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JEE Main & Advanced JEE Main Paper (Held On 9 April 2014)

  • question_answer
    Water is flowing at a speed of 1.5 \[\text{m}{{\text{s}}^{\text{-1}}}\] through a horizontal tube of cross-sectional area \[\text{1}{{\text{0}}^{\text{-2}}}{{\text{m}}^{\text{2}}}\]and you are trying to stop the flow by your palm. Assuming that the water stops immediately after hitting the palm, the minimum force that you must exert should be (density of water \[\text{=1}{{\text{0}}^{3}}\text{kg}{{\text{m}}^{-3}}\])   [JEE Main Online Paper ( Held On 09 Apirl  2014  )

    A) 15 N      

    B) 22.5 N

    C) 33.7 N                  

    D) 45 N

    Correct Answer: A

    Solution :

                    For 1 m length of horizontal tube Mass of water M = density × volume \[={{10}^{3}}\times \text{area}\times \text{length}\] \[={{10}^{3}}\times {{10}^{-2}}\times 1=10\,\text{kg}\] Therefore minimum force\[=\frac{\Delta p}{\Delta t}\](rate of change of momentum)= 10 × 1.5 = 15 N                


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