General Physics Question 14
Question 14
- 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 $(60 \pm 1) \mathrm{mm}$ and $(10 \pm 1) \mathrm{mm}$, respectively. In five successive measurements, the time period is found to be $0.52 \mathrm{~s}, 0.56$, $0.57 \mathrm{~s}, 0.54 \mathrm{~s}$ and $0.59 \mathrm{~s}$. The least count of the watch used for the measurement of time period is $0.01 \mathrm{~s}$. Which of the following statement(s) is (are) true?
(2016 Adv.) (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 measurement of $g$ is $11 %$
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Answer:
Correct Answer: 14. $(a, b, d)$
Solution:
- Mean time period $=\frac{0.52+0.56+0.57+0.54+0.59}{5}$
$$ =0.556 \cong 0.56 \mathrm{sec} \text { as per significant figures } $$
Error in reading $=\left|T_{\text {mean }}-T_{1}\right|=0.04$
$$ \begin{aligned} \left|T_{\text {mean }}-T_{2}\right| & =0.00 \ \left|T_{\text {mean }}-T_{3}\right| & =0.01 \ \left|T_{\text {mean }}-T_{4}\right| & =0.02 \ \left|T_{\text {mean }}-T_{5}\right| & =0.03 \end{aligned} $$
Mean error $=0.1 / 5=0.02$
$%$ error in $T=\frac{\Delta T}{T} \times 100=\frac{0.02}{0.56} \times 100=3.57 %$
$%$ error in $r=\frac{0.001 \times 100}{0.010}=10 %$
$%$ error in $R=\frac{0.001 \times 100}{0.60}=1.67 %$
$%$ error in $\frac{\Delta g}{g} \times 100=\frac{\Delta(R-r)}{R-r} \times 100+2 \times \frac{\Delta T}{T}$
$=\frac{0.002 \times 100}{0.05}+2 \times 3.57=4 %+7 %=11 %$
15. For vernier calipers
$1 \mathrm{MSD}=\frac{1}{8} \mathrm{~cm}$
$5 \mathrm{VSD}=4 \mathrm{MSD}$
$\therefore 1 \mathrm{VSD}=\frac{4}{5} \mathrm{MSD}=\frac{4}{5} \times \frac{1}{8}=\frac{1}{10} \mathrm{~cm}$ Least count of vernier calipers $=1 \mathrm{MSD}-1 \mathrm{VSD}$
$$ \begin{aligned} & =\frac{1}{8} \mathrm{~cm}-\frac{1}{10} \mathrm{~cm} \ & =0.025 \mathrm{~cm} \end{aligned} $$
(a) and (b)
Pitch of screw gauge $=2 \times 0.025=0.05 \mathrm{~cm}$
Least count of screw gauge $=\frac{0.05}{100} \mathrm{~cm}=0.005 \mathrm{~mm}$
(c) and (d)
Least count of linear scale of screw gauge $=0.05$
Pitch $=0.05 \times 2=0.1 \mathrm{~cm}$
Least count of screw gauge $=\frac{0.1}{100} \mathrm{~cm}$
$$ =0.01 \mathrm{~mm} $$
