A pendulum clock (fitted with a small heavy bob that is connected with a metal rod) is 5 seconds fast each day at a temperature of 15° C and 10 seconds slow at a temperature of 30° C. The temperature at which it is designed to give correct time, is

Question

A pendulum clock (fitted with a small heavy bob that is connected with a metal rod) is 5 seconds fast each day at a temperature of 15° C and 10 seconds slow at a temperature of 30° C. The temperature at which it is designed to give correct time, is (A) 18° C (B) 20° C (C) 24° C (D) 25° C

Tardigrade Admin · Accepted Answer

Fractional loss of time per second =(1/2) α Δ T Therefore (1/2) α(T0-15) ×(24 hrs )=5 and (1/2) α(30-T0) ×(24 hrs )=10 on solving T0=20° C

Q. A pendulum clock (fitted with a small heavy bob that is connected with a metal rod) is 5 seconds fast each day at a temperature of $1 5^{\circ} C$ and 10 seconds slow at a temperature of $3 0^{\circ} C$ . The temperature at which it is designed to give correct time, is

$1 8^{\circ} C$

$2 0^{\circ} C$

$2 4^{\circ} C$

$2 5^{\circ} C$

Fractional loss of time per second $= \frac{1}{2} α Δ T$
Therefore $\frac{1}{2} α (T_{0} - 15) \times (24 h rs) = 5$
and $\frac{1}{2} α (30 - T_{0}) \times (24 h rs) = 10$
on solving $T_{0} = 2 0^{\circ} C$

Q. A pendulum clock (fitted with a small heavy bob that is connected with a metal rod) is 5 seconds fast each day at a temperature of 15∘C and 10 seconds slow at a temperature of 30∘C. The temperature at which it is designed to give correct time, is

18∘C

20∘C

24∘C

25∘C