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QJEEM20Ph5730: JEE Main 2020 Physics - Class 12 JEEM20Ph5730 Solution: A solid sphere of radius R carries a charge Q + q distributed…

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 v when it has fallen through a vertical height y (see figure), then :
(assume the remaining portion to be spherical). JEE Main 2020 (Online) 5th September Morning Slot Physics - Electrostatics Question 148 English

Multiple Choice Options

  • A v2 = y[qQ4πε0R(R+y)m+g]y\left[ {{{qQ} \over {4\pi {\varepsilon _0}R\left( {R + y} \right)m}} + g} \right]
  • B v2 = 2y[qQR4πε0(R+y)3m+g]2y\left[ {{{qQR} \over {4\pi {\varepsilon _0}{{\left( {R + y} \right)}^3}m}} + g} \right]
  • C v2 = 2y[qQ4πε0R(R+y)m+g]2y\left[ {{{qQ} \over {4\pi {\varepsilon _0}R\left( {R + y} \right)m}} + g} \right]
  • D v2 = y[qQ4πε0R2ym+g]y\left[ {{{qQ} \over {4\pi {\varepsilon _0}{R^2}ym}} + g} \right]

Step-by-Step Verified Solution

JEE Main 2020 (Online) 5th September Morning Slot Physics - Electrostatics Question 148 English Explanation

Applying mechanical energy conservation,

KA+UA=KB+UB{K_A} + {U_A} = {K_B} + {U_B}

⇒\Rightarrow 0+mgy+qVA=12mv2+0+(+qVB)0 + mgy + q{V_A} = {1 \over 2}m{v^2} + 0 + ( + q{V_B})

⇒\Rightarrow mgy+qVA=12mv2+q(VB)mgy + q{V_A} = {1 \over 2}m{v^2} + q({V_B})

⇒\Rightarrow mgy+qk(Q)R=12mv2+qk(Q)R+ymgy + {{qk(Q)} \over R} = {1 \over 2}m{v^2} + {{qk(Q)} \over {R + y}}

⇒\Rightarrow 12mv2=−kq(Q)R+y+kq(Q)R+mgy{1 \over 2}m{v^2} = - {{kq(Q)} \over {R + y}} + {{kq(Q)} \over R} + mgy

⇒\Rightarrow mv22=−kq(Q)R+kq(Q)(R+y)R(R+y)+mgy{{m{v^2}} \over 2} = {{ - kq(Q)R + kq(Q)(R + y)} \over {R(R + y)}} + mgy

⇒\Rightarrow v2=2m[−kqQR+kqQR+kqQR(R+y)+mgy]{v^2} = {2 \over m}\left[ {{{ - kqQR + kqQR + kqQ} \over {R(R + y)}} + mgy} \right]

⇒\Rightarrow v2=2m[kq(Q)yR(R+y)+mgy]{v^2} = {2 \over m}\left[ {{{kq(Q)y} \over {R(R + y)}} + mgy} \right]

⇒\Rightarrow v2=2y[q(Q)4πε0R(R+y)m+g]=2y[qQ4πε0R(R+y)m+g]{v^2} = 2y\left[ {{{q(Q)} \over {4\pi {\varepsilon _0}R(R + y)m}} + g} \right] = 2y\left[ {{{qQ} \over {4\pi {\varepsilon _0}R(R + y)m}} + g} \right]

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