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Question
A reservoir in the form of the frustum of a right circular cone contains 44 × 107 litres of water which fills it completely. The radii of the bottom and top of the reservoir are 50 metres and 100 metres respectively. Find the depth of water and the lateral surface area of the reservoir. (Take: π = 22/7)
Solution
Let the depth of the frustum cone like reservoir is hm. The radii of the top and bottom circles of the frustum cone like reservoir are r1 =100m and r2 =50m respectively.
The volume of the reservoir is
`V=1/3pi(r_1^2+r_1r_2+r_2^2)xxh`
`=1/3pi(100^2+100xx50+50^2)xxh`
`=1/3xx22/7xx17500xxh`
`=1/3xx22xx2500xxh m^3`
`=1/3xx22xx2500xxhxx10^6 cm^3`
`=1/3xx22xx2500xxhxx10^3` liters
Given that the volume of the reservoir is44x107. Thus, we have
`1/3xx22xx2500xxhxx10^3=44xx10^7`
⇒ `h=(3xx44xx10^7)/(22xx2500xx10^3)`
⇒ h = 24
Hence, the depth of water in the reservoir is 24 m
The slant height of the reservoir is
`l=sqrt((r_1-r_2)^2+h^2)`
`=sqrt((100-50)^2+24^2)`
`=sqrt(3076)`
= 55.46169 meter
The lateral surface area of the reservoir is
`S_1=pi(r_1+r_2)xxl`
= π x (100+50) x 55.46169
= π x 150 x 55.46169
= 26145.225 m2
Hence, the lateral surface area is 26145.225 m2
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