A simple pendulum of length $2 \mathrm{~m}$ is given a horizontal push through angular displacement of $60^{\circ}$. If the mass of bob is 200 gram, the angular velocity of the bob will be (Take Acceleration due to gravity $=10 \mathrm{~m} / \mathrm{s}^2$ ) $\left(\sin 30^{\circ}=\cos 60^{\circ}=0.5, \cos 30^{\circ}=\sin 60^{\circ}=\sqrt{3} / 2\right)$

• A

  $2 \sqrt{2} \mathrm{rad} / \mathrm{s}$
• B

  $3 \sqrt{2} \mathrm{rad} / \mathrm{s}$
• Correct answer
  C

  $2 \sqrt{2.5} \mathrm{rad} / \mathrm{s}$
• D

  $3 \sqrt{2.5} \mathrm{rad} / \mathrm{s}$

Solution

Given: $l=2 \mathrm{~m}, \theta=60^{\circ}, \mathrm{m}=200 \mathrm{~g}=2 \mathrm{~g}$
From the figure, $\mathrm{T}=\mathrm{mr} \omega^2$... (i) also $\mathrm{T} \cos \theta=\mathrm{mg}$... (ii) putting (i) into (ii) $\mathrm{mr} \omega^2-\cos \theta=\mathrm{mg}$... (iii) putting the given values into equation (iii) $\begin{aligned} & 2 \times 2 \times \omega^2 \frac{1}{2}=2 \times 10 \quad \ldots .\left(\because \cos 60=\frac{1}{2}\right) \\ & \omega^2=10 \\ & \Rightarrow \omega=\sqrt{10} \quad \ldots .(\because \sqrt{10}=\sqrt{2 \times 2 \times 2.5}) \\ & \quad=2 \sqrt{2.5} \mathrm{rad} / \mathrm{s} \end{aligned}$

Hence, the correct answer is option: 'C'.