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Physics Question on Thermal Expansion

A clock pendulum made of invar has a period of 0.5 s0.5 \, s, at 20∘C20^{\circ}C. If the clock is used in a climate where the temperature averages to 30∘C30^{\circ}C , how much time does the clock lose in each oscillation? (For invar, a=9×10−7/∘C,ga = 9 \times 10^{-7} /^{\circ} C, g = constant)

A

2.25×10−6s2.25 \times 10^{-6} s

B

2.5×10−7s2.5 \times 10^{-7} s

C

5×10−7s5\times 10^{-7} s

D

1.125×10−6s1.125 \times 10^{-6} s

Answer

2.25×10−6s2.25 \times 10^{-6} s

Explanation

Solution

Time period of oscillation,
T=2π1gT = 2\pi \frac{1}{g}
⇒dTT=12dll\Rightarrow \frac{dT}{T} = \frac{1}{2} \frac{dl}{l}
As, dll=adt\frac{dl}{l} = adt
⇒dTT=12adt\Rightarrow \frac{dT}{T} = \frac{1}{2} adt
=12×9×10−7×30−20= \frac{1}{2} \times 9 \times 10^{-7} \times 30 - 20
∴\therefore Loss in time =4.5×10−6 = 4.5\times 10^{-6}
=4.5×10−6×0.5= 4.5 \times 10^{-6}\times 0.5
=2.25×10−6s= 2.25 \times 10^{-6} s