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Q.

The square of the resultant of two forces 4N and 3N exceeds the square of the resultant of the two forces by 12 when they are mutually perpendicular. The angle between the vectors is

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a

90°

b

30°

c

60°

d

120°

answer is B.

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Detailed Solution

P2+Q2+2PQcosθ

R12=42+32+2.4.3.cosθ

=25+24cosθ

if F1⊥F2     R22=42+32=25

R12=R22+12⇒R12−R22=12

24cosθ=12

cosθ=12⇒θ=60°

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