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Indian National Physics Olympiad (INPhO)-2023 — Question Paper · 29 януари 2023 г. · 16 т.

Указания

Date: 29 January 2023

Time: 09:00-12:00 (3 hours) Maximum Marks: 60

Instructions Roll No.:

This booklet consists of 6 pages and total of 5 questions. Write roll number at the top wherever asked.
Booklet to write the answers is provided separately. Instructions to write the answers are on the Answer Booklet.
Non-programmable scientific calculators are allowed. Mobile phones cannot be used as calculators.
Please submit the Answer Sheet at the end of the examination. You may retain the Question Paper.

Table of Constants

Speed of light in vacuum	$c$	$3.00 \times 10^{8}\ \mathrm{m{\cdot}s^{-1}}$
Planck’s constant	$h$	$6.63 \times 10^{-34}\ \mathrm{J{\cdot}s}$
	$\hbar$	$h/2\pi$
Universal constant of Gravitation	$G$	$6.67 \times 10^{-11}\ \mathrm{N{\cdot}m^2{\cdot}kg^{-2}}$
Magnitude of electron charge	$e$	$1.60 \times 10^{-19}\ \mathrm{C}$
Rest mass of electron	$m_e$	$9.11 \times 10^{-31}\ \mathrm{kg}$
Value of $1/4\pi\epsilon_0$		$9.00 \times 10^{9}\ \mathrm{N{\cdot}m^2{\cdot}C^{-2}}$
Avogadro’s number	$N_A$	$6.022 \times 10^{23}\ \mathrm{mol^{-1}}$
Acceleration due to gravity	$g$	$9.81\ \mathrm{m{\cdot}s^{-2}}$
Universal Gas Constant	$R$	$8.31\ \mathrm{J{\cdot}K^{-1}{\cdot}mol^{-1}}$
	$R$	$0.0821\ \mathrm{l{\cdot}atm{\cdot}mol^{-1}{\cdot}K^{-1}}$
Boltzmann constant	$K_B$	$1.3806 \times 10^{-23}\ \mathrm{J{\cdot}K^{-1}}$
Permeability constant	$\mu_0$	$4\pi \times 10^{-7}\ \mathrm{H{\cdot}m^{-1}}$
1 Angstrom unit	1Å	$1 \times 10^{-10}\ \mathrm{m}$
1 micro unit	$1\mu$	$1 \times 10^{-6}$ units
1 electron volt	1 eV	$1.6 \times 10^{-19}\ \mathrm{J}$

Q No	1	2	3	4	5	Total
Maximum Marks	6	6	16	16	16	60

Указания

Date: 29 January 2023

Time: 09:00-12:00 (3 hours) Maximum Marks: 60

Instructions Roll No.:

This booklet consists of 19 pages and total of 5 questions. Write roll number at the top wherever asked.
Booklet to write the answers is provided separately. Instructions to write the answers are on the Answer Booklet.
Non-programmable scientific calculators are allowed. Mobile phones cannot be used as calculators.
Please submit the Answer Sheet at the end of the examination. You may retain the Question Paper.

Table of Constants

Speed of light in vacuum	$c$	$3.00 \times 10^{8}\ \mathrm{m{\cdot}s^{-1}}$
Planck’s constant	$h$	$6.63 \times 10^{-34}\ \mathrm{J{\cdot}s}$
	$\hbar$	$h/2\pi$
Universal constant of Gravitation	$G$	$6.67 \times 10^{-11}\ \mathrm{N{\cdot}m^2{\cdot}kg^{-2}}$
Magnitude of electron charge	$e$	$1.60 \times 10^{-19}\ \mathrm{C}$
Rest mass of electron	$m_e$	$9.11 \times 10^{-31}\ \mathrm{kg}$
Value of $1/4\pi\epsilon_0$		$9.00 \times 10^{9}\ \mathrm{N{\cdot}m^2{\cdot}C^{-2}}$
Avogadro’s number	$N_A$	$6.022 \times 10^{23}\ \mathrm{mol^{-1}}$
Acceleration due to gravity	$g$	$9.81\ \mathrm{m{\cdot}s^{-2}}$
Universal Gas Constant	$R$	$8.31\ \mathrm{J{\cdot}K^{-1}{\cdot}mol^{-1}}$
	$R$	$0.0821\ \mathrm{l{\cdot}atm{\cdot}mol^{-1}{\cdot}K^{-1}}$
Boltzmann constant	$K_B$	$1.3806 \times 10^{-23}\ \mathrm{J{\cdot}K^{-1}}$
Permeability constant	$\mu_0$	$4\pi \times 10^{-7}\ \mathrm{H{\cdot}m^{-1}}$
1 Angstrom unit	1Å	$1 \times 10^{-10}\ \mathrm{m}$
1 micro unit	$1\mu$	$1 \times 10^{-6}$ units
1 electron volt	1 eV	$1.6 \times 10^{-19}\ \mathrm{J}$

