The gravitational potential of two homogeneous spherical shells $$A$$ and $$B$$ of same surface density at their respective centres are in the ratio $$3 : 4.$$  If the two shells coalesce into single one such that surface charge density remains same, then the ratio of potential at an internal point of the view shell to shell $$A$$ is equal to

Solution :
$${M_A} = \sigma 4\pi R_A^2,{M_B} = \sigma 4\pi R_B^2,$$
where $$\sigma $$ is surface density
$$\eqalign{ & {V_A} = \frac{{ - G{M_A}}}{{{R_A}}},{V_B} = \frac{{ - G{M_B}}}{{{R_B}}} \cr & \frac{{{V_A}}}{{{V_B}}} = \frac{{{M_A}}}{{{M_B}}}\frac{{{R_B}}}{{{R_A}}} = \frac{{\sigma 4\pi R_A^2}}{{\sigma 4\pi R_B^2}}\frac{{{R_B}}}{{{R_A}}} = \frac{{{R_A}}}{{{R_B}}} \cr & {\text{Given}}\,\,\frac{{{V_A}}}{{{V_B}}} = \frac{{{R_A}}}{{{R_B}}} = \frac{3}{4}\,\,{\text{then}}\,\,{R_B} = \frac{4}{3}{R_A} \cr} $$
for new shell of mass $$M$$ and radius $$R$$
$$\eqalign{ & M = {M_A} + {M_B} = \sigma 4\pi R_A^2 + \sigma 4\pi R_B^2 \cr & \sigma 4\pi {R^2} = \sigma 4\pi \left( {R_A^2 + R_B^2} \right) \cr} $$
then
$$\eqalign{ & \frac{V}{{{V_A}}} = \frac{M}{R}\frac{{{R_A}}}{{{R_B}}} = \frac{{\sigma 4\pi \left( {R_A^2 + R_B^2} \right)}}{{{{\left( {R_A^2 + R_B^2} \right)}^{\frac{1}{2}}}}} = \frac{{{R_A}}}{{\sigma 4\pi R_A^2}} \cr & = \frac{{\sqrt {R_A^2 + R_B^2} }}{{{R_A}}} = \frac{5}{3} \cr} $$