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MCQEasyJEE 2023Young's Modulus, Bulk & Rigidity Modulus

JEE Physics 2023 Question with Solution

The Young's modulus of a steel wire of length 6 m6 \, \text{m} and cross-sectional area 3 mm23 \, \text{mm}^2, is 2×1011 N/m22 \times 10^{11} \, \text{N/m}^2. The wire is suspended from its support on a given planet. A block of mass 4 kg4 \, \text{kg} is attached to the free end of the wire. The acceleration due to gravity on the planet is 14\frac{1}{4} of its value on the earth. The elongation of wire is (Take gg on the earth = 10 m/s210 \, \text{m/s}^2):

  • A

    1 cm1 \, \text{cm}

  • B

    1 mm1 \, \text{mm}

  • C

    0.1 mm0.1 \, \text{mm}

  • D

    0.1 cm0.1 \, \text{cm}

Answer

Correct answer:A

Step-by-step solution

Standard Method

Given: length of wire L=6 mL = 6 \, \text{m}, cross-sectional area A=3 mm2=3×10−6 m2A = 3 \, \text{mm}^2 = 3 \times 10^{-6} \, \text{m}^2, Young's modulus Y=2×1011 N/m2Y = 2 \times 10^{11} \, \text{N/m}^2, mass m=4 kgm = 4 \, \text{kg}, and gravity on the planet gp=g4=104=2.5 m/s2g_p = \frac{g}{4} = \frac{10}{4} = 2.5 \, \text{m/s}^2.

Find: elongation ΔL\Delta L of the wire.

The tension in the wire is the weight of the block on that planet:

F=mgp=4×2.5=10 NF = mg_p = 4 \times 2.5 = 10 \, \text{N}

For a wire under tensile load, elongation is given by:

ΔL=FLAY\Delta L = \frac{FL}{AY}

Substituting the values:

ΔL=10×63×10−6×2×1011\Delta L = \frac{10 \times 6}{3 \times 10^{-6} \times 2 \times 10^{11}} ΔL=606×105=10−4 m\Delta L = \frac{60}{6 \times 10^5} = 10^{-4} \, \text{m} 10−4 m=0.1 mm10^{-4} \, \text{m} = 0.1 \, \text{mm}

Therefore, the elongation of the wire is 0.1 mm0.1 \, \text{mm}. The solution marks A, but that conflicts with the computed value and the listed options. The correct option corresponding to 0.1 mm0.1 \, \text{mm} is C.

Unit Conversion Check

Given: the area is written as 3 mm23 \, \text{mm}^2. A common place to make an error is unit conversion.

Since

1 mm=10−3 m1 \, \text{mm} = 10^{-3} \, \text{m}

we get

1 mm2=(10−3)2 m2=10−6 m21 \, \text{mm}^2 = (10^{-3})^2 \, \text{m}^2 = 10^{-6} \, \text{m}^2

Hence,

3 mm2=3×10−6 m23 \, \text{mm}^2 = 3 \times 10^{-6} \, \text{m}^2

Using this correctly gives

ΔL=10×63×10−6×2×1011=10−4 m=0.1 mm\Delta L = \frac{10 \times 6}{3 \times 10^{-6} \times 2 \times 10^{11}} = 10^{-4} \, \text{m} = 0.1 \, \text{mm}

So the defensible answer from the working is C.

Common mistakes

  • Using earth's gravity directly as 10 m/s210 \, \text{m/s}^2 is wrong because the planet's gravity is only 14\frac{1}{4} of that. First compute gp=2.5 m/s2g_p = 2.5 \, \text{m/s}^2, then find the tension.

  • Converting 3 mm23 \, \text{mm}^2 to 3×10−3 m23 \times 10^{-3} \, \text{m}^2 is wrong because area conversion squares the length factor. Use 3×10−6 m23 \times 10^{-6} \, \text{m}^2 instead.

  • Substituting into ΔL=FLAY\Delta L = \frac{FL}{AY} without checking powers of ten can produce an answer off by a factor of 1010. After calculation, convert 10−4 m10^{-4} \, \text{m} carefully to 0.1 mm0.1 \, \text{mm}.

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