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Byju's Answer

Standard XI

Physics

Vector Addition

Two forces 3N...

Question

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

$30^{\circ}$

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$60^{\circ}$

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$90^{\circ}$

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$120^{\circ}$

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Solution

The correct option is D
$120^{\circ}$

$A = 3 N, B = 2 N t h e n R = \sqrt{A^{2} + B^{2} + 2 A B c o s θ} R = \sqrt{9 + 4 + 12 c o s θ}$
Now A =6 N, B =2 N then
$2 R = \sqrt{36 + 4 + 24 c o s θ}$
From (i) and (ii) we get $c o s θ = - \frac{1}{2} ∴ θ = 120^{\circ}$

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

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

The correct option is D
$120^{\circ}$

$A = 3 N, B = 2 N t h e n R = \sqrt{A^{2} + B^{2} + 2 A B c o s θ} R = \sqrt{9 + 4 + 12 c o s θ}$
Now A =6 N, B =2 N then
$2 R = \sqrt{36 + 4 + 24 c o s θ}$
From (i) and (ii) we get $c o s θ = - \frac{1}{2} ∴ θ = 120^{\circ}$