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Two forces P and Q, of magnitude 2F and 3F, respectively, are at an angle $θ$ with each other. If the force Q is doubled, then their resultant also gets doubled. Then, the angle $θ$ is

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$120^{o}$

$60^{o}$

$90^{o}$

$30^{o}$

answer is A.

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Detailed Solution

$R^{2} = P^{2} + Q^{2} + 2 PQcosθ$

$4 F^{2} + 9 F^{2} + 12 F^{2} \cos θ = R^{2}$

$When force Q is doubled$

$4 F^{2} + 36 F^{2} + 24 F^{2} cosθ = 4 R^{2}$

$40 F^{2} + 24 F^{2} cosθ = 4 (13 F^{2} + 12 F^{2} \cos θ)$

$\cos θ = \frac{- 12 F^{2}}{24 F^{2}} = \frac{- 1}{2}; θ = 120^{o}$

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