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The square of resultant of two equal for...

The square of resultant of two equal forces is equal to three times their product. Angle between the forces in radian is

A

`pi`

B

`pi//2`

C

`pi//3`

D

`pi//4`

Text Solution

Verified by Experts

The correct Answer is:
C
By law of parallelogram of vectors,
`R^(2) = P^(2) + Q^(2) + 2PQ cos theta`
Here P = Q
`R^(2) = 3PQ`
`therefore 3PQ = P^(2) + P^(2) + 2P^(2) cos theta`
`therefore 3P^(2) - 2P^(2) = 2P^(2) cos theta`
`therefore cos theta = 1//2 rArr theta = 60^(@) = (pi)/(3)` radian
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