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JEE Mains Previous Years Questions with Solutions

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JEE Main 2015 (Offline)

MCQ (More than One Correct Answer)
A uniformly charged solid sphere of radius $$R$$ has potential $${V_0}$$ (measured with respect to $$\infty $$) on its surface. For this sphere the equipotential surfaces with potentials $${{3{V_0}} \over 2},\,{{5{V_0}} \over 4},\,{{3{V_0}} \over 4}$$ and $${{{V_0}} \over 4}$$ have radius $${R_1},\,\,{R_2},\,\,{R_3}$$ and $${R_4}$$ respectively. Then
A
$${R_1} = 0$$ and $${R_2} < \left( {{R_4} - {R_3}} \right)$$
B
$$2R < {R_4}$$
C
$${R_1} = 0$$ and $${R_2} > \left( {{R_4} - {R_3}} \right)$$
D
$${R_1} \ne 0$$ and $$\left( {{R_2} - {R_1}} \right) > \left( {{R_4} - {R_3}} \right)$$

Explanation

$$(a,b)$$ We know, $${V_0} = {{Kq} \over R} = Vsurface$$

Now, $${V_i} = {{Kq} \over {2{R^3}}}\left( {3{R^2} - {R^2}} \right)\,\,\,\,\,$$ [For $$r < R$$]

At the center of sphere $$r=0.$$ Here

$$V = {3 \over 2}{V_0}$$

Now, $${5 \over 4}{{Kq} \over R} = {{Kq} \over {2{R^3}}}\left( {3{R^2} - {r^2}} \right)$$

$${R_2} = {R \over {\sqrt 2 }}$$

$${3 \over 4}{{Kq} \over R} = {{Kq} \over {{R^3}}}$$

$${1 \over 4}{{Kq} \over R} = {{Kq} \over {R{}_4}}$$

$${R_4} = 4R$$

Also, $${R_1} = 0$$ and $${R_2} < \left( {{R_4} - R{}_3} \right)$$

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