Vectors Question 32
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
[HP PMT 2000]
Options:
A) $ 30{}^\circ $
B) $ 60{}^\circ $
C) $ 90{}^\circ $
D) $ 120{}^\circ $
Correct Answer: D $ A=3N $ , $ B=2N $ then $ R=\sqrt{A^{2}+B^{2}+2AB\cos \theta } $ $ R=\sqrt{9+4+12\cos \theta } $ ?(i) Now $ A=6N $ , $ B=2N $ then $ 2R=\sqrt{36+4+24\cos \theta } $ ?(ii) from (i) and (ii) we get $ \cos \theta =-\frac{1}{2} $ Therefore $ $ \theta =120{}^\circ $
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Solution:
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