A clock which keeps correct time at 20-C- is subjected to 40-C- If coefficient of linear expansion of the pendulum is 12-10-6 per -C- How much will it gain or loose in time-

The correct option is A 10.3 seconds/day T = 2π√(l/g) T/T = 2π×1/2(l/g)^ 1/2× l/l (∵ T + T = 2π√(l+ l/g)) ∴ T/T = 1/2 l/l = 1/2∝θ ...

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Byju's Answer

Standard VIII

Physics

Thermal Expansion of Solids

A clock which...

Question

A clock which keeps correct time at $20^{\circ} C$ , is subjected to $40^{\circ} C$ . If coefficient of linear expansion of the pendulum is $12 \times 10^{- 6}$ per $^{\circ} C$ . How much will it gain or loose in time?

10.3 s e c o n d s / d a y

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20.6 s e c o n d s / d a y

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5 s e c o n d s / d a y

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20 m i n u t e s / d a y

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Solution

The correct option is A $10.3 s e c o n d s / d a y$
$T = 2 π \sqrt{\frac{l}{g}}$
$\frac{△ T}{T} = 2 π \times \frac{1}{2} (l / g)^{- 1 / 2} \times △ l / l$
$(∵ T + △ T = 2 π \sqrt{\frac{l + △ l}{g}})$
$∴ \frac{△ T}{T} = \frac{1}{2} \frac{△ l}{l} = \frac{1}{2} \propto △ θ$
$= \frac{1}{2} \times 12 \times 10^{- 6} \times (40 - 20) = 12 \times 10^{- 5}$
$△ T = T \times 12 \times 10^{- 5} = 24 \times 60 \times 60 \times 10^{- 5} = 10.3 s / d a y$

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A clock which keeps correct time at 20∘C, is subjected to 40∘C. If coefficient of linear expansion of the pendulum is 12×10−6 per ∘C. How much will it gain or loose in time?

The correct option is A 10.3 seconds/dayT=2π√lg△TT=2π×12(l/g)−1/2×△l/l(∵T+△T=2π√l+△lg)∴△TT=12△ll=12∝△θ=12×12×10−6×(40−20)=12×10−5△T=T×12×10−5=24×60×60×10−5=10.3 s/day

A clock which keeps correct time at $20^{\circ} C$ , is subjected to $40^{\circ} C$ . If coefficient of linear expansion of the pendulum is $12 \times 10^{- 6}$ per $^{\circ} C$ . How much will it gain or loose in time?