Three uniform spheres of mass M and radius R each are kept in such a way that each touches the other two. The magnitude of the gravitational force on any of the spheres due to the other two is

The correct option is √3 GM²/4R².

Step-by-Step Explanation:

1. Determine the Distance between the Centers of the Spheres:
Let there be three identical, uniform spheres of mass M and radius R. When three such spheres touch each other, the distance between the center of any sphere and the center of another sphere is equal to the sum of their radii:

$d=R+R=2R$

Thus, the centers of the three spheres form an equilateral triangle with a side length of $2R$.

2. Calculate the Gravitational Force between any Two Spheres:
According to Newton's law of universal gravitation, the magnitude of the force of attraction between two spheres of mass M separated by a distance d is given by:

$F=\frac{G{M}^2}{d^2}$

Substituting $d=2R$ into the equation:

$F=\frac{G{M}^2}{{\left(2R\right)}^2}=\frac{G{M}^2}{4R^2}$

3. Determine the Net Gravitational Force on one Sphere:
Let us consider one sphere. It experiences two gravitational forces of equal magnitude, say $F_1$ and $F_2$, due to the other two spheres:

$F_1=F_2=F=\frac{G{M}^2}{4R^2}$

Since the centers of the spheres form an equilateral triangle, the angle between the lines of action of these two forces is:

$\theta =60°$

The magnitude of the resultant force $F_{\text{net}}$ can be calculated using the vector addition formula:

$F_{\text{net}}=\sqrt{F_1^2+F_2^2+2F_1{F}_2\cos \theta }$

Substituting $F_1=F_2=F$ and $\cos (60°)=\frac{1}{2}$:

$F_{\text{net}}=\sqrt{F^2+F^2+2F^2\left(\frac{1}{2}\right)}$

$F_{\text{net}}=\sqrt{2F^2+F^2}=\sqrt{3F^2}=\sqrt{3}F$

4. Substitute the value of F:

$F_{\text{net}}=\sqrt{3}\left(\frac{G{M}^2}{4R^2}\right)=\frac{\sqrt{3}G{M}^2}{4R^2}$