JEE Advanced · 2010 · 11 Apr 2010 · Paper 1 · Units & Measurement

A student uses a simple pendulum of exactly 1m length to determine

g

, the acceleration due to gravity. He uses a stop watch with the least count of 1 sec for this and records 40 seconds for 20 oscillations. For this observation, which of the following statement(s) is (are) true ?

(A)Error $Δ T$ in measuring T, the time period, is 0.05 seconds✓ Correct
(B)Error $Δ T$ in measuring T , the time period, is 1 second
(C)Percentage error in the determination of $g$ is $5%$ ✓ Correct
(D)Percentage error in the determination of $g$ is 2.5%

Answer: (A), (C)

Solution

Time for 20 oscillations is

t = 40

s, so time period

T = 40/20 = 2

s. The least count of the stopwatch is 1 s, so the error in measuring

t

Δ t = 1

s. The error in

T

Δ T = Δ t /20 = 1/20 = 0.05

s (option A is correct). For a pendulum,

g = 4 π^{2} L / T^{2}

. The percentage error in

g

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

. With

L

exact (

Δ L = 0

\frac{Δ g}{g} = 2 \times \frac{0.05}{2} = 2 \times 0.025 = 0.05 = 5%

(option C is correct). Correct answers: (A) and (C).

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JEE Advanced · 2010 · 11 Apr 2010 · Paper 1 · Units & Measurement

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