A particle is moving along a circular path with a constant speed `10 ms^-1`.What is the magnitude of the change in velocity of the particle ,when it moves through an angle of `60^@` around the center of the circle ?
A
zero
B
10m/s
C
`10sqrt2` m/s
D
`10sqrt3`m/s
Text Solution
Verified by Experts
The correct Answer is:
B
`|DeltavecV)=|vecV_F-vecV_i|=sqrt(10^2+10^2 -2xx10xx10 cos 60^@)=sqrt(100)` m/s = 10 m/s
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VMC MODULES ENGLISH-JEE Main Revision Test-6 | JEE-2020 -PHYSICS
