A reservoir in the form of the frustum of a right circular cone contains 44*10^7 litres of water which fills it completely.The radii of the bottom and top of the reservoir are 50 m and 100 m respectively .Find the depth of water and the lateral surface area of the reservoir.TakeΠ=22/7
Answers
Answer:
25928.57 m²
Step-by-step explanation:
Let the depth of the reservoir = h
Reservoir Capacity = 44 x 10 ^ 7 liter
Volume of reservoir = 44 x 10 ^ 4 m³ ( 1000 l = 1 m³)
Volume of frustum = Volume of reservoir
1/3 Π h ( r1² + r2² + r1.r2) = 44 x 10 ^ 4 ...... (1)
Here r1 = 100, r2 = 50
Substituting in (1) r1, r2, we get h = 24
Depth of reservoir = 24
Slant Height of frustum = √ ( h² + (r1 - r2)²) = √ 24² + (100 - 50)² = √3076 = 55 m
Lateral area of reservoir = Π l (r1 + r2) = 22/7 x 55 x (100 + 50)
= 25928.57 m²
Answer:
Depth (h) of the reservoir is 24 m and Lateral Surface area of the reservoir is 26,145.42 m².
Step-by-step explanation:
SOLUTION :
GIVEN :
Let ‘h’ be the height of the reservoir which is in the form of frustum of cone.
Radius of the top of the reservoir, R = 100 m
Radius of the bottom of the reservoir, r= 50 m
Volume of the reservoir = 44 × 10^7 litres
= 44 × 10^7 × 10^-3 = 44 × 10⁴ m³
[1 litres = 10^-3 m³]
Volume of the reservoir (frustum of Cone) = π/3 (R² + r² + Rr) h
= ⅓ × π (100² + 50² + 100× 50)× h
= ⅓ π (10000 + 2500 + 5000)× h
= ⅓ × 22/7 × 17500 × h
= (⅓ × 22 × 2500 × h)
(44 × 10⁴) m³ = (⅓ × 22 × 2500 × h)
h = (44 × 10⁴ × 3) / (22 × 2500 )
h = 12 × 10⁴ / 5000
h = 12 × 10⁴ / 5 × 10³
h = 12 × 10 / 5 = 120/5 = 24 m
Depth (h) of the reservoir = 24 m
Slant height of a reservoir , l = √(R - r)² + h²
l =√(100 - 50)² + 24²
l = √50² + 576
l = √2500 + 576
l = √3076
l = 55.46 m
Lateral Surface area of the reservoir = π(R + r)l
= π(100 + 50) × 55.46
= π × 150 × 55.46
= 22/7 × 150 × 55.46
= 183,018/7
= 26,145.42 m²
Lateral Surface area of the reservoir = 26,145.42 m²
Hence, Depth (h) of the reservoir is 24 m and Lateral Surface area of the reservoir is 26,145.42 m².
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