
Integrating these equations, with the initial conditions taken into account, we get $x=\left(v_{0} / \omega\right)(1-\cos \omega t), y=\left(v_{0} / \omega\right) \sin \omega t$. Hence $\left(x-v_{0} / \omega\right)^{2}+y^{2}=\left(v_{0} / \omega\right)^{2}$. This is the equation of a circle of radius $v_{0} / \omega$ with the centre at the point $x_{0}=v_{0} / \omega, y_{0}=0$.
4.43. Will increase $\sqrt{1+2 / 5(R / l)^{2}}$ times. It is taken into account here that the water (when in liquid phase) moves translationwise, and the system behaves as a mathematical pendulum.
4.44. $\omega=\sqrt{\frac{3 g}{2 l}\left(1+\frac{2 x l}{m g}\right)}$.
4.45. (a) $T=2 \pi V \overline{l / 3 g}=1.1 \mathrm{~s}$;
(b) $E=\frac{1}{2} m g l \alpha^{2}=0.05 \mathrm{~J}$.
4.46. $\varphi_{m}=\varphi_{0} \sqrt{1+m R^{2} \dot{\varphi}_{0}^{2} / 2 k \varphi_{0}^{2}}, E={ }_{1 / 2} k \varphi_{m}^{2}$.
4.47. $\langle T\rangle=1 / 8 m g l \theta_{0}^{2}+1 / 12 m l^{2} \dot{\theta}_{0}^{2}$.
4.48. $T=4 \pi / \omega$.
4.49. $I=m l^{2}\left(\omega_{2}^{2}-g / l\right) /\left(\omega_{1}^{2}-\omega_{2}^{2}\right)=0.8 \mathrm{~g} \cdot \mathrm{m}^{2}$.
4.50. $\omega=\sqrt{\left(I_{1} \omega_{1}^{2}+I_{2} \omega_{2}^{2}\right) /\left(I_{1}+I_{2}\right)}$.
4.51. $x=l / 2 V \overline{3}, T_{\min }=2 \pi \sqrt{l / g V \overline{3}}$.
4.52. $T=\pi \sqrt{2 h / g}, l_{\text {red }}=h / 2$.
4.53. $\omega_{0}=\sqrt{3 a \omega^{2} / 2 l}$.
4.54. $\omega_{0}=\sqrt{x /\left(m+I / R^{2}\right)}$.
4.55. $\omega_{0}=\sqrt{\frac{2 m g \cos \alpha}{M R+2 m R(1+\sin \alpha)}}$.
4.56. $T=2 \pi V \overline{3(R-r) / 2 g}$.
4.57. $T=\pi \sqrt{3 m / 2 x}$.
4.58. $\omega_{0}=\sqrt{x / \mu}$, where $\mu=m_{1} m_{2} /\left(m_{1}+m_{2}\right)$.
4.59. (a) $\omega=\sqrt{x / \mu}=6 \mathrm{~s}^{-1}$; (b) $E=1 / 2 \mu v_{1}^{2}=5 \mathrm{nIJ}, a=v_{1} / \omega=$ $=2 \mathrm{~cm}$. Here $\mu=m_{1} m_{2} /\left(m_{1}+m_{2}\right)$.
4.60. $T=2 \pi V \overline{I^{\prime} / k}$, where $I^{\prime}=I_{1} I_{2} /\left(I_{1}+I_{2}\right)$.
4.61. $\omega_{2} / \omega_{1}=\sqrt{1+2 m_{\mathrm{O}} / m_{\mathrm{C}}} \approx 1.9$, where $m_{0}$ and $m_{\mathrm{C}}$ are the masses of oxygen and carbon atoms.
4.62. $\omega=S \sqrt{2 \gamma p_{0} / m V_{0}}$, where $\gamma$ is the adiabatic exponent.
4.63. $q=4 h V \overline{\pi \varepsilon_{0} m g\left(\eta^{2}-1\right)}=2.0 \mu \mathrm{C}$.
4.64. The induction of the field increased $\eta^{2}=25$ times.
4.65. $x=\left(v_{0} / \omega\right) \sin \omega t$, where $\omega=l B / \sqrt{m L}$.
4.66. $x=(1-\cos \omega t) g / \omega^{2}$, where $\omega=l B / \sqrt{m L}$.
4.67. (a) $a_{0}$ and $a_{0} \omega ;$ (b) $t_{n}=\frac{1}{\omega}\left(\arctan \frac{\omega}{\beta}+n \pi\right)$, where $n=0,1,2, \ldots$
4.68. (a) $\dot{\varphi}(0)=-\beta \varphi_{0}, \quad \ddot{\varphi}(0)=\left(\beta^{2}-\omega^{2}\right) \varphi_{0}$; (b) $t_{n}=$ $=\frac{1}{\omega}\left(\arctan \frac{\omega^{2}-\beta^{2}}{2 \beta \omega}+n \pi\right)$, where $n=0,1,2, \ldots$
4.69. (a) $a_{0}=\frac{\left|\dot{x}_{0}\right|}{\omega}, \quad \alpha=\left\{\begin{array}{lll}-\pi / 2, & \text { when } & \dot{x}_{0}>0, \\ +\pi / 2, & \text { when } & \dot{x}_{0}<0 ;\end{array}\right.$
(b) $a_{0}=$ $=\left|x_{0}\right| \sqrt{1+(\beta / \omega)^{2}}, \quad \alpha=\arctan (-\beta / \omega)$, with $-\pi / 2<\alpha<0$, if $x_{0}>0$ and $\pi / 2<\alpha<\pi$, if $x_{0}<0$.
4.70. $\beta=\omega \sqrt{\eta^{2}-1}=5 \mathrm{~s}^{-1}$.
4.71. (a) $v(t)=a_{0} \sqrt{\omega^{2}+\beta^{2}} \mathrm{e}^{-\beta t}$; (b) $v(t)=\left|\dot{x}_{0}\right| \sqrt{1+(\beta / \omega)^{2}} \mathrm{e}^{-\beta t}$.
4.72. The answer depends on what is meant by the given question. The first oscillation attenuates faster in time. But if one takes the natural time scale, the period $T$, for each oscillation, the second oscillation attenuates faster during that period.
4.73. $\lambda=n \lambda_{0} / \sqrt{1+\left(1-n^{2}\right)\left(\lambda_{0} / 2 \pi\right)^{2}}=3.3, n^{\prime}=\sqrt{1+\left(2 \pi / \lambda_{c}\right)^{2}}=$ $=4.3$ times.
4.74. $T=\sqrt{\left(4 \pi^{2}+\lambda^{2}\right) \Delta x / g}=0.70 \mathrm{~s}$.
4.75. $Q=\pi n / \ln \eta=5 \cdot 10^{2}$.
4.76. $s \approx l\left(1+\mathrm{e}^{-\lambda / 2}\right) /\left(1-\mathrm{e}^{-\lambda / 2}\right)=2 \mathrm{~m}$.
4.77. $Q=1 / 2 \sqrt{\frac{4 g \tau^{2}}{l \ln ^{2} \eta}-1}=1.3 \cdot 10^{2}$.
4.79. $\omega=\sqrt{\frac{2 \alpha}{m R^{2}}-\left(\frac{\pi \eta R^{2}}{m}\right)^{2}}$.
4.80. $\eta=2 \lambda h I / \pi R^{4} T$.
4.81. $\tau=2 R I / a^{4} B^{2}$.
4.82. (a) $T=2 \pi \sqrt{m / x}=0.28 \mathrm{~s}$; (b) $n=\left(x_{0}-\Delta\right) / 4 \Delta=3.5$ oscillations, here $\Delta=k m g / x$.
4.83. $x=\frac{F_{0} / m}{\omega^{2}-\omega_{0}^{2}}\left(\cos \omega_{0} t-\cos \omega t\right)$.
4.84. The motion equations and their solutions:
\[
\begin{array}{ll}
t \leqslant \tau, \ddot{x}+\omega_{0}^{2} x=F / m, & x=\left(1-\cos \omega_{0} t\right) F / k, \\
t \geqslant \tau, \ddot{x}+\omega_{0}^{2} x=0, & x=a \cos \left[\omega_{0}(t-\tau)+\alpha\right],
\end{array}
\]
where $\omega_{0}^{2}=k / m, a$ and $\alpha$ are arbitrary constants. From the continuity of $x$ and $\dot{x}$ at the moment $t=\tau$ we find the sought amplitude:
\[
a=(2 F / k)\left|\sin \left(\omega_{0} t / 2\right)\right| .
\]
4.85. $\omega_{\text {res }}=\sqrt{\frac{1-(\lambda / 2 \pi)^{2}}{1+(\lambda / 2 \pi)^{2}} \frac{g}{\Delta l}}, a_{\text {res }}=\frac{\lambda F_{0} \Delta l}{4 \pi m g}\left(1+\frac{4 \pi^{2}}{\lambda^{2}}\right)$.
4.86. $\omega_{\text {res }}=\sqrt{\left(\omega_{1}^{2}+\omega_{2}^{2}\right) / 2}=5.1 \cdot 10^{2} \mathrm{~s}-1$.
4.87. (a) $\omega_{0}=\sqrt{\omega_{1} \omega_{2}}$;
\[
=\sqrt{\omega_{1} \omega_{2}-\left(\omega_{2}-\omega_{1}\right)^{2} / 12} \text {. }
\]
\[
\beta=\left|\omega_{2}-\omega_{1}\right| / 2 \sqrt{3}, \quad \omega=
\]
4.88. $\eta=\left(1+\lambda^{2} / 4 \pi^{2}\right) \pi / \lambda=2.1$.
4.89. $A=\pi a F_{0} \sin \varphi$.
4.90. (a) $Q=1 / 2 \sqrt{\frac{4 \omega^{2} \omega_{0}^{2}}{\left(\omega^{2}-\omega_{0}^{2}\right)^{2} \tan ^{2} \varphi}-1}=2.2$; (b) $A=\pi m a^{2}\left(\omega_{0}^{2}-\right.$ $\left.-\omega^{2}\right) \tan \varphi=6 \mathrm{~mJ}$. Here $\omega_{0}=\sqrt{x / m}$.
4.91. (a) $\langle P\rangle=\frac{F_{0}^{2} \beta \omega^{2} / m}{\left(\omega_{0}^{2}-\omega^{2}\right)^{2}+4 \beta^{2} \omega^{2}}$;
(b) $\omega=\omega_{0},\langle P\rangle_{\max }=F_{0}^{2} / 4 \beta m$.
4.92. $\frac{\langle P\rangle_{\max }-\langle P\rangle}{\langle P\rangle_{\max }}=\frac{100}{\eta^{2}-1} \%$.
4.43. (a) $A=-\pi \varphi_{m} N_{m} \sin \alpha$;
(b) $Q=\frac{\sqrt{\left(\cos \alpha+2 \omega^{2} I \varphi_{m} / N_{m}\right)^{2}-1}}{2 \sin \alpha}$.
4.94. $\omega=\sqrt{n e^{2} / \varepsilon_{0} m}=1.65 \cdot 10^{16} \mathrm{~s}^{-1}$.
4.95. $V^{2}+I^{2} L / C=V_{m}^{2}$.
4.96. (a) $I=I_{m} \sin \omega_{0} t$, where $I_{m}=V_{m} \sqrt{C / L}, \quad \omega_{0}=1 / \sqrt{L C}$;
(b) $\mathscr{E}_{s}=V_{m} / \sqrt{2}$.
4.97. $A=\left(\eta^{2}-1\right) W$.
4.98. (a) $T=2 \pi \sqrt{\bar{L}\left(C_{1}+C_{2}\right)}=0.7 \mathrm{~ms}$;
(b) $I_{m}=V \sqrt{\left(C_{1}+C_{2}\right) / L}=8$ A.
4.99. $V=1 / 2(1 \pm \cos \omega t) V_{0}$, where the plus sign refers to the left-hand capacitor, and the minus sign to the right-hand one; $\omega=\sqrt{2 / L C}$.
4.100. $I=\frac{\Phi}{L} \cos (t / \sqrt{L C})$.
4.101. (a) $t_{n}=\frac{\pi n}{\omega}$;
(b) $t_{n}=\frac{1}{\omega}\left[\arctan \left(-\frac{\beta}{\omega}\right)+\pi n\right]$. Here $n=0,1,2, \ldots$
4.102. $V_{0} / V_{m}=\sqrt{1-\frac{R^{2} C}{4 L}}$.
4.103. $V_{C}=I_{m} \sqrt{L / C} \mathrm{e}^{-\beta t} \sin (\omega t+\alpha)$ with $\tan \alpha=\omega / \beta ; V_{C}(0)=$ $=I_{m} \sqrt{\frac{L}{C\left(1+\beta^{2} / \omega^{2}\right)}}$.
4.104. $W_{L} / W_{C}=L / C R^{2}=5$.
4.105. $L=L_{1}+L_{2}, R=R_{1}+R_{2}$.
4.106. $t=\frac{Q}{\pi v} \ln \eta=0.5 \mathrm{~s}$.
4.107. $n=\frac{1}{2 \pi} \sqrt{\frac{4 L}{C R^{2}}-1}=16$.
4.108. $\frac{\omega_{0}-\omega}{\omega_{0}}=1-\frac{1}{\sqrt{1+1 /(2 Q)^{2}}} \approx \frac{1}{8 Q^{2}}=0.5 \%$.
4.109. (a) $W_{0}=1 / 2 \mathscr{E}^{2}\left(L+C R^{2}\right) /(r+R)^{2}=2.0 \mathrm{~mJ}$; (b) $W=$ $=W_{0} \mathrm{e}^{-t R / L}=0.10 \mathrm{~mJ}$.
4.110. $t \approx \frac{Q}{2 \pi v_{0}} \ln \eta=1.0 \mathrm{~ms}$.
4.111. (a) $\omega=\sqrt{\frac{1}{L C}-\frac{1}{4 R^{2} C^{2}}}$; (b) $Q=\frac{1}{2} \sqrt{\frac{4 R^{2} C}{L}-1}$.

