Browse by Stream
Discuss Doubts
Home
Search for Exam, Articles, Questions
get app
Go To Your Study Dashboard
  1. Home
  2. From a solid sphere of mass M and radius R, a spherical portion of radius  is removed, as shown in the figure. Taking gravitati
# Engineering

15501352265161439601867.jpg

From a solid sphere of mass M and radius R, a spherical portion of radius \frac{R}{2} is removed, as shown in the figure. Taking

gravitational potential v=0 at  r=\infty, the potential at the centre of the cavity thus formed is :

(G = gravitational constant)

Answers (1)
S safeer

@Mannika

Outside the surface

V=-\frac{GM}{r}

on the surface

V_{surface}=-\frac{GM}{R}

Inside the surface 

r<R

V=-\frac{GM}{2R}\left[3-\left ( \frac{r}{R} \right )^{2}\right]

\left (3R^{2}-\left ( \frac{R}{2} \right )^{2} \right )

=\frac{-11GM}{8R}

Mass of spherical portion to be removed

=\frac{M}{8}

Potential at point P due to spehrical portion to be removed =\frac{-3GM}{8R}

=\frac{-11GM}{8R}-(\frac{-3GM}{8R})=\frac{-GM}{R}

Similar Questions

Preparation Products

Knockout JEE Main April 2021

An exhaustive E-learning program for the complete preparation of JEE Main..

₹ 22999/- ₹ 14999/-
Buy Now
Knockout JEE Main April 2022

An exhaustive E-learning program for the complete preparation of JEE Main..

₹ 34999/- ₹ 24999/-
Buy Now
Test Series JEE Main April 2021

Take chapter-wise, subject-wise and Complete syllabus mock tests and get in depth analysis of your test..

₹ 6999/- ₹ 4999/-
Buy Now
JEE Main Rank Booster 2021

This course will help student to be better prepared and study in the right direction for JEE Main..

₹ 13999/- ₹ 9999/-
Buy Now
Test Series JEE Main April 2022

Take chapter-wise, subject-wise and Complete syllabus mock tests and get an in-depth analysis of your test..

₹ 6999/-
Buy Now
Boost your Preparation for JEE Main 2021 with Personlized Coaching
Start Now
 
Exams
Articles
Questions