A pendulum clock gains 5s a day if the temperature is 15° C and loses 10s a day if the temperature is 30° C. What will be value of x if α =(x/T) . here α is the coefficient of linear expansion of the metal of the pendulum shaft and (.T.) is the time period of the pendulum.

Question

A pendulum clock gains 5s a day if the temperature is 15° C and loses 10s a day if the temperature is 30° C. What will be value of x if α =(x/T) . here α is the coefficient of linear expansion of the metal of the pendulum shaft and (.T.) is the time period of the pendulum. (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

Fractional change in time period of pendulum is,

\frac{Δ T}{T} = \frac{1}{2} α Δ t

Given, when

t_{1} = 1 5^{\circ} C, Δ T = 5 s

For linear expansion,

\frac{Δ T}{T} = \frac{1}{2} α (t_{0} - t_{1})

∴ \frac{5}{T} = \frac{1}{2} α (t_{0} - 15)

\dots . (i)

Similarly, when

t_{2} = 3 0^{\circ} C, Δ T = 10 s

∴ \frac{10}{T} = \frac{1}{2} α (30 - t_{0}) .... (ii)

Dividing equation

(ii)

by

(i)

,

\frac{10}{5} = \frac{( 30 - t _{0} )}{( t _{0} - 15 )}

∴ 2 t_{0} - 30 = 30 - t_{0}

∴ t_{0} = 2 0^{\circ} C

Substituting it in equation

(i)

,

\frac{5}{T} = \frac{1}{2} α (20 - 15)

∴ α = \frac{2}{T}

∴ x = 2

Q. A pendulum clock gains 5s a day if the temperature is 15∘C and loses 10s a day if the temperature is 30∘C. What will be value of x if α=Tx​ . here α is the coefficient of linear expansion of the metal of the pendulum shaft and (T) is the time period of the pendulum.