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\[\begin{align}

& \text{A}.\text{ }2.5 \\

& \text{B}.\text{ }5 \\

& \text{C}.10 \\

& \text{D}.\text{ }5\sqrt{2} \\

\end{align}\]

Answer
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Time period of a simple pendulum,

$T=2\pi \sqrt{\dfrac{l}{g}}$ .

In the question we are given the length of the pendulum,$l$ = 1 m

Amplitude of the pendulum, A = 2cm

And the time period of this pendulum, T = 5 seconds.

We have to find the time period of this pendulum, when its amplitude becomes 4 cm.

We know, time period of a simple pendulum is given by the equation,

$T=2\pi \sqrt{\dfrac{l}{g}}$ , where ‘$l$’ is the length of the pendulum and ‘g’ is acceleration due to gravity.

From this equation, it is clear that the time period of a simple pendulum does not depend on its amplitude.

In the question, we change the amplitude of the pendulum from 2 cm to 4 cm. Length of the pendulum remains the same and acceleration due to gravity; ‘g’ is a constant.

Therefore, the time period of the pendulum when its amplitude = 4cm, length ‘$l$’=1 m will be 5 seconds.

Time period of simple pendulum

For a simple pendulum, we know its angular frequency $\omega $ is given by

$\omega =\sqrt{\dfrac{g}{l}}$ , where ‘$l$’ is the length of the pendulum and ‘g’ is acceleration due to gravity.

Time period of an oscillation is generally expressed as,

$T=\dfrac{2\pi }{\omega }$ , where ‘T’ is the time period and ‘$\omega $’ is the angular frequency of the pendulum.

By substituting the value of angular frequency (ω) in the above equation, we get

$T=2\pi \sqrt{\dfrac{l}{g}}$

Therefore the time period of a pendulum is, $T=2\pi \sqrt{\dfrac{l}{g}}$