When solving the problem, it should be taken into account that $d q / d t=I-I^{\prime}$, where $q$ is the charge of the capacitor, $I$ is the current in the coil winding, $I^{\prime}$ is the leakage current $\left(I^{\prime}=V / R\right)$.
4.112. $Q=\frac{V^{2} m}{2\langle P\rangle} \sqrt{\frac{C}{L}}=1.0 \cdot 10^{2}$.
4.113. $\langle P\rangle=R\left\langle I^{2}\right\rangle=1 / 2 R I_{m}^{2}=20 \mathrm{~mW}$.
4.114. $\langle\mathrm{P}\rangle=1 / 2 R C V_{m}^{2} / L=5 \mathrm{~mW}$.
4.115. $\omega=\sqrt{\frac{1}{L C}-\frac{1}{4 R^{2} C^{2}}} ; R<\frac{1}{2} \sqrt{\frac{L}{C}}$.
4.116. $\frac{1}{L_{1}}+\frac{1}{L_{2}}=\frac{1}{L}$ and $\frac{1}{R_{1}}+\frac{1}{R_{2}}=\frac{1}{R}$.
4.117. $I=\frac{V_{0}}{L} t \mathrm{e}^{-t / \sqrt{L C}} ; \quad I=I_{\max }=\frac{V_{0}}{\mathrm{e}} \sqrt{\frac{\bar{C}}{L}}$ at the moment $t_{m}=\sqrt{L C}$.
4.118. $I=\frac{V_{m}}{\sqrt{R^{2}+\omega^{2} L^{2}}}\left[\cos (\omega t-\varphi)-\cos \varphi \cdot \mathrm{e}^{-t R / L}\right], \tan \varphi=\omega L / R$.
4.119. $I=\frac{V_{m}}{\sqrt{R^{2}+1 /(\omega C)^{2}}}\left[\cos (\omega t-\varphi)-\cos \varphi \cdot \mathrm{e}^{-t / R C}\right], \quad \tan \varphi=$ $=-\frac{1}{\omega R C}$.
4.120. The current lags behind the voltage by phase angle $\varphi$, defined by the equation $\tan \varphi=\frac{\mu_{0} \pi^{2} v a}{4 n \rho}$.
4.121. The current is ahead of the voltage by the phase angle $\varphi=60^{\circ}$, defined by the equation $\tan \varphi=\sqrt{\left(V_{m} / R I_{m}\right)^{2}-1}$.
4.122. (a) $V^{\prime}=V_{0}+V_{m} \cos (\omega t-\alpha)$, where $V_{m}=V_{0} / \sqrt{1+(\omega R C)^{2}}$, $\alpha=\arctan (\omega R C)$; (b) $R C=\sqrt{\eta^{2}-1} / \omega=22 \mathrm{~ms}$.
4.123. See Fig. 31 .
4.124. (a) $I_{m}=V_{m} / \sqrt{R^{2}+(\omega L-1 / \omega C)^{2}}=4.5 \mathrm{~A}$; (b) $\tan \varphi=$ $=\frac{\omega L-1 / \omega C}{R}, \varphi=-60^{\circ}$ (the current is ahead of the voltage); (c) $V_{C}=I_{m} / \omega C=0.65 \mathrm{kV}, V_{L}=I_{m} \sqrt{R^{2}+\omega^{2} L^{2}}=0.50 \mathrm{kV}$.
4.125. (a) $\omega=\sqrt{\omega_{0}^{2}-2 \beta^{2}}$; (b) $\omega=\omega_{0}^{2} / \sqrt{\omega_{0}^{2}-2 \beta^{2}}$, where $\omega_{0}^{2}=$ $=1 / L C, \beta=R / 2 L$.
4.126. For $C=\frac{1}{\omega^{2} L}=28 \mu \mathrm{F} ; V_{L}=V_{m} \sqrt{1+(\omega L / R)^{2}}=0.54 \mathrm{kV}$; $V_{C}=V_{m} \omega L / R=0.51 \mathrm{kV}$.
4.127. $I=I_{m} \cos (\omega t+\varphi)$, where $\quad I_{m}=\frac{V_{m}}{R} \sqrt{1+(\omega R C)^{2}} \quad$ and $\tan \varphi=\omega R C$.
4.128. $\omega_{0}=\sqrt{\frac{L_{2}}{C\left(L_{1} L_{2}-L_{12}^{2}\right)}}$.
4.129. $Q=\sqrt{n^{2}-1 / 4}$.
4.130. $Q=\sqrt{\frac{\eta^{2}-1}{(n-1)^{2}}-\frac{1}{4}}$.
4.131. (a) $\omega_{0}=V \overline{\omega_{1} \omega_{2}}$;
(b) $Q=\sqrt{\frac{\omega_{1} \omega_{2}\left(n^{2}-1\right)}{\left(\omega_{2}-\omega_{1}\right)^{2}}-\frac{1}{4}}$.
4.133. $I_{0} / I=\sqrt{1+\left(Q^{2}+1 / 4\right)\left(\eta^{2}-1\right)^{2} / \eta^{2}}, 2$ and 19 respectively.
4.134. $t=1 / 2 \pi t_{0}$.
4.135. (a) $I=\frac{2}{\sqrt{3}} I_{0} \approx 1.15 I_{0}$;
(b) $I=\frac{\pi}{\sqrt{8}} I_{0} \approx 1.11 I_{0}$.
4.136. $v=\frac{R}{2 \pi L} \sqrt{\eta-1}=2 \mathrm{kHz}$.
4.137. The current lags behind the voltage by the phase angle $\varphi=\arccos \sqrt{1-\left(X_{L} / Z\right)^{2}} \approx 37^{\circ}, P=\frac{V^{2}}{Z^{2}} \sqrt{Z^{2}-X_{L}^{2}}=0.16 \mathrm{~kW}$.
4.138. For $R=\omega L-r=0.20 \mathrm{k} \Omega ; P_{\max }=\frac{V^{2}}{2 \omega L}=0.11 \mathrm{~kW}$.
4.139. Increased by $\sqrt{n}-1 \approx 30 \%$.
4.140. For $Q \gg 1$ the ratio is $\Delta \omega / \omega_{0} \approx 1 / 2 V \overline{n-1} / Q=0.5 \%$.
4.141. $P_{2}=1 / 2\left(V^{2}-V_{1}^{2}-V_{2}^{2}\right) / R=30 \mathrm{~W}$.
4.142. $P_{1}^{2}=1 / 2\left(I^{2}-I_{1}^{2}-I_{2}^{2}\right)^{2} R=2.5 \mathrm{~W}$.
4.143. $Z=R / \sqrt{1+(\omega C R)^{2}}=40 \Omega$.
4.144. See Fig. 32.
4.145. (a) $\omega_{\text {res }}=\sqrt{\frac{1}{L C}-\frac{R^{2}}{L^{2}}}=3 \cdot 10^{4} \mathrm{rad} / \mathrm{s}$;
(b) $I=V R C / L=$ $=3 \mathrm{~mA}, I_{L}=V \sqrt{C / L}=1.0 \mathrm{~A}, I_{C}=V \sqrt{\frac{C}{L}-\left(\frac{R C}{L}\right)^{2}}=1.0 \mathrm{~A}$.
4.146. $\tan \varphi=\frac{\omega C\left(R^{2}+\omega^{2} L^{2}\right)-\omega L}{R}$.
4.147. $Z=\sqrt{\frac{R^{2}+\omega^{2} L^{2}}{(\omega C R)^{2}+\left(1-\omega^{2} C L\right)^{2}}}$.
4.149. $\left\langle F_{x}\right\rangle=\frac{\omega^{2} L_{2} L_{12} I_{0}^{2}}{2\left(R^{2}+\omega^{2} L_{2}^{2}\right)} \frac{\partial L_{12}}{\partial x}$.
4.150. $t=\frac{2 l}{\alpha\left(\sqrt{T_{1}}+\sqrt{T_{2}}\right)}$.
4.151. $\Delta \varphi=\frac{\omega}{v}\left|\left(x_{1}-x_{2}\right) \cos \alpha+\left(y_{1}-y_{2}\right) \cos \beta+\left(z_{1}-z_{2}\right) \cos \gamma\right|$.
4.152. $\mathbf{k}=\omega\left(\frac{\mathbf{e}_{x}}{v_{1}}+\frac{\mathbf{e}_{y}}{v_{2}}+\frac{\mathbf{e}_{z}}{v_{3}}\right)$.
4.153. $\xi=a \cos \left[(1-V / v) \omega t-k x^{\prime}\right]$, where $v=\omega / k$.
4.155. (a) $a / \lambda=5.1 \cdot 10^{-5}$; (b) $v_{m}=11 \mathrm{~cm} / \mathrm{s}, \quad 3.2 \cdot 10^{-4}$; (c) $(\partial \xi / \partial x)_{m}=3.2 \cdot 10^{-4},(\partial \xi / \partial t)_{m}=v(\partial \xi / \partial x)_{m}$, where $v=0.34 \mathrm{~km} / \mathrm{s}$ is the velocity of the wave.
330
4.156. See Fig. 33 .
4.157. $\Delta \varphi=-\frac{2 \pi}{\gamma \lambda} \ln (1-\eta) \approx \frac{2 \pi \eta}{\gamma \lambda}=0.3 \mathrm{rad}$.
4.158. $\mathrm{r}=\left(a_{1} \mathrm{r}_{1}+a_{2} \mathrm{r}_{2}\right) /\left(a_{1}+a_{2}\right)$.
4.159. (a) $\gamma=\frac{\ln \left(\eta r_{0} / r\right)}{r-r_{0}}=0.08 \mathrm{~m}^{-1}$;
(b) $v_{m}=\frac{2 \pi v a_{0}}{\eta}=15 \mathrm{~cm} / \mathrm{s}$.
4.160. (a) See Fig. 34a. The particles of the medium at the points lying on the solid straight lines $(y=x \pm n \lambda, n=0,1,2, \ldots)$
Fig. 33.
oscillate with maximum amplitude, those on the dotted lines do not oscillate at all.
(b) See Fig. $34 b$. The particles of the medium at the points lying on the straight lines $y=x \pm n \lambda, y=x \pm(n \pm 1 / 2) \lambda$ and $y=$ $=x \pm(n \pm 1 / 4) \lambda$ oscillate respectively along those lines, at
Fig. 34 .
right angles to them, or move along the circles (here $n=0,1,2, \ldots$ ). At all other points the particles move along the ellipses.
4.161. $\langle w\rangle=2 / 3 w_{0}$.
4.162. $\langle\Phi\rangle=2 \pi l^{2} I_{0}\left(1-\frac{1}{\sqrt{1+(R / l)^{2}}}\right)=20 \mu \mathrm{W}$.
4.163. $\langle\Phi\rangle=P / V \overline{1+(2 R / h)^{2}}=0.07 \mathrm{~W}$.
4.164. Fee Fig. 35, for (a) and (b); see Fig. 36 for (c).
4.165. (a) $w_{p}=1 / 2 \rho a^{2} \omega^{2} \sin ^{2} k x \cdot \cos ^{2} \omega t$;
(b) $w_{k}=1 / 2 \rho a^{2} \omega^{2} \times$ $\times \cos ^{2} k x \cdot \sin ^{2} \omega t$. See Fig. 37 .
Fig. 35.
4.166. $a_{\max }=5 \mathrm{~mm}$; to the third overtone.
4.167. $\frac{v_{2}}{v_{1}}=\sqrt{\frac{\eta_{2}\left(1+\eta_{1}\right)}{\eta_{1}\left(1+\eta_{2}\right)}}=1.4$.
4.168. Will increase $\eta=\frac{\sqrt{1-\Delta T / T}}{1+\Delta l / l}=2$ times.
Fig. 36.
Fig. 37.
4.169. $v=2 l v=0.34 \mathrm{~km} / \mathrm{s}$.
4.170. (a) $v_{n}=\frac{v}{4 l}(2 n+1)$, six oscillations; (b) $v_{n}=\frac{v}{2 l}(n+1)$, also six oscillations. Here $n=0,1,2, \ldots$
4.171. $v_{n}=\frac{2 n+1}{2 l} \sqrt{\frac{E}{\rho}}=3.8(2 n+1) \mathrm{kHz}$; four oscillations with frequencies $26.6,34.2,41.8$, and $49.4 \mathrm{kHz}$.
4.172. (a) $T_{\max }=1 / 4 m \omega^{2} a_{\max }^{2}$; (b) $\langle T\rangle=1 / 8 m \omega^{2} a_{\text {max }}^{2}$.
4.173. $W=1 / 4 \pi S \rho \omega^{2} a^{2} / k$.
4.174. $v=2 v_{0} v u /\left(v^{2}-u^{2}\right) \approx 2 v_{0} u / v=1.0 \mathrm{~Hz}$.
4.175. $u=\frac{v v_{0}}{v}\left(\sqrt{1+\left(v / v_{0}\right)^{2}}-1\right) \approx \frac{v v}{2 v_{0}}=0.5 \mathrm{~m} / \mathrm{s}$.
4.176. $\omega=\frac{v_{0} v}{a \Delta v}\left(\sqrt{1+\left(\Delta v / v_{0}\right)^{2}}-1\right)=34 \mathrm{~s}^{-1}$.
4.177. $v=v_{0} / \sqrt{1+2 w t / v}=1.35 \mathrm{kHz}$.
4.178. (a) $v=v_{0} /\left(1-\eta^{2}\right)=5 \mathrm{kHz}$; (b) $r=l \sqrt{1+\eta^{2}}=0.32 \mathrm{~km}$.
4.179. Decreases by $2 u /(v+u)=2.0 \%$.
4.180. $v=2 v_{0} u /(v+u)=0.60 \mathrm{~Hz}$.
4.181. $\gamma=\frac{\ln \left(\eta r_{1}^{2} / r_{2}^{2}\right)}{2\left(r_{2}-r_{1}\right)}=6 \cdot 10^{-3} \mathrm{~m}^{-1}$.
4.182. (a) $L^{\prime}=L-20 \gamma x \log \mathrm{e}=50 \mathrm{~dB}$; (b) $x=0.30 \mathrm{~km}$.
4.183. (a) $L=L_{0}+20 \log \left(r_{0} / r\right)=36 \mathrm{~dB}$; (b) $r>0.63 \mathrm{~km}$.
4.184. $\beta=\ln \left(r_{B} / r_{A}\right) /\left[\tau+\left(r_{B}-r_{A}\right) / v\right]=0.12 \mathrm{~s}^{-1}$.
4.185. (a) Let us consider the motion of a plane element of the medium of thickness $d x$ and unit area of cross-section. In accordance with Newton’s second law $\rho d \dot{x} \dot{\xi}=-d p$, where $d p$ is the pressure increment over the length $d x$. Recalling the wave equation $\ddot{\xi}=$ $=v^{2}\left(\partial^{2} \xi / \partial x^{2}\right)$, we can write the foregoing equation as
\[
\rho v^{2} \frac{\partial^{2} \xi}{\partial x^{2}} d x=-d p \text {. }
\]

Integrating this equation, we get
\[
\Delta p=-\rho v^{2} \frac{\partial \xi}{\partial x}+\text { const. }
\]

In the absence of a deformation (a wave) the surplus pressure is $\Delta p=0$. Hence, const $=0$.
4.186. $\langle\Phi\rangle=\pi R^{2}(\Delta p)_{m}^{2} / 2 \rho v \lambda=11 \mathrm{~mW}$.
4.187. (a) $(\Delta p)_{m}=\sqrt{\rho v P / 2 \pi r^{2}}=5 \mathrm{~Pa}$,
(b) $a=(\Delta p)_{m} / 2 \pi v \rho v=3 \mu \mathrm{m}, a / \lambda=5 \cdot 10^{-6}$.
\[
(\Delta p)_{m} / p=5 \cdot 10^{-5} ;
\]
4.188. $P=4 \pi r^{2} \mathrm{e}^{2 \gamma r} I_{0} \cdot 10^{L}=1.4 \mathrm{~W}$, where $L$ is expressed in bels.
4.189. $\Delta \lambda=(1 / V \bar{\varepsilon}-1) c / v=-50 \mathrm{~m}$.
4.190. $t=2\left(\sqrt{\varepsilon_{1}}-\sqrt{\varepsilon_{2}}\right) l / c \ln \left(\varepsilon_{1} / \varepsilon_{2}\right)$.
4.191. $j / j_{d t s}=\sigma / 2 \pi v \varepsilon \varepsilon_{0}=2$.
4.192. $\mathbf{H}=\frac{1}{k} \sqrt{\varepsilon_{0} / \mu_{0}}\left[\mathbf{k E}_{m}\right] \cos (c k t)$, where $c$ is the velocity of the wave in vacuum.
\[
\text { 4.193. (a) } \mathbf{H}=\mathbf{e}_{z} E_{m} \sqrt{\varepsilon_{0} / \mu_{0}} \cos k x=-0.30 \mathbf{e}_{z} \text {; }
\]
(b) $\mathbf{H}=\mathbf{e}_{z} E_{m} \sqrt{\varepsilon_{0} / \mu_{0}} \cos \left(c k t_{0}-k x\right)=0.18 \mathbf{e}_{\boldsymbol{z}}$. Here $\mathbf{e}_{\boldsymbol{z}}$ is the unit vector of the $z$ axis, $H$ is expressed in $\mathrm{A} / \mathrm{m}$.
4.194. $\varepsilon_{m}=2 \pi
u l^{2} E_{m} / c=13 \mathrm{mV}$.
4.196. $\langle\mathrm{S}\rangle=1 / 2 \mathrm{k} \varepsilon_{0} c^{2} E_{m}^{2} / \omega$.
4.197. (a) $j_{d t s}=\pi V \overline{2} \varepsilon_{0} v E_{m}=0.20 \quad \mathrm{~mA} / \mathrm{m}^{2}$;
(b) $\langle S\rangle=$ $=1 / 2 \varepsilon_{0} c E_{m}^{2}=3.3 \mu \mathrm{W} / \mathrm{m}^{2}$.
4.198. Since $t \gg T$, where $T$ is the period of oscillations, $W=$ $=1 / 2 V \overline{\varepsilon \varepsilon_{0} / \mu_{0}} E^{2} m \pi R^{2} t=5 \mathrm{~kJ}$.
4.199. $\mathbf{B}=\mathbf{B}_{m} \sin k x \cdot \sin \omega t$, where $\mathbf{B}_{m} \perp \mathbf{E}_{m}$, with $B_{m}=E_{m} / c$.
4.200. $S_{x}=1 /{ }_{4} \varepsilon_{0} c E_{m}^{2} \sin 2 k x \cdot \sin 2 \omega t,\left\langle S_{x}\right\rangle=0$.
4.201. $W_{m} / W_{e}=1 / 8 \varepsilon_{0} \mu_{0} \omega^{2} R^{2}=5.0 \cdot 10^{-15}$.
4.202. $W_{e} / W_{m}^{e}=1 / 8 \varepsilon_{0} \mu_{0} \omega^{2} R^{2}=5.0 \cdot 10^{-15}$.
4.204. $\Phi_{S}=I^{2} R$.
4.205. $S=I^{2} V \overline{m / 2 e U} / 4 \pi^{2} \varepsilon_{0} r^{2}$.
4.207. To the left.
4.208. $\Phi=V I$.
4.209. $\langle\Phi\rangle=1 / 2 V_{0} I_{0} \cos \varphi$.
4.211. The electric dipole moment of the system is $\mathbf{p}=\sum e \mathbf{r}_{i}=$ $=(e / m) M \mathbf{r}_{C}$, where $M$ is the mass of the system, $\mathbf{r}_{C}$ is the radius vector of its centre of inertia. Since the radiation power $P \propto \ddot{p}^{2} \propto$ $\propto \ddot{\mathbf{r}}_{C}^{2}$, and in our case $\ddot{\mathbf{r}_{C}}=0, P=0$ too.
4.212. $\langle P\rangle=\frac{1}{4 \pi \varepsilon_{0}} \frac{e^{2} a^{2} \omega^{4}}{3 c^{3}}=5 \cdot 10^{-15} \mathrm{~W}$.
4.213. $P=\frac{1}{\left(4 \pi \varepsilon_{0}\right)^{3}} \frac{2}{3 c^{3}}\left(\frac{q e^{2}}{m R^{2}}\right)^{2}$.
4.214. $\Delta W \approx \frac{1}{\left(4 \pi \varepsilon_{0}\right)^{3}} \frac{\pi e^{4} q^{2}}{3 c^{3} m^{2} v b^{3}}$.
4.215. $\Delta W / T=1 / 3 e^{3} B / \varepsilon_{0} c^{3} m^{2}=2 \cdot 10^{-18}$.
4.216. $T=T_{0} \mathrm{e}^{-\alpha t}$, where $\alpha=\frac{1}{3} e^{4} B^{2} / \pi \varepsilon_{0} c^{3} m^{3}$. After $t_{0}=\frac{1}{\alpha}=$ $=\left\{\begin{array}{ll}2.5 \mathrm{~s} & \text { for the electron, } \\ 1.6 \cdot 10^{10} \mathrm{~s}=0.5 \cdot 10^{3} & \text { years for the proton. }\end{array}\right.$
4.217. $S_{1} / S_{2}=\tan ^{2}(\omega l / c)=3$.
4.218. (a) Suppose that $t$ is the moment of time when the particle is at a definite point $x, y$ of the circle, and $t^{\prime}$ is the moment when the information about that reaches the point $P$. Denoting the observed values of the $y$ coordinate at the point $P$ by $y^{\prime}$ (see Fig. 4.40), we shall write
\[
t^{\prime}=t+\frac{l-x(t)}{c}, \quad y^{\prime}\left(t^{\prime}\right)=y(t) .
\]

