JEE MAIN - Physics (2019 - 12th January Morning Slot - No. 28)
There is a uniform spherically symmetric surface charge density at a distance R0 from the origin. The charge distribution is initially at rest and starts expanding because of mutual repulsion. The figure that represents best the speed V (R(t)) of the distribution as a function of its instantaneous radius R (t) is :
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Explanation
At any instant 't'
Total energy of charge distribution is constant
i.e. $${1 \over 2}m{V^2} + {{K{Q^2}} \over {2R}} = 0 + {{K{Q^2}} \over {2{R_0}}}$$
$$ \therefore $$ $${1 \over 2}m{V^2} = {{K{Q^2}} \over {2{R_0}}} - {{K{Q^2}} \over {2R}}$$
$$ \therefore $$ V = $$\sqrt {{2 \over m}{{K{Q^2}} \over 2}.\left( {{1 \over {{R_0}}} - {1 \over R}} \right)} $$
$$ \therefore $$ V = $$\sqrt {{{K{Q^2}} \over m}\left( {{1 \over {{R_0}}} - {1 \over R}} \right)} $$ = C$$\sqrt {{1 \over {{R_0}}} - {1 \over R}} $$
Also the slope of v-s curve will go on decreasing
Total energy of charge distribution is constant
i.e. $${1 \over 2}m{V^2} + {{K{Q^2}} \over {2R}} = 0 + {{K{Q^2}} \over {2{R_0}}}$$
$$ \therefore $$ $${1 \over 2}m{V^2} = {{K{Q^2}} \over {2{R_0}}} - {{K{Q^2}} \over {2R}}$$
$$ \therefore $$ V = $$\sqrt {{2 \over m}{{K{Q^2}} \over 2}.\left( {{1 \over {{R_0}}} - {1 \over R}} \right)} $$
$$ \therefore $$ V = $$\sqrt {{{K{Q^2}} \over m}\left( {{1 \over {{R_0}}} - {1 \over R}} \right)} $$ = C$$\sqrt {{1 \over {{R_0}}} - {1 \over R}} $$
Also the slope of v-s curve will go on decreasing
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