question_answer

The gravitational filed, 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):             JEE Main Online Paper (Held On 09 April 2013)         

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) \[\frac{6}{7}\frac{GM}{{{x}^{2}}}\]                

Correct Answer: D

Solution :

Mass of the small part                 \[=\frac{M}{\frac{4}{3}\pi {{R}^{3}}}\times \frac{4}{3}\pi {{\left( \frac{R}{2} \right)}^{3}}\]          \[=\frac{M\times {{R}^{3}}}{{{R}^{3}}\times 8}\]                 \[M'=\frac{M}{8}\]                                  g (rest part) = g (complete sphere) ? g (small part)                 \[=\frac{GM}{{{X}^{2}}}-\frac{G\frac{M}{8}}{{{\left( \frac{R}{2}+X \right)}^{2}}}\]                From far point \[X>>\frac{R}{2}\] , so neglect \[\frac{R}{2}\]                 \[=\frac{GM}{{{X}^{2}}}\left( 1-\frac{1}{8} \right)\]                 \[=\frac{7\,GM}{8\,{{X}^{2}}}\]