Question

The pendulum of a clock is made of brass. If the clock keeps correct time at ${20}^{\circ }C$ .Calculate how many seconds per day will it loose at ${35}^{\circ }C$ $(\alpha$ for brass =$2\times$${10}^{-5}$${}^{o}C$)

• A. 12.96 s
• B. 1.29 s
• C. 129.6 s
• D. 8.64 s

Answer 1

The correct option is A 12.96 s

Variation in time period w.r.t. temperature is given by,

Let, $T=2\pi \sqrt{\frac{L_o}{g}}$ at temperature ${\theta }_o$

and, $T^{\prime }=2\pi \sqrt{\frac{L}{g}}$ at temperature $\theta$

$\frac{T^{\prime }}{T}=\sqrt{\frac{L^{\prime }}{L}}=\sqrt{\frac{L(1+\alpha △\theta )}{L}}=1+\frac{1}{2}\alpha △\theta$

Therefore, loss or gain per unit time lapsed is

$\frac{T^{\prime }-T}{T}=\frac{1}{2}\alpha △\theta$

gain or loss in time in duration of $t$ in

So, $△t=\frac{1}{2}\alpha △\theta t\left(1\right)$

Now, in the question $△\theta ={15}^{o}C\alpha =2\times {10}^{-5}{/}^{o}Ct=oneday=24\times 60\times 60s$

So, putting values in equation $\left(1\right)$

$△t=\frac{1}{2}\times 2\times {10}^{-5}\times 15\times 24\times 60\times 60△t=12.96s$

So, it will gain $12.96s$ per day at ${35}^{o}C$.