Mechanical Properties of Solids

JEE Main level, single correct, medium

35 of 118

Question 35 of 118

An aluminium rod with Young’s modulus Y=7.0×1010Nm−2Y=7.0\times {10}^{10}N{m}^{-2} undergoes elastic strain of 0.040.04%. The energy per unit volume stored in the rod in SISI unit

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Answer: (C) 56005600

The data given is
Y=7×1010Nm−2and∆ll=0.04100Y=7\times {10}^{10}N{m}^{-2}\text{and} \dfrac{∆l}{l}=\dfrac{0.04}{100}
Young's modulus is given by the ratio of stress and strain,
Y=FA∆ll=FlA∆lY=\dfrac{\dfrac{F}{A}}{\dfrac{∆l}{l}}=\dfrac{Fl}{A∆l}
The energy stored is given as
E=(YA2l)∆l2⇒E=(YA2)(∆ll)2×l...(i)E=\left(\dfrac{YA}{2l}\right)∆{l}^{2} \Rightarrow E=\left(\dfrac{YA}{2}\right){\left(\dfrac{∆l}{l}\right)}^{2}\times l...(i)
The energy per unit volume can be written from equation (i)
EV=Y2×(∆ll)2...(ii)\dfrac{E}{V}=\dfrac{Y}{2}\times {\left(\dfrac{∆l}{l}\right)}^{2}...(ii)
Substituting the values in equation (ii)
EV=12×7×1010×0.04×0.04104=56×102\dfrac{E}{V}=\dfrac{1}{2}\times 7\times {10}^{10}\times \dfrac{0.04\times 0.04}{{10}^{4}}=56\times {10}^{2}

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