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Properties of Matter question

2019 · 8 Apr · Shift 2 · Q69
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Properties of Matter question

2019 · 8 Apr · Shift 2 · Q69

JEE MainPhysicsProperties of MatterMCQ+4 / −1
Young's moduli of two wires A and B are in the ratio 7 : 4. Wire A is 2 m long and has radius R. Wire B is 1.5 m long and has radius 2 mm. If the two wires stretch by the same length for a given load, then the value of R is close to :-
  1. A
    1.7 mm
  2. B
    1.9 mm
  3. C
    1.3 mm
  4. D
    1.5 mm
View written solutionFree

Correct answer: A

  1. Use the extension formula for a wire

For a wire under a load FFF, the extension is

ΔL=FLYA\Delta L = \frac{F L}{Y A}ΔL=YAFL​

where:

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

For wire AAA:

  • YA:YB=7:4Y_A : Y_B = 7:4YA​:YB​=7:4
  • LA=2 mL_A = 2\,\text{m}LA​=2m
  • radius =R= R=R
  • area AA=πR2A_A = \pi R^2AA​=πR2

For wire BBB:

  • LB=1.5 mL_B = 1.5\,\text{m}LB​=1.5m
  • radius =2 mm= 2\,\text{mm}=2mm
  • area AB=π(2)2=4π mm2A_B = \pi (2)^2 = 4\pi\,\text{mm}^2AB​=π(2)2=4πmm2

Also, both wires have the same extension under the same load.

So,

FLAYAAA=FLBYBAB\frac{F L_A}{Y_A A_A} = \frac{F L_B}{Y_B A_B}YA​AA​FLA​​=YB​AB​FLB​​

Cancel FFF:

LAYAAA=LBYBAB\frac{L_A}{Y_A A_A} = \frac{L_B}{Y_B A_B}YA​AA​LA​​=YB​AB​LB​​
  1. Substitute values
2YAπR2=1.5YB⋅4π\frac{2}{Y_A \pi R^2} = \frac{1.5}{Y_B \cdot 4\pi}YA​πR22​=YB​⋅4π1.5​

Cancel π\piπ:

2YAR2=1.54YB\frac{2}{Y_A R^2} = \frac{1.5}{4Y_B}YA​R22​=4YB​1.5​

Cross-multiply:

8YB=1.5YAR28Y_B = 1.5 Y_A R^28YB​=1.5YA​R2

Thus,

R2=8YB1.5YAR^2 = \frac{8Y_B}{1.5Y_A}R2=1.5YA​8YB​​

Using

YAYB=74  ⟹  YBYA=47\frac{Y_A}{Y_B} = \frac{7}{4} \implies \frac{Y_B}{Y_A} = \frac{4}{7}YB​YA​​=47​⟹YA​YB​​=74​

So,

R2=81.5⋅47R^2 = \frac{8}{1.5}\cdot \frac{4}{7}R2=1.58​⋅74​ R2=163⋅47=6421R^2 = \frac{16}{3}\cdot \frac{4}{7} = \frac{64}{21}R2=316​⋅74​=2164​ R=6421=821≈1.75 mmR = \sqrt{\frac{64}{21}} = \frac{8}{\sqrt{21}} \approx 1.75\,\text{mm}R=2164​​=21​8​≈1.75mm
  1. Closest option
R≈1.75 mmR \approx 1.75\,\text{mm}R≈1.75mm

This is closest to 1.7 mm1.7\,\text{mm}1.7mm.

So the correct option is A.

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