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A particle is released from rest from a tower of height 3h. The ratio of time intervals for fall of equal height h i.e. `t_(1):t_(2):t_(3)` is :A. `sqrt(3):sqrt(2):1`B. `3:2:1`C. `9:4:1`D. `1:(sqrt(2)-1):(sqrt(3)-sqrt(2))` |
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Answer» Correct Answer - D A particle ………. `h = (1)/(2)"gt"_(1)^(2) 2h=(1)/(2)g(t_(1)+t_(2))^(2)3h` `= (1)/(2)g(t_(1)+t_(2)+t_(3))^(2)` `:. T_(1):(t_(1)+t_(2)): (t_(1)+t_(2)+t_(3)) = 1: sqrt(2): sqrt(3)` `implies t_(1) : t_(2) : t_(3) = 1: (sqrt(2)-1) : (sqrt(3)-sqrt(2))` |
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