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← JEE Advanced 2024 · Paper 2
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JEE Advanced 2024 · Paper 2, Physics question 5

A small electric dipole p⃗0\vec p_0, having a moment of inertia II about its center, is kept at a distance rr from the center of a spherical shell of radius RR. The surface charge density σ\sigma is uniformly distributed on the spherical shell. The dipole is initially oriented at a small angle θ\theta as shown in the figure. While staying at a distance rr, the dipole is free to rotate about its center.

If released from rest, then which of the following statement(s) is(are) correct?

[ε0\varepsilon_0 is the permittivity of free space.]

A spherical shell of radius R (circle, radius arrow drawn from its centre). A dashed vertical line rises from the centre; at distance r above the centre (r marked with a vertical double arrow) sits a small dipole p₀, drawn as an arrow tilted at a small angle θ from the dashed radial line.
Choose one or more options
Answer and worked solution

Official answer: B, D

Field of the shell.

  • Inside (r<Rr<R) the field is zero.
  • Outside it is radial: E=σR2ε0r2E=\dfrac{\sigma R^2}{\varepsilon_0r^2}.

Small oscillations. A dipole tilted by a small θ\theta from the radial field feels a restoring torque p0Esin⁡θ≈p0Eθp_0E\sin\theta\approx p_0E\theta. So
ω=p0EI=σp0R2ε0Ir2.\omega=\sqrt{\frac{p_0E}{I}}=\sqrt{\frac{\sigma p_0R^2}{\varepsilon_0Ir^2}}.

  • (A) For r<Rr<R there is no field and no torque, so no oscillation. ✗
  • (B) For any r>Rr>R there is a radial field, so it oscillates. ✓
  • (C) At r=2Rr=2R, ω=σp04ε0I\omega=\sqrt{\frac{\sigma p_0}{4\varepsilon_0I}}, not 2σp0ε0I\sqrt{\frac{2\sigma p_0}{\varepsilon_0I}}. ✗
  • (D) At r=10Rr=10R, ω=σp0100ε0I\omega=\sqrt{\frac{\sigma p_0}{100\varepsilon_0I}}. ✓