Question

A uniform heavy rod of length L and area of cross section 'A' is hanging from a fixed support. It young's modulus of the material of the rod is Y, the increase the length of rod is ( p is density of the mateiral of the rod)

• A. $\frac{L^2y}{2pg}$
• B. $\frac{L^2pg}{2y}$
• C. $\frac{L^2g}{2Yg}$
• D. $\frac{L^2g}{3y\,p}$

Answer 1

Answer: (B)

$\frac{L^2pg}{2y}$

Answer 2

Consider an element $dx$ at a distance $x$ from the top. Let the wt. of rod is W. The force acting on this because of a part which is below $dx$ is given as $\frac{( L - x ) W }{ L }$ The elongation in element $dx$ is $=\left(\frac{ L - x }{ L }\right) \frac{ W d x }{ AY }=\frac{( L - x ) W dx }{ LAY }$ Total elongation, $\Delta L =\int_{ O }^{ L } \frac{( L - x ) W dx }{ LAY }=\frac{ WL }{2 AY }$ $W = ALpg$ $\Delta L =\frac{ AL \rho g L }{2 AY }=\frac{ L ^{2} \rho g }{2 Y }$