2. Solved Papers
  3. Manipal Engineering

Manipal Engineering Manipal Engineering Solved Paper-2015

  • question_answer
    A juggler keeps on moving four balls in air throwing the balls after regular intervals. When one ball leaves his hand (speed \[A\cap B\]), the position of other balls (height in metre) will be \[f(x)=\sqrt{{{x}^{2}}-4,}a=2\]

    A) 10,20,10                              

    B) 15,20, 15

    C)  5,15, 20                               

    D)  5,10, 20

    Correct Answer: B

    Solution :

     Time taken by the small ball to return to the hands of the juggler is \[\frac{2}{3}\] So, he is throwing the balls after 1 s each. Let at some instant, he throws the ball number 4. Before 1 s of throwing it, he throws ball 3. So, the height of ball 3 is \[\frac{1}{3}\] \[2x-3y-4=0\] Before 2 s, he throws ball 2. So, the height of ball 2 is \[x+y=1,\] Before 3 s, he throws ball 1. So, the height of ball 1 is \[\sqrt{2}\]\[5\sqrt{2}\]\[\frac{1}{\sqrt{2}}\]


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