Radius of Spherical part (r₁) = 4.25 cm Height of cylindrical part (h) = 8 cm Radius of cylindrical part (r₂) = 1 cm Volume of vessel = Volume of spherical part + Volume of cylindrical part = 4/3πr₁³ + πr₂²h = 4/3π(8.5/2) ² + π(1) ²(8) = 4/3 × 3.14 × 76.77 + 8 × 3.14 = 321.392 + 25.12 = 346. 51 cm³Read more
Radius of Spherical part (r₁) = 4.25 cm
Height of cylindrical part (h) = 8 cm
Radius of cylindrical part (r₂) = 1 cm
Volume of vessel = Volume of spherical part + Volume of cylindrical part
= 4/3πr₁³ + πr₂²h = 4/3π(8.5/2) ² + π(1) ²(8)
= 4/3 × 3.14 × 76.77 + 8 × 3.14 = 321.392 + 25.12 = 346. 51 cm³
Hence, her answer is incorrect.
Radius of metallic sphere (r₁) = 4.2 cm, radius of cylinder (r₂) = 4.2 cm Let, the height of cylinder = h According to question, volume of sphere = Volume of cylinder ⇒ 4/3πr₁³ = πr₂²h ⇒ 4/3π(4.2)³ = π(6)²h ⇒ h = 4/3 × (4.2 × 4.2 × 4.2/36) = 1.4 × 1.4 × 1.4 = 2.74 cm Hence, the height of cylinder isRead more
Radius of metallic sphere (r₁) = 4.2 cm, radius of cylinder (r₂) = 4.2 cm
Let, the height of cylinder = h
According to question, volume of sphere = Volume of cylinder
⇒ 4/3πr₁³ = πr₂²h ⇒ 4/3π(4.2)³ = π(6)²h
⇒ h = 4/3 × (4.2 × 4.2 × 4.2/36) = 1.4 × 1.4 × 1.4 = 2.74 cm
Hence, the height of cylinder is 2.74 cm.
Height of vessel (h) = 8 cm Radius of vessel (r₁) 5 cm Radium of lead shots (r₂) = 0.5 cm Let the number of lead short dropped = n Therefore, volume of water flows out = n × Volume of 1 lead shot ⇒ 1/4 × Volume of water flow out = n × Volume of 1 lead shot ⇒ 1/4 × 1/3πr₁²h = n × 4/3π³₂ ⇒ r²₁h = n ×Read more
Height of vessel (h) = 8 cm
Radius of vessel (r₁) 5 cm
Radium of lead shots (r₂) = 0.5 cm
Let the number of lead short dropped = n
Therefore, volume of water flows out = n × Volume of 1 lead shot
⇒ 1/4 × Volume of water flow out = n × Volume of 1 lead shot
⇒ 1/4 × 1/3πr₁²h = n × 4/3π³₂
⇒ r²₁h = n × 16 ×r³₂
⇒ 5² × 8 = n × 16 × (0.5)³
⇒ n = (25×8)/(16× (0.5)³ = 100
Hence, the number of lead shots dropped in vessel is 100.
Radius of larger cylinder (r₁) = 12 cm Height of larger cylinder (h₁) = 220 cm Radius of smaller cylinder (r₂) = 8 cm Height of smaller cylinder (h₂) = 60 cm Volume of pole = Volume of larger sylinder + Volume of smaller cylinder = πr₁ ²h₁ + πr₂² h₂ =π(12)² 220 + π(8)² × 60 = π[144 × 220 + 64 × 60]Read more
Radius of larger cylinder (r₁) = 12 cm
Height of larger cylinder (h₁) = 220 cm
Radius of smaller cylinder (r₂) = 8 cm
Height of smaller cylinder (h₂) = 60 cm
Volume of pole = Volume of larger sylinder + Volume of smaller cylinder
= πr₁ ²h₁ + πr₂² h₂
=π(12)² 220 + π(8)² × 60
= π[144 × 220 + 64 × 60] = 3.14 35520 = 111532.8 cm³
Mass of 1 cm³ of iron = 8 g
Therefore, the mass of 111532.8 cm³ of iron
= 111532.8 × 8 g = 892262.4g = 892.262 kg
A spherical glass vessel has a cylindrical neck 8 cm long, 2 cm in diameter; the diameter of the spherical part is 8.5 cm. By measuring the amount of water it holds, a child finds its volume to be 345 cm3. Check whether she is correct, taking the above as the inside measurements, and A spherical glass vessel has a cylindrical neck 8 cm long, 2 cm in diameter; the diameter of the spherical part is 8.5 cm. By measuring the amount of water it holds, a child finds its volume to be 345 cm3. Check whether she is correct, taking the above as the inside measurements, and π= 3.14.
Radius of Spherical part (r₁) = 4.25 cm Height of cylindrical part (h) = 8 cm Radius of cylindrical part (r₂) = 1 cm Volume of vessel = Volume of spherical part + Volume of cylindrical part = 4/3πr₁³ + πr₂²h = 4/3π(8.5/2) ² + π(1) ²(8) = 4/3 × 3.14 × 76.77 + 8 × 3.14 = 321.392 + 25.12 = 346. 51 cm³Read more
Radius of Spherical part (r₁) = 4.25 cm
See lessHeight of cylindrical part (h) = 8 cm
Radius of cylindrical part (r₂) = 1 cm
Volume of vessel = Volume of spherical part + Volume of cylindrical part
= 4/3πr₁³ + πr₂²h = 4/3π(8.5/2) ² + π(1) ²(8)
= 4/3 × 3.14 × 76.77 + 8 × 3.14 = 321.392 + 25.12 = 346. 51 cm³
Hence, her answer is incorrect.
