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Standard XII

Physics

Radial & Tangential Acceleration for Non Uniform Circular Motion

A clock, with...

Question

A clock, with a brass pendulum, keeps correct time at $20^{\circ} C$ , but losses $8.212 s$ per day, when the temperature rises to $30^{\circ} C$ . The coefficient of linear expansion of brass is:-

25 \times 10^{- 6} /^{\circ} C

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19 \times 10^{- 6} /^{\circ} C

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20 \times 20^{- 6} /^{\circ} C

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11 \times 10^{- 6} /^{\circ} C

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Solution

The correct option is B $19 \times 10^{- 6} /^{\circ} C$
Loss of time in a day= $\frac{1}{2} \times α \times (t_{2} - t_{1}) \times 86400$
So, $8.212 = 1 / 2 \times α \times 10 \times 86400$
Answer is $19 \times 10^{- 16} /^{o} C$

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Similar questions

Q. A clock, with a brass pendulum, keeps correct time at

20^{0} C

, but loses

8.212 s

per day, when the temperature rises to

30^{C}

. Calculate the coefficient of linear expansion of brass:

Q. A thin brass sheet at

10^{\circ} C

and thin steel sheet at

20^{\circ} C

have the same surface area.The common temperature at which both would have the same area : (Coefficient of linear expansion for brass and steel are respectively

19 \times 10^{- 6} /^{\circ} C

and

11 \times 10^{- 6} /^{\circ} C .

)

Q. A clock while keeps correct time at

30^{\circ} C

has a pendulum rod made of brass. The number of seconds it gains (or) looses per second when the temperature falls to

10^{\circ} C

[α

of brass =

18 \times 10^{- 6} /^{\circ} C]

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

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 $\frac{12 \times 10^{- 6}}{^{\circ} C}$ . How much time will it gain or lose per day

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A clock, with a brass pendulum, keeps correct time at 20∘C, but losses 8.212s per day, when the temperature rises to 30∘C. The coefficient of linear expansion of brass is:-

The correct option is B 19×10−6/∘CLoss of time in a day= 12×α×(t2−t1)×86400So, 8.212=1/2×α×10×86400Answer is 19×10−16/oC

A clock, with a brass pendulum, keeps correct time at $20^{\circ} C$ , but losses $8.212 s$ per day, when the temperature rises to $30^{\circ} C$ . The coefficient of linear expansion of brass is:-

The correct option is B $19 \times 10^{- 6} /^{\circ} C$
Loss of time in a day= $\frac{1}{2} \times α \times (t_{2} - t_{1}) \times 86400$
So, $8.212 = 1 / 2 \times α \times 10 \times 86400$
Answer is $19 \times 10^{- 16} /^{o} C$