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

30^{0}

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60^{0}

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90^{0}

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120^{0}

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Solution

The correct option is D $120^{0}$
$A = 3 N, B = 2 N$ then $R = \sqrt{A^{2} + B^{2} + 2 A B cos θ}$
$R = \sqrt{9 + 14 + 12 cos θ}$ ....(i)
Now $A = 6 N, B = 2 N$ then
$2 R = \sqrt{36 + 4 + 24 cos θ}$ ....(ii)
from (i) and (ii) we get $cos θ = - \frac{1}{2} ∴ θ = 120^{o}$

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

The correct option is D 1200A=3 N,B=2 N then R=√A2+B2+2ABcosθR=√9+14+12cosθ ....(i)Now A=6 N,B=2 N then 2R=√36+4+24cosθ ....(ii)from (i) and (ii) we get cosθ=−12∴θ=120o

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

The correct option is D $120^{0}$
$A = 3 N, B = 2 N$ then $R = \sqrt{A^{2} + B^{2} + 2 A B cos θ}$
$R = \sqrt{9 + 14 + 12 cos θ}$ ....(i)
Now $A = 6 N, B = 2 N$ then
$2 R = \sqrt{36 + 4 + 24 cos θ}$ ....(ii)
from (i) and (ii) we get $cos θ = - \frac{1}{2} ∴ θ = 120^{o}$