Test Index

Thermodynamics Part 1

Section: Physics

Question : 43 of 100

Marks: +1, -0

Solution:

First Method:

Time period T

=2\pi\sqrt{\frac{l}{g}}=2\pi\sqrt{\frac{l_0}{g(1+\alpha\Delta\theta)}}

T=T_0\left[1+\frac{1}{2}\alpha\Delta\theta\right]

N=\frac{2\times86400}{T}\left(\frac{2\times86400}{T_0}\right)\left(1+\frac{1}{2}\alpha\Delta\theta\right)=N\left(1+\frac{\alpha\Delta\theta}{2}\right)

\Delta N=N-N_0=\frac{1}{2}\alpha\Delta\theta N_0\Rightarrow\Delta N\propto\Delta\theta

\Rightarrow\theta_0=25^{\circ}

Putting

\theta_0

,we get

\alpha=18.5\times10^{-5}/

°C

Second MethodAccording to given conditions

86412=2\pi\sqrt{\frac{l_40}{g}}

.....(i)

86396=2\pi\sqrt{\frac{l_20}{g}}

.....(ii)

86400=2\pi\sqrt{\frac{l}{g}}

.....(iii)

From equation (i) and (iii)

12=\frac{2\pi}{\sqrt{g}}\left[\sqrt{l_40}-\sqrt{l}\right]

...(iv)

and

4=\frac{2\pi}{\sqrt{g}}\left[\sqrt{l}-\sqrt{l_20}\right]

.....(v)

on dividing (iv) and (v)

3=\frac{\sqrt{1+\alpha(40-\theta)}-1}{1-\sqrt{1+\alpha(20-\theta)}}

\Rightarrow3=\frac{40-\theta}{\theta-20}

(by Binomial theorem)

\Rightarrow\theta=25^{\circ}

on using in θ (i) and (iii)

\frac{86412}{86400}=\sqrt{1+\alpha15}\Rightarrow\alpha=1.85\times10^{-5}/

°C

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