### 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 ${R_2} < \left( {{R_4} - R{}_3} \right)$

On those following papers in MCQ (Multiple Correct Answer)
Number in Brackets after Paper Indicates No. of Questions
JEE Main 2015 (Offline) (1)
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