Ask Questions, Get Answers

Want to ask us a question? Click here
Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class11  >>  Thermodynamics
0 votes

A pendulum clock having copper rod keeps correct at $\;20^{0} C\;.$ It gains 15 seconds per day if cooled to $\;0^{0}C\;.$ Calculate the coefficient of linear expansion of copper :

$(a)\;{2.6 \times 10^{-5}} ^{0}{C^{-1}}\qquad(b)\;{1.6 \times 10^{-5}} ^{0}{C^{-1}}\qquad(c)\;{2.7 \times 10^{-5}} ^{0}{C^{-1}}\qquad(d)\;{1.7 \times 10^{-5}} ^{0}{C^{-1}}$

Can you answer this question?

1 Answer

0 votes
Answer : (d) $\;{1.7 \times 10^{-5}} ^{0}{C^{-1}}$
Explanation :
$T_{0}=2 \pi \sqrt{\large\frac{L_{0}}{g}}$
$T=2 \pi \sqrt{\large\frac{L}{g}}=2 \pi \sqrt{\large\frac{l_{0}(1+\alpha \bigtriangleup \theta)}{g}} \approx 2 \pi \sqrt{\large\frac{l_{0}}{g}}(1+\large\frac{\alpha}{2}\bigtriangleup \theta)$
For $\; \alpha \bigtriangleup \theta \to 0$
$T=T_{0}(1+\large\frac{\alpha \bigtriangleup theta}{2})-----(1)$
It gains 15 sec/24 hrs =>$\; 15 sec / 360\times 24$
From $\;(1) \to \large\frac{\bigtriangleup T}{T_{0}}=\large\frac{15}{3600 \times 24 } \times \large\frac{\alpha \bigtriangleup \theta}{2}\;,\bigtriangleup \theta=20$
$\alpha=\;{1.7 \times 10^{-5}} ^{0}{C^{-1}}$
answered Mar 12, 2014 by yamini.v

Related questions

Ask Question
student study plans
JEE MAIN, CBSE, NEET Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App