JEE Mains · Physics · STD 11 - Units,dimensions and measurement
- JEE Mains 2026
- Medium
- MCQ
The time period of a simple harmonic oscillator is \(T =2 \pi \sqrt{\frac{ k }{ m }}\). The measured value of mass (m) of the object is 10 g with an accuracy of 10 mg , and time for 50 oscillations of the spring is found to be 60 s using a watch of 2 s resolution. Percentage error in determination of spring constant( \(k\) ) is ___________ %.
Pick an option to check your answer.
Answer & Solution
Correct Answer
(D) 6.76
Step-by-step Solution
Detailed explanation
\(\frac{\Delta K}{K}=\frac{2\Delta T}{T}+\frac{\Delta m}{m}\)
\(T=\frac{60}{50}=1.2~sec\)
\(\Delta T=\frac{2}{50}\)
\(\therefore \frac{\Delta K}{K}=\frac{2\times2}{50\times1.2}+\frac{10\times10^{-3}}{10}=0.0676\)
\(\therefore \%Error=6.76\%\)
\(T=\frac{60}{50}=1.2~sec\)
\(\Delta T=\frac{2}{50}\)
\(\therefore \frac{\Delta K}{K}=\frac{2\times2}{50\times1.2}+\frac{10\times10^{-3}}{10}=0.0676\)
\(\therefore \%Error=6.76\%\)
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