A solid sphere of radius R/2 is cut out of a solid sphere of radius ,R such that the spherical cavity so formed touches the surface on one side and the centre of the sphere on the other side as shown in fig. The initial
mass of the solid sphere was M. If a particle of mass m is placed at a distance 2.5 R from the centre of the cavity, then the gravitational potential at the mass ,n will be

see full answer

Start JEE / NEET / Foundation preparation at rupees 99/day !!

21% of IItians & 23% of AIIMS delhi doctors are from Sri Chaitanya institute !!

An Intiative by Sri Chaitanya

$\frac{GMm}{4 R^{2}}$

$\frac{GMm}{R^{2}}$

$\frac{23}{100} \frac{GMm}{R^{2}}$

$\frac{GMm}{8 R^{2}}$

answer is D.

(Unlock A.I Detailed Solution for FREE)

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

Detailed Solution

Let M' be the mass removed. If p be the density, then
$ρ = \frac{M}{(4 / 3 \cdot {πR}^{3})}$
Now $M^{'} = ρ [\frac{4}{3} π {(\frac{R}{2})}^{2}]$
$= \frac{M}{(4 / 3 {πR}^{3})} \times [\frac{4}{3} π {(\frac{R}{2})}^{2}] = \frac{M}{8}$
So, gravitational force =force due to solid sphere of mass M - force due to mass removed to produce the cavity
$\begin{array}{l} ∴ \frac{GMm}{(2 R)^{2}} - \frac{GMm}{8 \times (2 \cdot 5 R)^{2}} \\ = \frac{GmM}{R^{2}} [\frac{1}{4} - \frac{1}{50}] \\ = \frac{23}{100} \times \frac{GMm}{R^{2}} \end{array}$

Watch 3-min video & get full concept clarity

tricks from toppers of Infinity Learn

Result Producing online coaching for JEE/NEET of south India

Largest All India Test Series for JEE/NEET

Best Books for IIT JEE

Best Books for NEET

🔥 Start Your JEE/NEET Prep at Just ₹1999 / month - Limited Offer! Check Now!

Get Expert Academic Guidance – Connect with a Counselor Today!

Result Producing online coaching for JEE/NEET of south India

Largest All India Test Series for JEE/NEET

Best Books for IIT JEE

Best Books for NEET