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NTA Abhyas 2020
Exam
NTA Abhyas
Year
2020
Subject
Physics
Chapter
Units and Dimensions

NTA Abhyas 2020 · Physics · Units and Dimensions

A student performs an experiment to determine young's modulus of a...

A student performs an experiment to determine young's modulus of a wire, exactly 2m2m long, by searle's method. In a particular reading, the student measures the extension in the length of the wire to be 0.8mm0. 8mm with an uncertainty of ±0.05mm\pm 0.05mm at a load of exactly 1.0kg1. 0kg. The student also measures the diameter of the wire to be0.4mm0. 4mm with an uncertainty of ±0.01mm.\pm 0.01mm. The young's modulus obtained from the reading is [Take[Takeg = 9.8 {ms}^{- 2} ]

Options

  1. A(2.0±0.3)×1011Nm−2\left(2 .0 \pm 0 .3\right) \times {10}^{11}{Nm}^{- 2}
  2. B(2.0±0.2)×1011Nm−2\left(2 .0 \pm 0 .2\right) \times {10}^{11}{Nm}^{- 2}
  3. C(2.0±0.1)×1011Nm−2\left(2 . 0 \pm 0 .1\right) \times {10}^{11}{Nm}^{- 2}
  4. D(2.0±0.05)×1011Nm−2\left(2 .0 \pm 0 .05\right) \times {10}^{11}{Nm}^{- 2}

Correct answer

B. (2.0±0.2)×1011Nm−2\left(2 .0 \pm 0 .2\right) \times {10}^{11}{Nm}^{- 2}

Step-by-step solution

Young's modulus Y=FLAl=4FL\pid2lY = \dfrac{FL}{Al} = \dfrac{4FL}{\pid^{2}l}

=(4)(1.0×9.8)(2)π(0.4×10−3)2(0.8×10−3)= \dfrac{\left(4\right)\left(1.0 \times 9.8\right)\left(2\right)}{\pi{\left(0.4 \times {10}^{- 3}\right)}^{2}\left(0.8 \times {10}^{- 3}\right)}

=2.0×1011Nm−2= 2.0 \times {10}^{11}{Nm}^{- 2}

Further, ∆YY=2(∆dd)+(∆ll)<br/><br/>\dfrac{∆ Y}{Y} = 2 \left(\dfrac{∆ d}{d}\right) +\left(\dfrac{∆ l}{l}\right)<br /><br />∴ ∆ Y = \left(\{2\left(\dfrac{∆ d}{d}\right) + \left(\dfrac{∆ l}{l}\right)\right\}YY

= \left\{2\times \dfrac{0.01}{0.4} + \dfrac{0.05}{0.8}\right \} \times2.0 \times {10}^{11}

=0.2= 0.2\times {10}^{11}{Nm}^{- 2}

Or \left(Y + ∆ Y\right) = \left(2 + 0.2\right)\times {10}^{11}{Nm}^{- 2}$
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