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)

Time period
$T=2\pi \sqrt{\frac{7(R-r)}{5g}}$
Time period
$0.52\sec Av⟨T⟩=\frac{0.52+0.56+0.57+0.54+0.59}{5}$
II $0.56\sec$
III $0.57\sec$
IV $0.54\sec ⟨T⟩=\frac{2.78}{5}$
V $0.59\sec ⟨T⟩=0.556\sec$

$T^{2}g(R-r{)}^{-1}=$ const
$\frac{2\Delta T}{T}+\frac{\Delta g}{g}-\frac{\Delta (R-r)}{R-r}=0$
$\frac{2\Delta T}{T}+\frac{\Delta g}{g}-\frac{1}{(R-r)}(\Delta R-\Delta r)=0$
$\frac{\Delta g}{g}=\left(\frac{2\times 3.57}{100}+\frac{2}{50}\right)$
$\%$ error $=\left(\frac{2\times 3.57}{100}+\frac{2}{50}\right)\times 100\approx 11\%$