Draw a schematic diagram of Cavendish's experiment for the determination of the universal gravitational constant $G$ and obtain the formula used in it.

(N/A) The value of the gravitational constant $G$ can be determined experimentally,which was first done by Cavendish in $1798$. The apparatus used is schematically shown in the figure.
The bar $AB$ has two small lead spheres attached at its ends. The bar is suspended from a rigid support by a fine wire.
Two large lead spheres are brought close to the small ones but on opposite sides. The big spheres attract the nearby small ones by equal and opposite forces. There is no net force on the bar,but there is a torque equal to $F \times L$,where $F$ is the force of attraction between a big sphere and a small sphere,and $L$ is the length of the bar. As a result,the bar rotates about the wire $OM$. The wire gets twisted until the restoring torque of the wire equals the gravitational torque.
In this equilibrium position,the angle of twist $\theta$ is measured. If $k$ is the torsion constant of the wire,the restoring torque is $\tau = k\theta$.
Gravitational force on each small sphere is $F = \frac{GMm}{d^2}$,where $M$ is the mass of the large sphere,$m$ is the mass of the small sphere,and $d$ is the distance between their centers.
The gravitational torque is $\tau_g = F \times L = \frac{GMm}{d^2} L$.
At equilibrium,$\tau_g = \tau$,so $\frac{GMm}{d^2} L = k\theta$.
Thus,$G = \frac{k\theta d^2}{MLm}$.