You are designing a pendulum clock to have a period of $1.0\,s$. How long should the pendulum be ?
A. 0.25 cm
B. 0.5 cm
C. 0.25 m
D. 0.25 mm

Hint:A pendulum is a weight that is suspended from a pivot and may freely swing. When a pendulum is moved sideways from its resting, equilibrium position, gravity acts as a restorative force, accelerating the pendulum back to its equilibrium position.

Formula used:
${\mathbf{T}} = 2\pi \sqrt {\dfrac{{\mathbf{L}}}{{\text{g}}}} $
$L$ = length of the bob and $G$ = acceleration due to gravity.

Complete step by step answer:
A simple pendulum may be built by tying a one-metre-long thread to a tiny metal ball (called a bob) and suspending it from a sturdy support, allowing the bob to swing freely. The bob of a pendulum is in the mean position when it is at rest.

A simple pendulum is made up of a mass m suspended on a string with a length $L$ and a pivot point $P$. The pendulum will swing back and forth with periodic motion when moved to a starting angle and released, with being the natural frequency of the motion. The time period of a pendulum is the amount of time it takes to complete one oscillation.

Hence, we find the answer using the formula
$T = 2\pi \sqrt {\dfrac{L}{g}} $
Squaring on both sides we get
${T^2} = 4{\pi ^2} \times \dfrac{L}{g}$
Substituting the value of T
$1 = 4{\pi ^2} \times \dfrac{L}{g}$
$L = \dfrac{g}{{4{\pi ^2}}}
\Rightarrow L= \dfrac{{9.8}}{{4 \times 3.14 \times 3.14}}m$
Hence the value of $L$ becomes
$L = \dfrac{{9.8}}{{39.44}}\,m \\
\Rightarrow L= 0.2484\,m \\
\Rightarrow L= 0.25\,m{\text{ (approx}}{\text{.) }}$
$ \therefore L = 0.25{\text{ m}}$

Hence option C is correct.

Note:The number of times a repeated event occurs per unit of time is known as frequency. It's also known as temporal frequency, which stresses the difference between spatial and angular frequency. The unit of frequency is hertz (Hz), which equals one occurrence per second.