JEE Main & Advanced JEE Main Solved Paper-2014

  • question_answer
    A bird is sitting on the top of a vertical pole 20 m high and its elevation from a point O on the ground is \[45{}^\circ\]. It flies off horizontally straight away from the point O. After one second, the elevation of the bird from O is reduced to 30°. Then the speed (in m/s) of the bird is:   JEE Main  Solved  Paper-2014

    A) \[40\left( \sqrt{2}-1 \right)\]                      

    B) \[40\left( \sqrt{3}-\sqrt{2} \right)\]

    C)  \[20\sqrt{2}\]                  

    D) \[20\left( \sqrt{3}-1 \right)\]

    Correct Answer: D

    Solution :

    Here, \[AP=QB=20m\] \[\angle POA={{45}^{o}},\angle QOB={{30}^{o}}\] \[\Rightarrow \]\[OA=20;OB=20\sqrt{3}\] \[\Rightarrow \]\[OB-OA=20(\sqrt{3}-1)\] Hence distance covered in one second by the bird is \[AB=20(\sqrt{3}-1)\] Thus, speed of bird \[=20(\sqrt{3}-1)m/s\]

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