StreakPeaked· Practice

ExamsJEE AdvancedPhysics

Two smooth non-conducting spherical shells, A (charge +Q, mass m) and B (charge -Q, mass 2m), each of radius R, are released from rest on a frictionless surface with their centres 5R apart. Find the speed of shell A just before they collide. (K = 1/(4*pi*epsilon0))

  1. sqrt(2*K*Q² / (5*m*R))
  2. sqrt(4*K*Q² / (5*m*R))
  3. sqrt(8*K*Q² / (5*m*R))
  4. sqrt(16*K*Q² / (5*m*R))

Correct answer: sqrt(2*K*Q² / (5*m*R))

Solution

PE released = KQ²*(1/2R - 1/5R) = 3KQ²/10R. With vA = 2vB, total KE = 3m*vB². Solving gives vA = sqrt(2KQ²/5mR).

Related JEE Advanced Physics questions

⚔️ Practice JEE Advanced Physics free + battle 1v1 →