Gravitation Question 343
Question: The gravitational field, due to the ’left over part* of a uniform sphere (from which a part as shown, has been ‘removed out’), at a very far off point, P, located as shown, would be (nearly):
Options:
A) $ [\frac{5}{6}\frac{GM}{x^{2}}] $
B) $ [\frac{8}{9}\frac{GM}{x^{2}}] $
C) $ [\frac{7}{8}\frac{GM}{x^{2}}] $
D) $ [\frac{6}{7}\frac{GM}{x^{2}}] $
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Answer:
Correct Answer: C
Solution:
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Let mass of smaller sphere (which has to be removed) is m Radius
$ [=\frac{R}{2}] $ $ [\frac{M}{\frac{4}{3}\pi R^{3}}=\frac{m}{\frac{4}{3}\pi {{\left( \frac{R}{2} \right)}^{3}}}\Rightarrow m=\frac{M}{8}] $
Mass of the left over part of the sphere
$ [M’=M-\frac{M}{8}=\frac{7}{8}M] $
Therefore gravitational field due to the left over part of the sphere
$ [=\frac{GM’}{x^{2}}=\frac{7}{8}\frac{GM}{x^{2}}] $
BETA
