The period of oscillation of a simple pendulum is $T=2\pi \sqrt{\dfrac{L}{g}}$. Measured value of L is 10 cm known to 1 mm accuracy and time for 100 oscillations of the pendulum is found to be 50s using a wrist watch of 1 s resolution. What is the accuracy of the determination of g?
A. $2\%$
B. $3\%$
C. $4\%$
D. $5\%$

Hint: The accuracy of measurement of a quantity $x$ is the value of the absolute error $\Delta x$. While measuring with an instrument it is usually the least count of the instrument.

Formula used:
If, $X={{Y}^{m}}\times {{Z}^{n}}$
$\dfrac{\Delta X}{X}=m\dfrac{\Delta Y}{Y}\pm n\dfrac{\Delta Z}{Z}$
For maximum possible relative error, + or – has to be used accordingly depending on the values of m and n.

Complete Step-by-Step solution:
Let us first analyze the information given to us. We are given that the period of oscillation of a pendulum is $T=2\pi \sqrt{\dfrac{L}{g}}$, where L is the length of the string of the pendulum and g is the acceleration due to gravity.
Given, the measured value of L is 10 cm. ---------------(1)
Also, given it is known upto 1mm accuracy. $\therefore \Delta L=1mm=0.1cm$ --$\left( 1mm=0.1cm \right)$ ----------------(2)
Give, time period for 100 oscillations (n) is 50 s (t)
Therefore, time period for one oscillation $T=\dfrac{t}{n}=\dfrac{50}{100}=0.5s$
$\therefore T=0.5s$ -------------------(3)
For 100 oscillations, the absolute error is 1 s, since the resolution of the wrist watch is 1 s.
$\therefore \text{Error for one oscillation time period = }\Delta \text{T =}\dfrac{\Delta t}{n}=\dfrac{1s}{100}=0.01s$ -----------------(4)
Now, $T=2\pi \sqrt{\dfrac{L}{g}}$
squaring both sides,
${{T}^{2}}=4{{\pi }^{2}}\dfrac{L}{g}$
$g=4{{\pi }^{2}}\dfrac{L}{{{T}^{2}}}$ ----------(5)
Now, the accuracy of measurement of a quantity $x$ is the value of the absolute error$\Delta x$. While measuring with an instrument it is usually the least count of the instrument. The relative error of a quantity $x$ is $\dfrac{\Delta x}{x}$
Now,
If, ${{X}^{p}}={{Y}^{m}}\times {{Z}^{n}}$
$p\dfrac{\Delta X}{X}=m\dfrac{\Delta Y}{Y}\pm n\dfrac{\Delta Z}{Z}$ --------------(6)
for, maximum possible relative error, + or – has to be used accordingly depending on the values of m and n.
Using (6) for (5)
$\dfrac{\Delta g}{g}=\dfrac{\Delta L}{L}+2\dfrac{\Delta T}{T}$ ---------(7)
In the above equation the term $4{{\pi }^{2}}$ does not appear since it is a constant and its relative error is zero. Also, we write \[+2\dfrac{\Delta T}{T}\] and not \[-2\dfrac{\Delta T}{T}\]to make the relative error maximum which is the rule of the process.
Therefore, putting (1), (2), (3) and (4) in (7), we get,
$\dfrac{\Delta g}{g}=\dfrac{0.1}{10}+2\dfrac{0.01}{0.5}=0.01+0.04=0.05$
Writing this as a percentage error,
$\dfrac{\Delta g}{g}\times 100=0.05\times 100=5%$
hence, the correct option is D) 5%.

Note: Students tend to make mistakes in this topic since they do not fully understand whether to add or subtract the relative errors of individual quantities while subtracting. However, it should be kept in mind that the rule is always to add the positive magnitude of individual relative errors and never to write in such a way that the total error gets decreased.
By making this mistake, students could have arrived at the answer of 3% in the above question. However, such errors should be avoided and it is to be always kept in mind that the addition of the individual relative errors has to be done in such a way so as to maximize the total relative error.