KVPY SX SB Exam 2016 Question Paper

Question : 95 of 120

Marks: +1, -0

Consider two masses with

m_1 > 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

$√{{m_1-m_2}/{m_1+m_2+1/R^2}}$
$√{{(m_1+m_2)(m_1-m_2)}/{m_1+m_2+1/R^2}}$
$√{{m_1+m_2+1/R^2}/{m_1-m_2}}$
$√{{1/R^2}/{m_1+m_2}}$

Validate

Solution:

The total mechanical energy of system = conserved
Hence

KE_i+PE_i=KE_f+PE_f

0-m_2g×2h=1/2m_2v^2+1/2m_1v^2+1/2Iω^2-m_1gh-m_2gh

Also

ω=v/R

(m_1-m_2)gh=1/2(m_1+m_2)v^2+1/2I(v/R)^2

(m_1-m_2)gh=v^2/2[m_1+m_2+I/R^2]

v∝

Go to Question:

Prev Question

Next Question