JEE PYQ: Units & Measurements - Question ID 99a39f47ef60 (JEE Main 2024)

ID: 99a39f47ef60JEE Main 2024Single Correct MCQ

The measured value of the length of a simple pendulum is 20 cm20 \mathrm{~cm} with 2 mm2 \mathrm{~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%\mathrm{N} \%. The value of N\mathrm{N} is:

Select Option

Step-by-step Explanation

Core Formula & Concept:

A simple pendulum’s time period TT is related to its length LL and the acceleration due to gravity gg by the formula T=2πLg.T = 2\pi \sqrt{\frac{L}{g}}. Rearranging for gg gives g=4π2LT2.g = \frac{4\pi^2 L}{T^2}. When quantities are measured with uncertainties, the relative (percentage) uncertainty in gg is found by combining the relative uncertainties in LL and TT using the rules of error propagation for multiplication/division: Δgg=ΔLL+2ΔTT.\frac{\Delta g}{g} = \frac{\Delta L}{L} + 2\,\frac{\Delta T}{T}. The percentage accuracy in gg is then 100×Δgg100 \times \frac{\Delta g}{g}.

Step-by-Step Derivation:

1. Identify measured values and their uncertainties
Length L=20 cm=0.20 mL = 20\ \text{cm} = 0.20\ \text{m}, with accuracy ΔL=2 mm=0.002 m\Delta L = 2\ \text{mm} = 0.002\ \text{m}.
Time for 50 oscillations t50=40 st_{50} = 40\ \text{s}, with resolution Δt50=1 s\Delta t_{50} = 1\ \text{s}.
Therefore the period of one oscillation is T=t5050=4050=0.80 s,T = \frac{t_{50}}{50} = \frac{40}{50} = 0.80\ \text{s}, and its uncertainty is ΔT=Δt5050=150=0.02 s.\Delta T = \frac{\Delta t_{50}}{50} = \frac{1}{50} = 0.02\ \text{s}.

2. Compute relative uncertainties
Relative uncertainty in length: ΔLL=0.0020.20=0.01=1%.\frac{\Delta L}{L} = \frac{0.002}{0.20} = 0.01 = 1\%. Relative uncertainty in period: ΔTT=0.020.80=0.025=2.5%.\frac{\Delta T}{T} = \frac{0.02}{0.80} = 0.025 = 2.5\%.

3. Propagate uncertainties to gg
From the formula g=4π2LT2g = \frac{4\pi^2 L}{T^2}, the relative uncertainty in gg is Δgg=ΔLL+2ΔTT=0.01+2×0.025=0.01+0.05=0.06.\frac{\Delta g}{g} = \frac{\Delta L}{L} + 2\,\frac{\Delta T}{T} = 0.01 + 2\times 0.025 = 0.01 + 0.05 = 0.06.

4. Convert to percentage accuracy
Percentage accuracy in gg is 100×Δgg=100×0.06=6%.100 \times \frac{\Delta g}{g} = 100 \times 0.06 = 6\%. Hence N=6N = 6, which corresponds to option A.

Common Traps & Exam Tip:

1. Unit mismatch: Students often forget to convert all lengths to the same unit (meters or centimeters). Here 20 cm20\ \text{cm} and 2 mm2\ \text{mm} must both be in meters or both in centimeters before computing ΔL/L\Delta L/L. 2. Period uncertainty: The resolution of 1 s1\ \text{s} applies to the total time for 50 oscillations, not to one oscillation. Divide by 50 to get ΔT\Delta T. 3. Error propagation factor: Because TT appears squared in the formula for gg, its relative uncertainty is multiplied by 2, not 1. 4. Rounding too early: Keep intermediate values (like 0.0250.025) in decimal form until the final percentage to avoid rounding errors.

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