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Elasticity

A metallic ro...

Question

A metallic rod undergoes a strain of $0.05$ %. The energy stored per unit volume is ( $Y$ = $2 \times 10$ $^{11}$ Nm $^{- 2}$ ):

0.5 \times 10

^{4}

^{- 3}

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0.5 \times 10

^{5}

^{- 3}

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2.5 \times 10

^{5}

^{- 3}

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2.5 \times 10

^{4}

^{- 3}

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Solution

The correct option is C $2.5 \times 10$ $^{4}$ Jm $^{- 3}$
$S t r e s s = Y \times s t r a i n$
$P o t e n t i a l e n e r g y = \frac{1}{2} \times s t r e s s \times s t r a i n \times v o l u m e$
$\frac{P o t e n t i a l e n e r g y}{v o l u m e} = \frac{1}{2} \times s t r e s s \times s t r a i n$
$= \frac{1}{2} \times Y s t r a i n \times s t r a i n$
$= \frac{1}{2} \times 2 \times 10^{11} \times (\frac{0.05}{100})^{2}$
$= 2.5 \times 10^{4} J / m^{3}$

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