JEE Main · 2019 · 12 Apr 2019 · Shift 2 · Elasticity & SHM

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 longitudinal compressional force F , its length remains unchanged. The coefficient of volume expansion, of the material of the rod, is (nearly) equals to :

(A) $9 F / (π r^{2} YT)$
(B) $F / (3 π r^{2} YT)$
(C) $3 F / (π r^{2} YT)$ ✓ Correct
(D) $6 F / (π r^{2} YT)$

Answer: (C)

Solution

Since the length is held unchanged, the mechanical compression exactly cancels the thermal expansion. The compressive strain from the force is

\frac{F}{π r ^{2} Y}

and the thermal strain is

α T

, so

\frac{F}{π r ^{2} Y} = α T

, giving the linear coefficient

α = \frac{F}{π r ^{2} Y T}

. The volume coefficient is three times the linear one, so

γ = 3 α = \frac{3 F}{π r ^{2} Y T}

JEE Main · 2019 · 12 Apr 2019 · Shift 2 · Single correct option

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JEE Main · 2019 · 12 Apr 2019 · Shift 2 · Elasticity & SHM

Solution

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