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Two forces 3 N and 2 N are at an angle $θ$ such that the resultant is R. The first force is now increased to 6N and the resultant becomes 2R. The value of $θ$ is

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$60^{\circ}$

$30^{\circ}$

$90^{\circ}$

$120^{\circ}$

answer is D.

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Detailed Solution

A=3, B=2N then

$\begin{array}{l} R = \sqrt{A^{2} + B^{2} + 2 ABcosθ} \\ R = \sqrt{9 + 4 + 12 cosθ} \end{array}$

Now $A = 6 N, B = 2 N$ then

$2 R = \sqrt{36 + 4 + 24 cosθ}$

From Eq.s (i) and (ii), we get $cosθ = \frac{1}{2}$ $\Rightarrow$ $θ = 120^{0}$

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