Question

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 new shell A is equal to –

• A. 3 : 2
• B. 4 : 3
• C. 5 : 3
• D. 5 : 6

Answer 1

Answer: (C)

5 : 3

Answer 2

M_A = s4p, M_B = s4p

where s is surface density

V_A =, V_B =

$\frac { 3 } { 4 }$

then R_B = $\frac { 4 } { 3 }$ R_A

For new shell of mass M and Radius R -

M = M_A + M_B = $\frac { \sigma 4 \pi \left[ \mathrm { R } _ { \mathrm { A } } ^ { 2 } + \mathrm { R } _ { \mathrm { B } } ^ { 2 } \right] } { \left( \mathrm { R } _ { \mathrm { A } } ^ { 2 } + \mathrm { R } _ { \mathrm { B } } ^ { 2 } \right) ^ { 1 / 2 } }$ $\frac { \mathrm { R } _ { \mathrm { A } } } { \sigma 4 \pi \mathrm { R } _ { \mathrm { A } } ^ { 2 } }$

= $\frac { \sqrt { \mathrm { R } _ { \mathrm { A } } ^ { 2 } + \mathrm { R } _ { \mathrm { B } } ^ { 2 } } } { \mathrm { R } _ { \mathrm { A } } }$ = $\frac { 5 } { 3 }$