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Question
A torsional pendulum consists of a solid disc connected to a thin wire (α = 2.4 × 10–5°C–1) at its centre. Find the percentage change in the time period between peak winter (5°C) and peak summer (45°C).
Solution
Given:
Coefficient of linear expansion of the wire, α = 2.4 × 10–5 °C–1
Let I0 be the moment of inertia of the torsional pendulum at 0 °C.
If K is the torsional constant of the wire, then time period of torsional pendulum (T):
`T = 2pi sqrt(1/K)`...(1)
Here, I = moment of inertia after change in temperature
When the temperature is changed by Δθ, moment of inertia (I),
l = l0(1+2αΔθ)
On substituting the value of I in equation(1), we get:
`T = 2 pi sqrt( I_0(1+2αΔθ))/sqrt K`
In winter , Δθ = 5 °C
∴ Time period , (T1)
= `2pi sqrt( I_0(1+2αΔθ))/sqrt K`
In summer , Δθ = 45 °C
Time period (T2)
`=2pi sqrt( I_0(1+2αΔθ))/sqrt K`
So,
`T_2/T_1 = sqrt(1+90α)/sqrt(1+10α)`
=`sqrt(1+90 xx 2.4 xx 10^-5)/sqrt(1+10 xx 2.4 xx 10^-5)`
`=> T_2/T_1 = sqrt(1.00216)/sqrt(1.00024)`
% change = `(T_2/T_1 -1) xx 100`
= 0.0959 %
⇒ % change in time period ≈ 9.6 × 10-2 %
Therefore, the percentage change in time period of a torsional pendulum between peak winters and peak summers is 9.6 × 10–2 % .
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