The length of second's hand in watch is `1 cm`. The change in Velocity of its tip in 15 seconds is

Options

A. Zero
B. `(pi)/(30sqrt(2))cm//sec`
C. `(pi)/(30) cm //sec`
D. `(pi sqrt(2))/(30) cm //sec`

Answer

D

Explanation

For the secong hand , T = 60 s
` therefore `In 15 s , it will rotate through moves through `90^(@) `
In U .C.M when the particle moves through an angle ` theta `
the change in velocity ` | d vecv | = 2v sin ((theta )/(2))`
in this case ` theta = 90^(@) ` and ` v= r omega and sin ((theta )/( 2)) = sin 45^(@)`
` therefore dv = 2 omega sin 45^(@) = 2r ((2pi)/(T)) sin 45^(@)`
`= 2xx1 xx(2pi)/(60 ) xx (1) / ( sqrt(2)) [ :' T = 60 s]`
` therefore dv = ( sqrt( 2 pi ))/(30) cm //s`