JEE Main · 2021 · 26 Aug 2021 · Shift 2 · Elasticity & SHM

If the length of the pendulum in pendulum clock increases by

0.1%

, then the error in time per day is

Answer: (B)

Solution

The time period is

T = 2 π \frac{ℓ}{g}

, so

T \propto ℓ

and

\frac{Δ T}{T} = \frac{1}{2} \frac{Δ ℓ}{ℓ} = \frac{1}{2} (\frac{0.1}{100}) = 5 \times 1 0^{- 4}

The clock loses this fraction of every interval it measures, so over one day of

86400

s the accumulated error is

Δ t = 5 \times 1 0^{- 4} \times 86400 = 43.2 s

JEE Main · 2021 · 26 Aug 2021 · Shift 2 · Single correct option