Radius of upper part of container (r₁) = 20 cm Radius of lower part of container (r₂) = 8 cm Height of container (h) = 16 cm Slant height of container = √((r₁ - r₂)² + h²) = √((20 - 8)² + 16²) = √(144+256) = √400 = 20 cm Capacity of container = Volume of frustum = 1/3πh(r²₁ + r²₂ + r₁r₂) = 1/3 × 3.1Read more
Radius of upper part of container (r₁) = 20 cm
Radius of lower part of container (r₂) = 8 cm
Height of container (h) = 16 cm
Slant height of container = √((r₁ – r₂)² + h²)
= √((20 – 8)² + 16²) = √(144+256) = √400 = 20 cm
Capacity of container = Volume of frustum = 1/3πh(r²₁ + r²₂ + r₁r₂)
= 1/3 × 3.14 × 16 × (20² + 8² + 20 × 8 ) = 1/3 × 3.14 × 16 × (400 + 64 + 160)
= 1/3 × 3.14 × 16 × 624 = 104449.92 cm³ = 10.45 litres.
Cost of 1 litre of milk = Rs 20
Therefore, the cost of 10.45 litres of milk = 10.45 × Rs 20 = Rs 209
The area of metal sheet used to make the container
= π(r₁+ r₂)l + π(r₂)²
= π(20 + 8) 20 + π(8)²
= 560π + 64 π = 624 π cm²
Cost of 100 cm² metal sheet = Rs 8
Cost of cm² metal sheet = Rs 8/100
Therefore, the cost of 624 π cm² metal sheet = Rs 8/100 × 624π = Rs 8/100 × 624 × 3.14 = Rs 156.75
Hence, the cost of the milk which can completely fill the container is Rs 209 and the cost of metal sheet used to make the container is Rs 156.75.
Radius of upper part of glass (r₁) = 4/2 = 2 cm Radius of lower part of glass (r₂) = 2/2 = 1 cm Height = 14 cm Capacity of glass = Volume of frustum = 1/3πh(r₁²+ r₂² + r₁r₂) = 1/3πh(2² + 1² + 2 × 1) = 1/3 × 22/7 × 14 × (7) = 1/3 × 22 × 14 = 308/3 = 102(2/3)cm³. Hence, the capacity of glass is 102(2/Read more
Radius of upper part of glass (r₁) = 4/2 = 2 cm
Radius of lower part of glass (r₂) = 2/2 = 1 cm
Height = 14 cm
Capacity of glass = Volume of frustum
= 1/3πh(r₁²+ r₂² + r₁r₂) = 1/3πh(2² + 1² + 2 × 1) = 1/3 × 22/7 × 14 × (7)
= 1/3 × 22 × 14 = 308/3 = 102(2/3)cm³.
Hence, the capacity of glass is 102(2/3)cm³.
Radius of pipe (r₁) = 20 / 200 = 0.1 m Area of cross- section of pipe = πr₁² = π(0.1)² = 0.01π m³ Speed of water = 3 km/h = 3000/60 = 50 m/min Volume of water flows through in 1 minute = 50 × 0.01π = 0.5πm³ Volume of water flows through in t = t × 0.5π m³ Radius of cylindrical tank (r₂) = 5 m HeightRead more
Radius of pipe (r₁) = 20 / 200 = 0.1 m
Area of cross- section of pipe = πr₁² = π(0.1)² = 0.01π m³
Speed of water = 3 km/h = 3000/60 = 50 m/min
Volume of water flows through in 1 minute = 50 × 0.01π = 0.5πm³
Volume of water flows through in t = t × 0.5π m³
Radius of cylindrical tank (r₂) = 5 m
Height of cylindrical tank (h₂) = 2 m
Let the tank will fill completely in t minutes.
Therefore,
The volume of water flows out in t minutes = volume of cylindrical tank
⇒ t × 0.5π = π ×(r₂)² ×h₂
⇒ t × 0.5 = 5² × 2
⇒ t = 100
Hence, the tank will completely fill in 100 minutes.
Radius of silver coin (r) = 1.75/2 = 0.875 cm, height of silver coin (h₁) = 0.2 cm Lenght of cuboid = 5.5 cm, breadth of cuboid = 10 cm Height of cuboid = 3.5 cm, Let the number of coins = n Therefore, n × volume of 1 silver coin = volume of cuboid ⇒ n(πr²h₁) = lbh ⇒ n × 22/7 × (0.875)² × 0.2 = 5.5Read more
Radius of silver coin (r) = 1.75/2 = 0.875 cm, height of silver coin (h₁) = 0.2 cm
Lenght of cuboid = 5.5 cm, breadth of cuboid = 10 cm Height of cuboid = 3.5 cm,
Let the number of coins = n
Therefore, n × volume of 1 silver coin = volume of cuboid
⇒ n(πr²h₁) = lbh ⇒ n × 22/7 × (0.875)² × 0.2 = 5.5 × 10 × 3.5
⇒ n = (5.5 × 10 × 3.5 × 7/0.875 × 0.875 × 0.2 × 22) ⇒ n = 400
Hence, the number of coins is 400.
