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. The initial mass of the solid sphere was M. lf a particle of mass m is placed at a distance 2.5R from the centre of the cavity, then what is the gravitational attraction on the mass m?

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

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

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

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

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

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 mass of the cavity $= M'$

Density of the sphere $=M/ (4 / 3 π R^{3})$

Mass of the cavity cut out $= M' = \frac{4}{3} π \frac{R^{3}}{8} \times \frac{M}{\frac{4}{3} {πR}^{3}}$

$\begin{array}{l} ∴ M' = \frac{M}{8} \Rightarrow F_{net} = F_{Mm} - F_{M' m} \\ = \frac{GMm}{4 R^{2}} - \frac{GM' m}{{(\frac{5}{2} R)}^{2}} = \frac{GMm}{4 R^{2}} - \frac{GMm}{50 R^{2}} \\ F_{net} = \frac{23}{100} \frac{GMm}{R^{2}} \end{array}$

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