Question

A pendulum clock shows correct time at 20°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–1. At what temperature will the clock show correct time? Coefficient of linear expansion of steel = 12 × 10–6 °C–1.

Answer 1

$\huge{\boxed{\mathfrak\pink{\fcolorbox{red}{purple}{Mr\:Awesome}}}}$

❤.option B is correct

Answer 2

The clock show correct time  at temperatures 82℃

Explanation:

The temperature at which the pendulum shows the proper time , $T_1$ = 20 °C

Linear expansion coefficient for steel,

$\alpha=12 \times 10^{-6} \cdot \mathrm{C}^{-1}$

Let  $\mathrm{T}_{2}$  be the temperature the value of g is  9.788$ms^{-2 }$ and ΔT be  Temperature change.

Thus, the pendulum time periods at different g values will be $t_1$ and $t_2$;

$T_{1}=2 \pi \frac{\sqrt{l_{1}}}{g_{1}}$

$T_{2}=2 \pi \frac{\sqrt{l_{2}}}{g_{2}}$

$=2 \pi \frac{\sqrt{l_{1}(1+\Delta T)}}{g}$

$= 2 \pi \frac{\sqrt{l_{2}}}{g_{2}}$

$=2 \pi \frac{\sqrt{l_{1}(1+\Delta T)}}{g}$

$\frac{l_{1}}{g_{1}}=\frac{l_{1}(1+\Delta T)}{g_{2}}$

$\frac{1}{9.8}=\frac{\left(1+12 \times 10^{-6} \times \Delta T\right)}{9.788}$

$\frac{9.788}{9.8}=1+12 \times 10^{-6} \times \Delta T$

$\frac{9.788}{9.8}-1=12 \times 10^{-6} \times \Delta T$

$0.9987-1=12 \times 10^{-6} \times \Delta T$

$-0.00122=12 \times 10^{-6} \times \Delta T$

$\Delta T=\frac{12 \times 10^{-6}}{-0.00122}$

$T_{2}-20=-101.6$

$T_{2}=-101.6+20$

$T_{2}=-81.6 \approx 82^{\circ} \mathrm{C}$