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MCQEasyJEE 2023Elasticity & Stress-Strain Curve

JEE Physics 2023 Question with Solution

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 \, \text{mm}. If same force is applied to another steel wire ‘B’ of double the length and a diameter 2.42.4 times that of the wire ‘A’, the elongation in the wire ‘B’ will be (wires having uniform circular cross sections):

  • A

    6.06×10−26.06 \times 10^{-2} mm\text{mm}

  • B

    2.77×10−22.77 \times 10^{-2} mm\text{mm}

  • C

    3.0×10−23.0 \times 10^{-2} mm\text{mm}

  • D

    6.9×10−26.9 \times 10^{-2} mm\text{mm}

Answer

Correct answer:B

Step-by-step solution

Standard Method

Given: Two steel wires are under the same applied force. For wire AA, elongation is Δl1=0.2 mm\Delta l_1 = 0.2 \, \text{mm}. Wire BB has length l2=2l1l_2 = 2l_1 and diameter d2=2.4d1d_2 = 2.4d_1.

Find: Elongation Δl2\Delta l_2 of wire BB.

Using Young's modulus,

Y=(l/Δl)F/AY = \frac{(l/\Delta l)}{F/A}

So,

F=(YAl)ΔlF = \left(\frac{YA}{l}\right) \Delta l

Since the material and applied force are the same for both wires, the solution uses

(lAΔl)1=(lAΔl)2\left(\frac{l}{A\Delta l}\right)_1 = \left(\frac{l}{A\Delta l}\right)_2

Therefore,

Δl1Δl2=A2A1×l1l2\frac{\Delta l_1}{\Delta l_2} = \frac{A_2}{A_1} \times \frac{l_1}{l_2}

Now, for circular cross-section, area is proportional to the square of diameter, so

A2A1=(d2d1)2=(2.4)2\frac{A_2}{A_1} = \left(\frac{d_2}{d_1}\right)^2 = (2.4)^2

and

l1l2=12\frac{l_1}{l_2} = \frac{1}{2}

Substituting in the relation shown in the solution,

0.2Δl2=2.4×2.41×12\frac{0.2}{\Delta l_2} = \frac{2.4 \times 2.4}{1} \times \frac{1}{2}

Hence,

Δl2=6.9×10−2 mm\Delta l_2 = 6.9 \times 10^{-2} \, \text{mm}

Therefore, the elongation of wire BB is 6.9×10−2 mm6.9 \times 10^{-2} \, \text{mm}. The solution concludes that the correct option is B.

Option Discrepancy Note

The solution working gives

Δl2=6.9×10−2 mm\Delta l_2 = 6.9 \times 10^{-2} \, \text{mm}

This numerical value matches option D in the listed options, but the solution explicitly states "The Correct Option is B" and also writes the value 6.9×10−2 mm6.9 \times 10^{-2} \, \text{mm}. Hence there is a mismatch between the option label and the option value on the solution's.

the answer is recorded as B.

Common mistakes

  • Using diameter directly instead of cross-sectional area is incorrect. Elongation depends on area AA, and for a circular wire A∝d2A \propto d^2. Always square the diameter ratio before substituting.

  • Assuming elongation is directly proportional only to length is incomplete. For the same force and material, Δl∝lA\Delta l \propto \frac{l}{A}. Both the doubled length and increased diameter must be accounted for together.

  • Confusing the option label with the numerical result can lead to a wrong final choice here. The working gives 6.9×10−2 mm6.9 \times 10^{-2} \, \text{mm}, so compare the computed value carefully with the listed options and note any source mismatch.

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