The sought acceleration is found by means of the double differentiation of $y^{\prime}$ with respect to $t^{\prime}$ :
\[
\frac{d y^{\prime}}{d t^{\prime}}=\frac{d y}{d t^{\prime}}=\frac{d y}{d t} \frac{d t}{d t^{\prime}}, \quad \frac{d^{2} y}{d t^{\prime 2}}=\frac{d t}{d t^{\prime}} \frac{d}{d t}\left(\frac{d y^{\prime}}{d t^{\prime}}\right)=\frac{v^{2}}{R} \frac{v / c-y / R}{(1-v y / c R)^{3}},
\]
where the following relations are taken into account: $x=R \sin \omega t$, $y=R \cos \omega t$, and $\omega=v / R$.
(b) Energy flow density of electromagnetic radiation $S$ is proportional to the square of the $y$ projection of the observed acceleration of the particle. Consequently, $S_{1} / S_{2}=(1+v / c)^{4} /(1-v / c)^{4}$.
4.219. $\langle P\rangle=8 / 3 \pi r^{2} S_{0}$.
4.220. $\langle w\rangle=3 /{ }_{8}^{3} P_{0} / \pi r^{2} c$.
4.221. $P=1 / 6 p^{2} \omega^{4} / \pi \varepsilon_{0} c^{3}$.
334
4.222. $\langle P\rangle /\langle S\rangle=\left(e^{2} / m\right)^{2} \mu_{0}^{2} / 6 \pi$.
4.223. $\langle P\rangle /\langle S\rangle=\frac{\mu_{0}^{2}}{6 \pi} \frac{\left(e^{2} / m\right)^{2} \omega^{4}}{\left(\omega_{0}^{2}-\omega^{2}\right)^{2}}$.
4.224. $R=3 P / 16 \pi c \gamma \rho M_{C} \approx 0.6 \mu \mathrm{m}$.
5.1. (a) 3 and $9 \mathrm{~mW}$; (b) $\Phi=1 / 2\left(V_{1}+V_{2}\right) \Phi_{e} / A=1.6 \mathrm{~lm}$, where $A=1.6 \mathrm{~mW} / \mathrm{lm}, V_{1}$ and $V_{2}$ are the values of relative spectral response of an eye for the given wavelengths.
5.2. $E_{m}^{2}=V \overline{\mu_{0} / \varepsilon_{0}} A \Phi / 2 \pi r^{2} V_{\lambda}$, hence $E_{m}=1.1 \mathrm{~V} / \mathrm{m}, \quad H_{m}=$ $=3.0 \mathrm{~mA} / \mathrm{m}$. Here $A=1.6 \mathrm{~mW} / \mathrm{lm}, V_{\lambda}$ is the relative spectral response of an eye for the given wavelength.
5.3
(a) $\langle E\rangle=1 / 2 E_{0}$;
(b) $\langle E\rangle=\frac{1-\sqrt{1-(R / l)^{2}}}{1-R / l} \frac{I}{R^{2}}=50 \mathrm{~lx}$.
5.4. $M=2 / 3 \pi L_{0}$.
5.5. (a) $\Phi=\pi L \Delta S \sin ^{2} \theta$; (b) $M=\pi L$.
5.6. $h \approx R, E=L S / 4 R^{2}=40 \mathrm{~lx}$.
5.7. $I=I_{0} / \cos ^{3} \theta, \Phi=\pi I_{0} R^{2} / h^{2}=3 \cdot 10^{2} \mathrm{~lm}$.
5.8. $E_{\max }=(9 / 16 \pi V \overline{3}) \rho E S / R^{2}=0.21 \mathrm{~lx}$, at the distance $R / \sqrt{3}$ from the ceiling.
5.9. $E=\pi L$.
5.10. $E=\pi L$.
5.11. $M=E_{0}\left(1+h^{2} / R^{2}\right)=7 \cdot 10^{2} \mathrm{~lm} / \mathrm{m}^{2}$.
5.12. $E_{0}=\pi L R^{2} / h^{2}=25 \mathrm{~lx}$.
5.13. $\mathrm{e}^{\prime}=\mathbf{e}-2(\mathrm{en}) \mathrm{n}$.
5.14. Suppose $n_{1}, n_{2}, n_{3}$ are the unit vectors of the normals to the planes of the given mirrors, and $\mathbf{e}_{0}, \mathbf{e}_{1}, \mathbf{e}_{2}, \mathbf{e}_{3}$ are the unit vectors of the incident ray and the rays reflected from the first, second, and the third mirror. Then (see the answer to the foregoing problem): $\mathbf{e}_{1}=e_{0}-2\left(\mathbf{e}_{0} \mathbf{n}_{1}\right) \mathbf{n}_{1}, e_{2}=e_{1}-2\left(e_{1} n_{2}\right) n_{2}, \mathbf{e}_{3}=e_{2}-2\left(e_{2} n_{3}\right) n_{3}$. Summing termwise the left-hand and right-hand sides of these expressions, it can be readily shown that $\mathbf{e}_{3}=-\mathbf{e}_{c}$.
5.15. $\theta_{1}=\arctan n=53^{\circ}$.
5.16. $n_{1} / n_{2}=1 / \sqrt{\eta^{2}-1}=1.25$.
5.17. $x=\left[1-\sqrt{\left(1-\sin ^{2} \theta\right) /\left(n^{2}-\sin ^{2} \theta\right)}\right] d \sin \theta=3.1 \mathrm{~cm}$.
5.18. $h^{\prime}=\left(h n^{2} \cos ^{3} \theta\right) /\left(n^{2}-\sin ^{2} \theta\right)^{3 / 2}$.
5.21. $\Theta=83^{\circ}$.
5.22. From 37 to $58^{\circ}$.
5.23. $\alpha=8.7^{\circ}$.
5.24. $\Delta \alpha=\frac{2 \sin (\theta / 2)}{\sqrt{1-n^{2} \sin ^{2}(\theta / 2)}} \Delta n=0.44^{\circ}$.
5.27. (a) $f=l \beta /\left(1-\beta^{2}\right)=10 \mathrm{~cm}$;
(b) $f=l \beta_{1} \beta_{2} /\left(\beta_{2}-\beta_{1}\right)=$ $=2.5 \mathrm{~cm}$.
5.28. $I^{\prime}=\rho I_{0} f^{2}(f-s)^{2}=2.0 \cdot 10^{3} \mathrm{~cd}$.
5.29. Suppose $S$ is a point source of light and $S^{\prime}$ its image (Fig. 38). According to Fermat’s principle the optical paths of all rays originating at $S$ and converging at $S^{\prime}$ are equal. Let us draw circles with the centres at $S$ and $S^{\prime}$ and radii $S O$ and $S^{\prime} M$. Consequently, the optical paths $(D M)$ and $(O B)$ must be equal:
\[
n \cdot D M=n^{\prime} \cdot O B \text {. }
\]

However, in the case of paraxial rays $D M \approx A O+O C$, where $A O \approx h^{2} /(-2 s)$ and $O C \approx h^{\prime 2} / 2 R$. Besides, $O B=O C-B C \approx$ $\approx h^{\prime 2} / 2 R-h^{\prime 2} / 2 s^{\prime}$. Substituting these expressions into $(*)$ and taking into account that $h^{\prime} \approx h$, we obtain $n^{\prime} / s^{\prime}-n / s=\left(n^{\prime}-n\right) / R$.
Fig. 38.
5.30. $x=\frac{n f}{n+1}\left(1-\sqrt{1-\frac{(n+1) r^{2}}{(n-1) f^{2}}}\right), \quad r_{\max }=f \sqrt{(n-1) /(n+1)}$.
5.31. $6.3 \mathrm{~cm}$.
5.32. (a) $\beta=1-d(n-1) / n R=-0.20$; (b) $E=\pi n^{2} D^{2} L / 4 d^{2}=$ $=42 \mathrm{~lx}$.
5.33. (a) $\Phi=\Phi_{0}\left(n-n_{0}\right) /(n-1)=2.0 \mathrm{D}, f^{\prime}=-f=n_{0} / \Phi=$ $=85 \mathrm{~cm}$; (b) $\Phi=1 / 2 \Phi_{0}\left(2 n-n_{0}-1\right) /(n-1)=6.7 \mathrm{D}, \quad f=$ $=1 / \Phi \approx 15 \mathrm{~cm} \cdot f^{\prime}=n_{0} / \Phi \approx 20 \mathrm{~cm}$. Here $n$ and $n_{0}$ are the refractive indices of glass and water.
5.35. $\Delta x \approx \Delta l f^{2} /(l-f)^{2}=0.5 \mathrm{~mm}$.
5.36. (a) $f=\left[l^{2}-(\Delta l)^{2}\right] / 4 l=20 \mathrm{~cm}$;
(b) $f=l \sqrt{\eta} /(1+\sqrt{\eta})^{2}=20 \mathrm{~cm}$.
5.37. $h=\sqrt{h^{\prime} h^{n}}=3.0 \mathrm{~mm}$.
5.38. $E=(1-\alpha) \pi L D^{2} / 4 f^{2}=15 \mathrm{~lx}$.
5.39. (a) Is independent of $D$; (b) is proportional to $D^{2}$.
5.40. $f=n_{0} R / 2\left(n_{1}-n_{2}\right)=35 \mathrm{~cm}$, where $n_{0}$ is refractive index of water.
5.41. $f=R / 2(2 n-1)=10 \mathrm{~cm}$.
5.42. (a) To the right of the last lens at the distance $3.3 \mathrm{~cm}$ from it; (b) $l=17 \mathrm{~cm}$.
5.43. (a) 50 and $5 \mathrm{~cm}$; (b) by a distance of $0.5 \mathrm{~cm}$.
5.44. $\Gamma=D / d$.
5.45. $\psi=\psi^{\prime} / \sqrt{\eta}=0.6^{\prime}$.
5.46. $\Gamma^{\prime}=(\Gamma+1) \frac{n-n_{0}}{n_{0}(n-1)}-1=3.1$, where $n_{0}$ is the refractive index of water.
336
5.47. $\Gamma \leqslant D / d_{0}=20$.
5.48. $\Gamma=60$.
5.49. (a) $\Gamma=2 \alpha l_{0} / d_{0}=15$, where $l_{0}$ is the distance of the best vision $(25 \mathrm{~cm})$; (b) $\Gamma \leqslant 2 \alpha l_{0} / d_{0}$.
5.50. The principal planes coincide with the centre of the lens. The focal lengths in air and water: $f=-1 / \Phi=-11 \mathrm{~cm}, f^{\prime}=$ $=n_{0} / \Phi=+15 \mathrm{~cm}$. Here $\Phi=\left(2 n-n_{0}-1\right) / R$, where $n$ and $n_{0}$ are the refractive indices of glass and water. The nodal points coincide and are located in water at the distance $x=f^{\prime}+f=3.7 \mathrm{~cm}$ from the lens.
5.51. See Fig. 39.
5.54. (a) The optical power of the system is $\Phi=\Phi_{1}+\Phi_{2}-$ $-d \Phi_{1} \Phi_{2}=+4 \mathrm{D}$, the focal length is $25 \mathrm{~cm}$. Both principal planes
Fig. 39.
are located in front of the converging lens: the front one at a distance of $10 \mathrm{~cm}$ from the converging lens, and the rear one at a distance of $10 \mathrm{~cm}$ from the diverging lens $\left(x=d \Phi_{2} / \Phi\right.$ and $\left.x^{\prime}=-d \Phi_{1} / \Phi\right)$; (b) $d=5 \mathrm{~cm}$; about $4 / 3$.
5.55. The optical power of the given lens is $\Phi=\Phi_{1}+\Phi_{2}-$ $-(d / n) \Phi_{1} \Phi_{2}, x=d \Phi_{2} / n \Phi=5.0 \mathrm{~cm}, x^{\prime}=-d \Phi_{1} / n \Phi=2.5 \mathrm{~cm}$, i.e. both principal planes are located outside the lens from the side of its convex surface.
5.56. $f=\frac{f_{1} f_{2}}{f_{1}+f_{2}-d}$. The lens should be positioned in the front principal plane of the system, i.e. at a distance of $x=$ $=f_{1} d /\left(f_{1}+f_{2}-d\right)$ from the first lens.
5.57. $\Phi=2 \Phi^{\prime}-2 \Phi^{\prime 2} l / n_{0}=3.0 \mathrm{D}$, where $\Phi^{\prime}=\left(2 n-n_{0}-1\right) / R$, $n$ and $n_{0}$ are the refractive indices of glass and water.
5.58. (a) $d=n \Delta R /(n-1)=4.5 \mathrm{~cm}$; (b) $d=3.0 \mathrm{~cm}$.
5.59. (a) $\Phi=d(n-1)^{2} / n R^{2}>0$, the principal planes are located on the side of the convex surface at a distance of $d$ from each other, with the front principal plane being removed from the convex surface of the lens by a distance of $R /(n-1)$; (b) $\Phi=\left(1 / R_{2}-1 / R_{1}\right) \times$ $\times(n-1) / n<0$; both principal planes pass through the common curvature centre of the surfaces of the lens.
5.60. $d=1 / 2 n\left(R_{1}+R_{2}\right) /(n-1)=9.0 \mathrm{~cm}, \Gamma=R_{1} / R_{2}=5.0$.
5.61. $\Phi=2\left(n^{2}-1\right) / n^{2} R=37 \mathrm{D}$.
5.63. $\rho=3 \cdot 10^{7} \mathrm{~m} ;|
abla n|=1.6 \cdot 10^{-7} \mathrm{~m}^{-1}$.
5.65. 1.9a.
5.66. Let us represent the $k$ th oscillation in the complex form
\[
\xi_{k}=a \mathrm{e}^{i[\omega t+(k-1) \varphi]}=a_{k}^{*} \mathrm{e}^{i \omega t},
\]
where $a_{k}^{*}=a \mathrm{e}^{i(k-1) \varphi}$ is the complex amplitude. Then the complex amplitude of the resulting oscillation is
\[
\begin{aligned}
A^{*}=\sum_{k=1}^{N} a \mathrm{e}^{i(k-1) \varphi} & =a\left[1+\mathrm{e}^{\mathrm{i} \varphi}+\mathrm{e}^{\mathrm{i} 2 \varphi}+\ldots+\mathrm{e}^{i(N-1) \varphi}\right]= \\
& =a\left(\mathrm{e}^{\mathrm{i} \varphi N}-1\right) /\left(\mathrm{e}^{\mathrm{i} \varphi}-1\right) .
\end{aligned}
\]

Multiplying $A^{*}$ by the complex conjugate value and extracting the square root, we obtain the real amplitude
\[
A=a \sqrt{\frac{1-\cos N \varphi}{1-\cos \varphi}}=a \frac{\sin (N \varphi / 2)}{\sin (\varphi / 2)} .
\]
5.67. (a) $\cos \theta=(k-\varphi / 2 \pi) \lambda / d, \quad k=0, \pm 1, \pm 2, \ldots$; (b) $\varphi=\pi / 2, d / \lambda=k+1 / 4, k=0,1,2, \ldots$
5.68. $\Delta \varphi=2 \pi[k-(d / \lambda) \sin (\omega t+\alpha)]$, where $k=0, \quad \pm \mathbf{1}$, $\pm 2, \ldots$
5.69. $\lambda=2 \Delta x \Delta h / l(\eta-1)=0.6 \mu \mathrm{m}$.
5.71. (a) $\Delta x=\lambda(b+r) / 2 \alpha r=1.1 \mathrm{~mm}, 9$ maxima; (b) the shift is $\delta x=(b / r) \delta l=13 \mathrm{~mm}$; (c) the fringe pattern is still sharp when $\delta x \leqslant \Delta x / 2$, hence $\delta_{\max }=(1+r / b) \lambda / 4 \alpha=43 \mu \mathrm{m}$.
5.72. $\lambda=2 \alpha \Delta x=0.64 \mu \mathrm{m}$.
5.73. (a) $\Delta x=\lambda f / a=0.15 \mathrm{~mm}, 13$ maxima; (b) the fringes are still sufficiently sharp when $\delta x \leqslant \Delta x / 2$, where $\delta x$ is the shift of the fringes from the extreme elements of the slit, hence, $\delta_{\max }=$ $=\lambda f^{2} / 2 a b=37 \mu \mathrm{m}$.
5.74. $\lambda=2 a \Theta(n-1) \Delta x /(a+b)=0.6 \mu \mathrm{m}$.
5.75. $\Delta x \approx \lambda / 2 \Theta\left(n-n^{\prime}\right)=0.20 \mathrm{~mm}$.
5.76. The fringes are displaced toward the covered slit over the distance $\Delta x=h l(n-1) / d=2.0 \mathrm{~mm}$.
5.77. $n^{\prime}=n+N \lambda / l=1.000377$.
5.78. (a) Let $\mathbf{E}, \mathbf{E}^{\prime}$, and $\mathbf{E}^{\prime \prime}$ be the electric field vectors in the incident, reflected and transmitted waves. Select the $x$-, $y$-axes at the interface so that they coincide in direction with $\mathbf{E}$ and $\mathbf{H}$ in the incident wave.

