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A right circular cone has height 9 cm and radius of the base 5 cm. it is inverted and water is poured into it. If an any instant the water level rises at the rate of(pi/A) cm /sec, where A is the area of the water surface at the instant , show that the vessel will be full in 75 seconds |
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Answer» Solution :LET r be the radius of the water surface and h be the HEIGHT of the water at time t. area of the water surface `A = pir^(2)` sqcm. Since height of the right cicrular cone is 9 cm and radius of the base is 5 cm. ` r/h = 5/9therefore r = 5/9 h` area of water surface i.e., A` = pi (5/9 h)^2` ` A = (25 pih^(2))/(81)` The water level, i.,e, the RATE of change of h is `(DH)/(dt) ` RISES at the rate of `(pi/2)` cm/sec. ` (dh)/(dt) = pi/A = ( pi xx 81)/( 25 pih^(2))` ` (dh)/(dt) = 81/ (25h^(2))"" therefore h^(2)dh = (81)/25 dt` On integrating we get ` int h^(2) dh = (81)/25 int dt +c"" therefore h^(3)/3 81/25 .t +c` Initially ,i.e. when t = 0 , h=0 ` therefore0=0+ctherefore c=0` ` h^(3)/3 = 81/25 t` when the vessel will be full h = 9` therefore ((9)^(3))/3= 81/25 xx t` ` t = (81 xx 9 xx 25)/(3 xx 81) =75` Hence, the vessel will be full in 75 seconds `(NVT_21_MAT_XII_C16_SLV_043_S01)` |
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