Question

A body is of mass $M$ is kept at rest by means of string connected to a body of mass ${}^{\prime }{m}^{\prime }$ kept on a turn-table rotating with constant angular velocity ${}^{\prime }{\omega }^{\prime }$ as shown. Find the ratio $\frac{M}{m}$. (Neglect the force due to friction)

• A. $\frac{g}{l{\omega }^2}$
• B. $\frac{l{\omega }^2}{g}$
• C. $\sqrt{\frac{l\omega }{g}}$
• D. $\frac{2l{\omega }^2}{g}$

Answer 1

The correct option is B $\frac{l{\omega }^2}{g}$

By the free body diagram,

It is given that the mass are kept at rest, writing the force of equation in the equilibrium condition.

By FBD of block of mass $M$
$T=Mg...\left(i\right)$

By FBD of block of mass $m$
For the equilibrium condition , the centrifugal force balances the tension force.
$T=ml{\omega }^2...\left(ii\right)$
Equating equation $\left(i\right)\text{and}\left(ii\right)$, we get
$Mg=mL{\omega }^2$
$\frac{M}{m}=\frac{l{\omega }^2}{g}$
Hence option B is the correct answer