Q No	1	2	3	4	5	Total
Maximum Marks	6	6	16	16	16	60

Please note that alternate/equivalent methods and different way of expressing final solutions may exist. A correct method will be suitably awarded.

Условие

Consider a fixed infinite vertical thin rod (shown by the red color in the figure below) of linear charge density $\lambda$ along the $z$ -axis at the origin (see figure below). A uniformly charged ring of total charge $Q$ , mass $M$ , and radius $a$ is placed with its center at the origin in the $x$ - $y$ plane. Point P is an arbitrary point on the ring. The projection of point P on $x$ - $y$ plane makes an angle $\theta$ with respect to the $x$ -axis in the anticlockwise direction as seen from the top.

The ring is now given an initial angular velocity $\omega_0$ about the $x$ -axis. We define the angle $\alpha$ which the plane of the ring makes with the $x$ - $y$ plane. This is illustrated by drawing line segment AB in the plane of the ring. Initially $\alpha = 0$ . Ignore gravity.

3D sketch: a red vertical rod along the z-axis through the origin; axes x (toward the viewer, lower left), y (right) and z (up); a tilted ring centred at the origin with point P on it; a dashed line A–B lies in the plane of the ring and makes angle α with the y-axis; angle θ is marked at the origin with a dashed projection of P.

You may find the following differentiation useful

D = \frac{d}{d\theta}[\tan^{-1}(q\tan\theta)] = \frac{1}{1 + (q\tan\theta)^2}[q(\sec^2\theta)] \qquad (4.1)

(a) [1 marks] State an expression for the electric field ( $\vec{E}_0$ ) due to the infinite rod at a point on the ring when $\alpha=0$ in terms of $x, y$ and $\theta$ , and related quantities.

(b) [2 marks] At some instant the ring makes an angle $\alpha$ . Derive an expression for the electric field $\vec{E}$ due to the infinite rod at a point on the ring in terms of $\theta$ , and $\alpha$ .

(c) [1 marks] Find the net force $\vec{F}$ acting on the ring.

(d) [5 marks] Find the net torque $\vec{\tau}$ acting on the ring in terms of $\alpha$ and the constants only. Qualitatively plot torque as a function of $\alpha$ .

(e) [2 marks] Let the ring is in equilibrium with respect to $\alpha = 0$ . Derive an expression for the time period $T$ of small oscillations of the ring in terms of $\lambda$ , and $Q$ . Take $\lambda = 0.1\mu\,\mathrm{C/m}$ , $Q = 2.0\,\mu\mathrm{C}$ , $M = 50.0\mathrm{g}$ , radius $a = 5.0\,\mathrm{cm}$ , and $\omega_0 = 1.0\,\mathrm{rad/s}$ . Calculate $T$ .

(f) [2.5 marks] Find an expression for the potential energy $U$ of the ring in terms of $\alpha$ . Qualitatively plot $U$ as a function of $\alpha$ . Take the zero of potential energy to be at $\alpha = 0$ .

(g) [2.5 marks] Obtain the expression of maximum value of $\alpha$ ( $\alpha_{\max}$ ) in terms of $\omega_0$ . Calculate $\alpha_{\max}$ .

Отговори

Решение

(a) Solution: The electric field on the ring due to the infinite rod is given by

\vec{E}_0 = \frac{\lambda(x\hat{x} + y\hat{y})}{2\pi\epsilon_0(x^2 + y^2)} \qquad (4.2)

Since the rod is infinite, the electric field will not depend on $z$ .

