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Properties of Matter question

2022 · 28 Jul · Shift 2 · Q59
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Properties of Matter question

2022 · 28 Jul · Shift 2 · Q59

JEE MainPhysicsProperties of MatterNumerical+4 / −1
A string of area of cross-section 4 mm24 \mathrm{~mm}^{2}4 mm2 and length 0.5 m0.5 \mathrm{~m}0.5 m is connected with a rigid body of mass 2 kg2 \mathrm{~kg}2 kg. The body is rotated in a vertical circular path of radius 0.5 m0.5 \mathrm{~m}0.5 m. The body acquires a speed of 5 m/s5 \mathrm{~m} / \mathrm{s}5 m/s at the bottom of the circular path. Strain produced in the string when the body is at the bottom of the circle is ‾\underline{\hspace{2cm}}​×10−5\times 10^{-5}×10−5. (use Young's modulus 1011 N/m210^{11} \mathrm{~N} / \mathrm{m}^{2}1011 N/m2 and g=10 m/s2\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^{2}g=10 m/s2)
Numerical answer
View written solutionFree

Correct answer: 30

  1. Given data
  • Cross-sectional area: A=4 mm2=4×10−6 m2A = 4\,\text{mm}^2 = 4 \times 10^{-6}\,\text{m}^2A=4mm2=4×10−6m2
  • Length of string: L=0.5 mL = 0.5\,\text{m}L=0.5m
  • Mass of body: m=2 kgm = 2\,\text{kg}m=2kg
  • Radius of circular path: r=0.5 mr = 0.5\,\text{m}r=0.5m
  • Speed at bottom: v=5 m/sv = 5\,\text{m/s}v=5m/s
  • Young's modulus: Y=1011 N/m2Y = 10^{11}\,\text{N/m}^2Y=1011N/m2
  • Acceleration due to gravity: g=10 m/s2g = 10\,\text{m/s}^2g=10m/s2
  1. Find tension in the string at the bottom

At the bottom of the vertical circle, centripetal force is directed upward toward the center.

So, T−mg=mv2rT - mg = \frac{mv^2}{r}T−mg=rmv2​

Substitute values: T−(2)(10)=(2)(52)0.5T - (2)(10) = \frac{(2)(5^2)}{0.5}T−(2)(10)=0.5(2)(52)​ T−20=2×250.5=100T - 20 = \frac{2 \times 25}{0.5} = 100T−20=0.52×25​=100 T=120 NT = 120\,\text{N}T=120N

  1. Use Young's modulus relation

Young's modulus is Y=stressstrain=T/AstrainY = \frac{\text{stress}}{\text{strain}} = \frac{T/A}{\text{strain}}Y=strainstress​=strainT/A​

Hence, strain=TAY\text{strain} = \frac{T}{AY}strain=AYT​

Substitute the values: strain=120(4×10−6)(1011)\text{strain} = \frac{120}{(4 \times 10^{-6})(10^{11})}strain=(4×10−6)(1011)120​

First compute denominator: (4×10−6)(1011)=4×105(4 \times 10^{-6})(10^{11}) = 4 \times 10^5(4×10−6)(1011)=4×105

Therefore, strain=1204×105=3×10−4\text{strain} = \frac{120}{4 \times 10^5} = 3 \times 10^{-4}strain=4×105120​=3×10−4

  1. Express in the required form

We need strain=‾×10−5\text{strain} = \underline{\hspace{1cm}} \times 10^{-5}strain=​×10−5

Now, 3×10−4=30×10−53 \times 10^{-4} = 30 \times 10^{-5}3×10−4=30×10−5

So the required integer is 30\boxed{30}30​

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