Question

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

• A.
• B.
• C.
• D.

Answer 1

The correct option is B

The gain/loss of time will be $(T^{\prime }-T)$ seconds. For example, what the clock was measuring to be 1 second will now be shown as $1(1+\frac{1}{2}a\Delta \theta )$ seconds. Which is, to say, a gain of $\left[1\right(1+\frac{1}{2}a\Delta \theta )-1]$ seconds, or, $\frac{1}{2}a\Delta \theta$seconds, has occurred. Thus, for $T$ seconds,the clock will show a gain of $T\times \frac{1}{2}a\Delta \theta$ seconds.

$T^{\prime }$ = $T(1+\frac{1}{2}a\Delta \theta )$

$\Rightarrow$ $T^{\prime }$ = $T+\frac{1}{2}aT\Delta \theta$

Now, a day has $24\times 60\times 60seconds$ = $86,400seconds$.

You might have figured the answer now.

Since, we are taking the measure at a temoerature ${\theta }_{final}$ = $t^{\circ }C$, and the clock is known to give a correct reading at ${\theta }_{initial}$ = $0^{\circ }C$, $\Delta \theta$ = $t^{\circ }C$. Also, $T$ = $86,400seconds$.