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The length of a seconds hand in watch is...

The length of a seconds hand in watch is `1 cm.` The change in velocity of its tip in `15 s` is

A

`(pi)/(15) cm//s`

B

`(pi)/(15sqrt(2)) cm//s`

C

`(pi)/(sqrt(15)) cm//s`

D

`(pi)/(sqrt(2)) cm//s`

Text Solution

Verified by Experts

The correct Answer is:
B

`v=|vecv|=(2pi)/(60)(1)=(pi)/(30) cm//s.` In 15 s, the radius vector and `vecv` rotates through an anlge `theta=90^(@).` Therefore, the angle between `vecv_(2) and -vecv_(1)` is `180^(@)-theta = 90^(@)." "therefore|Delta vecv|^(2)=|vecv_(2)-vecv_(1)|^(2)=|vecv_(2)+(-vecv_(1))|^(2)=2v^(2) cos (180-theta)=2v^(2) cos (180- theta)=2v^(2)(1-cos theta)=2v^(2)(2 sin^(2)""(theta)/(2)).`
`therefore" "|Delta vecv|=2v sin ""(theta)/(2).`
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