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PhysicsSHMMiscellaneousEasy2 minPYQ_2020

PhysicsEasysingle choice

A second's pendulum clock having steel wire is calibrated at $20 ° C$ . When the temperature is increased to $30 ° C$ , how much times does the clock lose or gain in the week? $[α_{s t e e l} = 1 .2 \times 10^{- 5} ° C^{- 1}]$

Options:

Answer:

Solution:

$T = 2 π \sqrt{\frac{l}{g}}$

$\frac{ΔT}{T} = \frac{1}{2} \frac{Δl}{l}$

$\frac{ΔT}{T} = \frac{1}{2} αΔθ$

$ΔT = \frac{1}{2} αΔθT$

$ΔT = \frac{1}{2} (1 .2 \times 10^{- 5}) (30 - 20) (3600 \times 24 \times 7)$

$ΔT = 36 .28 s$

Stream:NTA_ABHYAS_NEETSubject:PhysicsTopic:SHMSubtopic:Miscellaneous

⏱ 2mℹ️ Source: PYQ_2020

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