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Question

A body is of mass M is kept at rest by means of string connected to a body of mass m kept on a turn-table rotating with constant angular velocity ω as shown. Find the ratio Mm. (Neglect the force due to friction)

A
glω2
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B
lω2g
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C
lωg
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D
2lω2g
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Solution

The correct option is B lω2g
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 ...(i)

By FBD of block of mass m
For the equilibrium condition , the centrifugal force balances the tension force.
T=mlω2 ...(ii)
Equating equation (i) and (ii), we get
Mg=mLω2
Mm=lω2g
Hence option B is the correct answer

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