A clock is calibrated at a temperature of $20^{0} C$ . Assume that the pendulum is a thin brass rod of negligible mass with a heavy bob attached to the end $(α_{b r a s s} = 19 \times 10^{- 6} / K)$

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On a hot day at $30^{0} C$ the clock gains 8.2 s

On a hot at $30^{0} C$ the clock loses 8.2 s

On a cold day at $10^{0} C$ the clock gains 8.2 s

On a cold day at $10^{0} C$ the clock loses 8.2 s

answer is B, C.

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Detailed Solution

$T = 2 π \sqrt{\frac{l}{g}} = 2 π \sqrt{\frac{l_{0} + α l_{0} Δ θ_{0}}{g}}$

$= T_{0} (1 + \frac{1}{2} α Δ θ)$

At $30^{0} C$ , fraction loss of time $= \frac{T_{30^{0}} - T_{20^{0}}}{T_{20^{0}}}$
$= 5 α = 5 \times 19 \times 10^{- 6}$
Time lost in 24 hours $= 86400 \times 95 \times 10^{- 6} = 8.2 s$
On a cold day at $10^{0} C$ , fraction gain of time $= \frac{T_{10^{0}} - T_{20^{0}}}{T_{20^{0}}} = 5 α$
Time gained in 24 hours = 8.2 s