Question

In an 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 $(10\pm 1)mm$, respectively. In five successive measurements, the time period is found to be $0.52s,0.56s,0.57s,0.54s$ 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 error in the determined value of $g$ is $11$%

The observed values of time period, $T_i=0.52s,$ $0.56s,$ $0.57s,$ $0.54s$ and $0.59s$
The mean value of time period, $T=\frac{\sum T_i}{5}=\frac{2.78}{5}=0.56s$
Magnitude of absolute error in each observation, $|\Delta T_1|=|0.56-0.52|=|0.04|s$
$|\Delta T_2|=|0.56-0.56|s$ =$|0.00|s$
Similarly,
$|\Delta T_3|=|0.01|s$
$|\Delta T_4|=|0.02|s$ $|\Delta T_5|=|0.03|s$


Mean absolute error in time period, $\Delta T_m=\frac{0.04+0.00+0.01+0.02+0.03}{5}=0.02s$
$\therefore$ Error in $T$, $\frac{\Delta T_m}{T}\times 100=\frac{0.02}{0.56}\times 100=3.57$ %


Error in the measurement of $r$: $\frac{\Delta r}{r}\times 100=\frac{1}{10}\times 100=10$ %


From the equation given, we get: $g=\frac{28{\pi }^2(R-r)}{5T^2}$
$\therefore$ Error in the measurement of $g$: $\frac{\Delta g}{g}\times 100=\frac{\Delta R+\Delta r}{(R-r)}\times 100+2\frac{\Delta T_m}{T}\times 100$
$⟹$ $\frac{\Delta g}{g}\times 100=\frac{1+1}{(60-10)}\times 100+2\left(3.57\right)$ % $=11.14$ %

Thus options A, B and D are correct.