Question

A metallic rod of length l and cross sectional area a is made of a material of young modulus y. If the rod is elongated by an amount y then work done is proportional to

Answer 1

We know,
workdone = 1/2 × stress × strain × volume
also we know, Young's modulus = stress/strain
so, stress = Young's modulus × strain.

Here given,
Young's modulus = Y
original length of rod = l
change in length of rod = y
Corsa section area = a
so, strain = change in length/original length
strain = y/l
so, stress = Y × y/l

now, work done = 1/2 × (Y × y/l) × y/l × al [because volume = area × length so, v = al]
where A is cross section area of rod.

workdone = 1/2 × Yy²a/l

here you can see that workdone is directly proportional to cross section area, square of change in Length and inversely proportional to length of rod.

Answer 2

Y = (F L) / (A ΔL)

Volume = A* l

V= Al

Strain = elongation / natural length = y/l

We know that work done = 1/2 * stress * strain * volume

W = 1/2 * Y * [y/2]^2 * Al

= 1/2 [YA/l] y^2

= w ∝ y^2