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8.Mechanical Properties of Solids
medium
The work done per unit volume to stretch the length of area of cross-section $2 \,mm ^2$ by $2 \%$ will be ....... $MJ / m ^3$ $\left[Y=8 \times 10^{10} \,N / m ^2\right]$
A
$40$
B
$32$
C
$64$
D
$16$
Solution
(b)
Work done $=$ Force $\times$ elongation
$W=F \cdot \Delta x \ldots .$ $\left\{F=\Delta x \cdot \frac{A Y}{L}\right\}$
$\Rightarrow W=\frac{A Y}{L} \times \Delta x^2$
Multiply and divide by $L$
We get
$W=\frac{\text { Volume } \cdot Y}{L^2} \Delta x^2$
Cross multiply ( $L \cdot A)$ volume
$\frac{W}{L \cdot A}=Y \cdot\left(\frac{\Delta x}{L}\right)^2$
Work done per unit volume
Substitute values
$=8 \times 10^{10} \cdot\left(\frac{2}{100}\right)^2$
$=32 \,MJ / m ^3$
Standard 11
Physics
