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Physics · Units, Dimensions & Error Analysis

JEE Main 2024 — 31 January, Shift 2 — Question 31

The measured value of the length of a simple pendulum is 20 cm with 2 mm accuracy. The time for 50 oscillations was measured to be 40 seconds with 1 second resolution. From these measurements, the accuracy in the measurement of acceleration due to gravity is N%. The value of N is:

  1. Option A:

    4

  2. Option B:

    8

  3. Option C:

    6

    Correct
  4. Option D:

    5

Answer: C

Step-by-step solution

We know that the time period is

T=2πℓgT = 2\pi \sqrt{\frac{\ell}{g}}

So,

g=4π2ℓT2g = \frac{4\pi^{2}\ell}{T^{2}}

Now, error propagation gives

Δgg=Δℓℓ+2ΔTT\frac{\Delta g}{g} = \frac{\Delta \ell}{\ell} + \frac{2\Delta T}{T} =0.220+2(140)= \frac{0.2}{20} + 2\left(\frac{1}{40}\right) =1.220= \frac{1.2}{20}

Percentage change is

1.220×100=6%\frac{1.2}{20} \times 100 = 6\%

Answer key and solution verified before publishing.

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Exam
JEE Main 2024
Paper
31 January, Shift 2
Subject
Physics
Chapter
Units, Dimensions & Error Analysis
Topic
Significant Figures and Error Analysis

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