Question

# A solid sphere of uniform density and radius $$R$$ applies a gravitational force of attractionÂ equal to $$F_1$$ on a particle placed at a distanceÂ 3R from the centre of the sphere. A sphericalÂ cavity of radius $$\dfrac{R}{2}$$Â is now made in theÂ sphere, as shown in the figure. The sphereÂ with cavity now applies a gravitational forceÂ $$F_2$$ on the same particle. The ratio $$\dfrac{F_2}{F_1}$$Â is

A
950
B
4150
C
325
D
2225

Solution

## The correct option is A $$\dfrac{41}{50}$$ From super position principle ,  $$F_{1}=F_{r}+F_{c}$$ Here $$F_{r}=$$ force due to remaining part $$=F_{2}$$ $$F_{c}=$$ force due to mass on the cavity $$F_{1}=\dfrac{GMm}{9R^{2}}$$ $$\displaystyle {F_{c}=\dfrac{G(\dfrac{M}{8})m}{(\dfrac{5}{2}R)^{2}}=\dfrac{GMm}{50R^{2}}}$$ $$=\displaystyle \succ F_{2}=F_{1}-F_{c}=\dfrac{GMm}{9R^{2}}-\dfrac{GMm}{50R^{2}}=\dfrac{41GMm}{450R^{2}}$$ $$\Rightarrow \displaystyle \dfrac{F_{2}}{F_{1}}=\dfrac{41}{50}$$ Physics

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