Question

A highly rigid cubical block $A$ of small mass $M$ and side $L$ is fixed rigidly onto another cubical block $B$ of the 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 After the force is withdrawn block $A$ executes small oscillations. The time period of which is given by

• A. $2\pi \sqrt{\frac{M\eta }{L}}$
• B. $2\pi \sqrt{\frac{L}{M\eta }}$
• C. $2\pi \sqrt{\frac{ML}{\eta }}$
• D. $2\pi \sqrt{\frac{M}{\eta L}}$

Answer 1

The correct option is D $2\pi \sqrt{\frac{M}{\eta L}}$

We know that,
$\eta =\frac{F}{A}=\left[M{L}^{-1}{T}^{-2}\right]$
We can take,
$T\propto M^a{L}^b{\eta }^c$
$T=C{M}^a{L}^b{\eta }^c$
$\left[T\right]=\left[M{]}^a\right[L{]}^b[M{L}^{-1}{T}^{-2}{]}^c$
$\left[T\right]=\left[M{]}^a\right[L{]}^b\left[M^c{L}^{-c}{T}^{-2c}\right]$
$\left[T\right]=\left[M^{a+c}\right]\left[L^{b-c}\right]\left[T^{-2c}\right]$

Equating powers in both sides, we have:
$a+c=0b-c=0-2c=1\Rightarrow c=-1/2\Rightarrow b=-1/2\Rightarrow a=1/2$
So,
$T=C{M}^{1/2}{L}^{-1/2}{\eta }^{-1/2}$
$\Rightarrow T=C\sqrt{\frac{M}{\eta L}}$

So, assuming $C=2\pi$, we can have option $d$ as the right answer.