A metal rod has length, cross-sectional area and Young's modulus as ' L^ , ^ A^ and Y^ respectively. If the elongation in the rod produced is ' ^ then work done is proportional to

Correct answer: C. ^ 2 (C) V olume =A × L or V=Al strain = Elongation Original length = Y l Young's modulus = stress strain Work done, W=(1)/(2) × stress × strain × volume array l W=(1)/(2) ×(strain)^ 2 =(1)/(2) [(y)/(1)]^ 2 =(1)/(2)[(Y A)/(1)]y^ 2 W y^ 2 array

JEE Main

Exam: JEE Main
Subject: Physics
Chapter: Mechanical Properties of Solids

JEE Main · Physics · Mechanical Properties of Solids

A metal rod has length, cross-sectional area and Young's modulus as...

A metal rod has length, cross-sectional area and Young's modulus as '

\mathrm{L}^{\prime},{ }^{\prime} \mathrm{A}^{\prime}

and

\mathrm{Y}^{\prime}

respectively. If the elongation in the rod produced is '

\ell^{\prime}

then work done is proportional to

Options

A $\ell$
B $\ell^{4}$
C $\ell^{2}$
D $\ell^{3}$

Correct answer

C. $\ell^{2}$

Step-by-step solution

(C)

\mathrm{V}

olume

=\mathrm{A} \times \mathrm{L}

\mathrm{V}=\mathrm{Al}

strain

=\frac{\text { Elongation }}{\text { Original length }}=\frac{\mathrm{Y}}{\mathrm{l}}

Young's modulus

=\frac{\text { stress }}{\text { strain }}

Work done,

W=\frac{1}{2} \times

stress

\times

strain

\times

volume

\begin{array}{l} W=\frac{1}{2} \timesY \times(\operatorname{strain})^{2} \timesAl \\ =\frac{1}{2} \timesY \left[\frac{y}{1}\right]^{2} \timesAl=\frac{1}{2}\left[\frac{Y A}{1}\right]y^{2} \Rightarrow W \propto y^{2} \end{array}

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