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Properties of Matter question

2023 · 30 Jan · Shift 2 · Q48
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Properties of Matter question

2023 · 30 Jan · Shift 2 · Q48

JEE MainPhysicsProperties of MatterMCQ+4 / −1
A force is applied to a steel wire 'A', rigidly clamped at one end. As a result elongation in the wire is 0.2 mm0.2 \mathrm{~mm}0.2 mm. If same force is applied to another steel wire ' B\mathrm{B}B' of double the length and a diameter 2.42.42.4 times that of the wire 'A\mathrm{A}A', the elongation in the wire 'B\mathrm{B}B ' will be (wires having uniform circular cross sections)
  1. A
    6.9×10−2 mm6 .9 \times 10^{-2} \mathrm{~mm}6.9×10−2 mm
  2. B
    6.06×10−2 mm6.06 \times 10^{-2} \mathrm{~mm}6.06×10−2 mm
  3. C
    2.77×10−2 mm2.77 \times 10^{-2} \mathrm{~mm}2.77×10−2 mm
  4. D
    3.0×10−2 mm3.0 \times 10^{-2} \mathrm{~mm}3.0×10−2 mm
View written solutionFree

Correct answer: A

  1. Use the formula for elongation of a wire

For a wire under a force FFF, the elongation is

ΔL=FLAY\Delta L = \frac{F L}{A Y}ΔL=AYFL​

where:

  • LLL = length of wire
  • AAA = cross-sectional area
  • YYY = Young's modulus

Since both wires are made of steel, YYY is same. Also, the same force is applied, so FFF is same.

Thus,

ΔL∝LA\Delta L \propto \frac{L}{A}ΔL∝AL​

And for circular cross-section,

A∝d2A \propto d^2A∝d2

So,

ΔL∝Ld2\Delta L \propto \frac{L}{d^2}ΔL∝d2L​
  1. Compare wire B with wire A

Given:

  • For wire A, elongation ΔLA=0.2 mm\Delta L_A = 0.2\,\text{mm}ΔLA​=0.2mm
  • Wire B has double the length:
LB=2LAL_B = 2L_ALB​=2LA​
  • Diameter of wire B is 2.42.42.4 times that of A:
dB=2.4dAd_B = 2.4 d_AdB​=2.4dA​

Therefore,

ΔLBΔLA=LB/dB2LA/dA2\frac{\Delta L_B}{\Delta L_A} = \frac{L_B/d_B^2}{L_A/d_A^2}ΔLA​ΔLB​​=LA​/dA2​LB​/dB2​​

Substitute values:

ΔLBΔLA=2LA/(2.4dA)2LA/dA2\frac{\Delta L_B}{\Delta L_A} = \frac{2L_A/(2.4d_A)^2}{L_A/d_A^2}ΔLA​ΔLB​​=LA​/dA2​2LA​/(2.4dA​)2​ =2⋅1(2.4)2= 2 \cdot \frac{1}{(2.4)^2}=2⋅(2.4)21​ =25.76= \frac{2}{5.76}=5.762​ =0.3472= 0.3472=0.3472
  1. Find elongation of wire B
ΔLB=ΔLA×0.3472\Delta L_B = \Delta L_A \times 0.3472ΔLB​=ΔLA​×0.3472 ΔLB=0.2×0.3472\Delta L_B = 0.2 \times 0.3472ΔLB​=0.2×0.3472 ΔLB=0.06944 mm\Delta L_B = 0.06944\,\text{mm}ΔLB​=0.06944mm ΔLB≈6.9×10−2 mm\Delta L_B \approx 6.9 \times 10^{-2}\,\text{mm}ΔLB​≈6.9×10−2mm
  1. Match with options

This corresponds to:

Option A: 6.9×10−2 mm6.9 \times 10^{-2}\,\text{mm}6.9×10−2mm

  1. Comparison with stored correct answer

Stored correct answer: A

Derived answer: A

They match.

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