First slide

Universal law of gravitation

Question

A solid sphere of uniform density and radius R applies a gravitational force of attraction equal to F₁ on a particle placed at a distance 2R from the centre of the sphere. A spherical cavity of radius (R/2) is now made in the sphere as shown in Figure. The sphere with the cavity now applies a gravitational force F₂ on the same particle. The ratio (F₂/F₁) is :

Moderate

A

(1/2)
B

(3/4)
C

(7/8)
D

(7/9)

Solution

$\begin{array}{l} \Rightarrow F_{1} = \frac{GMm}{(2 R)^{2}}, F_{2} = \frac{GMm}{(2 R)^{2}} - \frac{G {(\frac{M}{8})}^{m}}{{(\frac{3 R}{2})}^{2}} = \frac{7 GMm}{36 R^{2}} \\ \Rightarrow \frac{F_{2}}{F_{1}} = \frac{7 / 36}{1 / 4} = \frac{7}{9} \end{array}$

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