Question

When the temperature increases the length of the pendulum also increases and hence its time period increases. Thus it losses the time when temperature is increased.

• A. True
• B. False

Answer 1

The correct option is A True

A pendulum clock keeps proper time at temperature ${\theta }_o$. If temperature is increased to $\theta (>{\theta }_o)$, then due to linear expansion, length of pendulum increases and hence its time period will increase.

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

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

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

$\therefore$ change in time per unit time lapsed

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

Fractional change in time period $\frac{△T}{T}=\frac{1}{2}\alpha △\theta$

So, as the temperature increases, length of pendulum increases and hence time period of pendulum increases. Due to increment in its time period, a pendulum dock become slow in summer and will lose time.