8.Mechanical Properties of Solids
medium

The length of a rod is $20\, cm$ and area of cross-section $2\,c{m^2}$. The Young's modulus of the material of wire is $1.4 \times {10^{11}}\,N/{m^2}$. If the rod is compressed by $5\, kg-wt$ along its length, then increase in the energy of the rod in joules will be

A

$8.57 \times {10^{ - 6}}$

B

$22.5 \times {10^{ - 4}}$

C

$9.8 \times {10^{ - 5}}$

D

$45.0 \times {10^{ - 5}}$

Solution

(a) Energy = $\frac{1}{2}Fl = \frac{1}{2} \times F \times \left( {\frac{{FL}}{{AY}}} \right) = \frac{1}{2} \times \frac{{{F^2}L}}{{AY}}$

$ = \frac{1}{2} \times \frac{{{{(50)}^2} \times 20 \times {{10}^{ – 2}}}}{{2 \times {{10}^{ – 4}} \times 1.4 \times {{10}^{11}}}}$$ = 8.57 \times {10^{ – 6}}J$

Standard 11
Physics

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