Question

A pendulum clock loses 12 s a day if the temperature is 40 °C and gains 4 s a day if the temperature is 20 °C. The temperature at which the clock will show correct time, and the coefficient of linear expansion (a) of the metal of the pendulum shaft are respectively:

(A) 60 πC, a = 1.85 × 10^-4/πC

(B) 30 πC, a = 1.85 × 10^-3/πC

(C) 55 πC, a = 1.85 × 10^-2/πC

(D) 25 πC, a = 1.85 × 10^-5/πC

Answer 1

(D) 25 πC, a = 1.85 × 10^-5/πC

Period of pendulum, T = 2π\(\sqrt{\frac{L}{g}}\)

∴ T ∝ \(\sqrt{L}\)

But, L = L₀ (1 + α∆t)

As L₀ is constant,

⇒ T ∝ (1 + α ∆t)

Calculating fractional change in time period of pendulum, \(\frac{\triangle T}{T}=\frac{1}{2}\) (αΔt)

For the given pendulum,

T = 24 × 60 × 60 = 864000 s

When t₁ = 40 °C, ∆T = 12 s,

Where, t0 is temperature at which the clock will show correct time.