A highly rigid cubical block `A` of small mass `M` and side `L` is fixed rigidly on the other cubical block of same dimensions and of modulus of rigidity `eta` such that the lower face of `A` completely covers the upper face of `B`. The lower face of `B` is rigidly held on a horizontal surface . `A` small force `F` is applied perpendicular to one of the side faces of `A`. After the force is withdrawn , block `A` executes faces of `A`. After the force is withdrawn , block `A` exceutes small oscillations , the time period of which is given by

Options

A. `2pi sqrt(M eta L)`
B. `2pi sqrt((M eta)/(L))`
C. `2pi sqrt((ML)/(eta))`
D. `2pi sqrt((M)/(eta L))`

Answer

D

Explanation

We know that `eta=(FL)/(Ax)` , where x is the displacement.
As, `A=L^(2)` So, `eta=(FL)/(L^(2)x)=(F)/(Lx)`, so shearing force
`F=eta L x` …….(i)
It means F `prop` x and this F is directed towards mean popsition, hence after the force is withdrawn, the block will execute liner S.H.M. Here spring factor `=eta L` and inertia factor = mass = M. ltBrgt As time period.
`T=2pi sqrt(("inertia factor")/("spring factor"))=2pi sqrt((M)/(eta L))`