A student performs an experiment for determination of $\mathrm{g}\left(=\frac{4 \pi^{2} \ell}{\mathrm{~T}^{2}}\right), \ell \approx 1 \mathrm{~m}$, and he commits an error of $\Delta \ell$. For T he takes the time of n oscillations with the stop watch of least count $\Delta \mathrm{T}$ and he commits a human error of 0.1sec. For which of the following data, the measurement of g will be most accurate? | | $\Delta \ell$ | $\Delta \mathrm{T}$ | n | Amplitude of oscillation | | --- | --- | --- | --- | --- | | (A) | 5 mm | 0.2 sec | 10 | 5 mm | | (B) | 5 mm | 0.2 sec | 20 | 5 mm | | (C) | 5 mm | 0.1 sec | 20 | 1 mm | | (D) | 1 mm | 0.1 sec | 50 | 1 mm |

JEE Advanced · 2006 · 09 Apr 2006 · Units & Measurement

A student performs an experiment for determination of

g (= \frac{4 π ^{2} ℓ}{T ^{2}}), ℓ \approx 1 m

, and he commits an error of

Δ ℓ

. For T he takes the time of n oscillations with the stop watch of least count

Δ T

and he commits a human error of 0.1sec. For which of the following data, the measurement of g will be most accurate?

	$Δ ℓ$	$Δ T$	n	Amplitude of oscillation
(A)	5 mm	0.2 sec	10	5 mm
(B)	5 mm	0.2 sec	20	5 mm
(C)	5 mm	0.1 sec	20	1 mm
(D)	1 mm	0.1 sec	50	1 mm

(A)5 mm, 0.2 sec, 10, 5 mm
(B)5 mm, 0.2 sec, 20, 5 mm
(C)5 mm, 0.1 sec, 20, 1 mm
(D)1 mm, 0.1 sec, 50, 1 mm✓ Correct

Answer: (D)

Solution

For

g = \frac{4 π ^{2} ℓ}{T ^{2}}

, the fractional error is:

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

where

T

is the time period for one oscillation. If we measure time for n oscillations,

T = \frac{t}{n}

where t is the total time measured.

The error in T is:

Δ T = \frac{Δ t}{n}

where

Δ t

includes both the least count and human error.

For most accurate measurement, we need:

Smallest $Δ ℓ$ (option D has 1 mm vs 5 mm in others)
Largest n to minimize $\frac{Δ T}{T}$ (option D has n=50)
Smallest amplitude to ensure SHM approximation (option D has 1 mm)
Smallest least count (options C and D have 0.1 sec)

Option (D) satisfies all criteria for maximum accuracy: smallest length error, largest number of oscillations, smallest amplitude, and smallest least count.

Therefore, the answer is (D).

JEE Advanced · 2006 · 09 Apr 2006 · Single correct option

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JEE Advanced · 2006 · 09 Apr 2006 · Units & Measurement

Solution

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