Two masses m and M are connected by a light string that passes through a smooth hole O at the centre of a table. Mass m lies on the table and M hangs vertically. m is moved round in a horizontal circle with O as the centre. If l is the length of the string from O to m then the frequency with which m should revolve so that M remains stationary is

The correct option is: (1/2π)√(Mg/ml)

Let us analyze the system to determine the frequency of revolution of the mass _m so that the mass _M remains stationary.

1. Force acting on the hanging mass _M:
For the hanging mass _M to remain stationary (in equilibrium), the upward tension _T in the string must balance the downward gravitational force acting on it:
$T\; =\; M\; *\; g$ (Equation 1)

2. Force acting on the mass _m on the table:
The mass _m moves in a horizontal circle of radius _l (the length of the string from the hole _O to _m). The tension _T in the string provides the necessary centripetal force for this circular motion:
$T\; =\; m\; *\; \omega 2*\; l$ (Equation 2)
where $\omega$ is the angular velocity of the revolving mass.

3. Equating the tension:
Since the string is light and passes through a smooth hole, the tension _T is the same throughout the string. Equating Equation 1 and Equation 2 gives:
$M\; *\; g\; =\; m\; *\; \omega 2*\; l$

Solve for the angular velocity $\omega$:
$\omega 2=\; (M\; *\; g)\; /\; (m\; *\; l)$
$\omega \; =\; \surd ((M\; *\; g)\; /\; (m\; *\; l))$

4. Determining the frequency:
The relationship between angular velocity $\omega$ and frequency of revolution _n is given by:
$\omega \; =\; 2\; *\; \pi \; *\; n$
Substituting this into the equation for $\omega$:
$2\; *\; \pi \; *\; n\; =\; \surd ((M\; *\; g)\; /\; (m\; *\; l))$

Solving for _n, we get:
$n\; =\; (1\; /\; (2\; *\; \pi ))\; *\; \surd ((M\; *\; g)\; /\; (m\; *\; l))$

Thus, the frequency with which _m should revolve is:
$(1\; /\; (2\; *\; \pi ))\; *\; \surd ((Mg)\; /\; (ml))$