Three students S _ 1 , S _ 2 and S _ 3 perform an experiment for...

Three students ${S}_{1}, {S}_{2}$ and ${S}_{3}$ perform an experiment for determining the acceleration due to gravity $( g )$ using a simple pendulum. They use different lengths of pendulum and record time for different number of oscillations. The observations are as shown in the table.

Student No.	Length of pendulum $( cm )$	Number of oscillations $( n )$	Total time for $n$ oscillations	Time period $( s )$
$1 .$	$64 . 0$	$8$	$128 . 0$	$16 . 0$
$2 .$	$64 . 0$	$4$	$64 . 0$	$16 . 0$
$3 .$	$20 . 0$	$4$	$36 . 0$	$9 . 0$

(Least count of length $= 0 . 1m$ , least count for time $= 0 . 1s$ )

If ${E}_{1}, {E}_{2}$ and ${E}_{3}$ are the percentage errors in $g$ for students $1, 2$ and $3$ , respectively, then the minimum percentage error is obtained by student no .

Correct answer

Step-by-step solution

$T = 2\pi\sqrt{\dfrac{ℓ}{g}} ⇒ g = \dfrac{4{\pi}^{2}ℓ}{{T}^{2}}$
$\dfrac{\Deltag}{g} = \dfrac{\Deltaℓ}{ℓ} + \dfrac{2 ∆ T}{T}$

$\DeltaT = \dfrac{leastcountoftime\left(\Delta{T}_{0}\right)}{numberofoscillations ( n )}$
$\dfrac{\Deltag}{g} = \dfrac{\Deltaℓ}{ℓ} + \dfrac{2\Delta{T}_{0}}{nT}$

As $\Deltaℓ$ and $\Delta{T}_{0}$ are same for all observations so
$\dfrac{\Deltag}{g}$ is minimum for highest value of $ℓ, n$ and $T$

$⇒$ Minimum percentage error in $g$ is for student number 1.

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