A spherical hole is made in a solid sphere of radius R . The mass of the sphere before hollowing was M . The gravitational field at the centre of the hole due to the remaining mass is -

Question

A spherical hole is made in a solid sphere of radius R . The mass of the sphere before hollowing was M . The gravitational field at the centre of the hole due to the remaining mass is - <img class="img-fluid question-image" alt="Question" src="https://cdn.tardigrade.in/q/nta/p-xnmkc0ejboud.png" /> (A) zero (B) (G M/8 R2) (C) (G M/2 R2) (D) (G M/R2)

Tardigrade Admin · Accepted Answer

By the principle of superposition of fields

E \to = E_{1} \to + E_{2} \to

Here,

E \to = n e t f i e l d a t t h e ce n t re o f h o l e d u e t o e n t i re ma ss

E_{1} \to = f i e l d d u e t o re mainin g ma ss

and

E_{2} \to = f i e l d d u e t o ma ss in h o l e = 0

∴ E_{1} \to = E \to = (\frac{GM}{R ^{3}}) . r

where

r = \frac{R}{2}

∴ E \to = \frac{GM}{2 R ^{2}}

Q. A spherical hole is made in a solid sphere of radius $R$ . The mass of the sphere before hollowing was $M$ . The gravitational field at the centre of the hole due to the remaining mass is -

zero

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

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

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

Q. A spherical hole is made in a solid sphere of radius R . The mass of the sphere before hollowing was M . The gravitational field at the centre of the hole due to the remaining mass is -

zero

8R2GM​

2R2GM​

R2GM​

By the principle of superposition of fields E→=E1​→​+E2​→​ Here, E→=netfieldatthecentreofholeduetoentiremass E1​→​=fieldduetoremainingmass and E2​→​=fieldduetomassinhole=0 ∴E1​→​=E→=(R3GM​).r where r=2R​ ∴E→=2R2GM​

Q. A spherical hole is made in a solid sphere of radius $R$ . The mass of the sphere before hollowing was $M$ . The gravitational field at the centre of the hole due to the remaining mass is -

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

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

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