A pendulum clock giving correct time at a place where $g = 9.8 m / s^{2}$ is taken to another place where it loses $24$ seconds during $24$ hours. Find the value of g at this new place.

Open in App

Solution

For the pendulum, $\frac{T_{1}}{T_{2}} = \sqrt{\frac{g_{2}}{g_{1}}}$
Given that, $T_{1} = 2 s e c, g_{1} = 9.8 m / s^{2}$
$T_{2} = \frac{24 \times 3600}{(\frac{24 \times 3600 - 24}{2})} = 2 \times \frac{3600}{3599}$

Now, $\frac{g^{2}}{g_{1}} = (\frac{T_{1}}{T_{2}})^{2}$

$∴ g_{2} = (9.8) (\frac{3599}{3600})^{2} = 9.795 m / s^{2}$

Suggest Corrections

Similar questions

Q. A pendulum clock is accurate at a place where

g = 9.8 m / s^{2}

. Find the value of g at another place where clock becomes slow by 24 seconds in a day (24 hrs).

Q. A seconds pendulum is shifted from a place where

g = 9.8 m / s^{2}

to another place where

g = 9.78 m / s^{2}

. The change in its length so that its time period of oscillation does not change is

Q. A pendulum clock gives correct time at

20^{o}

(at a place where

g = 10 m / s^{2}

) The pendulum consists of a light steel rod connected to a heavy ball. If it is taken to a different place

(w h e r e g = 10) 01 m / s^{2}

at what temperature the pendulum gives correct time

(α o f s t e e l i s 10^{- 5 /^{o} C})

[2007 E ]

Q. A pendulum gives correct time at

20^{o} C

at a place where

g = 9.8 m s^{- 2}

. The pendulum consists of a light steel rod connected to a heavy ball. If it is taken to a different place where

g = 9.788 m s^{- 2}

. At what temperature will it give correct time? Coefficient of linear expansion of steel is

{12 \times 10^{- 6}}^{o} C^{- 1}

A pendulum clock gives correct time at $20^{\circ} C$ at a place where g = 9.800 $m s^{- 2}$ . The pendulum consists of a light steel rod connected to a heavy ball. It is taken to a different place where g 9.788 $m s^{- 2}$ . At what temperature will it give correct time ? Coefficient of linear expansion of steel $= 12 \times 10^{- 6}^{\circ} C^{- 1}$

Join BYJU'S Learning Program

A pendulum clock giving correct time at a place where g=9.8m/s2 is taken to another place where it loses 24 seconds during 24 hours. Find the value of g at this new place.

For the pendulum, T1T2=√g2g1Given that, T1=2sec,g1=9.8m/s2T2=24×3600(24×3600−242)=2×36003599Now, g2g1=(T1T2)2∴g2=(9.8)(35993600)2=9.795m/s2

A pendulum clock giving correct time at a place where $g = 9.8 m / s^{2}$ is taken to another place where it loses $24$ seconds during $24$ hours. Find the value of g at this new place.