question_answer

A particle is moving along a circular path with a constant speed of \[10m{{s}^{-1}}\]. 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? [JEE Main Online Paper (Held On 11-Jan-2019 Morning]

A) zero                                          

B) \[10\sqrt{2}m/s\]

C) \[10\sqrt{3}m/s\]

D)                  \[10\,m/s\]

Correct Answer: D

Solution :

Let the particle be moving in counter clockwise direction. The initial velocity of the particle \[{{\vec{v}}_{1}}=v\hat{j}\] The velocity of the particle after time t is \[v(\cos {{60}^{0}}\hat{j}-\sin {{60}^{o}}\hat{i})=\frac{v}{2}(j-\sqrt{3}\hat{i})\] So, the change in velocity \[|\Delta v|=|{{\vec{v}}_{2}}-{{\vec{v}}_{1}}|=v\left| \frac{1}{2}(\hat{j}-\sqrt{3}\hat{i})-\hat{j} \right|\] \[=\frac{10}{2}\left| -\hat{j}-\sqrt{3}\hat{i} \right|=5\sqrt{{{(-1)}^{2}}+{{(-\sqrt{3})}^{2}}}=10m/s\]