When a force is applied on a wire of uniform cross-sectional area 3 × {10^{

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8.Mechanical Properties of Solids

medium

When a force is applied on a wire of uniform cross-sectional area $3 \times {10^{ - 6}}\,{m^2}$ and length $4m$, the increase in length is $1\, mm.$ Energy stored in it will be $(Y = 2 \times {10^{11}}\,N/{m^2})$

$6250\, J$

$0.177 \,J$

$0.075\, J$

$0.150 \,J$

Solution

(c) $U = \frac{1}{2} \times \frac{{YA{l^2}}}{L}$$ = \frac{1}{2} \times \frac{{2 \times {{10}^{11}} \times 3 \times {{10}^{ – 6}} \times {{(1 \times {{10}^{ – 3}})}^2}}}{4}$

$ = 0.075\;J$

Standard 11

Physics

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hard

When a force is applied on a wire of uniform cross-sectional area $3 \times {10^{ - 6}}\,{m^2}$ and length $4m$, the increase in length is $1\, mm.$ Energy stored in it will be $(Y = 2 \times {10^{11}}\,N/{m^2})$

Solution

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