A clock which keeps correct time at $25^{\circ} C$ has a pendulum made of brass whose coefficient of linear expansion is 0.000019. How many seconds a day will it gain if the temperature fall to $0^{\circ} C$ .

10.52 sec.

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20.52 sec.

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220.52 sec.

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202 sec.

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Solution

The correct option is B 20.52 sec.
$L e t L_{0} a n d L_{25} b e t h e l e n g t h o f p e n d u l u m a t 0^{\circ} C a n d 25^{\circ} C r e s p e c t i v e l y W e k n o w t h a t L_{25} = L_{0} (1 + α T) = L_{0} (1 + 0.000019 \times 25) = 1.000475 L_{0} I f T_{25} a n d T_{0} b e t h e t i m e p e r i o d s a t 25^{\circ} C a n d 0^{\circ} C r e s p e c t i v e l y, t h e n T_{25} = 2 π \sqrt{(\frac{L_{25}}{g})} a n d T_{0} = 2 π \sqrt{(\frac{L_{0}}{g})} ∴ \frac{T_{25}}{T_{0}} = \sqrt{(\frac{L_{25}}{g})} = \sqrt{(\frac{1.000475 L_{0}}{L_{0}})} = \sqrt{(1.000475)} = 1.000237 N o w \frac{T_{25} - T_{0}}{T_{0}} = 0.000237 ∴ G a i n i n t i m e f o r o n e v i b r a t i o n 2 \times 0.000237 s e c N u m b e r o f v i b r a t i o n i n o n e d a y = \frac{24 \times 60 \times 60}{2} s e c . (∵ T = 2 s e c) H e n c e, t h e g a i n i n t i m e i n o n e d a y = 2 \times 0.000237 \times \frac{24 \times 60 \times 60}{2} = 20.52 s e c$

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