JEE Main · 2021 · 24 Feb 2021 · Shift 2 · Units & Measurement

The period of oscillation of a simple pendulum is

T = 2 π \frac{L}{g}

. Measured value of '

L

' is 1.0 m from meter scale having a minimum division of 1 mm and time of one- complete oscillation is 1.95 s measured from stopwatch of 0.01 s resolution. The percentage error in the determination of '

g

' will be:

(A) $1.33%$
(B) $1.13%$ ✓ Correct
(C) $1.03%$
(D) $1.30%$

Answer: (B)

Solution

With

g = \frac{4 π ^{2} L}{T ^{2}}

the fractional errors combine as

\frac{Δ g}{g} = \frac{Δ L}{L} + 2 \frac{Δ T}{T}

The least count of the metre scale fixes

Δ L = 1 mm = 0.001

m against

L = 1.0

m, so

\frac{Δ L}{L} = 0.001

. The stopwatch resolution fixes

Δ T = 0.01

s against

T = 1.95

s, so

\frac{Δ T}{T} = 0.00513

. The time enters squared, so its error is counted twice:

\frac{Δ g}{g} = 0.001 + 2 (0.00513) = 0.01126

The percentage error in the determination of

g

1.13%

JEE Main · 2021 · 24 Feb 2021 · Shift 2 · Single correct option

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JEE Main · 2021 · 24 Feb 2021 · Shift 2 · Units & Measurement

Solution

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