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JEE Mains · Physics · STD 11 - 10.1, thermonetry,thermal expansion and calorimetry
- JEE Mains 2023
- Medium
- Numerical
A thin rod having a length of \(1\; m\) and area of cross-section \(3 \times 10^{-6}\,m ^2\) is suspended vertically from one end. The rod is cooled from \(210^{\circ}\,C\) to \(160^{\circ}\,C\). After cooling, a mass \(M\) is attached at the lower end of the rod such that the length of rod again becomes \(1\,m\). Young's modulus and coefficient of linear expansion of the rod are \(2 \times 10^{11} Nm ^{-2}\) and \(2 \times 10^{-5} K ^{-1}\), respectively. The value of \(M\) is \(.......kg .\left(\right.\) Take \(\left.g=10\,m s ^{-2}\right)\)
Pick an option to check your answer.
Answer & Solution
Correct Answer
(A) \(60\)
Step-by-step Solution
Detailed explanation
If \(\Delta l\) is decease in length of rod due to decease in temperature
\(\Delta l=l \alpha \Delta T\)
\(\alpha=2 \times 10^{-5}\,K ^{-1}, \Delta T =(210-160)=50\,K\)
\(\Delta l=1 \times 2 \times 10^{-5} \times 50=10^{-3}\,m\)
\(\text { Young Modulus }= Y =\frac{ F / A }{\Delta l / l} \quad A =3 \times 10^{-6}\,m ^2\)
\(2 \times 10^{11}=\frac{ Mg / 3 \times 10^{-6}}{10^{-3} / 1}\)
\(Mg =2 \times 10^{11} \times 3 \times 10^{-9}=6 \times 10^{-2}\)
\(M =60\,kg\)
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