Consider the situation shown in figure. The force F is equal to the `m_2g/2`. If the area of cross section of the string is A and its Young modulus Y, find the strain develolped in it. The string is lilght and there is no friction anywhere.

Options

A. `(m_(2)g(2m_(1)+m_(2)))/(2Ay(m_(1)+m_(2)))`
B. `(m_(1)g(2m_(1)+m_(2)))/(2Ay(m_(1)+m_(2)))`
C. `(m_(1)g(2m_(2)+m_(1)))/(2Ay(m_(1)+m_(20))`
D. `(m_(2)g(m_(2)+m_(1)))/(Ay(m_(1)+m_(2)))`

Answer

A

Explanation

Where `F=(m_(2)g)/2`
Writing equations from diagram
`T-F=m_(1)a`……..1
`m_(2)g-T=m_(2)a`…..2
Where T is tension a is accelerationn and g is gravitational acceleration
On adding 1 and 2
`m_(2)g-F=(m_(1)+m_(2))a`
`impliesa=(m_(2)g-F)/(m_(1)+m_(2))`
`impliesa=(m_(2)g-m_(2)g//2)/(m_(1)+m_(2))=1/2xx`
`((m_(1)g)/(m_(1)+m_(2)))`
Putting in eq. 1
`impliesT=F+m_(1)a`
`impliesT=(m_(2)g)/2+m_(1)(1/2)((m_(2)g)/(m_(1)+m_(2)))`
`impliesT=1/2[(m_(1)m_(2)+m_(2)^(2)+m_(1)m_(2))/(m_(1)+m_(2))]g`
`impliesT=1/2[(m_(2)^(2)+2m_(1)m_(2))/(m_(1)+m_(2))]g`
And we knwo that
Stress =Y strain
`implies`Strain`=("Stres")/Y=F/(AY)`
`implies` Strain `=1/2[(m_(2)^(2)+2m_(1)m_(2))/((m_(1)+m_(2))AY)]g`