Question

In an experiment to determine the acceleration due to gravity g, the formula used for the time period of a periodic motion is $T =2 \pi \sqrt{\frac{7( R - r )}{5\, g }} .$ 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 $0.52\, s,\, 0.56\, s$, $0.57\, s,\, 0.54\, s$ and $0.59\, s$. The least count of the watch used for the measurement of time period is $0.01\, s$. 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
$T =2 \pi \sqrt{\frac{7( R - r )}{5 g }}$
Time period
$0.52\, \sec\,\, Av \langle T \rangle=\frac{0.52+0.56+0.57+0.54+0.59}{5}$
II $0.56\, \sec$
III $0.57\, \sec$
IV $0.54\, \sec \langle T \rangle=\frac{2.78}{5}$
V $0.59\, \sec \langle T \rangle=0.556\, \sec$

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