NEET-XII-Physics
23: Heat and Temperature
- #19A pendulum clock shows correct time at 20°C at a place where g = 9.800 m s-2. The pendulum consists of a light steel rod connected to a heavy ball. It is taken to a different place where g = 9.788 m s-1. At what temperature will the clock show correct time? Coefficient of linear expansion of steel = 12 × 10-6 °C-1.Ans : Given:
The temperature at which the pendulum shows the correct time, T1 = 20 °C
Coefficient of linear expansion of steel, `` \alpha `` = 12 × 10-6 °C-1
Let T2 be the temperature at which the value of g is 9.788 ms-2 and `` \Delta ``T be the change in temperature.
So, the time periods of pendulum at different values of g will be t1 and t2 , such that
`` {t}_{1}=2\,\mathrm{\,\pi \,}\sqrt{\frac{{l}_{\mathit{1}}}{{g}_{1}}}``
`` {t}_{2}=2\,\mathrm{\,\pi \,}\sqrt{\frac{{l}_{2}}{{g}_{2}}}``
`` =\frac{2\,\mathrm{\,\pi \,}\sqrt{{l}_{1}\left(1+\alpha \Delta T\right)}}{\sqrt{{g}_{2}}}\left(\because {l}_{2}={l}_{1}\left(1+\alpha ∆T\right)\right)``
`` \,\mathrm{\,Given\,},{t}_{1}={t}_{2}``
`` \Rightarrow \frac{2\pi \sqrt{{l}_{1}}}{\sqrt{{g}_{1}}}=\frac{2\pi \sqrt{{l}_{1}\left(1+\alpha ∆T\right)}}{\sqrt{{g}_{2}}}``
`` \Rightarrow \sqrt{\left(\frac{{l}_{1}}{{g}_{1}}\right)}=\frac{\sqrt{{l}_{1}\left(1+\alpha ∆T\right)}}{\sqrt{{g}_{2}}}``
`` \Rightarrow \frac{1}{9.8}=\frac{1+12\times {10}^{-6}\times ∆T}{9.788}``
`` \Rightarrow \frac{9.788}{9.8}=1+12\times {10}^{-6}\times ∆T``
`` \Rightarrow \frac{9.788}{9.8}-1=12\times {10}^{-6}\times ∆T``
`` \Rightarrow ∆T=\frac{-0.00122}{12\times {10}^{-6}}``
`` \Rightarrow {T}_{2}-20=-102.4``
`` \Rightarrow {T}_{2}=-102.4+20``
`` =-82.4``
`` \Rightarrow {T}_{2}\approx -82°\,\mathrm{\,C\,}``
Therefore, for a pendulum clock to give correct time, the temperature at which the value of g is 9.788 ms-2 should be `` -`` 82 oC.
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