Correct Answer - C
Consider a small element of length dx at a distance x from the load as shown in the wire.
Tension in the wire at a distance x from the lower end is `T(x) = mugx + Mg`
Let dl be increase in length of the elemenet
Then `Y= (T(x)//A)/(dl//dx)`
`dl =(T(x)dx)/(YA) = ((mugx+Mg)/(YA))dx " "("using (i)")`
`therefore` Total extension produced in the wire is `l = underset(0)overset(L)int ((mugx+Mg)/(YA))dx = (1)/(Y) [(mugx)/(2) +Mgx]_(0)^(L) = (1)/(YA) [(mugL^(2))/(L)+Mgl] =(mugx^(2)+2MgL)/(2YA)`
