A solid sphere of uniform density and radius R exerts a gravitational force of attraction F 1 on a particle P, distant 2 R from the centre of the sphere. A spherical cavity of radius R / 2 is now formed in the sphere as shown in figure. The sphere with cavity now applies a gravitational force F 2 on the same particle P. Find the ratio F 2 / F 1.

Question

A solid sphere of uniform density and radius R exerts a gravitational force of attraction F 1 on a particle P, distant 2 R from the centre of the sphere. A spherical cavity of radius R / 2 is now formed in the sphere as shown in figure. The sphere with cavity now applies a gravitational force F 2 on the same particle P. Find the ratio F 2 / F 1. (A) (4/9) (B) (9/7) (C) (7/9) (D) (9/4)

Tardigrade Admin · Accepted Answer

F 1=( GMm /4 R 2), F 2 = force due to whole sphere - force due to the sphere forming the cavity =( GMm /4 R 2)-( GMm /18 R 2) ⇒ (7 GMm /36 R 2) ∴ ( F 2/ F 1)=(7/9)

$\frac{4}{9}$

$\frac{9}{7}$

$\frac{7}{9}$

$\frac{9}{4}$

$F_{1} = \frac{GM m}{4 R ^{2}}, F_{2}$
$=$ force due to whole sphere - force due to the sphere forming the cavity
$= \frac{GM m}{4 R ^{2}} - \frac{GM m}{18 R ^{2}}$
$\Rightarrow \frac{7 GM m}{36 R ^{2}}$
$∴ \frac{F _{2}}{F _{1}} = \frac{7}{9}$

94​

79​

97​

49​

F1​=4R2GMm​,F2​ = force due to whole sphere - force due to the sphere forming the cavity =4R2GMm​−18R2GMm​ ⇒36R27GMm​ ∴F1​F2​​=97​

$\frac{4}{9}$

$\frac{9}{7}$

$\frac{7}{9}$

$\frac{9}{4}$

$F_{1} = \frac{GM m}{4 R ^{2}}, F_{2}$
$=$ force due to whole sphere - force due to the sphere forming the cavity
$= \frac{GM m}{4 R ^{2}} - \frac{GM m}{18 R ^{2}}$
$\Rightarrow \frac{7 GM m}{36 R ^{2}}$
$∴ \frac{F _{2}}{F _{1}} = \frac{7}{9}$