Two blocks of masses `m_(1)` and `m_(2)` are hanging from a pulley with the help of a ropw of Young's modulus Y as shown in the diagram. The length of the wire is L and area of cross section is A. When system is released from rest, the elongation in the length of wire is.

Options

A. `(m_(1)m_(2)gL)/((m_(1)+m_(2)AY))`
B. `(m_(1)m_(2)gL)/(2(m_(1)+m_(2))AY)`
C. `(2m_(1)m_(2)gL)/((m_(1)+m_(2))AY)`
D. `(m_(2)gL)/((m_(1)+m_(2))AY)`

Answer

C

Explanation

Acceleration of system,
`a=("Net pulley force")/("Total mass")=((m_(2)m_(1))g)/((m_(1)+m_(2)))`
From FBD block `2,m_(2)g-T=m_(2)a`

`rArrT=m_(2)g-m_(2)a`
`=m_(2){g-((m_(2)-m_(1))g)/((m_(1)+m_(2)))}`
`=(2m_(1)m_(2)g)/((m_(1)+m_(2)))`
`:.` Stress in wire `=(T)/(A)=(2m_(1)m_(2)g)/((m_(1)+m_(2))A)`
`:.` Strain `=(l)/(L)=("Stress")/(Y)=(2m_(1)m_(2)g)/((m_(1)+m_(2))A)`
`rArrl=(2m_(1)m_(2)gL)/((m_(1)+m_(2))AY)`