Width of canal = 6 m, depth of canal = 1.5 m Speed of water = 10 km /h = 1000/60 m/min Volume of water in 1 minite = 6 × 1.5 × 1000/60 = 1500m³ Therefore, the volume of water in 30 minutes = 30 × 1500 = 45000 m³ Let the area of irrigated field = A m² Therefore, Volume of water from canal in 30 minutRead more
Width of canal = 6 m, depth of canal = 1.5 m
Speed of water = 10 km /h = 1000/60 m/min
Volume of water in 1 minite = 6 × 1.5 × 1000/60 = 1500m³
Therefore, the volume of water in 30 minutes
= 30 × 1500 = 45000 m³
Let the area of irrigated field = A m²
Therefore,
Volume of water from canal in 30 minutes = volume of water in the field
⇒ 45000 = (A × 8/100)
⇒ A = (45000 × 100/8) = 562500 m²
Hence, canal irrigate 562500 m² area is 30 minutes.
Radius of first sphere (r₁) = 6 cm, Radius of second sphere (r₂) = 8 cm Radius of third sphere (r₃) = 10 cm, Let the radius of the new sphere = r According to question, volumes of three spheres = Volume of new sphere ⇒ 4/3π(r₁³ + r₂³ + r₃³) 4/3πr ³ ⇒ 4/3π (6³+ 8³ + 10³) = 4/3πr³ ⇒ 216 + 512 + 1000 =Read more
Radius of first sphere (r₁) = 6 cm, Radius of second sphere (r₂) = 8 cm
Radius of third sphere (r₃) = 10 cm,
Let the radius of the new sphere = r
According to question, volumes of three spheres = Volume of new sphere
⇒ 4/3π(r₁³ + r₂³ + r₃³) 4/3πr ³ ⇒ 4/3π (6³+ 8³ + 10³) = 4/3πr³
⇒ 216 + 512 + 1000 = r³ ⇒ r³ = 1728 ⇒ r = 12 cm
Hence, the radius of resulting sphere is 12 cm.
Radius of well (r) = 7/2m, Height of well = 20 m Lenght of platfrom = 22 m, Width of platform = 14 m Let the height of the platform = H According to Question, volume of the earth dug out = Volume of earth of platform πr²h = 22 × 14 × H ⇒ 22/7 × (7/2)² × 20 = 22 × 14 × H ⇒ 11 × 7 × 10 = 22 × 14 × H ⇒Read more
Radius of well (r) = 7/2m, Height of well = 20 m
Lenght of platfrom = 22 m, Width of platform = 14 m
Let the height of the platform = H
According to Question, volume of the earth dug out = Volume of earth of platform
πr²h = 22 × 14 × H
⇒ 22/7 × (7/2)² × 20 = 22 × 14 × H
⇒ 11 × 7 × 10 = 22 × 14 × H
⇒ (11 × 7 × 10/22 × 14) = H ⇒ H = 2.5 m
Hence, the height of the platform is 2.5 m.
Radius of well (r) = 3/2 m, Depth of well h₁ = 14 m, Width of embankment = 14 m The embankment is in the form of hollow cylinder with internal radius r₂ = 3/2 m and Outer radius r₁ = 3/2 + 4 = 11/2 m Let the height of embankment = H According to question, the earth dug out from the well = volume ofRead more
Radius of well (r) = 3/2 m, Depth of well h₁ = 14 m, Width of embankment = 14 m
The embankment is in the form of hollow cylinder with internal radius r₂ = 3/2 m and Outer radius r₁ = 3/2 + 4 = 11/2 m
Let the height of embankment = H
According to question, the earth dug out from the well = volume of embankment
⇒ πr₁² = π (r₁² – r₂² )h
⇒ π(3/2)² 14 = π [(11/2)² – (3/2)²] × h ⇒ 9/4 14 = 112/4h ⇒ h = 9/8 = 1.125m
Hence, the height of embankment is 1.125 m.
Radius of cylinder (r₁) = 12/2 = 6 cm, height of cylinder (h₁) = 15 cm Height of conical part (h₂) = 12 cm, radius of conical part (r₂) = 6/2 = 3 cm Radius of hemisphere (r₂) = 3 cm, let the number of ice cream cones = n Therefore, volume of cylinder = n × [volume of cone + volume of hemisphere] ⇒ πRead more
Radius of cylinder (r₁) = 12/2 = 6 cm, height of cylinder (h₁) = 15 cm
Height of conical part (h₂) = 12 cm, radius of conical part (r₂) = 6/2 = 3 cm
Radius of hemisphere (r₂) = 3 cm, let the number of ice cream cones = n
Therefore, volume of cylinder = n × [volume of cone + volume of hemisphere]
⇒ πr₁²h₁ = n(1/3πr₂²h + 2/3πr₂³) ⇒ 6² × 15 = n(1/3 × 9 × 12 + 2/3 × 3³)
⇒ 36 × 15 = n (36 + 18) ⇒ n = (36 × 15)/54 = 10
Hence, the number of cones of filled with ice cream is 10.
Water in a canal, 6 m wide and 1.5 m deep, is flowing with a speed of 10 km/h. How much area will it irrigate in 30 minutes, if 8 cm of standing water is needed?
