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(D) Let the charge on the smaller sphere be $Q$. The potential of the smaller sphere is given by $V_1 = \frac{kQ}{R_1}$,where $R_1 = 2 \, m$ and $V_1

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$40$

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(D) Let the charge on the smaller sphere be $Q$. The potential of the smaller sphere is given by $V_1 = \frac{kQ}{R_1}$,where $R_1 = 2 \, m$ and $V_1 = 120 \, V$.Thus,$120 = \frac{kQ}{2}$,which implies $kQ = 240$.When the smaller sphere is placed inside the larger hollow sphere of radius $R_2 = 6 \, m$,the potential of the larger sphere is determined by the charge $Q$ on the inner sphere.The potential at any point on or inside the larger sphere due to the inner sphere is $V_2 = \frac{kQ}{R_2}$.Substituting the value of $kQ = 240$ and $R_2 = 6 \, m$:$V_2 = \frac{240}{6} = 40 \, V$.Therefore,the potential of the bigger sphere is $40 \, V$.

$A$ spherical conductor of radius $2 \, m$ is charged to a potential of $120 \, V$. It is now placed inside another hollow spherical conductor of radius $6 \, m$. Calculate the potential to which the bigger sphere would be raised. (in $, V$)

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