Three students S1, S2 and S3 perform an experiment for determining… | JEE Main 2021 (22 Jul, Shift 2)

Q84JEE Main 2021 · 22 JUL · SHIFT 2Difficult

Errors in Measurement

Three students S1, S2 and S3 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.

(Least count of length = 0.1 cm and Least count for time = 0.1 s)

If E1, E2 and E3 are the percentage errors in 'g' for students 1, 2 and 3 respectively, then the minimum percentage error is obtained by student no. ______________.

Numerical answer type. Work it out, then open the answer.

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Answer: 1

$g = \frac{4\pi^2 L}{T^2} \Rightarrow \frac{\Delta g}{g} = \frac{\Delta L}{L} + 2\frac{\Delta t}{t}$ ( $t$ = total time measured).
S1: $\frac{0.1}{64} + 2\left(\frac{0.1}{128}\right) = 0.156\% + 0.156\% = 0.31\%$
S2: $\frac{0.1}{64} + 2\left(\frac{0.1}{64}\right) = 0.156\% + 0.313\% = 0.47\%$
S3: $\frac{0.1}{20} + 2\left(\frac{0.1}{36}\right) = 0.5\% + 0.56\% = 1.06\%$
Minimum error: student 1. Answer: 1.

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Errors in Measurement question from JEE Main 2021 (22 Jul, Shift 2)

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