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 3R from the centre of the sphere. A spherical cavity of radius R/2 is now made in the sphere as shown in the figure. The sphere with cavity now applies a gravitational force F2 on the same particle . the ratio F2/F1 is 82/x. find x.​

Answer 1

$\huge\bold{\mathbb{\purple{ค{\pink{ภ{\blue{ร{\green{ฬ{\red{є{\orange{г}}}}}}}}}}}}}$

$\implies \: \frac{9}{7} = \frac{82}{x} \\ \implies \: 9x = 574 \\ \implies \: x = \frac{574}{9} = 63.77$