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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$90^{0}$

$60^{0}$

$120^{0}$

$30^{0}$

answer is A.

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Detailed Solution

Using $R^{2} = P^{2} + Q^{2} + 2 P Q \cos θ$

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

When forces Q is doubled,

$4 F^{2} + 36 F^{2} + 24 F^{2} \cos θ = 4 R^{2} 4 F^{2} + 36 F^{2} + 24 F^{2} \cos θ = 4 (13 F^{2} + 12 F^{2} \cos θ) = 52 F^{2} + 48 F^{2} \cos θ ∴ \cos θ = - \frac{12 F^{2}}{24 F^{2}} = - \frac{1}{2} \Rightarrow θ = 120^{0}$

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