T△T=21l△l=21αΔθ
⇒ΔT=21TαΔθ
ΔθΔT=2Tα
But,
T=2s
ΔθΔT=α
JEE Main 2016 — Physics Thermodynamics
A simple pendulum made of a bob of mass m and a metallic wire of a negligible mass has a time period of 2s at T=0∘C. If the temperature of the wire is increased, and the corresponding change in its time period is plotted against its temperature, the resulting graph is a line of slope S. If the coefficient of linear expansion of metal is α, then the value of S is
Held on 9 Apr 2016 · Verified 6 Jul 2026.
2α
2α
α
α1
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10 mole of an ideal gas is undergoing the process shown in the figure. The heat involved in the process from $P_{1}$ to $P_{2}$ is $\alpha$ Joule ($P_{1}=21.7 \mathrm{~Pa}$ and $\left.P_{2}=30 \mathrm{~Pa}, \mathrm{C}_{v}=21 \mathrm{~J} / \mathrm{K}. \mathrm{mol}, R=8.3 \mathrm{~J} / \mathrm{mol}. \mathrm{K}\right)$. The value of $\alpha$ is $\_\_\_\_$. 
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Work through every JEE Main Thermodynamics PYQ, year by year.