A spherical ball of mass $m$ is kept at the highest point in the space between two fixed, concentric spheres $\mathrm{A}$ and $\mathrm{B}$ (see fig.). The smaller sphere $\mathrm{A}$ has a radius $R$ and space between the two spheres has a width $d$. The ball has a diameter very slightly less than $d$. All surfaces are frictionless. The ball is given a gentle push (towards the right in the figure). The angle made by the radius vector of the ball with the upward vertical is denoted by $\theta$.  What is the total normal reaction force exerted by the spheres on the ball in terms of angle $\theta$?

• Correct answer
  A

  $mg (3\cos \theta –2 )$
• B

  $mg ( 2\cos \theta – 3 )$
• C

  $3mg ( 2\cos \theta – 1 )$
• D

  $2mg ( 3\cos \theta – 1 )$

   

Solution

$h=\left(R+\frac{d}{2}\right)\left(1-\text{cos}\mathrm{ \theta }\right)$
 

 
The velocity of the ball at an angle $\theta$ is

${\nu }^2=2gh=2\left(R+\frac{d}{2}\right)\left(1-\text{cos}\mathrm{ \theta }\right)g$  ...... (i)

Let N be the normal reaction (away from the centre) at angle $\theta$. Then, $\mathit{\text{mg }}\text{cos}\mathrm{ \theta }-N=\frac{m{\nu }^2}{\left(R+\frac{d}{2}\right)}$

Substituting the value of ${\nu }^2$ from equation (i), we get

$\mathit{\text{mg}}\text{ cos}\mathrm{ \theta }-N=\text{2}\mathit{\text{mg}}\left(1-\text{cos}\mathrm{ \theta }\right)$

$N=mg\left(\text{3 cos}\mathrm{ \theta }-2\right)$

Hence, the correct answer is option: 'A'.