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JEE Main 202116 MarchEvening Shift
Exam
JEE Main
Year
2021
Date
16 March
Shift
Evening Shift
Subject
Physics
Chapter
Mathematics in Physics

JEE Main 2021 · 16 March · Evening Shift · Physics · Mathematics in Physics

In order to determine the Young's Modulus of a wire of radius 0 ....

In order to determine the Young's Modulus of a wire of radius 0.2cm0 . 2cm (measured using a scale of least count =0.001cm= 0 . 001cm) and length 1m1m (measured using a scale of least count =1mm= 1mm), a weight of mass 1kg1kg (measured using a scale of least count =1g= 1g ) was hanged to get the elongation of 0.5cm0 . 5cm (measured using a scale of least count 0.001cm0 . 001cm). What will be the fractional error in the value of Young's Modulus determined by this experiment?

Options

  1. A0.140 . 14 %
  2. B0.90 . 9 %
  3. C99 %
  4. D1.41 . 4 %

Correct answer

D. 1.41 . 4 %

Step-by-step solution

Y=StressStrain=FLAl=mgLπR2lY = \dfrac{Stress}{Strain} = \dfrac{FL}{Al} = \dfrac{mgL}{\pi{R}^{2}l}

\DeltaYY=\Deltamm+\DeltaLL+2\DeltaRR+Δll\dfrac{\DeltaY}{Y} = \dfrac{\Deltam}{m} + \dfrac{\DeltaL}{L} + 2\dfrac{\DeltaR}{R} + \dfrac{\Delta l}{l}

\DeltaYY×100=100[11000+11000+2(0.0010.2)+0.0010.5]\dfrac{\DeltaY}{Y} \times 100 = 100\left[\dfrac{1}{1000} + \dfrac{1}{1000} + 2\left(\dfrac{0 . 001}{0 . 2}\right) + \dfrac{0 . 001}{0 . 5}\right]

=110+110+1+15=1410=1.4= \dfrac{1}{10} + \dfrac{1}{10} + 1 + \dfrac{1}{5} = \dfrac{14}{10} = 1 . 4 %

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