A container, opened from the top and made up of a metal sheet, is in the form of a frustum of a cone of height 16 cm with radii of its lower and upper ends as 8 cm and 20 cm, respectively. Find the cost of the milk which can completely fill the container, at the rate of ` 20 per litre. Also find the cost of metal sheet used to make the container, if it costs ` 8 per 100 cm2. (Take π = 3.14)
Radius of upper part of container (r₁) = 20 cm Radius of lower part of container (r₂) = 8 cm Height of container (h) = 16 cm Slant height of container = √((r₁ - r₂)² + h²) = √((20 - 8)² + 16²) = √(144+256) = √400 = 20 cm Capacity of container = Volume of frustum = 1/3πh(r²₁ + r²₂ + r₁r₂) = 1/3 × 3.1Read more
Radius of upper part of container (r₁) = 20 cm
See lessRadius of lower part of container (r₂) = 8 cm
Height of container (h) = 16 cm
Slant height of container = √((r₁ – r₂)² + h²)
= √((20 – 8)² + 16²) = √(144+256) = √400 = 20 cm
Capacity of container = Volume of frustum = 1/3πh(r²₁ + r²₂ + r₁r₂)
= 1/3 × 3.14 × 16 × (20² + 8² + 20 × 8 ) = 1/3 × 3.14 × 16 × (400 + 64 + 160)
= 1/3 × 3.14 × 16 × 624 = 104449.92 cm³ = 10.45 litres.
Cost of 1 litre of milk = Rs 20
Therefore, the cost of 10.45 litres of milk = 10.45 × Rs 20 = Rs 209
The area of metal sheet used to make the container
= π(r₁+ r₂)l + π(r₂)²
= π(20 + 8) 20 + π(8)²
= 560π + 64 π = 624 π cm²
Cost of 100 cm² metal sheet = Rs 8
Cost of cm² metal sheet = Rs 8/100
Therefore, the cost of 624 π cm² metal sheet = Rs 8/100 × 624π = Rs 8/100 × 624 × 3.14 = Rs 156.75
Hence, the cost of the milk which can completely fill the container is Rs 209 and the cost of metal sheet used to make the container is Rs 156.75.
A drinking glass is in the shape of a frustum of a cone of height 14 cm. The diameters of its two circular ends are 4 cm and 2 cm. Find the capacity of the glass.
A drinking glass is in the shape of a frustum of a cone of height 14 cm. The diameters of its two circular ends are 4 cm and 2 cm. Find the capacity of the glass.
Radius of upper part of glass (r₁) = 4/2 = 2 cm Radius of lower part of glass (r₂) = 2/2 = 1 cm Height = 14 cm Capacity of glass = Volume of frustum = 1/3πh(r₁²+ r₂² + r₁r₂) = 1/3πh(2² + 1² + 2 × 1) = 1/3 × 22/7 × 14 × (7) = 1/3 × 22 × 14 = 308/3 = 102(2/3)cm³. Hence, the capacity of glass is 102(2/Read more
Radius of upper part of glass (r₁) = 4/2 = 2 cm
See lessRadius of lower part of glass (r₂) = 2/2 = 1 cm
Height = 14 cm
Capacity of glass = Volume of frustum
= 1/3πh(r₁²+ r₂² + r₁r₂) = 1/3πh(2² + 1² + 2 × 1) = 1/3 × 22/7 × 14 × (7)
= 1/3 × 22 × 14 = 308/3 = 102(2/3)cm³.
Hence, the capacity of glass is 102(2/3)cm³.
A farmer connects a pipe of internal diameter 20 cm from a canal into a cylindrical tank in her field, which is 10 m in diameter and 2 m deep. If water flows through the pipe at the rate of 3 km/h, in how much time will the tank be filled?
Radius of pipe (r₁) = 20 / 200 = 0.1 m Area of cross- section of pipe = πr₁² = π(0.1)² = 0.01π m³ Speed of water = 3 km/h = 3000/60 = 50 m/min Volume of water flows through in 1 minute = 50 × 0.01π = 0.5πm³ Volume of water flows through in t = t × 0.5π m³ Radius of cylindrical tank (r₂) = 5 m HeightRead more
Radius of pipe (r₁) = 20 / 200 = 0.1 m
See lessArea of cross- section of pipe = πr₁² = π(0.1)² = 0.01π m³
Speed of water = 3 km/h = 3000/60 = 50 m/min
Volume of water flows through in 1 minute = 50 × 0.01π = 0.5πm³
Volume of water flows through in t = t × 0.5π m³
Radius of cylindrical tank (r₂) = 5 m
Height of cylindrical tank (h₂) = 2 m
Let the tank will fill completely in t minutes.
Therefore,
The volume of water flows out in t minutes = volume of cylindrical tank
⇒ t × 0.5π = π ×(r₂)² ×h₂
⇒ t × 0.5 = 5² × 2
⇒ t = 100
Hence, the tank will completely fill in 100 minutes.
How many silver coins, 1.75 cm in diameter and of thickness 2 mm, must be melted to form a cuboid of dimensions 5.5 cm × 10 cm × 3.5 cm?
Radius of silver coin (r) = 1.75/2 = 0.875 cm, height of silver coin (h₁) = 0.2 cm Lenght of cuboid = 5.5 cm, breadth of cuboid = 10 cm Height of cuboid = 3.5 cm, Let the number of coins = n Therefore, n × volume of 1 silver coin = volume of cuboid ⇒ n(πr²h₁) = lbh ⇒ n × 22/7 × (0.875)² × 0.2 = 5.5Read more
Radius of silver coin (r) = 1.75/2 = 0.875 cm, height of silver coin (h₁) = 0.2 cm
See lessLenght of cuboid = 5.5 cm, breadth of cuboid = 10 cm Height of cuboid = 3.5 cm,
Let the number of coins = n
Therefore, n × volume of 1 silver coin = volume of cuboid
⇒ n(πr²h₁) = lbh ⇒ n × 22/7 × (0.875)² × 0.2 = 5.5 × 10 × 3.5
⇒ n = (5.5 × 10 × 3.5 × 7/0.875 × 0.875 × 0.2 × 22) ⇒ n = 400
Hence, the number of coins is 400.