Laws Of Motion Question 81

Question: Two forces are such that the sum of their magnitudes is 18 N and their resultant is perpendicular to the smaller force and magnitude of resultant is 12 N. Then the magnitudes of the forces are [AIEEE 2002]

Options:

A) 12 N, 6 N

B) 13 N, 5N

C) 10 N, 8 N

D) 16 N, 2 N

Show Answer

Answer:

Correct Answer: B

Solution:

$ A+B=18 $ -(i)

$ 12=\sqrt{A^{2}+B^{2}+2AB\cos \theta } $ -(ii)

$ \tan \alpha =\frac{B\sin \theta }{A+B\cos \theta }=\tan 90{}^\circ $

therefore $ \cos \theta =-\frac{A}{B} $ -(iii) By solving (i), (ii) and (iii), $ A=13N $

and $ B=5N $



जेईई के लिए मॉक टेस्ट

एनसीईआरटी अध्याय वीडियो समाधान

दोहरा फलक