According to the universal law of gravitation, gravitational force (\(F\)) acting between two objects is inversely proportional to the square of the distance (\(r\)) between them, i.e.,
\(𝐹∝ \frac{1 }{𝑟^2 }\)
If distance r becomes \(\frac{r}2\), then the gravitational force will be proportional to
\(\frac1{ (\frac𝑟{2})^2}\) = \(\frac4{𝑟^2}\)
Hence, if the distance is reduced to half, then the gravitational force becomes four times larger than the previous value.
A small point of mass \(m\) is placed at a distance \(2R\) from the center \(O\) of a big uniform solid sphere of mass \(M\) and radius \(R\). The gravitational force on \(m\) due to \(M\) is \(F_1\). A spherical part of radius \(R/3\) is removed from the big sphere as shown in the figure, and the gravitational force on \(m\) due to the remaining part of \(M\) is found to be \(F_2\). The value of the ratio \( F_1 : F_2 \) is: 
Match the LIST-I with LIST-II
\[ \begin{array}{|l|l|} \hline \text{LIST-I} & \text{LIST-II} \\ \hline \text{A. Gravitational constant} & \text{I. } [LT^{-2}] \\ \hline \text{B. Gravitational potential energy} & \text{II. } [L^2T^{-2}] \\ \hline \text{C. Gravitational potential} & \text{III. } [ML^2T^{-2}] \\ \hline \text{D. Acceleration due to gravity} & \text{IV. } [M^{-1}L^3T^{-2}] \\ \hline \end{array} \]
Choose the correct answer from the options given below:
∆ABC is an isosceles triangle in which AB = AC. Side BA is produced to D such that AD = AB (see Fig. 7.34). Show that ∠ BCD is a right angle.
