JEE Advanced · Physics · 1. Math in Physics
A student uses a simple pendulum of exactly \(1 \mathrm{~m}\) length to determine \(g\), the acceleration due to gravity. He uses a stop watch with the least count of \(1 \mathrm{~s}\) for this and records \(40 \mathrm{~s}\) for 20 oscillations. For this observation, which of the following statement(s) is/are true?
- A Error \(\Delta T\) in measuring \(T\), the time period, is \(0.05 \mathrm{~s}\)
- B Error \(\Delta T\) in measuring \(T\), the time period, is \(1 \mathrm{~s}\)
- C Percentage error in the determination of \(g\) is \(5 \%\)
- D Percentage error in the determination of \(g\) is \(2.5 \%\)
Answer & Solution
Correct Answer
(C) Percentage error in the determination of \(g\) is \(5 \%\)
Step-by-step Solution
Detailed explanation
\(T=\frac{40 \mathrm{~s}}{20}=2 \mathrm{~s}\).
Further, \(t=n T=20 T\) or \(\Delta t=20 \Delta T\)
\(\therefore \quad \frac{\Delta t}{t}=\frac{\Delta T}{T}\)
or \(\Delta T=\frac{T}{t} \cdot \Delta t=\left(\frac{2}{40}\right)(1)=0.05 \mathrm{~s}\)
Further, \(T=2 \pi \sqrt{\frac{l}{g}}\) or \(\quad T \propto g^{-1 / 2}\)
\(\therefore \frac{\Delta T}{T} \times 100=-\frac{1}{2} \times \frac{\Delta g}{g} \times 100\)
or \(\%\) error in determination of \(g\) is
\(\begin{aligned}\frac{\Delta g}{g} \times 100 & =-200 \times \frac{\Delta T}{T} \\& =-\frac{200 \times 0.05}{2}=-5 \%\end{aligned}\)
\(\therefore\) correct options are (a) and (c).
Further, \(t=n T=20 T\) or \(\Delta t=20 \Delta T\)
\(\therefore \quad \frac{\Delta t}{t}=\frac{\Delta T}{T}\)
or \(\Delta T=\frac{T}{t} \cdot \Delta t=\left(\frac{2}{40}\right)(1)=0.05 \mathrm{~s}\)
Further, \(T=2 \pi \sqrt{\frac{l}{g}}\) or \(\quad T \propto g^{-1 / 2}\)
\(\therefore \frac{\Delta T}{T} \times 100=-\frac{1}{2} \times \frac{\Delta g}{g} \times 100\)
or \(\%\) error in determination of \(g\) is
\(\begin{aligned}\frac{\Delta g}{g} \times 100 & =-200 \times \frac{\Delta T}{T} \\& =-\frac{200 \times 0.05}{2}=-5 \%\end{aligned}\)
\(\therefore\) correct options are (a) and (c).
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