The continuity of the tangential components across the interface yields
\[
\mathbf{E}+\mathbf{E}^{\prime}=\mathbf{E}^{\prime \prime} .
\]

The minus sign before $\mathbf{H}^{\prime}$ appears because $\mathbf{H}^{\prime} 1 \uparrow \mathbf{H}$.
Rewrite the second equation taking into account that $\mathbf{H} \propto n E$. Solving the obtained and the first equation find:
\[
\mathbf{E}^{\prime \prime}=2 \mathbf{E} n_{1} /\left(n_{1}+n_{2}\right) .
\]
338
Hence, we see that $\mathbf{E}^{\prime \prime}$ and $\mathbf{E}$ are collinear, that is. cophasal.
\[
\text { (b) } \mathbf{E}^{\prime}=\mathbf{E}\left(n_{1}-n_{1}\right) /\left(n_{1}+n_{2}\right) \text {, }
\]
that is at $n_{2}>n_{1}$ and $\mathbf{E}^{\prime} \downarrow \mid \mathbf{E}$ the phase abruptly changes by $\pi$ at the interface. If $n_{2}<n_{1}$ the phase jump does not occur.
5.79. $d=1 / 4 \lambda(1+2 k) / \sqrt{n^{2}-\sin ^{2} \theta_{1}}=0.14(1+2 k) \mu \mathrm{m}$, where $k=0,1,2, \ldots$
5.80. $d_{\text {min }}=0.65 \mu \mathrm{m}$.
5.81. $d=1 / 4 \lambda(1+2 k) / V \bar{n}$, where $k=0,1,2, \ldots$
5.82. $d=\lambda \frac{\sqrt{n^{2}-\sin ^{2} \theta}}{\sin 2 \theta \cdot 8 \theta}=15 \mu \mathrm{m}$.
5.83. $\lambda \approx \frac{d\left(r_{i}^{2}-r_{k}^{2}\right)}{4 n l^{2}(i-k)}$.
5.84. $\Delta x=\frac{\lambda \cos \theta_{1}}{2 \alpha \sqrt{n^{2}-\sin ^{2} \theta_{1}}}$.
5.85. (a) $\Theta=1 / 2 \lambda / n \Delta x=3^{\prime}$; (b) $\Delta \lambda / \lambda \approx \Delta x / l=0.014$.
5.86. $\Delta r \approx 1 / 4 \lambda R / r$.
5.87. $r^{\prime}=V \overline{r^{2}-2 R \Delta h}=1.5 \mathrm{~mm}$.
5.88. $r=\sqrt{r_{0}^{2}+(k-1 / 2) \lambda R}=3.8 \mathrm{~mm}$, where $k=6$.
5.89. $\lambda=1 / 4\left(d_{2}^{2}-d_{1}^{2}\right) / R\left(k_{2}-k_{1}\right)=0.50 \mu \mathrm{m}$, where $k_{1}$ and $k_{2}$ are the numbers of the dark rings.
5.90. $\Phi=2(n-1)(2 k-1) \lambda / d^{2}=2.4 \mathrm{D}$, where $k$ is the number of the bright ring.
5.91. (a) $r=\sqrt{2 k \lambda(n-1) / \Phi}=3.5 \mathrm{~mm}$, where $k=10$; (b) $r^{\prime}=$ $=r / \sqrt{n_{0}}=3.0 \mathrm{~mm}$, where $n_{0}$ is the refractive index of water.
5.93. $k_{\text {min }}=1 / 2 \lambda_{1} /\left(\lambda_{2}-\lambda_{1}\right)=140$.
5.94. The transition from one sharp pattern to another occurs if the following condition is met:
\[
(k+1) \lambda_{1}=k \lambda_{2},
\]
where $k$ is a certain integer. The corresponding displacement $\Delta h$ of the mirror is determined from the equation $2 \Delta h=k \lambda_{2}$. From these two equations we get
\[
\Delta h=\frac{\lambda_{1} \lambda_{2}}{2\left(\lambda_{2}-\lambda_{1}\right)} \approx \frac{\lambda^{2}}{2 \Delta \lambda}=0.3 \mathrm{~mm} .
\]
5.95. (a) The condition for maxima: $2 d \cos \theta=k \lambda$; hence, the order of interference $k$ diminishes as the angle $\theta$, i.e. the radius of the rings, increases (see Fig. 5.18). (b) Differenting both sides of the foregoing equation and taking into account that on transition from one maximum to another the value of $k$ changes by unity, we obtain $\delta \theta=1 / 2 \lambda / d \sin \theta$; this shows that the angular width of the fringes decreases with an increase of the angle $\theta$, i.e. with a decrease in the order of interference.
5.96.(a) $k_{\text {max }}=2 d \lambda=1.0 \cdot 10^{3}$, (b) $\Delta \lambda=\lambda / k=\lambda \cdot / 2 d=5$ pm.
5.97. $I_{0}=\frac{2}{6 N \lambda} \int r(r) r d r$.
5.98. $b=a r^{2} /\left(k \lambda a-r^{2}\right)=2.0 \mathrm{~m}$.
5.99. $\lambda=\left(r^{2}-r^{2}\right)(a+b) / 2 a b=0.60 \mu \mathrm{m}$.
5.100. (a) $I^{\prime} \approx 4 I_{6}, I \approx 2 I_{4} ;$ (b) $I \approx I_{6}$.
5.101. (a) $I \approx 0$; (b) $I \approx I, 2$.
5.102. (a) $I_{1} \approx \theta_{14} I_{6}, I_{3}=1 / 4 I_{6}, I_{3}=1 / 14 I_{6}, I_{4}=I_{2}, \quad I \approx$ $=I_{6} I \approx(1+\mp / 2 \pi)^{2} I_{6}$. Here $\varphi$ is the angle covered by the screen.
5.103. (a) $h=\lambda(k+3 / 8) /(n-1)=1.2(k+3 / 8) \mu \mathrm{m} ;$ (b) $h=$ $=1.2(k+7 / 8) \mu \mathrm{m}$, (c) $h=1.2 k$ or $1.2(k+3 / 4) \mu \mathrm{m}$. Here $k=$ $=0,1,2, \ldots$
5.104. $h=\lambda(k+3 / 4) /(n-1)$, where $k=0,1,2, \ldots$, (b) $I_{\mathrm{mat}} \approx 8 I_{\text {, }}$
5.105. $h_{\text {min }} \approx \lambda(k+5 / 8) /(n-1)=2.5 \mu \mathrm{m}$, where $k=2$.
5.106. $r=\sqrt{k \lambda f b /(b-n)}=0.90 \sqrt{k} \mathrm{~mm}$, where $k=1,3,5, \ldots$
5.107. $b^{\prime}=b / \eta^{2}-1.0 \mathrm{~m}$.
5.108. (a) $y^{\prime}=y b / a=9 \mathrm{~mm} ;$ (b) $h_{\min } \approx a b M D(a+b)=$ $=0.10 \mathrm{~mm}$.
5.109. $f=a b /(a+b)=0.6 \mathrm{~m}$. This value corresponds to the principal focal point, apart from which there are other points as well.
5.119. (a) $h^{h}=0.60(2 k+1) \mathrm{mm} ;$
(b) $h=0.30(2 k+1) \mu m$. Here $\boldsymbol{k}=0,1,2, \ldots$
5.111. (a) $I_{\max } j I_{\text {mis }} \approx 1.7, \quad$ (b) $\quad \lambda=2(\Delta x)^{2} / b\left(v_{1}-v_{2}\right)^{2}=$ $=0.7 \mu \mathrm{m}$, where $v_{2}$ and $v_{1}$ are the corresponding values of the parameter along Cornu’s spiral.
5.112. $I_{\text {equn }} / I_{\text {ett }} \approx 2.6$.
5.113. $\lambda=(\Delta h)^{f} / 2 b\left(v_{1}-v_{1}\right)^{2}=0.55 \mu \mathrm{m}$, where $v_{1}$ and $v_{2}$ are the corresponding values of the parameter along Cornu’s spiral.
5.114. $h \approx \lambda(k+3 / 4) /(n-1)$, where $k=0,1,2, \ldots$
5.115. $I_{4} I_{1} \approx 1.9$.
5.116. $I \approx 2.8 I_{8}$
5.117. $I_{1}: I_{2}: I_{3} \approx 1: 4: 7$.
5.118. $t \approx I_{0}$
5.119. $I_{0} \sim\left(\sin ^{2} \alpha\right) / \alpha^{2}$, where $a=(a b / \lambda) \sin \theta ; \quad b \sin \theta=k \lambda$, $k=1,2,3, \ldots$
5.120. The condition for a maximum leads to the transcendental equation $\tan \alpha=\alpha$, where $\alpha=(\pi b / \lambda) \sin \theta$. The solution of this equation (by means of plotting or selection) provides the following root values: $a_{1}=1.43 \pi, \alpha_{2}=2.46 \pi, \alpha_{2}=3.47 \pi$. Hence $b \sin \theta_{1}=$ $=1.43 \lambda, b \sin \theta_{1}=2.46 \lambda, b \sin \theta_{3}=3.47 \lambda$.
5.121. b $\left(\sin \theta-\sin \theta_{0}\right)=k \lambda_{i}$ for $k=+1$ and $k=-1$ the angles $\theta$ are equal to $33^{\circ}$ and $27^{\circ}$ respectively.
36
5.122. (a) $\Delta \theta=\arcsin (n \sin \theta)-\theta=7.9^{\circ}$; (b) from the condition $b\left(\sin \theta_{1}-n \sin \theta\right)= \pm \lambda$ we obtain $\Delta \theta=\theta_{+1}-\theta_{-1}=$ $=7.3^{\circ}$.
5.123. $\lambda \approx\left(a^{2}-a\right) d / 2 k=0.6 \mu \mathrm{m}$.
5.125. $55^{\circ}$.
5.126. $d=2.8 \mathrm{\mu m}$.
5.127. $\lambda=(d \sin \Delta \theta) / y \overline{5-4 \cos \Delta \theta}=0.54 \mu \mathrm{m}$.
5.128. (a) $45^{\circ} ;$ (b) $-64^{\circ}$.
5.129. $x=2 R /(\mathrm{n}-1) V(d / 2)^{2}-1=8 \mathrm{~cm}$.
5.130. From the condition $d\left[n \sin \theta-\sin \left(\theta+\theta_{2}\right)\right]=k \lambda=$ obtain $\theta_{4}=-18.5^{\circ}, \quad \theta_{41}=0^{\circ} ; k_{\text {mux }}=+6, \theta_{44}=+78.5^{\circ}$. See Fig. 40 .
5.131. $h_{k}-\lambda(k-1 / 2) /(n-1)$, where $k=1,2, \ldots$; $a \sin \theta_{1}-\lambda / 2$.
$5.132 v=\lambda v / / \Delta x=1.5 \mathrm{~km} / \mathrm{s}$.
5.133. Each star produces its own diffraction pattern in the objective’s focal plane, with their zeroth maxima being separated
Fig. 40.
ทis. 41.
by an angle \& (Fig. 41). As the distance $d$ decreases the angle $\theta$ between the neighbouring maxima in each diffraction pattern increases, and when $\theta$ becomes equal to $2 \%$, the first deterioration of visibility occurs: the maxima of one system of fringes coincide with the minima of the other system. Thus, from the condition $\theta=2 \mathrm{y}$ and the formule $\sin \theta=2 / d$ we obtain $\downarrow=\lambda / 2 d \approx 0.06^{\circ}$.
5.134. (a) $D=k / d \sqrt{1-(k N / d)^{2}}=6.5$ ang. min/nm, where $k=2$; (b) $D=k / d \sqrt{1-\left(k \lambda / d-\sin \theta_{4}\right)^{2}}=13$ ang. $\mathrm{min} / \mathrm{nm}$, where $k=4$. 5.135. $d \theta / d \lambda=(\tan \theta) / \lambda$.
5.136. $\Delta \theta-2 \lambda / N d \sqrt{1-(k N / \delta)^{1}}=11^{\circ}$.
5.139. $\theta=46^{\circ}$.
5.140. (a) In the fourth order; (b) $\delta \lambda_{\min } \approx \lambda^{1 / l}=7 \mathrm{pm}$.
5.141. (a) $d=0.05 \mathrm{~mm}$; (b) $l=6 \mathrm{~cm}$.
5.142. (a) 6 and 12 um: (b) not in the first order, yes in the secend order.
5.143. According to Rayleigh’s eriterion the maximum of the line of wavelength $\lambda$ must coincide with the first minimum of the line of wavelength $\lambda+8 \lambda$. Let $y$ s write both conditions for the least deviation angle in terms of the optical path differences for the extreme rays (see Fig. 5.28):
\[
b n-(D C+C E)=0, \quad b(n+8 n)-(D C+C E)-\lambda+8 \lambda .
\]

Hence, $b \delta n \approx \lambda$. What follows is obvious.
5.144. (a) $N / 8 \lambda-2 b B / \lambda^{3} ; 1.2 \cdot 10^{4}$ and $0.35 \cdot 10^{4}$
(b) $1.0 \mathrm{~cm}$.
5.145. Abeut $20 \mathrm{~cm}$.
5.146. $R=7.10^{4}, \Delta y_{\mathrm{mtn}} \approx 4 \mathrm{~cm}$.
5.147. About $50 \mathrm{~m}$.
5.148. Suppose $\Delta \psi$ and $\Delta \psi^{\prime}$ are the minimum angular separations resolved by the telescope’s objective and the eye respectively $\left(\Delta \psi=1.22 N D, \Delta \psi^{\prime}=1.222 / d_{0}\right)$. Then the sought magnification of the telescope is $\mathrm{r}_{\mathrm{Nin}}=\Delta \psi^{\prime} / \Delta \psi_{4}-D / d_{4}=13$.
5.149. $d_{\text {min }}=0.61 \lambda{ }^{2} \sin a=1.4 \mu \mathrm{m}$.
5.150. Suppose $d_{\text {sis }}$ is the minimum separation resolved by the microscope’s objective, $\Delta \psi$ is the angle subtended by the eye at the object over the distance of the best visibility $l_{0}(25 \mathrm{~cm})$, and $\Delta \psi^{\prime}$ is the minimum angular separation resolved by the eye $\left(\Delta \psi^{\prime}=\right.$ $\left.=1.22 \mathrm{~N} d_{0}\right)$. Then the sought magnification of the microscope is $\mathrm{r}_{\text {mtn }}=\Delta \psi^{\prime} / \Delta \psi=2\left(L_{0} / d_{0}\right) \sin a=30$.
5.151. 26, 60, 84, iof and $134^{\circ}$.
5.152. $a=0.28 \mathrm{~nm}, b=0.41 \mathrm{~nm}$.
5.153. Suppose $\alpha, \beta$, and $\gamma$ are the angles between the direction to the diffraction maximum and the directions of the array along the periods a, b, and e respectively. Then the values of these angles can be found from the following conditions: $a(1-\cos \alpha)=k_{1} \lambda$, $b \cos \beta=k_{1} \lambda$, and $c \cos \gamma=k_{2} \lambda$. Recalling that $\cos ^{2} \alpha+\cos ^{2} \beta+$ $+\cos ^{2} \gamma=1$, we obtain
\[
\lambda=\frac{2 k_{1} / a}{\left(k_{1} / a\right)^{2}+\left(k_{2} /\right)^{2}+\left(k_{2} / e\right)^{2}}
\]
5.154. $\lambda=\frac{2}{k} \sqrt[3]{\frac{\pi}{2}} \sin \alpha=244$ pm, where $k=2, m$ is the mass of a $\mathrm{NaCl}$ molecule.
5.155. $d=\frac{\lambda}{2 \sin (\alpha / 2)} \sqrt{R_{1}+R_{1}-2 k_{1} k_{2} \cos (\alpha / 2)}=0.28 \mathrm{pm}$, where $k_{1}$ and $k_{2}$ are the orders of relection.
5.156. $r=i \tan 2 \alpha=3.5 \mathrm{~cm}$, where $\alpha$ is the glancing angle found from the condition $2 d$ sin $a=k \lambda$.
5.157. $I_{0} / 4$.
5.158. (a) $I_{\mathrm{s}}$ (b) $2 I_{6}$.
5.159. $\boldsymbol{E}=\pi \Phi_{0} / \omega=0.6 \mathrm{~mJ}$.
5.160. $\eta=1 / 2(\cos q)^{2(N-1)}=0.12$
5.161. $I_{0} / I=\frac{2}{x^{5} \cos ^{6} \varphi} \approx 60$.
5.162. $I_{\text {pel }} / I_{\text {nat }}=P /(1-P)=0.3$.
5.163. $P=(\eta-1) /(1-\eta \cos 2 \varphi)=0.8$.
5.164. (a) Let us represent the natural light as a sum of two mutualIy perpendicular components with intensities $I_{0}$. Suppose that each polarizer transmits in its plane the fraction $\alpha_{1}$ of the light with oscillation plane parallel to the polarizer’s plane, and the fraction $\alpha_{3}$ with oscillation plane perpendicular to the polarizer’s plane. The intensity of light transmitted through the system of two polsrisers is then equal to
\[
I_{n}=\alpha_{4}^{3} I_{4}+\alpha_{*}^{*} I_{n}
\]
when their planes are parallel, and to
\[
I_{4}=a_{1} \alpha_{4} I_{6}+a_{4} a_{1} I_{6}
\]
when their planes are perpendicular; according to the condition, $I_{0} I_{i}=\eta$.

On the other hand, the degree of polarization produced separately by each polarizer is
\[
P_{0}=\left(a_{1}-a_{2}\right) /\left(a_{1}+a_{2}\right) \text {. }
\]

Eliminating $\alpha_{1}$ and $\alpha_{1}$ from these equations, we get
\[
P_{0}=\sqrt{(\eta-1) /(\eta+1)}=0.905 .
\]
(b) $P=V \overline{1-1 / \eta^{2}}=0.995$.
5.165. The relative intensity variations of both beams in the cases $A$ and $B$ are

Hence
\[
(\Delta I /)_{,}=4 \cot (\varphi / 2) \cdot \delta \varphi, \quad(\Delta I / I)=-4 \tan (\varphi / 2) \cdot 8 \varphi .
\]
\[
\eta=(\Delta / / I)_{A} /(\Delta / / I)_{n}=\cot ^{2}(\varphi / 2), \varphi=11.5^{\circ} .
\]
5.166. $90^{\circ}$.
5.167. (a) $\rho=1 / 2\left(n^{4}-1\right)^{*} /\left(n^{2}+1\right)^{2}=0.074$;
(b) $P=\rho /(1-\rho)=\frac{\left(1+n^{2}\right)^{2}-4 n^{2}}{\left(1+n^{2}\right)^{2}+4 n^{2}}=0.080$. Here $n$ is the refractive index of glass.
5.168. $I=I_{0}(1-\rho) / n=0.721 I_{6}$, where $n$ is the refractive index of water.
5.169. $p=\left[\left(n^{2}-1\right) /\left(n^{2}+1\right)\right] \sin ^{2} \varphi=0.038$, where $n$ is the refractive index of water.
5.170. $P_{1}=P_{3}=1, \quad P_{2}=\frac{\rho}{1-p}=0.087, \quad P_{4}=\frac{2 p(1-\rho)}{1-2 p(1-p)}=0.17$.
5.171. (a) In this case the coefficient of reflection from each surface of the plate is equal to $p=\left(n^{2}-1\right)^{2} /\left(n^{2}+1\right)^{2}$, and therefore $I_{4}=I_{0}(1-p)^{2}=16 I_{6} n^{4}\left(1+n^{2}\right)^{2}=0.725 I_{4}$
(b) $P=\frac{1-(1-\rho)^{2}}{1+\left(1-p^{2}\right)^{2}}=\frac{\left(1+n^{4}\right)^{2}-16 n^{4}}{\left(1+n^{4}+16 n^{4}\right.} \approx 0.16$, where $\rho^{\prime}$ is the coeflicient of refection for the component of light whose electric vector oscillates at right angles to the incidence plane.
5.172. (a) $P=\left(1-a^{4 N}\right) /\left(1+a^{i n}\right)$, where $a=2 n /\left(1+n^{2}\right)$, $n$ is the refractive index of glass; (b) $0.16,0.31,0.67$, and 0.92 respectively.
5.173. (a) $\rho=(n-1)^{2} /(n+1)^{2}=0.040$; (b) $\Delta \Phi / \Phi=1-$
– $(1-\rho)^{2 N}=0.34$, where $N$ is the number of lenses.
5.175. (a) 0.83 ; (b) 0.044 .
Fig. 42.
5.176. See Fig. 42, where $o$ and $e$ are the ordinary and extraordinary rays.
5.177. $\delta \approx 11^{\circ}$.
5.178. For the right-handed system of coordinates:
(1) circular anticlockwise polarization, when observed toward the incoming wave;
(2) elliptical clockwise polarization, when observed toward the incoming wave; the major axis of the ellipse coincides with the straight line $y=x$;
(3) plane polarization, along the straight line $y=-x$.
5.179. (a) $0.490 \mathrm{~mm}$; (b) $0.475 \mathrm{~mm}$.
5.180. $\lambda=4 d \Delta n /(2 k+1) ; 0.58,0.55$ and $0.51 \mu \mathrm{m}$ respectively at $k=15,16$ and 17 .
5.181. Four.
5.182. 0.69 and $0.43 \mu \mathrm{m}$.
5.183. $d=(k-1 / 2) \lambda_{1} / \Delta n=0.25 \mathrm{~mm}$, where $k=4$.
5.184. $\Delta n=\lambda / \Theta \Delta x=0.009$.
5.185. Let us denote the intensity of transmitted light by $I_{\perp}$ in the case of the crossed Polaroids, and by $I_{\mathrm{ll}}$ in the case of the parallel Polaroids. Then
\[
\begin{array}{l}
I_{1}=1 / 2 I_{0} \sin ^{2} 2 \varphi \cdot \sin ^{2}(\delta / 2), \\
I_{\|}=1 / 2 I_{0}\left[1-\sin ^{2} 2 \varphi \cdot \sin ^{2}(\delta / 2)\right] .
\end{array}
\]

