JEE Main & Advanced Physics Vectors JEE PYQ - Vectors

  • question_answer
    Two forces are such that the sum of their magnitudes is 18 N and their resultant which has magnitude 12 N, is perpendicular to the smaller force. Then, the magnitudes of the forces are                                     [AIEEE 2002]

    A) 12 N. 6 N

    B) 13 N, 5 N

    C) 10 N, 8 N

    D) 16 N, 2 N

    Correct Answer: B

    Solution :

    [b] Given, the sum of magnitude of forces
                \[A+B=18\]                   ... (i)
                The magnitude of resultant,
                \[12=\sqrt{{{A}^{2}}+{{B}^{2}}+2AB\cos \theta }\]           …. (ii)
                            \[\tan \alpha =\frac{B\sin \theta }{A+B\cos \theta }\]
                \[\Rightarrow \tan {{90}^{o}}=\frac{B\sin \theta }{A+B\cos \theta }\]
                \[\Rightarrow \cos \theta =\frac{-A}{B}\]                          ….. (iii) 
                Solving Eqs. (i), (ii) and (iii),
                \[A=5N,\,B=13N\]


You need to login to perform this action.
You will be redirected in 3 sec spinner