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MCQMediumJEE 2024Young's Modulus, Bulk & Rigidity Modulus

JEE Physics 2024 Question with Solution

Each of three blocks P, Q, and R (each 3 kg3 \, \text{kg}) is attached to a wire. Wires A and B each have a cross-sectional area of 0.005 cm20.005 \, \text{cm}^2 and Young’s modulus of 2×1011 N/m22 \times 10^{11} \, \text{N/m}^2. Neglecting friction, the longitudinal strain on wire B is x×10−4x \times 10^{-4}:

  • A

    11

  • B

    22

  • C

    33

  • D

    44

Answer

Correct answer:B

Step-by-step solution

Standard Method

Given: Each block has mass 3 kg3 \, \text{kg}. Cross-sectional area of wire B is 0.005 cm20.005 \, \text{cm}^2 and Young's modulus is 2×1011 N/m22 \times 10^{11} \, \text{N/m}^2.

Find: The value of xx if the longitudinal strain in wire B is x×10−4x \times 10^{-4}.

From the solution, the system acceleration is found using Newton's second law:

a=FnetMtotal=309=103 m/s2a = \frac{F_{net}}{M_{total}} = \frac{30}{9} = \frac{10}{3} \, \text{m/s}^2

For block R, taking downward direction along its motion:

mRg−TB=mRam_R g - T_B = m_R a 30−TB=3×10330 - T_B = 3 \times \frac{10}{3} 30−TB=1030 - T_B = 10 TB=20 NT_B = 20 \, \text{N}

Convert the cross-sectional area into SI units:

A=0.005 cm2=0.005×10−4 m2=5×10−7 m2A = 0.005 \, \text{cm}^2 = 0.005 \times 10^{-4} \, \text{m}^2 = 5 \times 10^{-7} \, \text{m}^2

Using Young's modulus,

Y=stressstrain⇒strain=TBAYY = \frac{\text{stress}}{\text{strain}} \quad \Rightarrow \quad \text{strain} = \frac{T_B}{AY}

Substitute the values:

strain=20(5×10−7)(2×1011)\text{strain} = \frac{20}{\left(5 \times 10^{-7}\right)\left(2 \times 10^{11}\right)} strain=2010×104=2×10−4\text{strain} = \frac{20}{10 \times 10^4} = 2 \times 10^{-4}

Therefore, x=2x = 2, so the correct option is B.

Using Stress-Strain Relation

Given: Tension in wire B from the motion analysis is 20 N20 \, \text{N}.

Find: Longitudinal strain in wire B.

Stress in wire B is:

stress=TBA\text{stress} = \frac{T_B}{A}

with

A=0.5×10−6 m2A = 0.5 \times 10^{-6} \, \text{m}^2

Now use:

strain=stressY=TBAY\text{strain} = \frac{\text{stress}}{Y} = \frac{T_B}{AY} strain=200.5×10−6×2×1011=2×10−4\text{strain} = \frac{20}{0.5 \times 10^{-6} \times 2 \times 10^{11}} = 2 \times 10^{-4}

Hence, the required value is x=2x = 2 and the correct option is B.

Common mistakes

  • Using the weight of block R directly as the tension in wire B is incorrect because block R is accelerating. First apply Newton's second law to block R and then find TBT_B from mRg−TB=mRam_R g - T_B = m_R a.

  • Forgetting to convert 0.005 cm20.005 \, \text{cm}^2 into m2\text{m}^2 gives a wrong strain by several powers of 1010. Convert area carefully: 0.005 cm2=5×10−7 m20.005 \, \text{cm}^2 = 5 \times 10^{-7} \, \text{m}^2.

  • Using strain=Ystress\text{strain} = \frac{Y}{\text{stress}} is wrong because Young's modulus is Y=stressstrainY = \frac{\text{stress}}{\text{strain}}. Therefore, strain must be calculated as stressY\frac{\text{stress}}{Y} or TAY\frac{T}{AY}.

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