Verified Solution • 2025–26 Session moderate

QJEEM23Ph1be6: JEE Main 2023 Physics - Class 11 JEEM23Ph1be6 Solution: A thin rod having a length of and area of cross-section is…

A thin rod having a length of 1 m1 \mathrm{~m} and area of cross-section 3×10−6 m23 \times 10^{-6} \mathrm{~m}^{2} is suspended vertically from one end. The rod is cooled from 210∘C210^{\circ} \mathrm{C} to 160∘C160^{\circ} \mathrm{C} . After cooling, a mass M\mathrm{M} is attached at the lower end of the rod such that the length of rod again becomes 1 m1 \mathrm{~m} . Young's modulus and coefficient of linear expansion of the rod are 2×1011 N m−22 \times 10^{11} \mathrm{~N} \mathrm{~m}^{-2} and 2×10−5 K−12 \times 10^{-5} \mathrm{~K}^{-1} , respectively. The value of M\mathrm{M} is __________ kg\mathrm{kg} .

(Take g=10 m s−2\mathrm{g=10~m~s^{-2}} )

Step-by-Step Verified Solution

When the rod is cooled from 210°C to 160°C, it will contract in length due to thermal contraction. The change in length of the rod is given by:

ΔL = L α\alpha ΔT

where L is the original length of the rod, α is the coefficient of linear expansion, and ΔT is the change in temperature.

When a mass M is attached to the lower end of the rod, it will stretch due to the weight of the mass. The elongation of the rod is given by:

ΔL=MgLAY\Delta L = {{MgL} \over {AY}}

where M is the mass, g is the acceleration due to gravity, A is the cross-sectional area of the rod, Y is the Young's modulus of the rod, and L is the original length of the rod.

∴\therefore LαΔT=MgLAYL\alpha \Delta T = {{MgL} \over {AY}}

⇒\Rightarrow αΔT=MgAY\alpha \Delta T = {{Mg} \over {AY}}

⇒\Rightarrow Mg=AYαΔTMg = AY\alpha \Delta T

⇒\Rightarrow M×10=2×1011×3×10−6×2×10−5×50\mathrm{M} \times 10=2 \times 10^{11} \times 3 \times 10^{-6} \times 2 \times 10^{-5} \times 50

⇒\Rightarrow M=60 kg\mathrm{M}=60 \mathrm{~kg}

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