Width of canal = 6 m, depth of canal = 1.5 m Speed of water = 10 km /h = 1000/60 m/min Volume of water in 1 minite = 6 × 1.5 × 1000/60 = 1500m³ Therefore, the volume of water in 30 minutes = 30 × 1500 = 45000 m³ Let the area of irrigated field = A m² Therefore, Volume of water from canal in 30 minutRead more
Width of canal = 6 m, depth of canal = 1.5 m
See lessSpeed of water = 10 km /h = 1000/60 m/min
Volume of water in 1 minite = 6 × 1.5 × 1000/60 = 1500m³
Therefore, the volume of water in 30 minutes
= 30 × 1500 = 45000 m³
Let the area of irrigated field = A m²
Therefore,
Volume of water from canal in 30 minutes = volume of water in the field
⇒ 45000 = (A × 8/100)
⇒ A = (45000 × 100/8) = 562500 m²
Hence, canal irrigate 562500 m² area is 30 minutes.
Metallic spheres of radii 6 cm, 8 cm and 10 cm, respectively, are melted to form a single solid sphere. Find the radius of the resulting sphere.
Radius of first sphere (r₁) = 6 cm, Radius of second sphere (r₂) = 8 cm Radius of third sphere (r₃) = 10 cm, Let the radius of the new sphere = r According to question, volumes of three spheres = Volume of new sphere ⇒ 4/3π(r₁³ + r₂³ + r₃³) 4/3πr ³ ⇒ 4/3π (6³+ 8³ + 10³) = 4/3πr³ ⇒ 216 + 512 + 1000 =Read more
Radius of first sphere (r₁) = 6 cm, Radius of second sphere (r₂) = 8 cm
See lessRadius of third sphere (r₃) = 10 cm,
Let the radius of the new sphere = r
According to question, volumes of three spheres = Volume of new sphere
⇒ 4/3π(r₁³ + r₂³ + r₃³) 4/3πr ³ ⇒ 4/3π (6³+ 8³ + 10³) = 4/3πr³
⇒ 216 + 512 + 1000 = r³ ⇒ r³ = 1728 ⇒ r = 12 cm
Hence, the radius of resulting sphere is 12 cm.
A 20 m deep well with diameter 7 m is dug and the earth from digging is evenly spread out to form a platform 22 m by 14 m. Find the height of the platform.
Radius of well (r) = 7/2m, Height of well = 20 m Lenght of platfrom = 22 m, Width of platform = 14 m Let the height of the platform = H According to Question, volume of the earth dug out = Volume of earth of platform πr²h = 22 × 14 × H ⇒ 22/7 × (7/2)² × 20 = 22 × 14 × H ⇒ 11 × 7 × 10 = 22 × 14 × H ⇒Read more
Radius of well (r) = 7/2m, Height of well = 20 m
See lessLenght of platfrom = 22 m, Width of platform = 14 m
Let the height of the platform = H
According to Question, volume of the earth dug out = Volume of earth of platform
πr²h = 22 × 14 × H
⇒ 22/7 × (7/2)² × 20 = 22 × 14 × H
⇒ 11 × 7 × 10 = 22 × 14 × H
⇒ (11 × 7 × 10/22 × 14) = H ⇒ H = 2.5 m
Hence, the height of the platform is 2.5 m.
A well of diameter 3 m is dug 14 m deep. The earth taken out of it has been spread evenly all around it in the shape of a circular ring of width 4 m to form an embankment. Find the height of the embankment.
Radius of well (r) = 3/2 m, Depth of well h₁ = 14 m, Width of embankment = 14 m The embankment is in the form of hollow cylinder with internal radius r₂ = 3/2 m and Outer radius r₁ = 3/2 + 4 = 11/2 m Let the height of embankment = H According to question, the earth dug out from the well = volume ofRead more
Radius of well (r) = 3/2 m, Depth of well h₁ = 14 m, Width of embankment = 14 m
See lessThe embankment is in the form of hollow cylinder with internal radius r₂ = 3/2 m and Outer radius r₁ = 3/2 + 4 = 11/2 m
Let the height of embankment = H
According to question, the earth dug out from the well = volume of embankment
⇒ πr₁² = π (r₁² – r₂² )h
⇒ π(3/2)² 14 = π [(11/2)² – (3/2)²] × h ⇒ 9/4 14 = 112/4h ⇒ h = 9/8 = 1.125m
Hence, the height of embankment is 1.125 m.
A container shaped like a right circular cylinder having diameter 12 cm and height 15 cm is full of ice cream. The ice cream is to be filled into cones of height 12 cm and diameter 6 cm, having a hemispherical shape on the top. Find the number of such cones which can be filled with ice cream.
Radius of cylinder (r₁) = 12/2 = 6 cm, height of cylinder (h₁) = 15 cm Height of conical part (h₂) = 12 cm, radius of conical part (r₂) = 6/2 = 3 cm Radius of hemisphere (r₂) = 3 cm, let the number of ice cream cones = n Therefore, volume of cylinder = n × [volume of cone + volume of hemisphere] ⇒ πRead more
Radius of cylinder (r₁) = 12/2 = 6 cm, height of cylinder (h₁) = 15 cm
See lessHeight of conical part (h₂) = 12 cm, radius of conical part (r₂) = 6/2 = 3 cm
Radius of hemisphere (r₂) = 3 cm, let the number of ice cream cones = n
Therefore, volume of cylinder = n × [volume of cone + volume of hemisphere]
⇒ πr₁²h₁ = n(1/3πr₂²h + 2/3πr₂³) ⇒ 6² × 15 = n(1/3 × 9 × 12 + 2/3 × 3³)
⇒ 36 × 15 = n (36 + 18) ⇒ n = (36 × 15)/54 = 10
Hence, the number of cones of filled with ice cream is 10.