Asked by | 2nd Mar, 2010, 12:20: PM
1000 lit = 1 m3
44x107 lit = 44x104 m3
Let R and H be the initial radius and height of the cone before the frustrum is created,
and r and h be the radius and height of the cone that was cut off to make the frustrum.
tan α = R/H = r/h ...... α - semivertical angle.
H = Rh/r = 100h/50 = 2h.
Volume of frustrum = Volume of Initial cone - Volume of cutoff cone
= (π/3)(R2H - r2h) = 44x104 m3
(π/3)(2R2 - r2)h = 44x104 m3
(π/3)(7)r2h = 44x104 m3 .........2R2 - r2 = 2(2r)2 - R2 = 8r2 - r2 = 7r2
(22/3)r2h = 44x104 m3
h = 3x44x104/(22x50x50) = 3x2x4 = 24 m
H = 2h = 48 m
Depth of water = 48 - 24 = 24 m
Lateral surface area = CSA of Initial cone - CSA of cutoff cone
= π(RS - rs)
= πr(2S - s)
= πr(2(R2 + H2) - (r2+h2))
= πr(4(r2 + h2) - (r2+h2))
= πr(3(r2 + h2)
= 26122.46 m2.
Answered by | 2nd Mar, 2010, 01:37: PM
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