Radius of bucket, (r₁) = 18/ 2 = 9 cm, Height of bucket (h₁) = 32 cm Height of conical heap (h₂) = 24 cm, Let the radius of conical heap = (r₂) Therefore, volume of sand in the bucket = volume of conical heap ⇒ πr₁²h₁ = 1/3πr₂²h₂ ⇒ π × 18² × 32 = 1/3π × r₂²× 24 ⇒ r₂² = (3 × 18 × 18 × 32/24 = 18 × 18Read more
Radius of bucket, (r₁) = 18/ 2 = 9 cm, Height of bucket (h₁) = 32 cm
Height of conical heap (h₂) = 24 cm,
Let the radius of conical heap = (r₂)
Therefore, volume of sand in the bucket = volume of conical heap
⇒ πr₁²h₁ = 1/3πr₂²h₂ ⇒ π × 18² × 32 = 1/3π × r₂²× 24
⇒ r₂² = (3 × 18 × 18 × 32/24 = 18 × 18 × 4 ⇒ r₂ = 36 cm
slant height = √(36² + 24²) = √(1296 + 576) = √1872 = 12√13 cm.
Hence, the radius of conical heap is 36 cm and its slant height is 12√13 cm.
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.
A cylindrical bucket, 32 cm high and with radius of base 18 cm, is filled with sand. This bucket is emptied on the ground and a conical heap of sand is formed. If the height of the conical heap is 24 cm, find the radius and slant height of the heap.
Radius of bucket, (r₁) = 18/ 2 = 9 cm, Height of bucket (h₁) = 32 cm Height of conical heap (h₂) = 24 cm, Let the radius of conical heap = (r₂) Therefore, volume of sand in the bucket = volume of conical heap ⇒ πr₁²h₁ = 1/3πr₂²h₂ ⇒ π × 18² × 32 = 1/3π × r₂²× 24 ⇒ r₂² = (3 × 18 × 18 × 32/24 = 18 × 18Read more
Radius of bucket, (r₁) = 18/ 2 = 9 cm, Height of bucket (h₁) = 32 cm
See lessHeight of conical heap (h₂) = 24 cm,
Let the radius of conical heap = (r₂)
Therefore, volume of sand in the bucket = volume of conical heap
⇒ πr₁²h₁ = 1/3πr₂²h₂ ⇒ π × 18² × 32 = 1/3π × r₂²× 24
⇒ r₂² = (3 × 18 × 18 × 32/24 = 18 × 18 × 4 ⇒ r₂ = 36 cm
slant height = √(36² + 24²) = √(1296 + 576) = √1872 = 12√13 cm.
Hence, the radius of conical heap is 36 cm and its slant height is 12√13 cm.
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.