Question

In the experiment to determine the acceleration due to gravity $g$ the formula used for the time period of a periodic motion is $T=2\pi \sqrt{\frac{7(R-r)}{5g}}$. The values of $R$ and $r$ are measured to be $(60\pm 1)mm$ and $(10\pm 1)mm$ respectivley. In five successive measurements, the time period is found to be $0.52s,0.56s,0.57s,0.54s\text{and}0.59s$. The least count of the watch used for the measurement of time period is $0.01s$. Which of the following statement(s) is(are) true?

• A. The error in the measurement of $r$ is $10\%$
• B. The error in the measurement of $T$ is $3.57\%$
• C. The error in the measurement of $T$ is $2\%$
• D. The error in the determined value of $g$ is $11\%$

Answer 1

Answer: (A), (B), (D)

The correct options are
A The error in the measurement of $r$ is $10\%$
B The error in the measurement of $T$ is $3.57\%$
D The error in the determined value of $g$ is $11\%$

$\begin{array}{ccc}\text{S.NO}& \text{T}& \text{Absolute error}=|T-T_{mean}|\\ 1& 0.52& 0.04\\ 2& 0.56& 0.00\\ 3& 0.57& 0.01\\ 4& 0.54& 0.02\\ 5& 0.59& 0.03\\ & T_{mean}=\frac{2.78}{5}& (\Delta T{)}_{mean}=0.02\\ & T_{mean}=0.56& \end{array}$

% error in $T=\frac{0.02}{0.56}\times 100=3.57\%$
% error in $r=\frac{dr}{r}\times 100=\frac{1}{10}\times 100=10\%$

According to the question
$g\propto \frac{T^2}{R-r}$
$\frac{dg}{g}=2\frac{dT}{T}+\frac{dR+dr}{R-r}$
$\frac{dg}{g}=2\left(3.57\%\right)+\frac{1+1}{60-10}\times 100\%$
$\frac{dg}{g}=11\%$