A pendulume clock loses 12;s a day if the tem | vedclass

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Question

$\mathrm{T}=2 \pi \sqrt{\frac{\ell}{\mathrm{g}}}$

$\frac{\Delta \mathrm{T}}{\mathrm{T}}=\frac{1}{2} \frac{\Delta \ell}{\ell}$

When clock gain $1

Vedclass · Accepted Answer

$25^o C$ ,$\;\alpha $$ = 1.85 	imes 10^{-5}/^o C$

Vedclass · Answer

$\mathrm{T}=2 \pi \sqrt{\frac{\ell}{\mathrm{g}}}$

$\frac{\Delta \mathrm{T}}{\mathrm{T}}=\frac{1}{2} \frac{\Delta \ell}{\ell}$

When clock gain $12\, sec$

$\frac{12}{\mathrm{T}}=\frac{1}{2} \alpha(40-	heta)$             $...(1)$

When clock lose $4\, sec.$

$\frac{4}{\mathrm{T}}=\frac{1}{2} \alpha(	heta-20)$             $...(2)$

From equation $( 1)\&(2)$

$3=\frac{40-	heta}{	heta-20}$

$3 	heta-60=40-	heta$

$4 	heta=100$

$	heta=25^{\circ} \mathrm{C}$

from equation $( 1 )$

$\frac{12}{\mathrm{T}}=\frac{1}{2} \alpha(40-25)$

$\frac{12}{24 	imes 3600}=\frac{1}{2} \alpha 	imes 15$

$\alpha=\frac{24}{24 	imes 3600 	imes 15}$

$\alpha=1.85 	imes 10^{-15} /^{\circ} \mathrm{C}$

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

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