A simple pendulum made of a bob of mass m and a metallic wire of a negligible mass has a time period of 2s at T = 0 ° C . If the temperature of the wire is increased, and the corresponding change in its time period is

Correct answer: C. α (△ T)/(T) = (1)/(2)(△ l)/(l) = (1)/(2)α ∆ θ ⇒∆ T = (1)/(2)Tα ∆ θ ∆ T ∆θ = (T)/(2)α But, T = 2s ∆T ∆θ = α

JEE Main 20169 AprilOnline

Exam: JEE Main
Year: 2016
Date: 9 April
Mode: Online
Subject: Physics
Chapter: Thermal Properties of Matter

JEE Main 2016 · 9 April · Online · Physics · Thermal Properties of Matter

A simple pendulum made of a bob of mass m and a metallic wire of a...

A simple pendulum made of a bob of mass m and a metallic wire of a negligible mass has a time period of

2s

T = 0 ° C

. If the temperature of the wire is increased, and the corresponding change in its time period is plotted against its temperature, the resulting graph is a line of slope

S

. If the coefficient of linear expansion of metal is

\alpha

, then the value of

S

Options

A $\dfrac{\alpha}{2}$
B $2\alpha$
C $\alpha$
D $\dfrac{1}{\alpha}$

Correct answer

C. $\alpha$

Step-by-step solution

$\dfrac{△ T}{T} = \dfrac{1}{2}\dfrac{△ l}{l} = \dfrac{1}{2}\alpha ∆ \text{\theta}$

$⇒∆ T = \dfrac{1}{2}T\alpha ∆ \text{\theta}$

$\dfrac{∆ T}{∆\text{\theta}} = \dfrac{T}{2}\alpha$

But,

$T = 2s$

$\dfrac{∆T}{∆\text{\theta}} = \alpha$

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