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Properties of Matter question

2023 · 31 Jan · Shift 2 · Q59
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  5. /2023 · 31 Jan · Shift 2 · Q59

Properties of Matter question

2023 · 31 Jan · Shift 2 · Q59

JEE MainPhysicsProperties of MatterMCQ+4 / −1
Under the same load, wire A having length 5.0 m5.0 \mathrm{~m}5.0 m and cross section 2.5×10−5 m22.5 \times 10^{-5} \mathrm{~m}^{2}2.5×10−5 m2 stretches uniformly by the same amount as another wire B of length 6.0 m6.0 \mathrm{~m}6.0 m and a cross section of 3.0×10−5m23.0 \times 10^{-5}\mathrm{m}^{2}3.0×10−5m2 stretches. The ratio of the Young's modulus of wire A to that of wire BBB will be :
  1. A
    1:21: 21:2
  2. B
    1:41: 41:4
  3. C
    1:11: 11:1
  4. D
    1:101: 101:10
View written solutionFree

Correct answer: C

  1. Use the formula for extension of a wire

For a wire under load FFF, the extension is

ΔL=FLAY\Delta L = \frac{FL}{AY}ΔL=AYFL​

where:

  • LLL = original length
  • AAA = cross-sectional area
  • YYY = Young's modulus
  1. Given condition: same load and same extension

For wire A and wire B:

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

Since the load FFF is same for both, it cancels:

LAAAYA=LBABYB\frac{L_A}{A_A Y_A} = \frac{L_B}{A_B Y_B}AA​YA​LA​​=AB​YB​LB​​

Rearranging,

YAYB=LAABLBAA\frac{Y_A}{Y_B} = \frac{L_A A_B}{L_B A_A}YB​YA​​=LB​AA​LA​AB​​
  1. Substitute the given values

For wire A:

  • LA=5.0 mL_A = 5.0\,\text{m}LA​=5.0m
  • AA=2.5×10−5 m2A_A = 2.5 \times 10^{-5}\,\text{m}^2AA​=2.5×10−5m2

For wire B:

  • LB=6.0 mL_B = 6.0\,\text{m}LB​=6.0m
  • AB=3.0×10−5 m2A_B = 3.0 \times 10^{-5}\,\text{m}^2AB​=3.0×10−5m2

So,

YAYB=5.0×3.0×10−56.0×2.5×10−5\frac{Y_A}{Y_B} = \frac{5.0 \times 3.0 \times 10^{-5}}{6.0 \times 2.5 \times 10^{-5}}YB​YA​​=6.0×2.5×10−55.0×3.0×10−5​

Cancel 10−510^{-5}10−5:

YAYB=5.0×3.06.0×2.5\frac{Y_A}{Y_B} = \frac{5.0 \times 3.0}{6.0 \times 2.5}YB​YA​​=6.0×2.55.0×3.0​ =1515=1= \frac{15}{15} = 1=1515​=1
  1. Final ratio
YA:YB=1:1Y_A : Y_B = 1:1YA​:YB​=1:1

So the correct option is C.

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