A block of mass M is situated on a smooth horizontal frictionless table. A thread tied to the block passes through hole H in the table and carries a mass m at its other end. If the length of thread above the table is 'l'. what should be the value of m so that it remains suspended at a constant height 'h' below the table when M is revolving in horizontal circle with angular speed co on the table ?

Options

A. `(MI)/(omega^(2))`
B. `(l)/(Momega^(2))`
C. `(Ml)/(gomega^(2))`
D. `(Momega^(2)I)/(g)`

Answer

D

Explanation

Here the weight mg of the supended mass must provide the centripetal force `Mtomega^(2)` or `Miomega^(2)=mg` or `m=(Mlomega^(2))/(g)`