\vec{E}_0 = \frac{\lambda(a\cos\theta\hat{x} + a\sin\theta\hat{y})}{2\pi\epsilon_0(a^2\cos^2\theta + a^2\sin^2\theta)} \qquad (4.3)

The above expression simplifies to

\vec{E}_0 = \frac{\lambda(\cos\theta\hat{x} + \sin\theta\hat{y})}{2\pi\epsilon_0 a} \qquad (4.4)

(b) Solution: The new coordinates of the ring are $x = \dfrac{a\cos\theta}{\sqrt{1 + \sin^2\theta\tan^2\alpha}}$ , $y = \dfrac{a\sin\theta}{\sqrt{1 + \sin^2\theta\tan^2\alpha}}$ and $z = \dfrac{a\sin\theta\tan\alpha}{\sqrt{1 + \sin^2\theta\tan^2\alpha}}$ .

\vec{E} = \frac{\lambda(x\hat{x} + y\hat{y})}{2\pi\epsilon_0(x^2 + y^2)} \qquad (4.5)

= \frac{\lambda}{2\pi\epsilon_0 a}\sqrt{(1 + \sin^2\theta\tan^2\alpha)}(\cos\theta\hat{x} + \sin\theta\hat{y}) \qquad (4.6)

(c) Solution: Charge $dQ$ in elementary length is $\dfrac{Q}{2\pi\cos\alpha}\dfrac{1}{(1 + \sin^2\theta\tan^2\alpha)}d\theta$ . Force on small element $ds$ is

d\vec{F} = \vec{E}dQ = \frac{\lambda}{2\pi\epsilon_0 a}\sqrt{(1 + \sin^2\theta\tan^2\alpha)}(\cos\theta\hat{x} + \sin\theta\hat{y})\frac{Q}{2\pi\cos\alpha}\frac{1}{(1 + \sin^2\theta\tan^2\alpha)}d\theta \qquad (4.7)

which simplifies to

d\vec{F} = \frac{\lambda Q}{4\pi^2\epsilon_0 a\cos\alpha}\frac{\cos\theta\hat{x} + \sin\theta\hat{y}}{\sqrt{(1 + \sin^2\theta\tan^2\alpha)}} \qquad (4.8)

where $C = \lambda Q/4\pi^2\epsilon_o a$ . Consider

dF_x = \frac{C}{\cos\alpha}\frac{\cos\theta}{\sqrt{(1 + \sin^2\theta\tan^2\alpha)}}d\theta \qquad (4.9)

F_x = \frac{C}{\cos\alpha}\int_{-\pi}^{\pi}\frac{\cos\theta}{\sqrt{(1 + \sin^2\theta\tan^2\alpha)}}d\theta \qquad (4.10)

The integrand is an even function, hence

F_x = \frac{2C}{\cos\alpha}\int_0^{\pi}\frac{\cos\theta}{\sqrt{(1 + \sin^2\theta\tan^2\alpha)}}d\theta \qquad (4.12)

using $\int_0^{2a} f(x)dx = \int_0^a f(x)dx + \int_0^a f(2a - x)dx$

F_x = \frac{2C}{\cos\alpha}\left[\int_0^{\pi/2}\frac{\cos\theta}{(1 + \sin^2\theta\tan^2\alpha)} + \int_0^{\pi/2}\frac{\cos(\pi - \theta)}{(1 + \sin^2(\pi - \theta)\tan^2\alpha)}\right] = 0 \qquad (4.14)

Similarly

F_y = 0 \qquad (4.15)

The total force acting on the ring is zero. Answers based on symmetric arguments will be also given credit.

