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JEE Main 20178 AprilOnline
Exam
JEE Main
Year
2017
Date
8 April
Mode
Online
Subject
Physics
Chapter
Thermal Properties of Matter

JEE Main 2017 · 8 April · Online · Physics · Thermal Properties of Matter

A compressive force, F is applied at the two ends of a long thin...

A compressive force, FF is applied at the two ends of a long thin steel rod. It is heated, simultaneously, such that its temperature increases by \DeltaT\DeltaT. The net change in its length is zero. Let ll be the length of the rod, A its area of cross-section, YY its Young's modulus, and α\alpha its coefficient of linear expansion. Then, FF is equal to:

Options

  1. AlAYα\DeltaTlAY\alpha\DeltaT
  2. BAYα\DeltaTAY\alpha\DeltaT
  3. CAYα\DeltaT\dfrac{AY}{\alpha\DeltaT}
  4. Dl2Yα\DeltaT{l}^{2}Y\alpha\DeltaT

Correct answer

B. AYα\DeltaTAY\alpha\DeltaT

Step-by-step solution

Net change in length =0= 0

Thermal expansion. \Deltal\Deltal=lα\Deltat= l\alpha\Deltat

Y=FA\DeltallY= \dfrac{\dfrac{F}{A}}{\dfrac{\Deltal}{l}}

\Deltall=FAY\dfrac{\Deltal}{l} = \dfrac{F}{AY}

\Deltal=FlAY\Deltal = \dfrac{Fl}{AY}

FlAY=lα\Deltat\dfrac{Fl}{AY} = l\alpha\Deltat

F=AYα\DeltatF = AY\alpha\Deltat
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