The conditions for the maximum and the minimum:

Here $\Delta$ is the optical path difference for the ordinary and extraordinary rays, $k=0,1,2, \ldots$
344
5.187. (a) The light with right-hand circular polarization (from the observer’s viewpoint) becomes plane polarized on passing through a quarter-wave plate. In this case the direction of oscillations of the electric vector of the electromagnetic wave forms an angle of $+45^{\circ}$ with the axis $O O^{\prime}$ of the crystal (Fig. $43 a$ ); in the case of lefthand polarization this angle will be equal to $-45^{\circ}$ (Fig. $43 \mathrm{~b}$ ).
(b) If for any position of the plate the rotation of the Polaroid (located behind the plate) does not bring about any variation in the intensity of the transmitted light, the initial light is natural;
Fig. 43.
if the intensity of the transmitted light varies and drops to zero, the initial light is circularly polarized; if it varies but does not drop to zero, then the initial light is composed of natural and circularly polarized light.
5.188. (a) $\Delta x=1 / 2 \lambda\left(n_{e}-n_{0}\right) \Theta$,
(b) $d\left(n_{o}^{\prime}-n_{e}^{\prime}\right)=$ $=-2\left(n_{e}-n_{0}\right) \Theta \delta x<0$.
5.189. $\Delta n=\alpha \lambda / \pi=0.71 \cdot 10^{-4}$, where $\alpha$ is the rotational constant.
5.190. $\alpha=\pi / \Delta x \tan \Theta=21$ ang. deg./mm, $I(x) \sim \cos ^{2}(\pi x / \Delta x)$, where $x$ is the distance from the maximum.
5.191. $d_{\min }=(1 / \alpha) \arcsin \sqrt{2 \eta}=3.0 \mathrm{~mm}$.
5.192. $8.7 \mathrm{~mm}$.
5.193. $[\alpha]=72$ ang. $\operatorname{deg} . /\left(\mathrm{dm} \cdot \mathrm{g} / \mathrm{cm}^{3}\right)$.
5.194. (a) $E_{\min }=1 / \sqrt{4 B l}=10.6 \mathrm{kV} / \mathrm{cm}$;
(b) $2.2 \cdot 10^{8}$ interruptions per second.
5.195. $\Delta n=2 c H V / \omega$, where $c$ is the velocity of light in vacuum.
5.196. $V=1 / 2\left(\varphi_{1}-\varphi_{2}\right) / l H=0.015$ ang. $\mathrm{min} / \mathrm{A}$.
5.197. If one looks toward the transmitted beam and counts the positive direction clockwise, then $\varphi=(\alpha-V N H) l$, where $N$ is the number of times the beam passes through the substance (in Fig. 5.35 the number is $N=5$ ).
5.198. $H_{\min }=\pi / 4 V l=4.0 \mathrm{kA} / \mathrm{m}$, where $V$ is the Verdet constant. The direction along which the light is transmitted changes to the opposite.
5.199. $t=m c \omega_{0} / \lambda I=12$ hours. Although the effect is very small, it was observed both for visible light and for SHF radiation.
is $v^{\prime}=a / d^{\prime}$. Due to the Doppler effect the observed frequency is
\[
v=v \frac{\sqrt{i-(v / \epsilon)^{T}}}{1-(v / \epsilon) \cos 6}=\frac{n d}{i-(v / \sigma) \cos \theta} .
\]

The corresponding wavelength is $\lambda=c / v=d(c / v-\cos \theta)$. When $\theta=45^{\circ}$ and $v \approx c$ the wavelength is $\lambda \approx 0.6 \mu \mathrm{m}$.
5.238. (a) Let $v_{x}$ be the projection of the velocity vector of the radiating atom on the obervation direction. The number of atoms with projections falling within the interval $v_{x}, v_{x}+d v_{x}$ is
\[
n\left(v_{x}\right) d v_{x} \sim \exp \left(-m v_{2} / 2 k T\right) \cdot d v_{x}
\]

The frequency of light emitted by the atoms moving with velocity $v_{s}$ is $\omega=\omega_{0}\left(1+v_{2} J c\right)$. From the expression the frequency distribution of atoms can be found: $n(\omega) d \omega=n\left(v_{n}\right) d v_{x}$. And finally it should be taken into account that the spectral radiation intensity
5.239. $u=\frac{e / n+V}{1+V / e n}$. If $V<c$, then $u \approx \frac{e}{n}+V\left(1-\frac{1}{n^{2}}\right)$.
5.240. $v=1 / 2 \mathrm{c} 6 \theta=\mathbf{3 0} \mathrm{km} / \mathrm{s}$.
5.24.2. $\theta^{\prime}=8^{\prime}$.
5.243. The field induced by a charged partiele moving with velocity $V$ excites the atoms of the medium turning them into sources of light waves. Let us consider two arbitrary points $A$ and $B$ along the path of the particle. The light waves emitted from these points when the particle passes them reach the point $P($ Fig. 4) simultaneously and amplify each other provided the time taken by the light wave to propagate from the point $A$ to the point $C$ is equal to that taken by the particle to dy over the distance $A B$. Hence, we obtain $\cos \theta=v / V$, where $v=c / n$ is the phase velocity of light. It is evident that the radiation is posible only if $V>v$, i.e. when the velocity of the particle exceeds the phase velocity of light in the medium.
5.244. $T_{\min }=\left(n / V \overline{n^{2}-1}-1\right) m e^{2} ; 0.14 \mathrm{MeV}$ and $0.26 \mathrm{GeV}$ respectively. For muons.
5.245. $T=\left(\frac{s \cos \theta}{V^{2} \cos x-1}-1\right) m e^{2}=0.23 \mathrm{MeV}$.
5.247. $T_{2}=b T_{1} /\left(b+T_{1} \Delta \lambda\right)=1.75 \mathrm{kK}$.
5.248. $\lambda_{\mathrm{m}}=3.4 \mu \mathrm{m}$.
$5.249 .5 \cdot 10^{\circ} \mathrm{kg} / \mathrm{s}$, about $10^{n}$ years.
5.250. $T=\sqrt[3]{3 \epsilon h \rho / \sigma M}=2.10^{\circ} \mathrm{K}$, where $A$ is the universal gas constant, $M$ is the molar mass of hydregen.
368
5.251. $t=\left(\eta^{2}-1\right)$ epd/18aT; -3 hours, where $c$ is the specific heat capacity of copper, $\rho$ is its density.
5 252. $T_{3}=T_{1} \sqrt{\text { dा2 }}=0.4 \mathrm{kK}$.
5.253. (a) $C_{v}=(\partial U / \partial T)_{v}=16 \sigma T^{2 V} / \epsilon=3 \mathrm{~nJ} / \mathrm{K}$, where $U=$ $=4 \sigma T^{*} V / e ;$ (b) $s=16 a T V / 3 e=1.0 \mathrm{~nJ} / \mathrm{K}$.
5.254. (a) $\omega_{p r}=3$ T/a $=7.85 \cdot 10^{14} \mathrm{s-1} ;$ (b) $\lambda_{p r}=2$ rea $/ 5 T=$ $-1.44 \mu \mathrm{m}$. $b$ is the constant in Wien’s displacement law.
5.258. (a) $1.1 \mu \mathrm{m} ;$ (b) 0.37 ; (c) $P_{2} / P_{1}=\left(T, / T_{1}\right)^{4}\left(1-y_{1}\right) /\left(1-y_{1}\right)=$ $=4.9$.
5.220. (a) (i) $=$ PN/8nefrr $=6 \cdot 10^{43} \mathrm{~cm}^{-1} \mathrm{~s}^{-1}$ :
(b) $r=\sqrt{P N / 2 \hbar / 2 a c}-9 \mathrm{~m}$.
5.251. $d p / d t=\Phi_{e} / c$.
5.352. (p) $-4(1+p)$ E/ad\”et $\approx 50 \mathrm{~atm}$.
5.263. $p=(E / c) \sqrt{1+\rho^{2}+2 p \cos 20}=35 \mathrm{nN} \cdot \mathrm{s}$.
5.364. $p=(I / c)(1+p) \cos ^{2} \theta=0.6 \mathrm{aN} / \mathrm{cm}^{2}$.
5.255. $F=a R^{2} / / e=0.18 \mu \mathrm{N}$.
5.266. $F=P / 2 e\left(1+\eta^{7}\right)$.
5.267. (a) $\Delta p=\frac{210}{\epsilon} \frac{\sqrt{1-\beta}}{1-\beta}$;
(b) $\Delta_{p}=\frac{210}{c} \frac{1}{1-\beta}$.
Here $\beta=$ $\sim$ V/e. It is evident that in the reference frame fixed to the mirror the latter obtains the smaller momentum.
5.268. sin $(\theta / 2) \approx E / m e V \bar{g}$, $\theta=0.5^{\circ}$.
5.259. $\Delta \omega_{0} / \omega_{0}=-\left(1-e^{-7 M / R e v}\right)<0$, i.e. the frequency of the photon decreases.
5.270. $V=2 \pi h e(1-1 / \eta) / e \Delta \lambda-16 \mathrm{kV}$.
5.271. $V=\pi$ theled $\sin \alpha=31 \mathrm{kV}$.
5.272. $\lambda_{\text {min }}=2 x h / m e(\gamma-1)=2.8 \mathrm{pm}$, where $\gamma=1 / \sqrt{1-(v / c)^{2}}$.
5.273. $332 \mathrm{~nm}, 6.6 \cdot 10^{\circ} \mathrm{m} / \mathrm{s}$.
5.274. $A=2 \pi c h \frac{\left(\eta^{2}-\lambda_{2} / \lambda_{1}\right)}{\lambda_{2}\left(\eta^{2}-1\right)}=1.9 \mathrm{eV}$.
5.275. $\varphi_{\max }=4 . \hat{4}^{2} \mathrm{~V}$.
5.276. $T_{\max }=\boldsymbol{\Lambda}\left(\omega_{0}+\omega\right)-A_{1}=0.38 \mathrm{eV}$
5.277. $w=2 \mathrm{reN} / \mathrm{ei}=0.020$.
5.278. $v_{\text {max }}=6.4 \cdot 10^{4} \mathrm{~m} / \mathrm{s}$.
5.279. $0.5 \mathrm{~V}_{\mathrm{i}}$ the polarity of the contact potential difference is opposite to that of external voltage.
5.280 . $\mathrm{A} / \mathrm{mc}$, the Compton wavelength for the given particle.
5.281. Let us write the energy and momentum conservation laws in the reference frame fixed to the electron for the moment preceding the collision with the photon: $A \in+m_{c} c^{2}=m c^{2}$, hele $=m v$, where $m=m_{0} y \overline{1-(v / c)^{2}}$. From this it follows that $v=0$ or $v=c$. The results have no physical meaning.
5.282. (a) Light is scattered by the free electrons; (b) the increase of the aumber of electrons that turn free (the free electrons have the binding energy much lower than the energy transferred to them by the photons); (c) the presence of a non-displaced component is due to scattering by the strongly bound electrons and the nuclei.
5.283. $\lambda=4 \pi \lambda_{c} \mid \sin \left(\theta_{1} / 2\right)-\eta \sin \left(\theta_{1} / 2\right) V(\eta-1)=1.2 \mathrm{pm}$.
5.284. $T=\operatorname{hen} /(1+\eta)=0.20 \mathrm{MeV}$.
5.285. (a) $\omega^{\prime}=2 \pi e /(\lambda+2 \pi / / m e)=2.2 \cdot 10^{\mathrm{m} n} \mathrm{rad} / \mathrm{s} ;$
(b) $T=\frac{2 \mathrm{men} / \lambda}{1+\lambda m / 2 \pi \mathrm{A}}=60 \mathrm{keV}$.
5.287. $\sin (\theta / 2)=V \overline{m c\left(p-p^{\prime}\right) / 2 p p^{\prime}}$. Hence $\theta=120^{\circ}$.
5.288. $\mathrm{A}_{\omega}=\left[1+V \overline{1+2 m c^{2} / T \sin ^{2}(\Theta / 2)}\right] T / 2=0.68 \mathrm{MeV}$.
5.289. $\lambda=(2 \pi \mathrm{n} / \mathrm{me})\left(\sqrt{1+2 \mathrm{me} / T_{\operatorname{mux}}}-1\right)=3.7 \mathrm{pm}$.
5.291. $\rho=\frac{2 n(1+v) m c}{(1+2 v) \epsilon B}=3.4 \mathrm{~cm}$.
5.292. $\Delta \lambda=(4 \mathrm{~h} / \mathrm{mc}) \sin ^{2}(\theta / 2)=1.2 \mathrm{pm}$.
6.1. $r=3 e^{2} / 2 E=0.16 \mathrm{~nm}, \lambda=(2 \pi / e) V \overline{m r}=0.24 \mu \mathrm{m}$.
6.2. $b=0.73 \mathrm{pm}$.
6.3. (a) $r_{\text {mis }}=0.59 \mathrm{pm} ;$ (b) $r_{\text {mis }}=(27 \mathrm{R} / T)\left(1+m_{\mathrm{a}} / m_{\text {Li }}\right)=$ $=0.034 \mathrm{pm}$.
6.4. (a) $P_{\text {ais }}-\left(Z e^{2} / T\right) \cot ^{2}(\theta / 2)=0.23 \mathrm{pm} ;$
(b) $r_{\text {mis }}=$ $-[1+\operatorname{cosec}(\theta / 2)] \mathrm{Ze}^{2} / T=0.56 \mathrm{pm}$.
6.5. $\left.p \approx 2 V 2 m T / 1+\left(2 b T / Z e^{2}\right)^{2}\right]$.
6.6. $T_{e}=m_{2} e^{4} / m_{s} b^{2} T=4$ eV.
6.7. $b=\frac{h_{n} \sin (\theta / 2)}{\sqrt{1+n^{2}-2 n \cos (4 / 2)}}$, where $n=\sqrt{1+U_{n} / T}$.
6.8. (a) $\cos (\theta / 2)=b /(R+r)$; (b) $d P=1 / 2 \sin \theta d \theta$; (c) $P=$ $=1 / 2$.
6.9. $3.3 \cdot 10^{-4}$
6.10. $d=\left(4 j r^{4} T^{4} / n I Z^{2} e^{4}\right) \sin ^{4}(\theta / 2)=1.5 \mu \mathrm{m}$, where $n$ is the concentration of nuclei.
6.11. $\mathrm{Z}_{\mathrm{pt}}-\mathrm{Z}_{\mathrm{Af}} \sqrt{\eta A_{\mathrm{pd}} / A_{\mathrm{Ag}}}=78$.
6.12. (a) $1.6 \cdot 10^{2}$; (b) $N=$ and $\left(Z e^{2} / T\right)^{2} \cot ^{2}(6, / 2) I_{0} \mathrm{x}=2.0 \cdot 10^{2}$, where $n$ is the concentration of nuclei.
6.13. $P=$ and $\left(\mathrm{Ze}^{3} / \mathrm{mu}^{h}\right)^{2}=0.006$, where $n$ is the concentration of nuelei.
6.14. $\Delta N / N=1-\pi n Z^{2} \mu^{4} / T^{2} \tan ^{2}(8,2)=0.6$.

