A-metal-sphere-has-radius-R-and-mass-m-A-spherical-hollow-of-diameter-R-is-made-in-this-sphere-such-that-its-surface-passes-through-the-centre-of-the-metal-sphere-and-touches-the-outside-surface-of-t

Question Number 173558 by MadhumitaSamanta last updated on 13/Jul/22

A metal sphere has radius R and mass m . A spherical

hollow of diameter R is made in this sphere such that its

surface passes through the centre of the metal sphere

and touches the outside surface of the metal sphere .

A unit mass is placed at a distance from the centre of the metal

sphere . The gravitational field at that point is

(a) \frac{G M}{R^{2}} (1 - \frac{1}{8 {(1 - \frac{2 R}{a})}^{2}})

(b) \frac{G M}{a^{2}} (1 - \frac{1}{8 {(1 - \frac{R}{2 a})}^{2}})

(c) \frac{G M}{{(R + a)}^{2}} (1 - \frac{1}{8 {(1 - \frac{R}{2 a})}^{2}})

(d) \frac{G M}{{(R - a)}^{2}} (1 - \frac{1}{8 {(1 - \frac{2 a}{R})}^{2}})

Commented by MadhumitaSamanta last updated on 13/Jul/22

Answered by aleks041103 last updated on 13/Jul/22

W e c a n s a y t h a t t h e m e t a l e l e m e n t i s

a s u p e r p o s i t i o n o f a s p h e r e o f r a d i u s R

a n d m a s s m a n d a s p h e r e o f d i a m e t e r R

a n d m a s s - \frac{m}{8}

\Rightarrow F = \frac{G m}{a^{2}} + \frac{G (- \frac{m}{8})}{{(a - R / 2)}^{2}} = \frac{G m}{a^{2}} (1 - \frac{1}{8 {(1 - \frac{R}{2 a})}^{2}})

\Rightarrow A n s . (b)

Commented by Tawa11 last updated on 13/Jul/22

Great sir

Facebook Tweet Pin