JEE Main & Advanced AIEEE Solved Paper-2002

  • 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  Solved  Paper-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 :

    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),    \[A=5N,\,B=13N\]


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