(d) Solution:

d\tau = \vec{r} \times d\vec{F}(\theta) \qquad (4.16)

Consider

d\tau_z = xdF_y - ydF_x \qquad (4.17)

= 0 \qquad (4.18)

\Longrightarrow \tau_z = 0 \qquad (4.19)

\tau_y = \int_{-\pi}^{\pi} zdF_x\tau_y = \frac{aC\tan\alpha}{2\cos\alpha}\int_{-\pi}^{\pi}\frac{\sin 2\theta}{(1 + \sin^2\theta\tan^2\alpha)} \qquad (4.20)

Integrand is odd function, hence

\tau_y = 0 \qquad (4.21)

\tau_x = \int_{-\pi}^{\pi} -zdF_y \qquad (4.22)

\tau_x = -\int_{-\pi}^{\pi}\frac{a\sin\theta\tan\alpha}{\sqrt{1 + \sin^2\theta\tan^2\alpha}}\frac{C\sin\theta}{\cos\alpha(1 + \sin^2\theta\tan^2\theta)} \qquad (4.23)

which simplifies to

\tau_x = -\,aC\int_{-\pi}^{\pi}\frac{\tan^2\theta\sin\alpha}{(\cos^2\alpha + \tan^2\theta)}d\theta \qquad (4.24)

The integrand is an even function, hence

\tau_x = -2aC\sin\alpha\int_0^{\pi}\frac{\tan^2\theta}{(\cos^2\alpha + \tan^2\theta)}d\theta \qquad (4.25)

using $\int_0^{2a} f(x)dx = \int_0^a f(x)dx + \int_0^a f(2a - x)dx$

\tau_x = -4aC\sin\alpha\int_0^{\pi/2}\frac{\tan^2\theta}{(\cos^2\alpha + \tan^2\theta)}d\theta \qquad (4.26)

\tau_x = -4aC\sin\alpha\sec^2\alpha\int_0^{\pi/2}\frac{\tan^2\theta}{(1 + \sec^2\alpha\tan^2\theta)}d\theta \qquad (4.27)

Substituting $\sec\alpha = u$ , in above equation, we get

\tau_x = -4aC\sin\alpha u^2\int_0^{\pi/2}\frac{\tan^2\theta}{(1 + u^2\tan^2\theta)}d\theta \qquad (4.28)

It is given that

D = \frac{d}{d\theta}(\tan^{-1}(u\tan\theta)) = \frac{u\sec^2\theta}{1 + u^2\tan^2\theta} \qquad (4.29)

= \frac{u(1 + \tan^2\theta)}{1 + u^2\tan^2\theta} \qquad (4.30)

D = \frac{u}{1 + u^2\tan^2\theta} + \frac{u\tan^2\theta}{1 + u^2\tan^2\theta} + u - u\frac{1 + u^2\tan^2\theta}{1 + u^2\tan^2\theta} \qquad (4.31)

Solving above equation, we get

\frac{\tan^2\theta}{1 + u^2\tan^2\theta} = \frac{D}{u - u^3} - \frac{1}{1 - u^2} \qquad (4.32)

Hence Substituting above equation in Eq.(4.28) , we get

\tau_x = -4Ca\sin\alpha u^2\int_0^{\pi/2}\left[\frac{D}{u - u^3} - \frac{1}{1 - u^2}\right]d\theta \qquad (4.33)

\tau_x = -4Ca\sin\alpha u^2\int_0^{\pi/2}\left[\frac{D}{u - u^3} - \frac{1}{1 - u^2}\right]d\theta \qquad (4.34)

Substituting value of D, we get

\tau_x = -4Ca\sin\alpha u^2\int_0^{\pi/2}\left[\frac{\frac{d}{d\theta}(\tan^{-1}(u\tan\theta))}{u - u^3} - \frac{1}{1 - u^2}\right]d\theta \qquad (4.35)

\tau_x = -4Ca\sin\alpha u^2\left[\frac{\tan^{-1}(u\tan\theta)}{u - u^3} - \frac{\theta}{1 - u^2}\right]_0^{\pi/2} \qquad (4.36)

Applying limits and solving further, we get

\tau_x = -\frac{\lambda Q}{2\pi\epsilon_0}\tan(\alpha/2) \qquad (4.38)

Working of $\tau_y, \tau_z$ is not required.

Qualitative plot of torque τ (vertical axis, from −1 to 1) against α (horizontal axis, from −π/2 to π/2 with ticks at −π/3, −π/6, 0, π/6, π/3): a decreasing curve through the origin, from 1 at −π/2 to −1 at π/2; dotted lines mark τ = 0 and α = 0.

