A simple pendulum made of a bob of mass $m$ and a metallic wire of negligible mass has time period $2 s$ at $T = 0$ . 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 $α$ then the value of $S$ is

2 α

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\frac{1}{α}

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α

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\frac{α}{2}

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Solution

The correct option is C $α$
Time period of a pendulum is given by
$T = 2 π \sqrt{\frac{l_{o}}{g}}$
where $l_{o} :$ length and $g :$ acceleration due to gravity
$T^{2} = \frac{4 π^{2} l_{o}}{g}$

At temperature $θ$ , length $= l_{o} (1 + α (θ - θ_{o}))$
$∴ T^{2} = \frac{4 π^{2} l_{o} (1 + α (θ - θ_{o}))}{g}$

Differentiating with respect to temperature,
$2 T \frac{d T}{d θ} = \frac{4 π^{2} l_{o} α}{g}$
$\frac{d T}{d θ} = α T^{2} / 2 T = α T / 2 = α$

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The correct option is C αTime period of a pendulum is given by T=2π√logwhere lo: length and g: acceleration due to gravityT2=4π2logAt temperature θ, length =lo(1+α(θ−θo)) ∴T2=4π2lo(1+α(θ−θo))gDifferentiating with respect to temperature,2TdTdθ=4π2loαgdTdθ=αT2/2T=αT/2=α