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Consider two masses with m(1) gt m(2) co...

Consider two masses with `m_(1) gt m_(2)` connected by a light inextensible string that passes over a pulley of radius R and moment of inertia I about its axis of rotation. The string does not slip on the pulley and the pulley turns without friction. The two masses are released from rest separated by a vertical distance 2h. When the two masses pass each other, the speed of the masses is proportional to

A

`sqrt((m_(1)-m_(2))/(m_(1) + m_(2) + (I)/(R^(2))))`

B

`sqrt(((m_(1) + m_(2))(m_(1) - m_(2)))/(m_(1) + m_(2) + (I)/(R^(2))))`

C

`sqrt((m_(1) + m_(2) + (I)/(R^(2)))/(m_(1)- m_(2)))`

D

`sqrt(((I)/(R^(2)))/(m_(1) + m_(2)))`

Text Solution

Verified by Experts

The correct Answer is:
A
`m_(1) g- T_(2)= m_(1) a` ..(i)
`T_(2)- m_(2)g = m_(2) a` ....(ii)
`(T_(1)- T_(2)) R= (I)/(R) a` ....(iii)
`a= ((m_(1) - m_(2))g)/(((I)/(R^(2)) + m_(1) + m_(2))) rArr v_("rel")^(2)= (8hg (m_(1)- m_(2)))/(((I)/(R^(2)) + m_(1) + m_(2)))`
As `v_("rel") = v- (-v) = 2v rArr v= (1)/(2) sqrt((8gh (m_(1)- m_(2)))/(((I)/(R^(2)) + m_(1) + m_(2))))`
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