A particle is kept at rest at the top of a sphere of diameter 42m. When disturbed slightly, it slides down. At what height h from the bottom, the particle will leave the sphere

The correct option is 35 m.

Let us analyze the problem step-by-step to find the height $h$ from the bottom of the sphere where the particle leaves its surface.

Step 1: Understand the setup and motion
Let $R$ be the radius of the sphere. The diameter of the sphere is given as $D=42\text{m}$, which means the radius is:
$R=\frac{D}{2}=\frac{42}{2}=21\text{m}$
Let the particle start from rest at the top of the sphere (height $2R$ from the bottom). When disturbed slightly, it slides down along the surface. Let $\theta$ be the angle made by the radius vector of the particle with the vertical when it reaches a certain point on the sphere.

Step 2: Apply the Principle of Conservation of Mechanical Energy
As the particle slides down by an angle $\theta$, its vertical height decreases. The vertical distance descended by the particle is:
$y=R-R\cos \theta =R(1-\cos \theta )$
Using conservation of energy, the loss in potential energy equals the gain in kinetic energy:
$mgR(1-\cos \theta )=\frac{1}{2}m{v}^2$
Simplifying this, we find the square of the velocity $v^2$ at angle $\theta$:
$v^2=2gR(1-\cos \theta )$

Step 3: Analyze the forces acting on the particle
The forces acting on the particle of mass $m$ at angle $\theta$ are:
1. The gravitational force $mg$ acting vertically downwards.
2. The normal reaction force $N$ acting radially outwards from the center of the sphere.

The component of gravity acting towards the center of the sphere is $mg\cos \theta$. Therefore, the net radial force providing the centripetal acceleration is:
$mg\cos \theta -N=\frac{m{v}^2}{R}$
Solving for the normal force $N$:
$N=mg\cos \theta -\frac{m{v}^2}{R}$

Step 4: Condition for leaving the sphere
The particle leaves the surface of the sphere when the normal contact force becomes zero ($N=0$):
$mg\cos \theta =\frac{m{v}^2}{R}$
Substituting the expression for $v^2$ from Step 2 into this equation:
$mg\cos \theta =\frac{m}{R}\left[2gR\right(1-\cos \theta \left)\right]$
Dividing both sides by $mg$:
$\cos \theta =2(1-\cos \theta )$
$\cos \theta =2-2\cos \theta$
$3\cos \theta =2$
$\cos \theta =\frac{2}{3}$

Step 5: Calculate the height from the bottom
The height $h$ of the particle from the center of the sphere is $R\cos \theta$. Therefore, the total height $h$ from the bottom of the sphere is:
$h=R+R\cos \theta =R(1+\cos \theta )$
Substitute the values of $R=21\text{m}$ and $\cos \theta =\frac{2}{3}$:
$h=21(1+\frac{2}{3})=21\left(\frac{5}{3}\right)=7\times 5=35\text{m}$
Thus, the particle will leave the sphere at a height of 35 m from the bottom.