If the length of a clock pendulum increase by $0.2 %$ due to atmospheric temperature rise, then the loss in time of clock per day is

86.4

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

43.2

No worries! We‘ve got your back. Try BYJU‘S free classes today!

72.5

No worries! We‘ve got your back. Try BYJU‘S free classes today!

32.5

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A $86.4$ s
$T = 2 π \sqrt{\frac{l}{g}}$

$\frac{Δ T}{T} \times 100 = \frac{1}{2} \frac{Δ l}{l} \times 100$

$\frac{Δ T}{24 \times 3600} \times 100 = \frac{1}{2} \frac{0.2}{100} \times 100$

$Δ T = 86.4 s$

Suggest Corrections

Similar questions

A pendulum clock keeps correct time at $0^{\circ}$ . Its mean coefficient of linear expansions is $α /^{\circ}$ , then the loss in seconds per day by the clock if the temperature rises by $t^{\circ}$ is

Q. 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:-

Q. A pendulum clock loses

12 s

a day if the temperature is

40^{\circ} C

and goes fast by

4 s

a day if the temperature is

20^{\circ} C

. Find the temperature in

^{\circ} C

at which the clock will show correct time.

Q. If length of a pendulum of wall clock is increased by

0.1 %

, then find error in time per day.

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:

Join BYJU'S Learning Program

If the length of a clock pendulum increase by 0.2% due to atmospheric temperature rise, then the loss in time of clock per day is

The correct option is A 86.4sT=2π√lgΔTT×100=12Δll×100ΔT24×3600×100=120.2100×100ΔT=86.4s

If the length of a clock pendulum increase by $0.2 %$ due to atmospheric temperature rise, then the loss in time of clock per day is

The correct option is A $86.4$ s
$T = 2 π \sqrt{\frac{l}{g}}$

$\frac{Δ T}{T} \times 100 = \frac{1}{2} \frac{Δ l}{l} \times 100$

$\frac{Δ T}{24 \times 3600} \times 100 = \frac{1}{2} \frac{0.2}{100} \times 100$

$Δ T = 86.4 s$