JEE Main · 2021 · 18 Mar 2021 · Shift 1 · Units & Measurement

The time period of a simple pendulum is given by

T = 2 π \frac{ℓ}{g}

. The measured value of the length of pendulum is 10 cm known to a 1 mm accuracy. The time for 200 oscillations of the pendulum is found to be 100 second using a clock of 1 s resolution. The percentage accuracy in the determination of '

g

' using this pendulum is '

x

'. The value of '

x

' to the nearest integer is,

(A) $4%$
(B) $3%$ ✓ Correct
(C) $2%$
(D) $5%$

Answer: (B)

Solution

Rearranging the given relation,

g = \frac{4 π ^{2} ℓ}{T ^{2}}

, so the fractional errors combine as

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

The length is

10

cm known to

1

mm, giving

\frac{Δ ℓ}{ℓ} = \frac{0.1}{10} = 1%

. The clock reads

100

s for

200

oscillations with a resolution of

1

s; the number of oscillations is counted exactly, so the fractional error in the period equals that in the total time,

\frac{Δ T}{T} = \frac{1}{100} = 1%

. Hence

\frac{Δ g}{g} = 1% + 2 \times 1% = 3%

The percentage accuracy in the determination of

g

3%

JEE Main · 2021 · 18 Mar 2021 · Shift 1 · Single correct option

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JEE Main · 2021 · 18 Mar 2021 · Shift 1 · Units & Measurement

Solution

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