Test Index

Units and Measurements Part 1

Section: Physics

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Question : 26 of 100

Marks: +1, -0

Three students

S_{1}, S_{2}

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 observationsare as shown in the table.

Student No.	Length of pendulum (cm	No. 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 = 0.1 m

least count for time = 0.1 s)

E_1, E_2

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._____.

[22 Jul 2021 Shift 2]

Your Answer:

Solution:

T = 2\pi \sqrt{\frac{l}{g}} \Rightarrow g = \frac{4\pi^{2} l}{T^{2}}

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

\Delta T = \frac{\text{least count of time }(\Delta T_0)}{\text{number of oscillations(n) }}

\frac{\Delta g}{g} = \frac{\Delta l}{l} + \frac{2 \Delta T_0}{n T}

As

\Delta l

and

\Delta T_0

are same for all observations so

\frac{\Delta g}{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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