the water container in the form of the frustum of a right circular cone contains 44×10×10×10×10×10×10×10 litres of water which fills it completely. the radii of the bottom and top of the container are 50m. and 100m. respectively. Find the depth of water and curved area of the container.(π=22/7)
Answers
12(234) 84
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 a frustum of a 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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