Question

Uniform rod of mass m, length l , area of cross-section A has Young’s modulus Y. If it is hanged vertically, elongation under its own weight will be(a) $\frac{mgl}{2AY}}$(b) $\frac{2mgl}{AY}}$(c) $\frac{mgl}{AY}}$(d) $\frac{mgY}{Al}}$

Answer 1

the option is d because it is tex som

Answer 2

Answer:

C) mgl/AY

Explanation:

Mass of the rod = m (Given)

Length of the rod = l (Given)

Area of cross section of the rod =A

Weight of the rod = mg

Thus,

Stress on the rod = mg/A

Let the change in length of the rod be = l'

Original length of the rod =l

Therefore, strain = l'/l

According to Young's modulus = stress/strain

Y = (mg/A)/(l'/l)

Y = mgl/Al'

l' = mgl/AY

Thus, if hanged vertically, under its own weight will be mgl/AY