A particle of mass M is situated at the centr | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) The total gravitational potential $V$ at a point $P$ inside the shell is the sum of the potential due to the particle at the centre and the potent

Vedclass · Accepted Answer

$\frac{-3GM}{a}$

Vedclass · Answer

(C) The total gravitational potential $V$ at a point $P$ inside the shell is the sum of the potential due to the particle at the centre and the potential due to the spherical shell.$1$. Potential due to the particle of mass $M$ at distance $r = \frac{a}{2}$ from the centre is $V_1 = \frac{-GM}{r} = \frac{-GM}{a/2} = \frac{-2GM}{a}$.$2$. Potential due to the spherical shell of mass $M$ and radius $a$ at any point inside it is constant and equal to the potential at its surface,which is $V_2 = \frac{-GM}{a}$.$3$. The total potential at point $P$ is $V = V_1 + V_2 = \frac{-2GM}{a} + \frac{-GM}{a} = \frac{-3GM}{a}$.

$A$ particle of mass $M$ is situated at the centre of a spherical shell of same mass $M$ and radius $a$. The gravitational potential at a point situated at a distance of $\frac{a}{2}$ from the centre will be:

Similar Questions

Acidified potassium permanganate solution is decolourised by

Keystone species deserve protection because these

Current is flowing through a conducting hollow pipe whose cross-section is shown in the figure. The value of magnetic induction will be zero at

From the top of a tower of height $40 \ m$,a ball is projected upwards with a speed of $20 \ m/s$ at an angle $30^{\circ}$ to the horizontal. The ball will hit the ground in time ......... $\sec$ (Take $g = 10 \ m/s^2$)

If $3x^2-11xy+10y^2-7x+13y+k=0$ denotes a pair of straight lines,then the point of intersection of the lines is

Vedclass Test Series

Exam Paper Generator

Online Exam Module