Laws Of Motion Ques 44
- A particle is moving along a circular path with a constant speed of $10 \mathrm{~ms}^{-1}$. What is the magnitude of the change in velocity of the particle, when it moves through an angle of $60^{\circ}$ around the centre of the circle?
(2019 Main, 11 Jan I)
(a) $10 \sqrt{2} \mathrm{~m} / \mathrm{s}$
(b) $10 \mathrm{~m} / \mathrm{s}$
(c) $10 \sqrt{3} \mathrm{~m} / \mathrm{s}$
(d) Zero
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Answer:
Correct Answer: 44.(b)
Solution:
Formula:
- Let $v_{1}$ be the velocity of the particle moving along the circular path initially, $v_{1}$ and $v_{2}$ be the velocity when it moves through an angle of $60^{\circ}$ as shown below.
From the figure,
$$ \begin{aligned} \Delta \mathbf{v} & =\mathbf{v}_ {2}-\mathbf{v}_ {1} \\ \Rightarrow \quad|\Delta \mathbf{v}| & =2 v \sin \frac{\theta}{2}=2 \mathbf{v} \sin 30^{\circ} \quad\left[\because\left|\mathbf{v}_ {1}\right|=\left|\mathbf{v}_ {2}\right|\right] \\ & =2 v \times \frac{1}{2}=v \quad(\text { Given, } v=10 \mathrm{~m} / \mathrm{s}) \\ \Rightarrow \quad|\Delta \mathbf{v}| & =10 \mathrm{~m} / \mathrm{s} \end{aligned} $$
Alternate method
$\because \Delta \mathbf{v}=\mathbf{v}_ {2}-\mathbf{v}_ {1}=\mathbf{v}_ {2}+\left(-\mathbf{v}_ {1}\right)$
$\therefore|\Delta \mathbf{v}|^{2}=v_ {1}^{2}+v_ {2}^{2}+2 v_ {1} v_ {2} \cos 120^{\circ}$
$$ =v^{2}+v^{2}+2 v \times v \times-\frac{1}{2} $$
$\Rightarrow|\Delta \mathbf{v}|=v=10 \mathrm{~m} / \mathrm{s}$.
BETA