(e) Solution: Under small angle approximation of $\alpha$ , $\tau_x$ becomes

\tau_x = \frac{-\lambda Q\alpha}{4\pi\epsilon_0} \qquad (4.39)

I\frac{d^2\alpha}{dt^2} = \frac{-\lambda Q\alpha}{4\pi\epsilon_0} \qquad (4.40)

\frac{Ma^2}{2}\frac{d^2\alpha}{dt^2} = \frac{-\lambda Q\alpha}{4\pi\epsilon_0} \qquad (4.41)

\frac{d^2\alpha}{dt^2} = -\frac{2\lambda Q}{4Ma^2\pi\epsilon_0}\alpha \qquad (4.42)

This is a differential equation of SHM, hence

T^2 = \frac{4\pi^2}{\frac{2\lambda Q}{4Ma^2\pi\epsilon_0}} \qquad (4.43)

\Longrightarrow T = 2\pi a\sqrt{\frac{2M\pi\epsilon_0}{Q\lambda}} \qquad (4.44)

T = 1.17\mathrm{s} \qquad (4.45)

This can be used as an electrostatic clock!

(f) Solution: We know that $\tau = -\dfrac{dU}{d\alpha}$

\Longrightarrow U = -\int \tau d\alpha \qquad (4.46)

= \frac{\lambda Q}{2\pi\epsilon_0}\int \tan(\alpha/2)d\alpha \qquad (4.47)

U = -\frac{\lambda Q}{2\pi\epsilon_0}2\log(\cos(\alpha/2)) + c \qquad (4.48)

where $c$ is the constant of integration. At $\alpha = 0$ , $U = 0$ , which implies that $c = 0$ .

\Longrightarrow U(\alpha) = -\frac{\lambda Q}{\pi\epsilon_0}\log(\cos(\alpha/2)) \qquad (4.49)

Qualitative plot of U(α) against α (ticks labelled −π/2, −π/6, −π/3, 0, π/6, π/3, π/2): a U-shaped curve with minimum zero at α = 0, rising symmetrically on both sides.

(g) Solution: Consider

I\frac{d^2\alpha}{dt^2} = \tau_x \qquad (4.50)

\frac{Ma^2}{2}\frac{d^2\alpha}{dt^2} = \frac{-\lambda Q}{2\pi\epsilon_0}\tan(\alpha/2) \qquad (4.51)

Multiplying both sides by $\frac{d\alpha}{dt}$

\frac{Ma^2}{2}\frac{d^2\alpha}{dt^2}\frac{d\alpha}{dt} = \frac{-\lambda Q}{2\pi\epsilon_0}\tan(\alpha/2)\frac{d\alpha}{dt} \qquad (4.52)

Integrating on both sides

\frac{Ma^2}{4}\left(\frac{d\alpha}{dt}\right)^2 = \frac{\lambda Q}{\pi\epsilon_0}\log(\cos(\alpha/2)) + c' \qquad (4.53)

\left(\frac{d\alpha}{dt}\right)^2 = \frac{32\pi^2}{T^2}\log(\cos(\alpha/2)) + c' \qquad (4.54)

where $c'$ is the constant of integration. At $t = 0$ , $\frac{d\alpha}{dt} = \omega_0$ , which implies that $c = \omega_0^2$ .

\left(\frac{d\alpha}{dt}\right)^2 = \frac{32\pi^2}{T^2}\log(\cos(\alpha/2)) + \omega_0^2 \qquad (4.55)

\frac{d\alpha}{dt} = \sqrt{\frac{32\pi^2}{T^2}\log(\cos(\alpha/2)) + \omega_0^2} \qquad (4.56)

For $\alpha = \alpha_{\max}$ , $\frac{d\alpha}{at} = 0$ , hence solving above equation, we get

\alpha_{\max} = 2\left[\cos^{-1}\left(\exp\left(-\frac{\omega_0^2 T^2}{32\pi^2}\right)\right)\right] \qquad (4.57)

\alpha_{\max} = 10.66^{\circ} \qquad (4.58)

Оригинал в Архива: INPhO2023-Question.pdf · официални решения: INPhO2023-Solution-20230215.pdf