A pendulume clock loses 12 ;s a day if the temperature is 40^oC a… - Vidyadip

MCQ

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 $ $= 1.85 \times 10^{-2}/^o C$
✓
$25^o C$ ,$\;\alpha $$ = 1.85 \times 10^{-5}/^o C$
D
$60^o $ $C$ ,$\;\alpha $ = $1.85 \times10^{-4}/^o C$

✓

Answer

Correct option: C.

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

c
$\mathrm{T}=2 \pi \sqrt{\frac{\ell}{\mathrm{g}}}$

$\frac{\Delta \mathrm{T}}{\mathrm{T}}=\frac{1}{2} \frac{\Delta \ell}{\ell}$

When clock gain $12\, sec$

$\frac{12}{\mathrm{T}}=\frac{1}{2} \alpha(40-\theta)$ $...(1)$

When clock lose $4\, sec.$

$\frac{4}{\mathrm{T}}=\frac{1}{2} \alpha(\theta-20)$ $...(2)$

From equation $( 1)\&(2)$

$3=\frac{40-\theta}{\theta-20}$

$3 \theta-60=40-\theta$

$4 \theta=100$

$\theta=25^{\circ} \mathrm{C}$

from equation $( 1 )$

$\frac{12}{\mathrm{T}}=\frac{1}{2} \alpha(40-25)$

$\frac{12}{24 \times 3600}=\frac{1}{2} \alpha \times 15$

$\alpha=\frac{24}{24 \times 3600 \times 15}$

$\alpha=1.85 \times 10^{-15} /^{\circ} \mathrm{C}$

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