A uniform heavy rod of mass 20 kg . Cross sectional area 0 . 4 m 2 and length 20 m is hanging from a fixed support. Neglecting the lateral contraction, the elongation in the rod due to its own weight is x × 10 - 9 m . The value of x is _____ . (Given. Young's modulus Y = 2 × 10 11 Nm - 2 and g = 10 m s - 2 )

Step-by-step solution

Tension at a distance x from lower end = m g l x If the elongation in the element is taken as d ∆ l then, using Hooke's law, Y = m g x d x A l d ∆ l ⇒ d ∆ l = m a x d x A l Y ⇒ ∫ 0 ∆ l d ∆ l = ∫ 0 l m g x d x A l Y ⇒ ∆ l = m g l 2 A Y ⇒ Δ l = 20 × 10 × 20 2 × 0 . 4 × 2 × 10 11 ⇒ Δ l = 25 × 10 - 9 ⇒ x = 25