JEE Main 201912 AprilEvening Shift

Exam: JEE Main
Year: 2019
Date: 12 April
Shift: Evening Shift
Subject: Physics
Chapter: Thermal Properties of Matter

JEE Main 2019 · 12 April · Evening Shift · Physics · Thermal Properties of Matter

A uniform cylindrical rod of length L and radius r, is made from a...

A uniform cylindrical rod of length

L

and radius r, is made from a material whose Young’s modulus of Elasticity equals

Y .

When this rod is heated by temperature T and simultaneously subjected to a net longutudinal compressional force F, its length remains unchanged. The coefficient of volume expansion, of the material of the rod, is (nearly) equal to:

Options

A $F / \left(3\pi{r}^{2}YT\right)$
B $9F / \left(\pi{r}^{2}YT\right)$
C $6F / \left(\pi{r}^{2}YT\right)$
D $3F / ( \pi{r}^{2}YT )$

Correct answer

D. $3F / ( \pi{r}^{2}YT )$

Step-by-step solution

Stress

= \dfrac{F}{A}

Y\left(\alphaT\right) = \dfrac{F}{A}

\alpha = \dfrac{F}{A\left(YT\right)}

\gamma = 3\alpha = \dfrac{3F}{AYT}

\gamma = \dfrac{3F}{\pi{r}^{2}YT}

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