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JEE Main & Advanced Physics Motion in a Straight Line / सरल रेखा में गति Question Bank Motion Under Gravity

  • question_answer
    From the top of a tower, a particle is thrown vertically downwards with a velocity of 10 m/s. The ratio of the distances, covered by it in the 3rd and 2nd seconds of the motion is (Take \[g=10m/{{s}^{2}}\])                             [AIIMS 2000; CBSE PMT 2002]

    A)             5 : 7    

    B)             7 : 5

    C)             3 : 6    

    D)             6 : 3

    Correct Answer: B

    Solution :

                    \[{{S}_{{{3}^{rd}}}}=10+\frac{10}{2}(2\times 3-1)=35\ m\]             \[{{S}_{{{2}^{nd}}}}=10+\frac{10}{2}(2\times 2-1)=25m\] Þ \[\frac{{{S}_{{{3}^{rd}}}}}{{{S}_{{{2}^{nd}}}}}=\frac{7}{5}\]


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