JEE MAIN - Physics (2020 - 5th September Morning Slot - No. 23)
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)._5th_September_Morning_Slot_en_23_1.png)
(assume the remaining portion to be spherical).
v2 = $$y\left[ {{{qQ} \over {4\pi {\varepsilon _0}R\left( {R + y} \right)m}} + g} \right]$$
v2 = $$2y\left[ {{{qQR} \over {4\pi {\varepsilon _0}{{\left( {R + y} \right)}^3}m}} + g} \right]$$
v2 = $$2y\left[ {{{qQ} \over {4\pi {\varepsilon _0}R\left( {R + y} \right)m}} + g} \right]$$
v2 = $$y\left[ {{{qQ} \over {4\pi {\varepsilon _0}{R^2}ym}} + g} \right]$$
Explanation
_5th_September_Morning_Slot_en_23_2.png)
Applying mechanical energy conservation,
$${K_A} + {U_A} = {K_B} + {U_B}$$
$$ \Rightarrow $$ $$0 + mgy + q{V_A} = {1 \over 2}m{v^2} + 0 + ( + q{V_B})$$
$$ \Rightarrow $$ $$mgy + q{V_A} = {1 \over 2}m{v^2} + q({V_B})$$
$$ \Rightarrow $$ $$mgy + {{qk(Q)} \over R} = {1 \over 2}m{v^2} + {{qk(Q)} \over {R + y}}$$
$$ \Rightarrow $$ $${1 \over 2}m{v^2} = - {{kq(Q)} \over {R + y}} + {{kq(Q)} \over R} + mgy$$
$$ \Rightarrow $$ $${{m{v^2}} \over 2} = {{ - kq(Q)R + kq(Q)(R + y)} \over {R(R + y)}} + mgy$$
$$ \Rightarrow $$ $${v^2} = {2 \over m}\left[ {{{ - kqQR + kqQR + kqQ} \over {R(R + y)}} + mgy} \right]$$
$$ \Rightarrow $$ $${v^2} = {2 \over m}\left[ {{{kq(Q)y} \over {R(R + y)}} + mgy} \right]$$
$$ \Rightarrow $$ $${v^2} = 2y\left[ {{{q(Q)} \over {4\pi {\varepsilon _0}R(R + y)m}} + g} \right] = 2y\left[ {{{qQ} \over {4\pi {\varepsilon _0}R(R + y)m}} + g} \right]$$
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