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JEE Main 2019 · Physics · Electrostatics · hard

A thin ring of radius 10 cm carries a uniformly distributed charge of +4 muC and is fixed with its axis vertical. A small bead of mass 9 g carrying a…

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A thin ring of radius 10 cm10\ \mathrm{cm} carries a uniformly distributed charge of +4 μC+4\ \mu\mathrm{C} and is fixed with its axis vertical. A small bead of mass 9 g9\ \mathrm{g} carrying a charge of −2.5 μC-2.5\ \mu\mathrm{C} is free to move along the axis (gravity is negligible) and is displaced slightly from the centre of the ring along the axis. Take 14πε0=9×109 N m2 C−2\dfrac{1}{4\pi\varepsilon_0}=9\times10^9\ \mathrm{N\,m^2\,C^{-2}}. The time period of its small oscillations is
  1. A. π100 s\dfrac{\pi}{100}\ \mathrm{s}
  2. B. π50 s\dfrac{\pi}{50}\ \mathrm{s}
  3. C. π25 s\dfrac{\pi}{25}\ \mathrm{s}
  4. D. π502 s\dfrac{\pi}{50\sqrt2}\ \mathrm{s}
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Answer: (B) π50 s\dfrac{\pi}{50}\ \mathrm{s}

On the axis at a small distance xx from the centre, the field of the ring is E=kQx(R2+x2)3/2≈kQxR3E=\dfrac{kQx}{(R^2+x^2)^{3/2}}\approx\dfrac{kQx}{R^3}, directed away from the centre. For the negative bead, the force is toward the centre: F=−kQqR3xF=-\dfrac{kQq}{R^3}x, which is simple harmonic. Then ω2=kQqmR3=9×109×4×10−6×2.5×10−69×10−3×10−3=104 s−2\omega^2=\dfrac{kQq}{mR^3}=\dfrac{9\times10^9\times4\times10^{-6}\times2.5\times10^{-6}}{9\times10^{-3}\times10^{-3}}=10^4\ \mathrm{s^{-2}}. So ω=100 rad s−1\omega=100\ \mathrm{rad\,s^{-1}} and T=2π/ω=π/50 s≈0.063 sT=2\pi/\omega=\pi/50\ \mathrm{s}\approx0.063\ \mathrm{s}.
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Original question written in the JEE Main 2019 pattern and checked twice for the correct answer. Not an official question paper. How we write and check questions · Corrections policy · Report a mistake