Nw
6.15. $\Delta N / N=\frac{\pi r^{4}}{4 r^{r}}\left(0.7 \frac{\eta}{M_{1}}+0.3 \frac{2}{M_{1}}\right) \rho d N_{A} \cot ^{2} \frac{\theta}{2}=1.4 \cdot 10^{-3}$, where $\mathbf{Z}_{4}$ and $\mathbf{Z}_{2}$ are the atomic numbers of copper and zibe. $M_{1}$ and $M_{2}$ are their molar masses, $N_{A}$ is Avegadro’s number.
6.16. $\Delta d=\pi\left(Z e^{2} / T\right)^{2} \cot ^{2}(\theta / / 2)=0.73 \mathrm{~kb}$.
6.17. (a) $0.9 \mathrm{MeV}$; (b) $d \sigma / d \Omega=\Delta \sigma / 4 \pi \sin ^{4}(\theta / 2)=0.64 \mathrm{~kb} / \mathrm{sp}$.
6.18. $t=\left(3 m e^{2} / 2 e^{i} \omega^{2}\right) \ln n=15$ ns.
6.19. $t \approx m^{2} c+1 / 4 \mu^{4} \approx 13$ ps.
6.21. $r_{0}=\sqrt{n \lambda / m \omega}, E_{n}=n \lambda \omega$, where $n=1,2, \ldots, \omega=$ $=\sqrt{k / m}$.
6.22.
6.23. $\omega=m e^{4} Z^{2} / n^{3} n^{3}=2.07 \cdot 10^{44} s^{-1}$,
6.24. $\mu_{n}=n e h / 2 m c, \mu_{n} / M_{n}=e / 2 m c, \mu_{1}=\mu_{n}$.
6.25. $B=m^{2} e^{2} / \mathrm{ch}^{3}=125 \mathrm{kG}$.
6.27. The Brackett series, $\lambda_{4-4}=2.63 \mu \mathrm{m}$.
6.28. (a) 657,487 and $434 \mathrm{~nm} ;$ (b) $2 / 82 \approx 1.5 \cdot 10^{\circ}$.
6.29. For $n \gg 1 \sin \theta \approx n^{2} \pi c i l R$, whence $\theta \approx 60^{\circ}$.
6.30. $\mathrm{He}^{+}$,
6.31. $\boldsymbol{N}=1 / 2 n(n-1)$.
6.32. $97.3,102.6$ and $121.6 \mathrm{~nm}$.
6.33. $n=5$.
6.34. $R=\frac{176 a t}{15243}=2.07 \cdot 104 \mathrm{~s}^{-1}$.
6.36. $\lambda=(2 \pi c / \Delta \omega)(2 V \overline{R / \Delta \omega}-1) /(2 z V \overline{R / \Delta \omega}-1)=0.47 \mu \mathrm{m}$.
6.37. $E_{b}=54.4 \circ \mathrm{V}\left(\mathrm{He}^{2}\right)$.
6.38. $E=E_{0}+4 h R=79 \cdot \mathrm{V}$.
6.40. $T_{\text {min }}=-1 / 2 n=20.5$ eV.
6.41. $y=3 M R / 4 m e=3.25 \mathrm{~m} / \mathrm{s}$, where $m$ is the mass of the atom.
6.42. $\left(\varepsilon-\varepsilon^{\prime}\right) / \varepsilon \approx 3 M R / 8 m c^{2}=0.55 \cdot 10^{-4 \%}$, where $m$ is the mass of the atom.
6.43. $v=2 \sqrt{\overline{h R / m}}-3.1 \cdot 10^{4} \mathrm{~m} / \mathrm{s}$, where $m$ is the mass of the electron.
6.44. $v=3 R \Delta M / 8 \pi \cos \theta=0.7 \cdot 10^{\mathrm{m}} \mathrm{m} / \mathrm{s}$.
6.45. (a) $E_{n}=n^{2} n^{2} n^{2} / 2 m l^{2}$; (b) $E_{n}=n^{2} n^{2} / 2 m r^{2}$;
(c) $E_{n}=$ $=n \hbar V \sqrt{\alpha / m} ;$ (d) $E_{n}=-m \alpha^{2} / 2 n^{2} n^{2}$.
6.46. $E_{\mathrm{b}}=\mu e^{4} / 2 n^{2}, R=\mu e^{4} / 2 n^{3}$, where $\mu$ is the reduced mass of the system. If the motion of the nucleus is not taken into account. these values (in the case of a hydrogen stom) are greater by $\mathrm{m} / \mathrm{M} \approx$ $\approx 0.055 \%$, where $m$ and $M$ are the masses of an electron and a proton.
6.47. $E_{\mathrm{p}}-E_{\mathrm{H}}=3.7 \mathrm{meV}, \lambda_{\mathrm{n}}-\lambda_{\mathrm{D}}-33 \mathrm{pm}$.
6.48. (a) $0.285 \mathrm{pm}, 2.53 \mathrm{keV}, 0.65 \mathrm{~nm}$; (b) $106 \mathrm{pm}, 6.8 \mathrm{eV}$, $0.243 \mu \mathrm{m}$.
6.49. $123,2.86$ and $0.186 \mathrm{pm}$.
6.50. $0.45 \mathrm{keV}$.
6.51. For both particles $\lambda=2 \pi h\left(1+m_{n} / m_{d}\right) / \sqrt{2 m_{n} T}=8.6 \mathrm{pm}$.
6.52. $\tau=2 \lambda_{1} \lambda_{2} / \sqrt{\lambda_{3}+\lambda_{2}^{*}}$.
6.53. $\lambda=2 \pi / / 2 \pi=128 \mathrm{pm}$.
6.54. First, let us find the distribution of molecules over de Broglie wavelengths. From the relation $f(v) d v=-\varphi(\lambda)$ di. where $f(v)$ is Maxwell’s distribution of velocities, we obtain
\[
\varphi(\lambda)=A \lambda-6-4 / 3, \quad \epsilon=2 \pi^{3} h^{2} / n k T \text {. }
\]

The condition $d q / d \lambda=0$ provides $\lambda_{y}=n N / V \overline{m K T}=0.09 \mathrm{~nm}$,
6.55. $\lambda=2 \pi N / \sqrt{2 m T\left(1+T / 2 m c^{2}\right)}, T \leqslant 4 m e^{2} \Delta \lambda / \lambda=20.4 \mathrm{keV}$ (for an electron) and $37.5 \mathrm{MeV}$ (for a proton).
6.56. $T=(\sqrt{2}-1) m e^{2}=0.21 \mathrm{MeV}$.
6.57. $\lambda=\lambda, N \sqrt{1+m e \lambda, \sqrt{\pi} \lambda}=3.3 \mathrm{pm}$.
6.58. $v=4 \mathrm{xhl} / \mathrm{mbs} \mathrm{x}=2.0 \cdot 10^{*} \mathrm{~m} / \mathrm{s}$.
6.59. $\Delta x=2 \mathrm{x} \Lambda l / d \sqrt{2 m e V}=4.9 \mu \mathrm{m}$.
6.60. $V_{0}=a^{2} n^{2} / 2 m e(\sqrt{\eta}-1)^{2} d^{2} \sin ^{2} \theta=0.15 \mathrm{keV}$.
6.61. $d=\pi h k / \sqrt{2 m T} \cos (\theta / 2)=0.21 \mathrm{~nm}$, where $k=4$.
6.62. $d=\pi A k / \sqrt{2 m T} \sin \theta=0.23 \pm 0.04 \mathrm{~nm}$, where $k=3$ and the angle $\theta$ is determined by the formula $\tan 2 \theta=D / 2 i$.
6.63. (a) $n=\sqrt{1+V_{i} V}=1.05$; (b) $V_{/} V_{1}>1 / \eta(2+\eta)=50$.
6.64. $E_{n}=n^{2} n^{2} n^{2} / 2 m l^{2}$, where $n=1,2, \ldots$
6.66. $1.10^{4}, 1.10$ and $1.10^{-10} \mathrm{~cm} / \mathrm{s}$.
6.67. $\Delta v \approx A / \mathrm{ml}=1 \cdot 10^{4} \mathrm{~m} / \mathrm{s} ; v_{1}=2.2 \cdot 10^{4} \mathrm{~m} / \mathrm{s}$.
6.69. $\Delta t \approx \eta m l^{2} / \mathrm{h} \approx 10^{-14} \mathrm{~s}$.
6.70. $T_{\text {pin }^{i n}} \approx A^{2} / 2 m l^{2}=1$ eV. Here we assumed that $p \approx \Delta p$ and $\Delta x=i^{i}$.
6.71. $\Delta v / v \sim M / I \sqrt{2 m T}=1 \cdot 10^{-4}$,
6.72. $F \approx A^{2} / m P$.
6.73. Taking into account that $p \sim \Delta p \sim M / \Delta x \sim M / x$, we get $E=T+U \approx n^{2} / 2 m x^{2}+k x^{2} / 2$. From the condition $d E / d x=0$ we find $x_{0}$ and then $E_{\min } \approx n V \sqrt{k / m}=n \omega$, where $\omega$ is the oscillator’s angular frequency. The rigorous calculations furnish the value ก $\omega / 2$.
6.74. Taking into account that $p \sim \Delta p \sim M / \Delta r$ and $\Delta r \sim r$, we get $E=p^{2} / 2 m-e^{2} / r \approx A^{3} / 2 m r^{3}-e^{2} / r$. From the condition
359
$d E / d r=0$ we find $r_{e t} \approx A^{4} / m e^{4}=53 \mathrm{pm}, E_{\text {min }} \approx-m e^{4} / 2 A^{3}=$ $=-13.6 \mathrm{ev}$.
6.75. The width of the image is $\Delta \approx \delta+\Delta^{\prime} \approx \delta+\Delta U_{p} \delta$, where $\Delta^{\prime}$ is an additional widening associated with the uncertainty of the momentum $\Delta p_{y}$ (when the hydrogen atoms pass through the slit), $p$ is the momentum of the incident hydrogen atoms. The funetion $\Delta(\delta)$ has the minimum when $\delta \approx \sqrt{\overline{h l} / m v}=0.01 \mathrm{~mm}$.
6.76. The solution of the Schroblinger equation should be sought in the form $\Psi=\Psi(x) \cdot f(t)$. The substitution of this function into the initial equation with subsequent separation of the variables $z$ and $t$ results in two equations. Their solutions are $\downarrow(x) \sim e^{\text {in } x}$, where $k=\sqrt{2 m E} / h, E$ is the energy of the particle, and $f(f) \sim$ tain constant.
6.77. $P=1 / 3+V / 3 / 2 \pi=0.61$.
6.78. $\leqslant=\left\{\begin{array}{l}A \cos (\pi n x / l), \text { if } n=1,3,5, \ldots, \\ A \sin (\pi n x / b), \text { if } n=2,4,6, \ldots\end{array}\right.$

Here $A=\sqrt{2 / 1}$.
6.80. $d N / d E=(l / a h) V / \overline{m / 2 E}$; if $E=1 \mathrm{eV}$, then $d N / d E=$ $=0.8 \cdot 10^{\prime}$ levels per eV.
6.81. (a) In this case the Schrodinger equation takes the form
\[
\frac{\partial^{2} v}{\partial r^{2}}+\frac{\partial^{2} v}{\partial v^{2}}+k^{2} q=0, k^{2}=2 m E / n^{2}
\]

Let us take the origin of coordinates at one of the corners of the well. On the sides of the well the function $\$(x, y)$ must turn into sero (according to the condition), and therefore it is convenient to seek this function inside the well in the form $\downarrow(x, y)=a \sin k_{1} x \times$ $X$ sin $k_{2} y$, since on the two sides $(x-0$ and $y=0), \psi=0$ automs. tically. The possible values of $k_{1}$ and $k_{2}$ are found from the condition of $\$$ turning inte zero on the opposite sides of the well:
\[
\begin{array}{ll}
\downarrow\left(l_{1}, y\right)=0, & k_{1}= \pm\left(\pi / l_{1}\right) n_{1}, \quad n_{1}=1,2,3, \ldots \\
\downarrow\left(x, l_{2}\right)=0, & k_{3}= \pm\left(\pi / l_{2}\right) n_{2}, \quad n_{3}=1,2,3, \ldots
\end{array}
\]

The substitution of the wave function into the Schrobinger equation leads to the relation $k_{i}+k_{i}=k^{3}$, whence
(b) $9.87,24.7,39.5$, and 49.4 units of $\mathrm{A}^{2} / \mathrm{ml}^{2}$.
6.82. $P=1 / 3-\sqrt{3} / 4 \pi=19.5 \%$.
6.83. (a) $E=\left(n_{1}^{2}+n_{3}^{2}+n_{3}^{2}\right) \pi^{2} n^{2} / 2 m a^{2}$, where $n_{3}, n_{3}, n_{3}$ are integers not equal to zere: (b) $A E=a^{2} n^{1} / \mathrm{ma}^{2}$; (c) for the b-th level $n_{i}^{2}+n_{j}^{2}+n_{3}^{2}-14$ and $E=7 n^{2} n^{2} / m a^{2}$; the number of states is equal o six (it is equal to the number of permutations of a triad 1, 2, 3.)
6.84. Let us integrate the Schrödinger equation over a small interval of the coordinate $x$ within which there is a discontinuity in $U(x)$, for example at the point $x=0$ :
\[
\frac{\partial \psi}{\partial x}(+\delta)-\frac{\partial \psi}{\partial x}(-\delta)=\int_{-\delta}^{+\delta} \frac{2 m}{\hbar^{2}}(E-U) \psi d x .
\]

Since the discontinuity $U$ is finite the integral tends to zero as $\delta \rightarrow 0$. What follows is obvious.
6.85. (a) Let us write the Schrödinger equation for two regions
\[
\begin{array}{l}
0<x<l, \quad \psi_{1}^{\prime \prime}+k^{2} \psi_{1}=0, k^{2}=2 m E / \hbar^{2}, \\
x>l, \quad \psi_{2}^{\prime \prime}-x^{2} \psi_{2}=0, \quad x^{2}=2 m\left(U_{0}-E\right) / \hbar^{2} .
\end{array}
\]

Their common solutions
\[
\psi_{1}(x)=a \sin (k x+\alpha), \quad \psi_{2}(x)=b \mathrm{e}^{-x x}+c \mathrm{e}^{x x}
\]
must satisfy the standard and boundary conditions. From the condition $\psi_{1}(0)=0$ and the requirement for the finiteness of the wave
Fig. 45.
function it follows that $\alpha=0$ and $c=0$. And finally, from the continuity of $\psi(x)$ and its derivative at the point $x=l$ we obtain $\tan k l=-k / x$, whence
\[
\sin k l= \pm k l \sqrt{\hbar^{2} / 2 m l^{2} U_{0}} .
\]

Plotting the left-hand and right-hand sides of the last equation (Fig. 45), we can find the points at which the straight line crosses the sine curve. The roots of the equation corresponding to the eigenvalues of energy $E$ are found from those intersection points $(k l)_{i}$ for which $\tan (k l)_{i}<0$, i.e. the roots of that equation are located in the even quadrants (these segments of the abscissa axis are shown heavy in the figure). It is seen from the plot that the roots of the equation, i.e. the bound states of the particle, do not always exist. The dotted line indicates the ultimate position of the straight line.
(b) $\left(l^{2} U_{0}\right)_{1 \text { min }}=\pi^{2} \hbar^{2} / 8 m, \quad\left(l^{2} U_{0}\right)_{n \text { min }}=(2 n-1) \pi^{2} \hbar^{2} / 8 m$.
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6.86. Suppose that $P_{a}$ and $P_{i}$ are the probabilities of the particle being outside and inside the well. Then
\[
\frac{P_{a}}{P_{i}}=\frac{\int_{i}^{\infty} b^{2} \mathrm{e}^{-2 x x} d x}{\int_{0}^{l} a^{2} \sin ^{2} k x d x}=\frac{2}{2+3 \pi},
\]
where the ratio $b / a$ can be found from the condition $\psi_{1}(l)=\psi_{2}(l)$. Now it remains to take into account that $P_{a}+P_{i}=1$; then $P_{a}=$ $=2 /(4+3 \pi)=14.9 \%$.

The penetration of the particle into the region where its energy $E<U$ is a purely quantum phenomenon. It occurs owing to the wave properties of the particle ruling out the simultaneous precise magnitudes of the coordinate and the momentum, and consequently the precise division of the total energy of the particle into the potential and the kinetic energy. The latter could be done only within the limits set by the uncertainty principle.
6.87. Utilizing the substitution indicated, we get
\[
\chi^{\prime \prime}+k^{2} \chi=0, \text { where } k^{2}=2 m E / \hbar^{2} .
\]

We shall seek the solution of this equation in the form $\chi=$ $=a \sin (k r+\alpha)$. From the finiteness of the wave function $\psi$ at the point $r=0$ it follows that $\alpha=0$. Thus, $\psi=(a / r) \sin k r$. From the boundary condition $\psi\left(r_{0}\right)=0$ we obtain $k r_{0}=n \pi$, where $n=1,2, \ldots$ Hence, $E_{n}=n^{2} \pi^{2} \hbar^{2} / 2 m r_{0}^{2}$.
6.88. (a) $\psi(r)=\frac{1}{\sqrt{2 \pi r_{0}}} \frac{\sin \left(n \pi r / r_{0}\right)}{r}, n=1,2, \ldots$;
(b) $r_{p r}=$ $=r_{0} / 2 ; 50 \%$.
6.89. (a) The solutions of the Schrödinger equation for the function $\chi(r)$ :
\[
\begin{array}{l}
r<r_{0}, \chi_{1}=A \sin (k r+\alpha), \text { where } k=\sqrt{2 m E} / \hbar, \\
r>r_{0}, \chi_{2}=B \mathrm{e}^{x r}+C \mathrm{e}^{-x r}, \text { where } x=\sqrt{2 m\left(U_{0}-E\right)} / \hbar .
\end{array}
\]

Since the function $\psi(r)$ is finite throughout the space, $\alpha=0$ and $B=0$. Thus,
\[
\psi_{1}=A \frac{\sin k r}{r}, \quad \psi_{2}=C \frac{\mathrm{e}^{-x r}}{r} .
\]

From the continuity of the function $\psi$ and its derivative at the point $r=r_{0}$ we get $\tan k r_{0}=-k / x_{3}$ or
\[
\sin k r_{0}= \pm \sqrt{\hbar^{2} / 2 m r_{0}^{2} U_{0}} k r_{0} .
\]

As it was demonstrated in the solution of Problem 6.85, this equation determines the discontinuous spectrum of energy eigenvalues. (b) $r_{0}^{2} U_{0}=\pi^{2} \hbar^{2} / 8 m$.
6.90. $\alpha=m \omega / 2 A, E=A \omega / 2$, where $\omega=V k / m$.
6.91. $E=-m e^{4} / 8 n^{3}$, i.e. the level with principal quantum number $n=2$.
6.92. (a) The probability of the electron being at the interval $r, r+d r$ from the nucleus is $d P=v^{2}(r) 4 y^{2} d r$. From the condition for the maximum of the function $d P / d r$ we $g e t r_{n}=r_{1}$ : (b) $(F)=2 e^{2} / r^{2},(c)(U)=-e^{2} / r_{\mathrm{r}}$.
6.93. $\varphi_{0}=\int(\rho / r) 4 \pi r^{2} d r=-e / r_{1}$, where $p=-q^{2}$ is the space charge density, $\phi$ is the normalized wave function.
6.94. (a) Let us write the solutions of the Schrobdinger equation to the left and to the right of the barrier in the following form:
\[
\begin{array}{l}
x<0, \hbar_{1}(x)=a_{1} e^{i k_{1} x}+b_{1} e^{-i \omega_{1} x} \text {, where } k_{1}=\sqrt{2 m E} / \hbar, \\
\left.x>0, \hbar_{2}(x)=a_{2} e^{i k_{4} x}+b_{2} e^{-i \omega_{y} x} \text {, where } k_{2}=\sqrt{2 m\left(E-U_{0}\right.}\right) / \hbar .
\end{array}
\]

