Question

A uniform heavy rope of weight W and cross-
sectional area A is hanging from rigid support. If
Young's modulus of material is Y then strain is

(1)W/4 AY
(2)W/AY
(3) W/2 AY
(4)W /3 AY​

Answer 1

answer : option (C) W/2AY

explanation : cut an element dx of rod at a distance x from the fixed end. The part below this small element has length (l - x).

here, tension of the rod at the element = weight of the rod below.

i.e., F = W × (l - x)/l

now Young's modulus = stress/strain

we know, strain = change in length/original length

so, strain = ∆l/dx

or, Y =(F/A)/(∆l/dx)

or, ∆l = Fdx/AY

={ W(l - x)/l}dx/AY

= {W/AYl} ∫(l - x)dx

= {W/AYl} $[lx - x^2/2]^l_0$

= W/AYl × [l² - l²/2]

= W/AYl × l²/2 = W/2AYl

or, ∆l/l = W/2AY

hence, strain of uniform rope is W/2AY

Answer 2

Explanation:

here's the solution best of luck