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Q.

A particle is moving on a circular path of radius r with uniform speed v. What is the displacement of the particle after it has described an angle of $60^{0}$ '?

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a

$r \sqrt{2}$

b

$r \sqrt{3}$

c

r

d

2r

answer is C.

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Detailed Solution

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According to cosine formula

Question Image

$\cos 60^{0} = \frac{r^{2} + r^{2} - x^{2}}{2 r^{2}}$

$∴ x = r$ $2 r^{2} \cos 60^{0} = 2 r^{2} - x^{2}$

$x^{2} = 2 r^{2} - 2 r^{2} \cos 60^{0}$

= $2 r^{2} [1 - \cos 60^{0}]$

= $2 r^{2} [2 \sin^{2} 30^{0}] = r^{2}$

$∴$ x = r

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