JEE Main20205 Sep 2020Morning ShiftPhysicsElectrostaticsActual
A solid sphere of radius R carries a charge Q + q distributed uniformly over its volume. A very small point like piece of it of mass m gets detached from the bottom of the sphere and falls down vertically under gravity. This piece carries charge q . If it acquires a speed ν when it has fallen through a vertical height y (see figure), then (assume the remaining portion to be spherical)
Options
- Av 2 = y q Q 4 π ϵ 0 R 2 y m + g
- Bv 2 = y q Q 4 π ϵ 0 R ( R + y ) m + g
- Cv 2 = 2 y Q q R 4 π ϵ 0 ( R + y ) 3 m + g
- Dv 2 = 2 y q Q 4 π ε 0 R ( R + y ) m + g
Correct answer
D. v 2 = 2 y q Q 4 π ε 0 R ( R + y ) m + g
Step-by-step solution
By using total energy conservation Δ KE + ( Δ PE ) electo   + ( Δ PE ) gravitation   = 0 1 2 m V 2 + k Q q R + y - k Q q R + ( - m g y ) = 0 1 2 m V 2 = m g y + k Q q 1 R - 1 R + y ;     V 2 = 2 g y + 2 k Q q m y R ( R + y ) V 2 = 2 y q Q 4 π ε 0 R ( R + y ) m + g