In the pulley system shown in the figure, the mass of A is half of that of rod B. The rod length is 500 cm. The mass of pulleys and the threads may be neglected The mass A is set at the same level as the lower end of the rod and then released After releasing the mass A, it would reach the top end of the rod B in time (Assume, `g = 10 m//s^2 ` )

Options

A. 2.0 s
B. 1.0 s
C. 3.0 s
D. 4.0 s

Answer

B

Explanation

According to the question,

Given, mass of body A. is half of mass of rod B.
l.e. `m_A = (m_B)/(2) rArr m_B = 2m_A`
and length of rod = 500 cm = 5m
Since, rod B and body C is in equilibrium, hence mass of rod B= mass of rod C
i.e., ` m_B = m_C`
By applying Newton's law of motion,
` 2T = m_A g = m_A a " " ....(i)`
`(m_B + m_C ) g - 2T = (m_B = m_C) a `
or ` 2m_B g - 2T = 2m_Ba `
or ` 4m_A g - 2T = 4m_A a " " ....(ii)`
Adding Eqs. (i) and (ii), we get
`3m_A g = 5 m_A a `
` rArr a = 3/5 g = ( 3 xx 10)/(5) = 6m//s^2`
If t be the time to cross the rod of length
500 cm = 5 m
` therefore ` From eqs., ` s = 0 + 1/2 at^2`
` rArr t = sqrt( (2s)/(a) ) = sqrt( (2 xx 5)/(6) ) = 1.28 s ~~ 1s`