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JEE Advanced 2021 Paper 1 · Physics · Electrostatics · medium

JEE Advanced 2021 Paper 1 Physics: Paragraph 1

A thin insulating ring of radius R=0.10 mR=0.10\ \mathrm{m} carries a uniformly distributed positive charge Q=4.0 μCQ=4.0\ \mu\mathrm{C} and is fixed in a vertical plane so that its axis is horizontal. A small bead of mass m=1.0 gm=1.0\ \mathrm{g} carrying a charge −q-q with q=1.0 μCq=1.0\ \mu\mathrm{C} is threaded on a smooth, thin, insulating horizontal rod along the axis of the ring, passing through the centre. Use 14πε0=9×109 N m2 C−2\frac{1}{4\pi\varepsilon_0}=9\times10^{9}\ \mathrm{N\,m^2\,C^{-2}}. The rod and the ring do not disturb the charge distribution, and gravity does not act along the rod.

Question 1

Numerical answer

The bead is displaced slightly along the axis and released. Find the period of its small oscillations in milliseconds. Give the answer correct to two decimal places.
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Answer: 33.12

On the axis of a uniformly charged ring the potential is V(x)=kQR2+x2V(x)=\frac{kQ}{\sqrt{R^2+x^2}}, so the axial force on a charge −q-q at distance xx is F=−kQq x(R2+x2)3/2F=-\frac{kQq\,x}{(R^2+x^2)^{3/2}}, directed towards the centre. For x≪Rx\ll R this is F≈−kQqR3xF\approx-\frac{kQq}{R^3}x, a linear restoring force, so the motion is simple harmonic with ω2=kQqmR3\omega^2=\frac{kQq}{mR^3}. Numerically kQq=9×109×4×10−6×10−6=3.6×10−2 N m2kQq=9\times10^9\times4\times10^{-6}\times10^{-6}=3.6\times10^{-2}\ \mathrm{N\,m^2} and mR3=10−3×10−3=10−6mR^3=10^{-3}\times10^{-3}=10^{-6}, giving ω2=3.6×104 s−2\omega^2=3.6\times10^{4}\ \mathrm{s^{-2}} and ω=189.7 s−1\omega=189.7\ \mathrm{s^{-1}}. The period is T=2π/ω=0.03312T=2\pi/\omega=0.03312 s =33.12=33.12 ms.

Question 2

Numerical answer

The bead is now released from rest at a distance x=Rx=R from the centre of the ring (not a small displacement). Find its speed, in m s−1\mathrm{m\,s^{-1}}, as it passes through the centre. Give the answer correct to two decimal places.
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Answer: 14.52

The force is conservative, so use energy conservation with the exact potential energy of the bead, U(x)=−kQqR2+x2U(x)=-\frac{kQq}{\sqrt{R^2+x^2}}. At x=Rx=R, U=−kQq2 RU=-\frac{kQq}{\sqrt2\,R}; at the centre, U=−kQqRU=-\frac{kQq}{R}. The kinetic energy gained is 12mv2=kQqR(1−12)\frac12mv^2=\frac{kQq}{R}\left(1-\frac1{\sqrt2}\right). With kQq=3.6×10−2kQq=3.6\times10^{-2}, R=0.10R=0.10 and 1−1/2=0.292891-1/\sqrt2=0.29289, we get 12mv2=0.36×0.29289=0.10544\frac12mv^2=0.36\times0.29289=0.10544 J, so v=2×0.10544/10−3=210.9=14.52 m s−1v=\sqrt{2\times0.10544/10^{-3}}=\sqrt{210.9}=14.52\ \mathrm{m\,s^{-1}}. The small-oscillation formula would give a different (wrong) value because x=Rx=R is not small.
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