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Water is dropped at the rate of `2m^(2)//s` into a cone of semivertical angel of `45^(@)`. The rate at which periphery of water surface changes when height of water in the cone is 2 m, isA. 2 m/sB. 1m/sC. 3 m/sD. 4 m/s |
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Answer» Correct Answer - a Given that, `(dV)/(dt) =2` ` rArr d/(dt)(1/3 pir^(3)) = 2 rArr pi r^(2) (dr)/(dt) = 2` ` rArr (dr)/(dt) = 2/(pir^(2))` ` rArr d/(dt) (2pir) = 4/(r^(2))" "` …(i) when H = 2m and r = 2m , then ` (dp)/(dt) = 4/4 = 1m//s ` [ where p =` 2 pi r` = perimeter ] |
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