Let us assume that the incident wave has an amplitude $a_{1}$ and the reflected wave an amplitude $b$, Since in the region $x>0$ there is only a travelling wave, $b_{1}=0$. The reflection coefficient $R$ is the ratio of the reflected stream of partieles to the incident stream, or, in other words, the ratio of the squares of amplitudes of corresponding waves. Due to the continuity of $\$$ and its derivative at the point $x=0$ we have $a_{1}+b_{1}=a_{2}$ and $a_{1}-b_{1}=\left(k_{2} / k_{1}\right) a_{3}$ whence
\[
\left.R=\left(b_{1} / a_{1}\right)^{2}=\left(k_{1}-k_{2}\right)\right)^{2} /\left(k_{1}+k_{2}\right)^{2} \text {. }
\]
(b) In the case of $E<U$, the solution of the Schrödinger equation to the right of the barrier takes the form
\[
\hbar_{1}(x)=a_{2} e^{x x}+b_{2} e^{-x} \text {, where } \mathrm{x}=\sqrt{2 m\left(U_{0}-B\right)} / \hbar .
\]

From the finiteness of $\$(x)$ it follows that $a_{3}=0$. The probability of finding the particle under the barrier has the density $P_{1}(x)=$ $=\phi^{\prime}(x) \sim e^{-4 x}$, Hence, $x_{e t}=1 / 2 x$.
6.95. (a) $D \approx \exp \left[-\frac{21}{\pi} \sqrt{2 m\left(U_{0}-E\right)}\right]$;
(b) $D \approx \exp \left[-\frac{8 \sqrt{2 m}}{30 U_{0}}\left(U_{0}-E\right)^{3 / 2}\right]$.
6.96. $D \approx \exp \left[-\frac{n l}{n} \sqrt{\frac{2 m}{U_{0}}}\left(U_{0}-E\right)\right]$;
6.97. -0.41 for an $S$ term and -0.04 for a $P$ term.
6.98. $a=\sqrt{\lambda /\left(E_{0}-e \varphi_{1}\right)}-3=-0.88$.
6.99. $E_{\mathrm{b}}=A R /\left(\sqrt{R \lambda_{1} \lambda_{2} / 2 \pi c \Lambda \lambda}-1\right)^{2}-5.3 \mathrm{eV}$.
6.100. $0.82 \mu \mathrm{m}(3 S \rightarrow 2 P)$ and $0.68 \mu \mathrm{m}(2 P \rightarrow 25)$.
6.101. $\Delta E=2 \pi \mathrm{A} \Delta \lambda \lambda^{2}=2.0 \mathrm{meV}$.
6.102. $\Delta_{\omega}=1.05 \cdot 10^{14} \mathrm{rad} / \mathrm{s}$.
6.103. $3 S_{v / 2}, 3 P_{1 / \mathrm{v}}, 3 P_{\mathrm{a} / \mathrm{s}} 3 D_{\mathrm{a} / \mathrm{p}}, 3 D_{\mathrm{s} / \mathrm{z}}$
6.104. (a) $1,2,3,4,5 ;$ (b) $0,1,2,3,4,5,6 ;$ (c) $1 / 2,3 / 2,5 / 2$, $7 / 2,9 / 2$.
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6.105. For the state ‘P: AV $\overline{3} / 2, A V \overline{15} / 2$, and $A V \overline{32} / 2$; for
6.106. (a) ${ }^{2} F_{1 / 2}, M_{\max }=\hbar V \frac{1}{2} / 2$, (b) $F_{6}, M_{\max }=2 n V 5$.
6.107. In the $F$ state $M,-M V \overline{6}$; for the $D$ state it can be only found that $M_{,}>\cap \boldsymbol{V} \overline{\text {. }}$
6.108. 3, 4, 5 .
6.109. (a) 1, 3, 5, 7, 9; (b) 2, 4, 6; (c) 5, 7, 9 .
6.110. $31^{\circ}$.
6.111. ‘ $D$,
6.113. The same as in the foregoing problem.
6.114. The second and the third term.
6.115. $g=4+6=10$.
6.116. 4,7 and 10.
6.117. ${ }^{*} F_{2}$.
6.118. As.
6.119. (a) ‘ $S_{\mathrm{ag}}$ (b) ( ${ }^{3} P_{\mathrm{z}}$.
6.121. (a) Two $d$ electrons; (b) five $p$ electrons; (c) five $d$ electrons,
6.122. (a) ${ }^{*} P_{0}$
(b) ${ } F_{3 / 2}$
6.123. $F_{\mathrm{av}} \mathrm{r}$
6.124. $\mu=\mu, \sqrt{35}\left(c S_{y_{1}}\right)$.
6.125. $\eta=n^{2}$-sear $=3 \cdot 10^{-17}$, where $\omega-R\left(1-1 / n^{2}\right)$.
6.126. $N / N_{0}=\left(\mathrm{g} / \mathrm{g}_{0}\right) \mathrm{e}^{-\mathrm{H} / \mathrm{kr}}=1.14 \cdot 10^{-1}$, where $g$ and $\mathrm{g}_{\mathrm{f}}$ are the statistical weights (degeneracy raties) of the levels $3 P$ and 35 respectively $\left(g=6, g_{0}-2\right)$.
6.127. $\mathrm{T}=V_{v} \ln n=1.3 \mu \mathrm{s}$.
6.128. $N=\lambda$ r. $P / 2$ ane $=7 \cdot 10^{\circ}$. are the degeneracy ratios of the resonant and the basic level.
6.130. (a) $P_{\text {in }} d P_{*}=1 /\left(e^{\operatorname{sos} / \omega T}-1\right) \approx 10^{-34}$, where $a=$ $=3 R_{i}$ (b) $T=1.7 \cdot 10^{0} \mathrm{~K}$.
6.131. Suppose that $I$ is the intensity of the passing ray. The decrease in this value on passing through the layer of the substance of thickness $d x$ is equal to
\[
-d I=x I d x=\left(N_{1} B_{14}-N_{4} B_{n}\right)(I / c) A_{\omega} d x \text {, }
\]
where $N_{1}$ and $N_{\text {, are }}$ the concentrations of atoms on the lower and upper levels, $B_{10}$ and $B_{n}$ a re the Einstein coefficients. Hence
\[
\text { ж }=(\mathrm{Ae} / \mathrm{e}) N_{1} B_{14}\left(1-\mathrm{g}_{2} N_{2} / \mathrm{g}_{2} N_{1}\right) .
\]

Next, the Boltrmann distribution should be taken into consideration, as well as the fact that $A_{\omega}>k T$ (in this case $N_{1}$ is approximately equal to $N_{6}$, the total concentration of the atems).
6.132. $\Delta \lambda_{\text {Dep }} / \Delta \lambda_{\text {sat }} \approx 4 \pi w_{p r} / \lambda \approx 10^{\circ}$, where $v_{p r}=V \sqrt{2 \pi T / M}$.
6.133. $\lambda=154 \mathrm{pm}$.
6.134. (a) $843 \mathrm{pm}$ for $\mathrm{Al}, 180 \mathrm{pm}$ for $\mathrm{Co}$; (b) $\approx 5 \mathrm{keV}$.
6.135. Three.
6.136. $V=15 \mathrm{kV}$.
6.137. Yes.
6.138. $Z=1+2 \sqrt{(n-1) e V_{1} / 3 \hbar R\left(n-V_{1} / V_{2}\right)}=29$.
6.139. $Z=1+\sqrt{4 \Delta \omega / 3 R}=22$, titanium.
6.140. $E_{b}=3 / 4 \hbar R(Z-1)^{2}+2 \pi c \hbar / \lambda_{L}=5.5 \mathrm{keV}$.
6.141. $E_{L}=\hbar \omega(2 \pi c / \omega \Delta \lambda-1) \approx 0.5 \mathrm{keV}$, where $\omega=$ $=3 / 4 R(Z-1)^{2}$.
6.142. $T=3 / 4 \hbar R(Z-1)^{2}-2 \pi c \hbar / \lambda_{K}=1.45 \mathrm{keV}, \quad v=$ $=2.26 \cdot 10^{7} \mathrm{~m} / \mathrm{s}$.
6.143. (a) $g=2$, with the exception of the singlet state, where $g=0 / 0$; (b) $g=1$.
6.144. (a) $-2 / 3$; (b) 0 ; (c) 1 ; (d) $5 / 2$; (e) $0 / 0$.
6.145. (a) $\sqrt{12} \mu_{B}$; (b) $2 \sqrt{3 / 5} \mu_{B}$; (c) $(8 / \sqrt{3}) \mu_{B}$.
6.146. $M_{s}=2 \sqrt{3} \hbar$.
6.147. $\mu=(8 / \sqrt{3}) \mu_{B}$.
6.148. $\mu=3 \sqrt{7 / 5} \mu_{B}$.
6.149. $\mu-(5 \sqrt{5} / 2) \mu_{B}$.
6.150. $M=\hbar \sqrt{3} / 2$.
6.151. ${ }^{5} F_{1}$.
6.152. $\omega=\mu_{B} / g B / \hbar=1.2 \cdot 10^{10} \mathrm{rad} / \mathrm{s}$, where $g$ is the Landé factor.
6.153. $F_{\max }=\mu_{B \max } \cdot|\partial B / \partial z|=(3 / \sqrt{8}) \pi I g J \mu_{B} / c r^{2}=$ $=4 \cdot 10^{-27} \mathrm{~N}$.
6.154. $F=2 I \mu_{B} / c r^{2}=3 \cdot 10^{-26} \mathrm{~N}$.
6.155. $\partial B / \partial z=2 T \delta / g J \mu_{B} l_{1}\left(l_{1}+2 l_{2}\right)=15 \mathrm{kG} / \mathrm{cm}$.
6.156. (a) It does not split; (b) splits into 6 sublevels; (c) does not split $(g=0)$.
6.157. (a) $58 \mu \mathrm{eV}$; (b) $\Delta E=2 g J \mu_{B} B=145 \mu \mathrm{eV}$.
6.158. (a) Normal; (b) anomalous; (c) normal; (d) normal (both terms have identical Landé factors).
6.159. $L=\Delta E / 2 \mu_{B} B=3 ;{ }^{1} F_{3}$.
6.160. $\Delta \lambda=\lambda^{2} e B / 2 \pi m c^{2}=35 \mathrm{pm}$.
6.161. $B_{m t n}=4.0 \mathrm{kG}$.
6.162. $B \stackrel{m i n}{=} \hbar \Delta \omega / g \mu_{B}=3 \mathrm{kG}$.
6.163. (a) 2.1 (the ratio of the corresponding Landé factors); (b) $B=2 \pi c \hbar \Delta \lambda / g \mu_{B} \eta \lambda^{2}=5.5 \mathrm{kG}$.
6.164. $\Delta \omega=( \pm 1.3, \pm 4.0, \pm 6.6) \cdot 10^{10} \mathrm{~s}^{-1}$, six components.
6.165. (a) Six (1) and four (2); (b) nine (1) and six (2).
6.166. $\Delta \omega=\left(m_{1} g_{1}-m_{2} g_{2}\right)_{m a x} e B / m c=1.0 \cdot 10^{11} \mathrm{~s}^{-1}$.
6.167. $\omega=4 \sqrt{2} \hbar / m d^{2}=1.57 \cdot 10^{11} \mathrm{~s}^{-1}$, where $m$ is the mass of the molecule.
6.168. 2 and 3 .
6.169. $M=\sqrt{m d^{2} E / 2}=3.5 \hbar$, where $m$ is the mass of the molecule.
6.170. $I=\hbar / \Delta \omega=1.93 \cdot 10^{-40} \mathrm{~g} \cdot \mathrm{cm}^{2}, d=112 \mathrm{pm}$.
6.171. 13 levels.
6.172. $N \approx \sqrt{2 I \omega / \hbar}:=33$ lines.
6.173. $d N / d E \approx \sqrt{I / 2 \hbar^{2} E}$, where $I$ is the moment of inertia of the molecule. In the case of $J=10 d N / d E=1.0 \cdot 10^{4}$ levels per eV.
6.174. $E_{v i b} / E_{\text {rot }}=\omega \mu d^{2} / \hbar$, where $\mu$ is the reduced mass of the molecule; (a) 36 ; (b) $1.7 \cdot 10^{2}$; (c) $2.9 \cdot 10^{3}$.
6.175. $N_{v t b} / N_{\text {rot }}=1 / 3 \mathrm{e}^{-\hbar(\omega-2 B) / k T}=3.1 \cdot 10^{-4}$, where $\quad B=$ $=\hbar / 2 I, I$ is the moment of inertia of the molecule.
6.176. According to the definition
\[
\langle E\rangle=\frac{\sum E_{v} \exp \left(-E_{v} / k T\right)}{\sum \exp \left(-E_{v} / k T\right)}=\frac{\sum E_{v} \exp \left(-\alpha E_{v}\right)}{\sum \exp \left(-\alpha E_{v}\right)} .
\]
where $E_{v}=\hbar \omega(v+1 / 2), \alpha=1 / k T$. The summation is carried out over $v$ taking the values from 0 to $\infty$ as follows:
\[
\begin{array}{c}
\langle E\rangle=-\frac{\partial}{\partial \alpha} \ln \sum \exp \left(-\alpha E_{v}\right)=-\frac{\partial}{\partial \alpha} \ln \frac{\exp (-\alpha \hbar \omega / 2)}{1-\exp (-\alpha \hbar \omega)}= \\
=\frac{\hbar \omega}{2}+\frac{\hbar \omega}{\exp (\hbar \omega / k T)-1} ; \\
C_{V v i b}=N \frac{\partial(E\rangle}{\partial T}=\frac{R(\hbar \omega / k T)^{2} \mathrm{e}^{\hbar \omega / k T}}{\left(\mathrm{e}^{\hbar \omega / k T}-1\right)^{2}}=0.56 R .
\end{array}
\]
where $R$ is the universal gas constant.
6.177. $d=\sqrt{2 \hbar / \mu \Delta \omega}=0.13 \mathrm{~nm}$, where $\mu$ is the reduced mass of the molecule.
6.178. $\lambda=\lambda_{0} /\left(1 \mp \omega \lambda_{0} / 2 \pi c\right)=423$ and $387 \mathrm{~nm}$.
6.179. $\omega=\pi c\left(\lambda_{r}-\lambda_{v}\right) / \lambda_{r} \lambda_{v}=1.37 \cdot 10^{14} \mathrm{rad} / \mathrm{s}, x=4.96 \mathrm{~N} / \mathrm{cm}$.
6.180. $I_{v} / I_{r}=\exp (-\hbar \omega / k T)=0.067$. Will increase 3.9 times.
6.181. (a) See Fig. $46 a$ in which the arrows indicate the motion directions of the nuclei in the molecule at the same moment. The oscillation frequencies are $\omega_{1}, \omega_{2}, \omega_{3}$, with $\omega_{3}$ being the frequency of two independent oscillations in mutually perpendicular planes. Thus, there are four different oscillations. (b) See Fig. $46 b$; there are seven different oscillations: three longitudinal ones $\left(\omega_{1}, \omega_{2}, \omega_{3}\right)$ and four transversal ones $\left(\omega_{4}, \omega_{5}\right)$, two oscillations for each frequency.
6.182. $d N_{\omega}=(l / \pi v) d \omega$.
6.183. $d N_{\omega}=\left(S / 2 \pi v^{2}\right) \omega d \omega$.
6.184. $d N_{\omega}=\left(V / \pi^{2} v^{3}\right) \omega^{2} d \omega$.
6.185. (a) $\Theta=(\hbar / k) \pi v n_{0}$;
(b) $\Theta=(\hbar / k) v \sqrt{4 \pi n_{0}}$;
(c) $\Theta=$ $=(\hbar / k) v \sqrt[3]{6 \pi^{2} n_{0}}$.
6.186. $\Theta=(\hbar / k) \sqrt[3]{18 \pi^{2} n_{0} /\left(v_{\|}^{-3}+2 v_{\perp}^{-3}\right)}=470 \mathrm{~K}$, where $n_{0}$ is the concentration of the atoms.
Fig. 46.
6.187. $v \approx k \Theta / \hbar \sqrt[3]{6 \pi^{2} n_{0}}=3.4 \mathrm{~km} / \mathrm{s}$, where $n_{0}$ is the concentration of the atoms. The tabulated values are: $v_{\|}=6.3 \mathrm{~km} / \mathrm{s}$, $v_{\perp}=3.1 \mathrm{~km} / \mathrm{s}$.
6.188. The oscillation energy of a mole of a \”crystal\” is
\[
U=R \Theta\left[\frac{1}{4}+\left(\frac{T}{\Theta}\right)^{2} \int_{0}^{\Theta / T} \frac{x d x}{\mathrm{e}^{x}-1}\right],
\]
where $x=\hbar \omega / k T$. Hence the molar heat capacity is
\[
C=R\left(\frac{2 T}{\Theta} \int_{0}^{\theta / T} \frac{x d x}{\mathrm{e}^{x}-1}-\frac{\theta / T}{\mathrm{e}^{\theta / T}-1}\right) .
\]

