A spherical marble of radius r is made to oscillate in a bowl of radius R. What is its period of oscillation ?

Options

A. `T=2pisqrt((R )/(g))`
B. `T=2pisqrt((R-r)/(g))`
C. `T=(1)/(2pi)sqrt((g)/(R-r))`
D. `T=2pisqrt((g)/(R ))`

Answer

B

Explanation

In this case, the equivalent length of the simple pendulum =Dis"tan"ce between the centre of the bowl and the centre of the marble =(R-r)=l
`therefore T=2pi sqrt((l)/(g))=2pi sqrt((R-r)/(g))`