A metallic sphere of radius 4.2 cm is melted and recast into the shape of a cylinder of radius 6 cm. Find the height of the cylinder.
Radius of metallic sphere (r₁) = 4.2 cm, radius of cylinder (r₂) = 4.2 cm Let, the height of cylinder = h According to question, volume of sphere = Volume of cylinder ⇒ 4/3πr₁³ = πr₂²h ⇒ 4/3π(4.2)³ = π(6)²h ⇒ h = 4/3 × (4.2 × 4.2 × 4.2/36) = 1.4 × 1.4 × 1.4 = 2.74 cm Hence, the height of cylinder isRead more
Radius of metallic sphere (r₁) = 4.2 cm, radius of cylinder (r₂) = 4.2 cm
See lessLet, the height of cylinder = h
According to question, volume of sphere = Volume of cylinder
⇒ 4/3πr₁³ = πr₂²h ⇒ 4/3π(4.2)³ = π(6)²h
⇒ h = 4/3 × (4.2 × 4.2 × 4.2/36) = 1.4 × 1.4 × 1.4 = 2.74 cm
Hence, the height of cylinder is 2.74 cm.
A pen stand made of wood is in the shape of a cuboid with four conical depressions to hold pens. The dimensions of the cuboid are 15 cm by 10 cm by 3.5 cm. The radius of each of the depressions is 0.5 cm and the depth is 1.4 cm. Find the volume of wood in the entire stand.
Height of depression (h) = 1.4 cm Radius of depression (r) = 0.5 cm Volume of wood in the entire stand = Volume of cuboid - 4 × Volume of depression = lbh - 4 × 1/3πr²h = 15 × 10 × 3.5 - 4 × 1/3 × 22 /7 × 1/2 × 1/2 × 1.4 = 525 - 1.47 = 523 .53 cm³
Height of depression (h) = 1.4 cm
See lessRadius of depression (r) = 0.5 cm
Volume of wood in the entire stand = Volume of cuboid – 4 × Volume of depression
= lbh – 4 × 1/3πr²h
= 15 × 10 × 3.5 – 4 × 1/3 × 22 /7 × 1/2 × 1/2 × 1.4
= 525 – 1.47
= 523 .53 cm³
A vessel is in the form of an inverted cone. Its height is 8 cm and the radius of its top, which is open, is 5 cm. It is filled with water up to the brim. When lead shots, each of which is a sphere of radius 0.5 cm are dropped into the vessel, one-fourth of the water flows out. Find the number of lead shots dropped in the vessel.
Height of vessel (h) = 8 cm Radius of vessel (r₁) 5 cm Radium of lead shots (r₂) = 0.5 cm Let the number of lead short dropped = n Therefore, volume of water flows out = n × Volume of 1 lead shot ⇒ 1/4 × Volume of water flow out = n × Volume of 1 lead shot ⇒ 1/4 × 1/3πr₁²h = n × 4/3π³₂ ⇒ r²₁h = n ×Read more
Height of vessel (h) = 8 cm
See lessRadius of vessel (r₁) 5 cm
Radium of lead shots (r₂) = 0.5 cm
Let the number of lead short dropped = n
Therefore, volume of water flows out = n × Volume of 1 lead shot
⇒ 1/4 × Volume of water flow out = n × Volume of 1 lead shot
⇒ 1/4 × 1/3πr₁²h = n × 4/3π³₂
⇒ r²₁h = n × 16 ×r³₂
⇒ 5² × 8 = n × 16 × (0.5)³
⇒ n = (25×8)/(16× (0.5)³ = 100
Hence, the number of lead shots dropped in vessel is 100.
A solid iron pole consists of a cylinder of height 220 cm and base diameter 24 cm, which is surmounted by another cylinder of height 60 cm and radius 8 cm. Find the mass of the pole, given that 1 cm3 of iron has approximately 8g mass.
Radius of larger cylinder (r₁) = 12 cm Height of larger cylinder (h₁) = 220 cm Radius of smaller cylinder (r₂) = 8 cm Height of smaller cylinder (h₂) = 60 cm Volume of pole = Volume of larger sylinder + Volume of smaller cylinder = πr₁ ²h₁ + πr₂² h₂ =π(12)² 220 + π(8)² × 60 = π[144 × 220 + 64 × 60]Read more
Radius of larger cylinder (r₁) = 12 cm
See lessHeight of larger cylinder (h₁) = 220 cm
Radius of smaller cylinder (r₂) = 8 cm
Height of smaller cylinder (h₂) = 60 cm
Volume of pole = Volume of larger sylinder + Volume of smaller cylinder
= πr₁ ²h₁ + πr₂² h₂
=π(12)² 220 + π(8)² × 60
= π[144 × 220 + 64 × 60] = 3.14 35520 = 111532.8 cm³
Mass of 1 cm³ of iron = 8 g
Therefore, the mass of 111532.8 cm³ of iron
= 111532.8 × 8 g = 892262.4g = 892.262 kg