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Q. A load of $1\, kg$ weight is a attached to one end of a steel wire of area of cross-section $ 3 \, mm^2$ and Young's modulus $10^{11} \, N/m^2$. The other end is suspended vertically from a hook on a wall, then the load is pulled horizontally and released. When the load passes through its lowest position the fractional change in length is $(g = 10 \, m/s^2)$

BITSATBITSAT 2008

Solution:

$Y=\frac{m g l}{A \Delta l}$
$\Rightarrow \frac{\Delta l}{l}=\frac{m g}{A Y}$
$\therefore \frac{\Delta l}{l}=\frac{1 \times 10}{3 \times 10^{-6} \times 10^{11}}$
$=0.3 \times 10^{-4}$