Question

A solid sphere of radius $R$ gravitationally attracts a particle placed at $3R$ from its centre with a force $F_1$. Now a spherical cavity of radius $\frac{R}{2}$ is made in the sphere (as shown in figure) and the force becomes $F_2$. The value of $F_1:F_2$ is:

• A. $41:50$
• B. $36:25$
• C. $50:41$
• D. $25:36$

Answer 1

The correct option is A

$41:50$

Step 1 : Given data

Radius of the solid sphere $=R$

The particle placed distance $=2R$

With acting force $=F$

Radius of spherical cavity $=\frac{R}{2}$

That time the force become changes to $\frac{F}{2}$

$g_1$ and $g_2$ are gravitational field intensities.

$M=$ Mass

$G=$Gravitational constant

Step 2 : To find the value of $F_1:F_2$

Gravitational field intensity , According to Universal law of Gravitation.

$$\begin{gathered} g_1=\frac{GM}{{\left(3R\right)}^2} \\ =\frac{GM}{9R^2} \end{gathered}$$

Gravitational field intensity

$$\begin{gathered} g_2=\frac{GM}{9R^2}-\frac{G\left({\displaystyle \frac{M}{8}}\right)}{{\left({\displaystyle \frac{5R}{2}}\right)}^2} \\ =\frac{GM}{9R^2}-\frac{GM}{R^{2}50} \\ =\frac{\left(\frac{41}{9\times 50}\right)GM}{R^2} \end{gathered}$$

Then we can write that

We know that $F=ma$ and $a=g$

$$\begin{gathered} \frac{g_1}{g_2}=\frac{41}{50} \\ \frac{F_1}{F_2}=\frac{m{g}_1}{m{g}_2} \\ =\frac{41}{50} \end{gathered}$$

Hence the value of $F_1:F_2$ is $41:50$

Therefore, correct answer is Option A.