Question

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JEE

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Answer 1

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GIVEN :

Length of the wire = 2m

Length of extension = 2mm

$\sf{= \: 2 \times {10}^{ - 3} m}$

Area of the Cross section $\sf{ = 4 {mm}^{2} }$

$\sf{ = 4 \times {10}^{ - 6} }$

Youngs Modulus, $\sf{ Y \: = \: 2 \times {10}^{11} {m}^{ - 2} }$

SOLUTION :

We've,

$\sf{ Strain = \frac{ 2 \times {10}^{ - 3} }{2}}$

$\sf{ Strain = \frac{ \Delta \: l}{l} }$

$\boxed{\sf{ Strain = {10}^{ - 3}}}$

Now, let us find out the Stress

$\sf{ Stress =Y \times \frac{ \Delta \: l}{l} }$

$\sf{ Stress =2 \times {10}^{11} \times ( \frac{ 2 \times {10}^{ - 3} }{2} )}$

$\boxed{\sf{ Stress =2 \times {10}^{8} \: { N m}^{ - 2}}}$

Now, By using the formula of Elastic Potential Energy we get,

$\sf{U = \: \frac{1}{2} \times Stress \times Strain \times Volume}$

$\sf{U = \: \frac{1}{2} \times 2 \times {10}^{8} \times {10}^{ - 3} \times ( A × l )}$

[Since, Volume = Area × Length]

$\sf{U = \: {10}^{5} \: \times (4 \times {10}^{ - 6} \times 2) }$

$\sf{U = \: 8 \: \times \: {10}^{ - 1} }$

$\boxed{\sf{U = \: 0.8J }}$

Hence, the required Elastic Potential Energy is 0.8J

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