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Properties of Matter question

2023 · 24 Jan · Shift 1 · Q48
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Properties of Matter question

2023 · 24 Jan · Shift 1 · Q48

JEE MainPhysicsProperties of MatterMCQ+4 / −1
A 100 m long wire having cross-sectional area 6.25×10−4 m2\mathrm{6.25\times10^{-4}~m^2}6.25×10−4 m2 and Young's modulus is 1010 Nm−2\mathrm{10^{10}~Nm^{-2}}1010 Nm−2 is subjected to a load of 250 N, then the elongation in the wire will be :
  1. A
    6.25×10−6 m\mathrm{6.25\times10^{-6}~m}6.25×10−6 m
  2. B
    4×10−3 m\mathrm{4\times10^{-3}~m}4×10−3 m
  3. C
    4×10−4 m\mathrm{4\times10^{-4}~m}4×10−4 m
  4. D
    6.25×10−3 m\mathrm{6.25\times10^{-3}~m}6.25×10−3 m
View written solutionFree

Correct answer: B

  1. Use the formula for elongation of a wire

For a wire under tensile load,

Y=FLA ΔLY = \frac{F L}{A\,\Delta L}Y=AΔLFL​

So,

ΔL=FLAY\Delta L = \frac{F L}{A Y}ΔL=AYFL​
  1. Substitute the given values

Given:

  • Length, L=100 mL = 100\,\text{m}L=100m
  • Cross-sectional area, A=6.25×10−4 m2A = 6.25\times10^{-4}\,\text{m}^2A=6.25×10−4m2
  • Young's modulus, Y=1010 N m−2Y = 10^{10}\,\text{N m}^{-2}Y=1010N m−2
  • Load, F=250 NF = 250\,\text{N}F=250N

Thus,

ΔL=250×100(6.25×10−4)(1010)\Delta L = \frac{250\times100}{(6.25\times10^{-4})(10^{10})}ΔL=(6.25×10−4)(1010)250×100​
  1. Simplify the denominator
(6.25×10−4)(1010)=6.25×106(6.25\times10^{-4})(10^{10}) = 6.25\times10^6(6.25×10−4)(1010)=6.25×106

So,

ΔL=250006.25×106\Delta L = \frac{25000}{6.25\times10^6}ΔL=6.25×10625000​
  1. Calculate
ΔL=4×10−3 m\Delta L = 4\times10^{-3}\,\text{m}ΔL=4×10−3m
  1. Match with the options

Option B is:

4×10−3 m4\times10^{-3}\,\text{m}4×10−3m

So the correct answer is B.

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