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NTA Abhyas 2020
Exam
NTA Abhyas
Year
2020
Subject
Physics
Chapter
Mechanical Properties of Solids

NTA Abhyas 2020 · Physics · Mechanical Properties of Solids

A load of 31. 4kg is suspended from a wire of radius 10 ^ - 3 m and...

A load of 31.4kg31. 4kg is suspended from a wire of radius 10−3m{10}^{- 3}m and density 9×103kgm−39 \times {10}^{3}kgm^{- 3}. Calculate the change in temperature of the wire if 7575 % of the stored elastic potential energy is converted into heat. The Young's modulus and heat capacity of the material of the wire is 9.8×1010Nm−29. 8 \times {10}^{10}Nm^{- 2} and 490Jkg−1K−1490J{kg}^{- 1}K^{- 1} respectively.

Options

  1. A4.33×10−2K4. 33 \times {10}^{- 2}K
    &
  2. B8.33×10−3K8. 33 \times {10}^{- 3}K
    &
  3. C2.44×10−5K2. 44 \times {10}^{- 5}K
    &
  4. D6.22×10−2K6. 22 \times {10}^{- 2}K
    &

Correct answer

B. 8.33×10−3K8. 33 \times {10}^{- 3}K

&

Step-by-step solution

As work done in stretching an elastic body per unit volume is given by

& & & WV=12stress×strain=12stress2Y[as Y=stressstrain and ]........(1)<br/><br/>SoW=12[MgA]2VY[as stress=FA=MgA]<br/><br/>NowaccordingtogivenproblemheatproducedH=(75/100)W<!−−MathType@Translator@5@5@MathML2(namespaceattr).tdl@MathML2.0(namespaceattr)@−−>=34.W<!−−MathType@End@5@5@−−><br/><br/><!−−MathType@Translator@5@5@MathML2(namespaceattr).tdl@MathML2.0(namespaceattr)@−−>\dfrac{\text{W}}{\text{V}} = \dfrac{1}{2} \text{stress} \times \text{strain} = \dfrac{1}{2} \dfrac{{stress}^{2}}{\text{Y}} \left[\text{as Y} = \dfrac{\text{stress}}{\text{strain} \text{ and } }\right] ........ (1)<br /><br />So \text{W} = \dfrac{1}{2} {\left[\dfrac{\text{Mg}}{\text{A}}\right]}^{2} \dfrac{\text{V}}{\text{Y}} \left[\text{as stress} = \dfrac{\text{F}}{\text{A}} = \dfrac{\text{Mg}}{\text{A}}\right]<br /><br />Now according to given problem heat produced H = (75/100) W<!-- MathType@Translator@5@5@MathML2 (namespace attr).tdl@MathML 2.0 (namespace attr)@ --> =\dfrac{3}{4} . \text{W} <!-- MathType@End@5@5@ --><br /><br /> <!-- MathType@Translator@5@5@MathML2 (namespace attr).tdl@MathML 2.0 (namespace attr)@ -->⇒<!-- MathType@End@5@5@ --> \text{ms} \Delta \theta = \dfrac{3}{4} \times \dfrac{1}{2} {\left[\dfrac{\text{Mg}}{\pi r^{2}}\right]}^{2} \dfrac{\text{V}}{\text{Y}} \left[\text{as H} = \text{ms} \Delta \theta\right]using equation (1)<br /><br />or \left(ρ \text{V}\right) \text{s} \Delta \theta = \dfrac{3}{4} \times \dfrac{1}{2} { \left[\dfrac{\text{Mg}}{\pi r^{2}}\right]}^{2}\dfrac{\text{V}}{\text{Y}}\left[\text{as m} =ρ \text{V}\right]<br /><br />∴ Δθ=38[31.4×9.8π×10−6]219.8×1010×19×103×490<br/><br/>i.e.,Δθ=1120K=8.33×10−3K\Delta \theta = \dfrac{3}{8} {\left[\dfrac{3 1 \text{.} 4 \times 9 \text{.} 8}{\pi \times 1 {0}^{- 6}}\right]}^{2} \dfrac{1}{9 \text{.} 8 \times 1 {0}^{1 0}} \times \dfrac{1}{9 \times 1 {0}^{3} \times 4 9 0}<br /><br />i.e., \Delta \theta = \dfrac{1}{1 2 0} \text{K} =8 \text{.}3 3 \times1 {0}^{- 3} \text{K}
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