If length of a pendulum of wall clock is increased by $0.1 %$ , then find error in time per day.

Open in App

Solution

Let, length of pendulum = L

Time period $T = \sqrt{\frac{L}{g}} = 1 sec$

Now, time in a day

$t = 1 \times 60 \times 60 \times 24$

Now, after increasing length o.1%

New length

$L^{'} = L + 0.001 L$

Now, new time period

$T^{'} = \sqrt{\frac{L^{'}}{g}}$

$T^{'} = \sqrt{\frac{1.001 L}{g}}$

$T^{'} = \sqrt{1.001} \times \sqrt{\frac{L}{g}}$

$T^{'} = \sqrt{1.001} s$

Now, time in a day

$t^{'} = \sqrt{(1.001)} \times 60 \times 60 \times 24$

Now, error in a day

$= \sqrt{(1.001)} \times 60 \times 60 \times 24 - 1 \times 60 \times 60 \times 24$

$= 60 \times 60 \times 24 (\sqrt{(1.001)} - 1)$

$= 60 \times 60 \times 24 (1.0005 - 1)$

$= 60 \times 60 \times 24 \times .0005$

$= 43.2 s$

Hence, the error in time per day is $43.2 s$

Suggest Corrections

Similar questions

Q. A pendulum clock shows accurate time. If the length increases by 0.1 %, deduce the error in time per day.

Q. 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

Q. A clock regulated by a seconds pendulum keeps correct time. During summer the length of the pendulum increases to

1.01 m

. How much will the clock gain or lose in one day?

(g = 9.8 m / s^{2})

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. If the length of seconds pendulum is increased by

2 %

, how many seconds it will lose per day

Join BYJU'S Learning Program

If length of a pendulum of wall clock is increased by 0.1%, then find error in time per day.

If length of a pendulum of wall clock is increased by $0.1 %$ , then find error in time per day.