JEE Mains · Physics · STD 11 - Units,dimensions and measurement
- JEE Mains 2024
- Hard
- MCQ
The measured value of the length of a simple pendulum is \(20 \mathrm{~cm}\) with \(2 \mathrm{~mm}\) 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 \(\mathrm{N} \%\). The value of \(\mathrm{N}\) is _______.
Pick an option to check your answer.
Answer & Solution
Correct Answer
(C) \(6\)
Step-by-step Solution
Detailed explanation
\( T = 2\pi\sqrt{\frac{l}{g}} \)
\( g = \frac{4\pi^2 l}{T^2} \)
The relative error in \( g \) is given by: \( \frac{\Delta g}{g} = \frac{\Delta l}{l} + 2\frac{\Delta T}{T} \)
Given values:
Length \( l = 20\text{ cm} \)
Error in length \( \Delta l = 2\text{ mm} = 0.2\text{ cm} \)
Total time \( t = 40\text{ s} \)
Error in time \( \Delta t = 1\text{ s} \)
Note: The relative error for the time period \( T \) is the same as the relative error for the total time \( t \),
so \( \frac{\Delta T}{T} = \frac{\Delta t}{t} = \frac{1}{40} \).
Substituting the values into the error equation:
\( \frac{\Delta g}{g} = \frac{0.2}{20} + 2\left(\frac{1}{40}\right) \)
Converting to decimals for clarity:
\( \frac{\Delta g}{g} = 0.01 + \frac{2}{40} \)
\( \frac{\Delta g}{g} = 0.01 + 0.05 \)
\( \frac{\Delta g}{g} = 0.06 \)
Percentage error = \( 0.06 \times 100\% = 6\% \)
\( g = \frac{4\pi^2 l}{T^2} \)
The relative error in \( g \) is given by: \( \frac{\Delta g}{g} = \frac{\Delta l}{l} + 2\frac{\Delta T}{T} \)
Given values:
Length \( l = 20\text{ cm} \)
Error in length \( \Delta l = 2\text{ mm} = 0.2\text{ cm} \)
Total time \( t = 40\text{ s} \)
Error in time \( \Delta t = 1\text{ s} \)
Note: The relative error for the time period \( T \) is the same as the relative error for the total time \( t \),
so \( \frac{\Delta T}{T} = \frac{\Delta t}{t} = \frac{1}{40} \).
Substituting the values into the error equation:
\( \frac{\Delta g}{g} = \frac{0.2}{20} + 2\left(\frac{1}{40}\right) \)
Converting to decimals for clarity:
\( \frac{\Delta g}{g} = 0.01 + \frac{2}{40} \)
\( \frac{\Delta g}{g} = 0.01 + 0.05 \)
\( \frac{\Delta g}{g} = 0.06 \)
Percentage error = \( 0.06 \times 100\% = 6\% \)
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