When $T \gg \Theta$, the heat capacity $C \approx R$.
6.189. (a) $d N / d \omega=2 l / \pi a \sqrt{\omega_{\max }^{2}-\omega^{2}} ; \quad$ (b) $\quad N=l / a, \quad$ i.e. is equal to the number of the atoms in the chain.
\[
\text { 6.190. } U_{0}=9 R \Theta / 8 \mu=48.6 \mathrm{~J} / \mathrm{g} \text {, }
\]
where $\mu$ is the molar mass of copper.
6.191 .
(a) $\Theta \approx 220 \mathrm{~K}$;
(b) $C \approx$
$10 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K})$;
$\times 10^{13} \mathrm{rad} / \mathrm{s}$.
(c) $\omega_{\max }=4.1 \times$
6.193. Yes, because the heat capacity is proportional to $T^{3}$ at these temperatures.
6.194. $\langle E\rangle=3 / 8 k \Theta$.
6.195. See Fig. 47.
6.196. $\hbar \omega_{\max }=28 \mathrm{meV}, \hbar k_{\max } \sim$ $\sim 10^{-19} \mathrm{~g} \cdot \mathrm{cm} / \mathrm{s}$.
Fig. 47.
(a) $T_{\max }=\left(3 \pi^{2} n\right)^{2 / 3} \hbar^{2} / 2 m$;
(b) $\langle T\rangle=3 / 5 T_{\max }$.
6.198. $\eta=1-2^{-3 / 2} \approx 65 \%$.
6.199. 0.93 .
6.200. $\approx 3 \cdot 10^{4} \mathrm{~K}$.
6.201. $\Delta E=2 \pi^{2} \hbar^{2} / m V\left(3 \pi^{2} n\right)^{1 / 3}=2 \cdot 10^{-22} \mathrm{eV}$.
6.202. (a) $d n_{v}=\left(m^{3} / \pi^{2} \hbar^{3}\right) v^{2} d v$; (b) $\langle v\rangle / v_{\max }=3 / 4$.
6.203. $d n_{\lambda}=8 \pi \lambda^{-4} d \lambda$.
6.204. $p=2 / 3 n\langle T\rangle=\left(\pi \sqrt[3]{9 \pi} \hbar^{2} / 5 m\right) n^{5 / 3} \approx 5 \cdot 10^{4}$ atm.
6.205. $A=k T(\eta T / \Delta T-2)=4.5 \mathrm{eV}$.
6.206. $n=\sqrt{1+U_{0} / T}=1.02$, where $U_{0}=T_{\max }+A, \quad T_{\max }=$ $=\left(3 \pi^{2} n\right)^{2 / 3} \hbar^{2 / 2 m}$. $A$ is the work function.
6.207. $E_{\text {min }}=\frac{2 k T_{1} T_{2}}{T_{2}-T_{1}} \ln \eta=0.33 \mathrm{eV}$.
6.208. $\alpha=\frac{1}{\rho} \frac{\partial \rho}{\partial T}=-\frac{\pi c \hbar}{k T^{2} \lambda_{T}}=-0.05 \mathrm{~K}^{-1}$, where $\rho \sim \mathrm{e}^{\Delta E_{0} / 2 k T}$, $\Delta E_{0}$ is the forbidden band width.
6.209. $\Delta E=-2 k \frac{\Delta \ln \sigma}{\Delta\left(T^{-1}\right)}=1.2$ and $0.06 \mathrm{eV}$.
6.210. $\tau=t / \ln \frac{\left(\rho-\rho_{1}\right) \rho_{2}}{\left(\rho-\rho_{2}\right) \rho_{1}}=0.01 \mathrm{~s}$.
6.211. $n=h B V / e l \rho V_{H}=5 \cdot 10^{15} \mathrm{~cm}^{-3}, u_{0}=l V_{H} / h B V=$ $=0.05 \mathrm{~m}^{2} /(\mathrm{V} \cdot \mathrm{s})$.
6.212. $\left|u_{0}^{-}-u_{0}^{+}\right|=1 / \eta B=0.20 \mathrm{~m}^{2} /(\mathrm{V} \cdot \mathrm{s})$.
6.213. $n^{+} / n^{-}=\eta^{2}=4.0$.
6.214. (a) $P=1-\exp (-\lambda t)$; (b) $\tau=1 / \lambda$.
6.215. About $1 / 4$.
6.216. $1.2 \cdot 10^{15}$.
6.217. $\tau \approx 16 \mathrm{~s}$.
6.218. $T=5.3$ days.
6.219. $4.6 \cdot 10^{2}$ part./min.
6.220. $\lambda=-(1 / t) \ln (1-\eta) \approx \eta / t=1.1 \cdot 10^{-5} \mathrm{~s}^{-1}, \tau=1 / \lambda=$ $=1.0$ years.
6.221. $T=4.5 \cdot 10^{9}$ years, $A=1.2 \cdot 10^{4} \mathrm{dis} / \mathrm{s}$.
6.222. $4.1 \cdot 10^{3}$ years.
6.223 . About $2.0 \cdot 10^{9}$ years.
6.224. $3.2 \cdot 10^{17}$ and $0.8 \cdot 10^{5} \mathrm{dis} /(\mathrm{s} \cdot \mathrm{g})$ respectively.
6.225. $V=\left(A / A^{\prime}\right) \exp (-t \ln 2 / T)=61$.
6.226. $0.19 \%$.
6.227. $T_{1}=1.6$ hours, $T_{2}=9.8$ hours; $N_{2} / N_{1}=\left(T_{2} / T_{1}\right) \times$ $\times \exp \left(\ln A_{2}-\ln A_{1}\right)=10$.
6.228. $t=-(T / \ln 2) \ln (1-A / q)=9.5$ days.
6.229. (a) $N_{2}(t)=N_{10} \frac{\lambda_{1}}{\lambda_{2}-\lambda_{1}}\left(\mathrm{e}^{-\lambda_{1} t}-\mathrm{e}^{-\lambda_{2} t}\right)$;
(b) $t_{m}=\frac{\ln \left(\lambda_{1} / \lambda_{2}\right)}{\lambda_{1}-\lambda_{2}}$.
6.230. (a) $N_{2}(t)=\lambda N_{10} t \exp (-\lambda t)$; (b) $t_{m}=1 / \lambda$.
6.231. $N_{3}(t)=N_{10}\left(1+\frac{\lambda_{1} \mathrm{e}^{-\lambda_{2} t}-\lambda_{2} \mathrm{e}^{-\lambda_{1} t}}{\lambda_{2}-\lambda_{1}}\right)$.
6.232. $\dot{N}_{\beta}=N_{0} \lambda_{1}$ exp $\left(-\lambda_{1} t\right)=0.72 \cdot 10^{11} \quad$ part./s, $\dot{N}_{\alpha}=$ $=N_{0}\left(\mathrm{e}^{-\lambda_{1} t}-\mathrm{e}^{-\lambda_{2} t}\right) \lambda_{1} \lambda_{2} /\left(\lambda_{2}-\lambda_{1}\right)=1.46 \cdot 10^{11}$ part./s. Here $N_{0}$ is the initial number of $\mathrm{Bi}^{210}$ nuclei.
6.233. (a) $\mathrm{Pb}^{206}$; (b) eight alpha decays and six beta decays.
6.234. $v=\sqrt{2 m_{\mathrm{g}} T_{\mathrm{a}}} / \mathrm{m}=3.4 \cdot 10^{4} \mathrm{~m} / \mathrm{s} ; 0.020$.
6.235. $1.6 \mathrm{MJ}$.
6.236. $0.82 \mathrm{MeV}$,
6.237. (a) $6.1 \mathrm{~cm}$; (b) $2.1 \cdot 10^{\circ}$ and $0.77-10^{\circ}$ respectively.
6.238. $Q=\left\{\begin{array}{l}\left(M_{p}-M_{d}\right) c^{2} \text { for } \beta^{-} \text {decay and } K \text {-eapture, } \\ \left(M_{p}-M_{4}-2 m\right) c^{2} \text { for } \beta^{+} \text {decay. }\end{array}\right.$
6.239. $0.56 \mathrm{MeV}$ and $47.5 \mathrm{eV}$.
6.240. $5 \mathrm{MJ}$.
6.241. 0.32 and $0.65 \mathrm{MeV}$.
6.242. $T \approx 1 / 2\left(Q+2 m c^{2}\right) / M \mathrm{nc}^{2}=0.11 \mathrm{keV}$, where $Q=$ $=\left(M_{N}-M_{C}-2 m\right) c^{2}, m$ is the wass of an electron.
6.243. $40 \mathrm{~km} / \mathrm{s}$.
6.244. $0.45 c$, where $c$ is the velocity of light.
6.245. $\Delta \varepsilon / \varepsilon^{2}=E / 2 m d^{2}-3.6 \cdot 10^{-1}$, where $m$ is the mass of the nucleas.
6.246. $v \approx d / m e=0.22 \mathrm{~km} / \mathrm{s}$, where $m$ is the mass of the nucleus.
6.267. $v=g h / e=65 \mu \mathrm{m} / \mathrm{s}$.
6.248. $h_{\text {min }}=\mathrm{Ac}^{2} / \mathrm{get}-4.6 \mathrm{~m}$.
6.249. $T=T_{0} / 11+(M-m)^{2 / 4 m} M \cos ^{2} \Theta \mathrm{l}=6.0 \mathrm{MeV}$, where $m$ and $M$ are the masses of an alpha particle and a lithium nucleus. 6.250. (a) $\eta=4 m M /(m+M)^{2}=0.89 ;(\mathrm{b}) \eta=2 m /(m+M)=$ $=2 / 3$. Here $m$ and $M$ are the masses of a neutron and a deuteron.
6.251. $\theta_{\text {max }}=\arcsin \left(m_{l} / m_{1}\right)=30^{\circ}$, where $m_{1}$ and $m_{1}$ are the masses of a proton and a deuteron.
6.252. $2 \cdot 10^{11} \mathrm{~kg} / \mathrm{cm}^{3}, 1 \cdot 10^{44}$ aucl. $/ \mathrm{cm}^{3}$.
6.253. (a) d; (b) $\mathrm{Fn}^{\prime}$; (c) a; (d) $\mathrm{Cl}^{n}$.
6.255. Be, $B_{1}=56.5 \mathrm{MeV}$.
6.256. (a) $8.0 \mathrm{MeV}$; (b) 11.5 and $8.7 \mathrm{MeV}$; (c) $14.5 \mathrm{MeV}$.
6.257. $E_{n}-E_{p}=0.22 \mathrm{MeV}$.
6.258. $R_{n}^{n}=20 \varepsilon_{\mathrm{Nn}}-2.4 \varepsilon_{\mathrm{s}}-12 \varepsilon_{\mathrm{C}}=11.9 \mathrm{MeV}$, where $\varepsilon$ is the binding energy per nucleon in the corresponding aucleus.
6.259. (a) 8.0225 a.m.u.; (b) 10.0135 a.m.u.
6.260. $\varphi=\left(E_{3}+E_{4}\right)-\left(E_{1}+B_{2}\right)$.
6.261. (a) $8.2 \cdot 10^{10} \mathrm{~kJ}, 2.7 \cdot 10^{\circ} \mathrm{kg}$ (b) $1.5 \mathrm{~kg}$.
6.362. $5.74 \cdot 10^{7} \mathrm{~kJ}, 2 \cdot 10^{4} \mathrm{~kg}$.
6.263. 2.79 MeV; $0.85 \mathrm{MeV}$.
6.264. $Q=8 \varepsilon_{\mathrm{s}}-7 \epsilon_{\mathrm{L}}=17.3 \mathrm{MeV}$.
6.265. $Q=\left(1+\eta_{p}\right) T_{p}-\left(1-\eta_{a}\right) T_{a}-2 V \overline{\eta_{p} \eta_{a} T_{p} T_{a}} \times$ $\times \cos \theta=-1.2 \mathrm{MeV}$, where $\eta_{p}=m_{2} / m_{0}, \eta_{c}=m_{0} / m_{0}$. 6.266.
(a) $-\mathbf{1 . 6 5} \mathrm{MeV}$ :
(b) $6.82 \mathrm{MeV}_{i}$
(c) $-2.79 \mathrm{MeV}$;
(d) $3.11 \mathrm{MeV}$.
6.267. $v_{\mathrm{C}}=0.92 \cdot 10^{1} \mathrm{~m} / \mathrm{s}, v_{\mathrm{L}}=0.53 \cdot 10^{1} \mathrm{~m} / \mathrm{s}$.
6.268. $1.9 \mathrm{MeV}$.
6.269. $T_{\mathrm{s}}=\frac{Q+\left(1-\mathrm{ma}_{\mathrm{a}} / \mathrm{m}_{\mathrm{c}}\right) T}{1+\mathrm{m}_{\mathrm{s}} / \mathrm{m}_{\mathrm{C}}}=8.5 \mathrm{MeV}$.
6.270. $9.1 \mathrm{MeV}, 170.5^{\circ}$.
6.272. $T \geqslant E_{\mathrm{b}}\left(m_{p}+m_{d}\right) / m_{4}=3.3 \mathrm{MeV}$.
6.273. Between 1.89 and $2.06 \mathrm{MeV}$.
6.274. $Q=-11 / 12 T_{15}=-3.7 \mathrm{MeV}$.
6.275. 1.88 and $5.75 \mathrm{MeV}$ respectively.
6.276. $4.4 \mathrm{MeV} ; 5.3 \cdot 10^{\circ} \mathrm{m} / \mathrm{s}$.
6.277. $T_{4}=\frac{1}{m_{0}+m_{4}}\left[\left(m_{4}-m_{1}\right) T-\frac{m_{1} m_{4}}{m_{1}+m_{4}} T_{t h}\right]=2.2 \mathrm{MeV}$, where $m_{4}, m_{2}, m_{2}, m_{4}$ are the masses of neutron, a $\mathrm{Cu}$ aucleus, a alpha particle, and a $\mathrm{Be}^{*}$ nueleus.
6.278. By $B_{v} / 2 m c^{2}=0.06 \%$, where $m$ is the mass of a deuteron.
6.279. $E=Q+2 / 3 T=6.5 \mathrm{MeV}$.
6.250. $E_{1}=E_{\mathrm{b}}+\frac{\mathrm{mc}}{\mathrm{C}_{\mathrm{c}}+\mathrm{mc}_{\mathrm{c}}} T_{1}=16.7,16.9,17.5$ and $17.7 \mathrm{MeV}$, where $B_{0}$ is the biading energy of a deuteron in the transitional aucleus.
6.281. $\sigma=(M / N \rho d)$ ln $\eta=2.5 \mathrm{~kb}$, where $M$ is the molar mass of cadmium, $N$ is the Avogadro number, $p$ is the density of cadmium.
6.282. $I_{d} I=\exp \left(\left(2 \sigma_{1}+o_{2}\right) n d\right]=20$, where $n$ is the concentration of heavy water molecules.
6.283. $w=\left\{1-\exp \left[-\left(\sigma_{*}+\sigma_{e}\right) n d\right]\right\} \sigma_{v}\left(\sigma_{s}+\sigma_{a}\right)=0.35$, where $n$ is the concentration of $\mathrm{Fe}$ auclel.
6.284. (a) $T=(w / k)$ is 2; (b) $w=$ ATe/It is $2=2 \cdot 10^{-4}$. $=1.0 \cdot 10^{3}$, where $N_{0}$ is the number of $\mathrm{Au}^{1 n^{n}}$ nuclei in the foil.
6.256. $\dot{N}=\left(1-e^{-1 /}\right) J_{n o} /$.
6.287. $J=\lambda e^{2} / \mathrm{N}_{6}\left(1-\mathrm{e}^{-\mathrm{M}}\right)=6 \cdot 10^{\circ} \mathrm{part} . /\left(\mathrm{cm}^{2} \cdot \mathrm{s}\right)$, where $\lambda$ is the decay constant, $N$, is the number of $\mathrm{Au}$ auclei in the foil.
6.288. $N=N_{0} k^{i-i}=1.3 \cdot 10^{3}$, where $t$ is the number of generstions.
6.289. $N=v P / E=0.8 \cdot 10^{10} \mathrm{~s}^{-1}$,
6.290. (a) $N / N_{4}=4 \cdot 10^{2}$; (b) $T^{\prime}=v /(k-1)=10 \mathrm{s.}$
6.291. $0.05,0.4$, and 9 GeV respectively.
6.292. $(D)=C \pi_{0} \sqrt{\eta(\eta+2)}=15 \mathrm{~m}$.
6.293. $\mathrm{r}_{4}-\operatorname{lm} / \sqrt{T\left(T+2 m c^{5}\right)}-26 \mathrm{~ns}$, where $m$ is the rest mass of a plon.
6.294. $J / J_{0}=\exp |-| m e / \mathrm{r}_{0} \sqrt{T\left(T+2 m c^{2}\right)} \mid=0.22$, where $m$ is the rest mass of a aegative pion.
6.295*. $T_{n}=\left(m_{n}-m_{n}\right)^{2} / 2 m_{n}=4.1 \mathrm{MeV}, \quad E_{v}=29.8 \mathrm{MoV}$.
6.296*. $T=1\left(m_{x}-m_{s}\right)^{2}-m \mathrm{H} / 2 m_{x}=19.5 \mathrm{MeV}$.
6.297*. $T_{\text {max }}=\left(m_{s}-m_{e}\right)^{2} / 2 m_{s}=52.5 \mathrm{MeV}$.
6.298*. $m=m_{p}+T+\sqrt{m+T\left(T+2 m_{p}\right)}=1115 \mathrm{MeV}$, a $\mathrm{A}$ particle.
6.299*, $R_{v}=1 /(m !-m !) /\left(m_{n}+T\right)=22 \mathrm{MeV}$.
– Is the assers to Problems 6.295-6.290 marked iby as asterisk the quatity met is abtiroviatiod as $=$.
\[
\text { 6.300*. } m=\sqrt{m_{\Sigma}^{2}+m_{\pi}^{2}-2\left(m_{\Sigma}+T_{\Sigma}\right)\left(m_{\pi}+T_{\pi}\right)}=0.94 \mathrm{GeV} \text {, }
\]
neutron.
6.301*. $T_{\pi}=m_{\pi}[\operatorname{cosec}(\Theta / 2)-1], E_{\gamma}=m_{\pi} / 2 \sin (\Theta / 2)$. For $\Theta=60^{\circ}$ the energy $T_{\pi}=E_{\gamma}=m_{\pi}$.
$6.303^{*} \cdot \cos (\Theta / 2)=1 / \sqrt{1+2 m / T}$, whence $\Theta=99^{\circ}$.
$6.304^{*}$. (a) $\varepsilon_{t h}=4 m_{e}=2.04 \mathrm{MeV}$; (b) $\varepsilon_{t h}=2 m_{\pi}\left(1+m_{\pi} / m_{p}\right)=$ $=320 \mathrm{MeV}$.
6.305*. (a) $T_{t h}=6 m_{p}=5.6 \mathrm{GeV}$; (b) $T_{t h}=m_{\pi}\left(4 m_{p}+m_{\pi}\right) / 2 m_{p}=$ $=0.28 \mathrm{GeV}$.
6.306. (a) $0.90 \mathrm{GeV}$; (b) $0.77 \mathrm{GeV}$.
6.307. $S=-2, Y=-1, \Xi^{0}$ particle.
6.308. Processes 1, 2, and 3 are forbidden.
6.309. Processes 2, 4, and 5 are forbidden.
6.310. Process 1 is forbidden in terms of energy; in other processes the following laws of conservation are broken: of baryon charge (2), of electric charge (3), of strangeness (4), of lepton charge (5), and of electron and muon charge (6).
* In the answers to Problems 6.300-6.305 marked by an asterisk the quantity $m c^{2}$ is abbreviated as $m$.