JEE Mains · Physics · STD 11 - 13. oscillations
A pendulume clock loses \(12\;s\) a day if the temperature is \(40^oC\) and gains \(4\;s\) a day if the temperature is \(20^oC\). The temperature at which the clock will show correct time, and the coeffecient of linear expansion \((\alpha)\) of the metal of the pendulum shaft are respectively
- A \(30^o \) \(C\) ,\(\;\alpha \) \(= 1.85 \times 10^{-3}/^o C\)
- B \(55^o C\) ,\(\;\alpha \) \(= 1.85 \times 10^{-2}/^o C\)
- C \(25^o C\) ,\(\;\alpha \)\( = 1.85 \times 10^{-5}/^o C\)
- D \(60^o \) \(C\) ,\(\;\alpha \) = \(1.85 \times10^{-4}/^o C\)
Answer & Solution
Correct Answer
(C) \(25^o C\) ,\(\;\alpha \)\( = 1.85 \times 10^{-5}/^o C\)
Step-by-step Solution
Detailed explanation
\(\mathrm{T}=2 \pi \sqrt{\frac{\ell}{\mathrm{g}}}\) \(\frac{\Delta \mathrm{T}}{\mathrm{T}}=\frac{1}{2} \frac{\Delta \ell}{\ell}\) When clock gain \(12\, sec\) \(\frac{12}{\mathrm{T}}=\frac{1}{2} \alpha(40-\theta)\) \(...(1)\) When clock lose \(4\, sec.\)…
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