Question

What is the gravitational field at Q?

• A. $\frac{31GM}{1024}$
• B. $\frac{GM}{1024}$
• C. $31GM$
• D. $GM$

Answer 1

Answer: (A)

Let us calculate field at Q in case the cavity weren't there, since mass is proportional to $R^3$

Gravitational field $g_0=GM\left(\frac{2^3}{4^3}\right)\frac{1}{2^2}=\frac{GM}{32}$

Also, field due to mass previously in right cavity will be 0 since the point is at its centre.

Gravitational field due to left sphere,

$g^{\prime }=GM\left(\frac{1^3}{4^3}\right)\frac{1}{4^2}=\frac{GM}{1024}$

Thus, actual magnetic field at Q is

$g=g_0-g^{\prime }=GM(\frac{1}{32}-\frac{1}{1024})=\frac{31GM}{1024}$

Answer is option A.