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JEE Main 20199 JanuaryMorning Shift
Exam
JEE Main
Year
2019
Date
9 January
Shift
Morning Shift
Subject
Physics
Chapter
Mechanical Properties of Solids

JEE Main 2019 · 9 January · Morning Shift · Physics · Mechanical Properties of Solids

A rod, of length L at room temperature and uniform area of cross...

A rod, of length LL at room temperature and uniform area of cross section A,A, is made of a metal having coefficient of linear expansion α/°C\alpha / ° C It is observed that an external compressive force FF is applied on each of its ends, prevents any change in the length of the rod when its temperature rises by ∆TK∆ TK Young's modulus, YY for this metal is:

Options

  1. A\dfrac{F}{A\alpha\left(∆ T - 273\right)}
  2. B2FAα∆T\dfrac{2F}{A\alpha ∆ T}
  3. CFAα∆T\dfrac{F}{A\alpha ∆ T}
  4. DF2Aα∆T\dfrac{F}{2A\alpha ∆ T}

Correct answer

C. FAα∆T\dfrac{F}{A\alpha ∆ T}

Step-by-step solution

Since the length of the rod does not change, the increment due to increase in temperature will be balanced by the compression due to applied force.

Y=FA\Deltall⇒\Deltall=FAYY = \dfrac{\dfrac{F}{A}}{\dfrac{\Deltal}{l}} ⇒ \dfrac{\Deltal}{l} = \dfrac{F}{AY}

Also \Deltall=α\DeltaT\dfrac{\Deltal}{l} = \alpha\DeltaT

∴FAY=α\DeltaT∴ \dfrac{F}{AY} = \alpha\DeltaT

Y=FAα\DeltaTY = \dfrac{F}{A\alpha\DeltaT}
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