Question

Find out elongation in a rod.

Given : Y=2×1011N/m2,ρ=104kg/m3Y=2×1011N/m2,ρ=104kg/m3

• A. $9mm$
• B. $10mm$
• C. $18mm$
• D. $3mm$

Answer 1

The correct option is A $9mm$

Mass of shaded portion $m^{\prime }=\frac{m}{l}(l-x)$ [where m = total mass = $\rho Al$]
$T=m^{\prime }{\omega }^2[\frac{l-x}{2}+x]$ $\Rightarrow T=\frac{m}{l}(l-x){\omega }^2\left(\frac{l+x}{2}\right)$ $T=\frac{m{\omega }^2}{2l}(l^2-x^2)$
$m^{\prime }\omega \{\frac{l-x}{2}+x\}$
this tension will be maximum at $A\left(\frac{m{\omega }^{2}l}{2}\right)$ and minimum at 'B' (zero) elongation in element of width 'dx' = $\frac{Tdx}{AY}$ Total elongation
$\delta =\int \frac{Tdx}{AY}={\int }_0^l\frac{m{\omega }^2(l^2-x^2)}{2lAY}dx$

$\delta =\frac{m{\omega }^2}{2lAY}{[l^{2}x-\frac{x^3}{3}]}_0^l=\frac{m{\omega }^2\times 2l^2}{2lAY\times 3}=\frac{m{\omega }^2{l}^2}{3AY}=\frac{\rho Al{\omega }^2{l}^2}{3AY}$

$\delta =\frac{\rho {\omega }^2{l}^3}{3Y}=\frac{{10}^4\times \left(400\right)\times (1.5{)}^3}{3\times 2\times {10}^{11}}=9\times {10}^{-3}=9mm$