A uniform metal rod is used as a bar pendulum. If the room temperature rises by $10 ° C$ , and the coefficient of linear expansion of the metal of the rod is $2 \times 10^{- 6}$ per $° C$ , the period of the pendulum will have percentage increase of

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$- 2 \times 10^{- 3}$

$- 1 \times 10^{- 3}$

$2 \times 10^{- 3}$

$1 \times 10^{- 3}$

answer is D.

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Detailed Solution

Fractional change in period
$\frac{Δ T}{T} = \frac{1}{2} α Δ θ = \frac{1}{2} \times 2 \times 10^{- 6} \times 10 = 10^{- 5}$
$% change = \frac{Δ T}{T} \times 100 = 10^{- 5} \times 100 = 10^{- 3} %$

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