Question

A light rod with uniform cross-section of ${10}^{-4}{m}^2$ is shown in the adjoining figure. The rod consists of three different materials whose lengths are 0.1 m, 0.2 m and 0.15 m respectively and whose Young's modulii are $2.5\times {10}^{10}\frac{N}{m^2},4\times {10}^{10}\frac{N}{m^2}$ and $1\times {10}^{10}\frac{N}{m^2}$ respectively.
The displacement of point B will be

• A. $24\times {10}^{-6}m$
• B. $9\times {10}^{-6}m$
• C. $4\times {10}^{-6}m$
• D. $1\times {10}^{-6}m$

Answer 1

The correct option is C $4\times {10}^{-6}m$

Increment in the length AB =
$\frac{MgL}{AY}=\frac{10\times 10\times 0.1}{{10}^{-4}\times 2.5\times {10}^{10}}=4\times {10}^{-6}m$
$\therefore$ Displacement of point B = $4\times {10}^{-6}$ m