In the arrangement shown below, pulleys are mass-less and friction-less and threads are in-extensible. Block of mass `m_(1)` will remain at rest if :

Options

A. `4/m_1=1/m_2+1/m_3`
B. `m_1=m_2=m_3`
C. `1/m_1=1/m_2+1/m_3`
D. `1/m_2=2/m_2+3/m_1`

Answer

A

Explanation

FBD of masses `m_1 and m_2 and m_3`

`T.=mg`
`m_2g-T/2=m_2a`……(i)
`(T.)/2-m_3g=m_3a`……(ii)
From eqns. (i) and (ii)
`a=((m_2-m_3)g)/(m_2+m_3)`
Putting value of a in eqn (i)
`T.=2m_2 (1+(m_3-m_2)/(m_2+m_3))g`
`m_1g= (2m_2 times 2m_3)/(m_2+m_3) g`
`4/m_1=1/m_2+1/m_3`