Gravitation Question 363
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 view shell to shell A is equal to
Options:
A) 3:2
B)4:3
C) 5:3
D)5:6
Show Answer
Answer:
Correct Answer: C
Solution:
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$ [M _{A}=\sigma 4\pi {R^{2}} _{A},M _{B}=\sigma 4\pi R] $ , where a is surface density
$ [{V _{A}}=\frac{-GM _{A}}{R _{A}},,V _{B}=\frac{-GM _{B}}{R _{B}}] $
$ [\frac{V _{A}}{V _{B}}=\frac{M _{A}}{M _{B}},,\frac{R _{B}}{R _{A}}=\frac{\sigma 4\pi R^{2} _{A}}{\sigma 4\pi R^{2} _{B}},\frac{R _{B}}{{R _{A}}}=\frac{{R _{A}}}{{R _{B}}}] $
Given $ [\frac{{V _{A}}}{{V _{B}}}=,\frac{R _{A}}{R _{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}}=\sigma 4\pi R^{2} _{A}+\sigma 4\pi R^{2} _{B}] $
$ [\sigma 4\pi R^{2}=\sigma 4\pi \left( R^{2} _{A}+{{R}^{2}} _{B} \right)] $
then $ [\frac{V}{{V _{A}}}=\frac{M}{M},,\frac{{R _{A}}}{{R _{B}}}=\frac{\sigma 4\pi \left( R^{2} _{A}+R^{2} _{B} \right)}{{{\left( R^{2} _{A}+R^{2} _{B} \right)}^{1/2}}},=\frac{{R _{A}}}{\sigma 4\pi R^{2} _{A}}] $
$ [\frac{\sqrt{R^{2} _{A}+R^{2} _{B}}}{R _{A}}=\frac{5}{3}] $
