Radius of hemispherical part = radius of cylindrical part (r) = 3.5 cm Height of cylindrical part (h) = 10 m The total surface area of article = CSA of cylindrical part + CSA of two hemispherical part = 2πrh + 2 × 2πr² = 2π × 3.5 × 10 + 2 × 2π × 3.5 ×3.5 = 70π + 49π = 119π = 119 × (22/7) = 17 × 22 =Read more
Radius of hemispherical part = radius of cylindrical part (r) = 3.5 cm
Height of cylindrical part (h) = 10 m
The total surface area of article
= CSA of cylindrical part + CSA of two hemispherical part
= 2πrh + 2 × 2πr²
= 2π × 3.5 × 10 + 2 × 2π × 3.5 ×3.5
= 70π + 49π = 119π
= 119 × (22/7) = 17 × 22 = 374 cm²
Hence, the total surface area of the article is 374 cm²
Height of conical part (h) = dadius of conical part (r) = 1 cm Radius of conical part (r) = radius of hemispherical part (r) = 1 cm Volume of Solid = Volume of conical part + volume of hemispherical part = 1/3πr²h + 2/3πr³ = 1/3π. 1² . 1 + 2/3π. 1³ = πcm³
Height of conical part (h) = dadius of conical part (r) = 1 cm
Radius of conical part (r) = radius of hemispherical part (r) = 1 cm
Volume of Solid = Volume of conical part + volume of hemispherical part
= 1/3πr²h + 2/3πr³
= 1/3π. 1² . 1 + 2/3π. 1³ = πcm³
Height of conical part (h₁) = 2 cm Radius of conical part (r) = Radius of cylindrical part (r) = 3/2 cm Height of cylindrical part (h₂) = 12 - 2 × Height of conical part = 12 - 2 × 2 = 8 cm The volume of air contained in the modal = Volume of cylindrical part + 2 × Volume of conical part = πr²h₂ + 2Read more
Height of conical part (h₁) = 2 cm
Radius of conical part (r) = Radius of cylindrical part (r) = 3/2 cm
Height of cylindrical part (h₂) = 12 – 2 × Height of conical part
= 12 – 2 × 2 = 8 cm
The volume of air contained in the modal
= Volume of cylindrical part + 2 × Volume of conical part
= πr²h₂ + 2 × 1/3πr²h₁
= π(3/2)² . 8 + 2 × 1/3π(3/2)².2
= 18π + 3π = 21π = 21 × 22/7 = 66 cm³.
Radium of hemispherical part (r) = Radius of cylindrical part (r) = 2.8/2 = 1.4 cm Height of Hemispherical part = Radius of hemispherical part = 1.4 cm Height of cylindrical part (h2) = 5 -2 × height of hemispherical part = 5 - 2 × 1.4 = 2.2 cm Volume of 1 Gulab jamun = Volume of cylindrical part +Read more
Radium of hemispherical part (r) = Radius of cylindrical part (r) = 2.8/2 = 1.4 cm
Height of Hemispherical part = Radius of hemispherical part = 1.4 cm
Height of cylindrical part (h2) = 5 -2 × height of hemispherical part = 5 – 2 × 1.4 = 2.2 cm
Volume of 1 Gulab jamun = Volume of cylindrical part + 2 × Volume of hemispherical part.
= πr²h + 2 × 2/3πr³ = πr²h + 4/3πr³
= π(1.4)² (2.2) + 4/3π(1.4)³
= 22/7 × 1.4 × 1.4 × 2.2 + 4/3 × 22/7 × 1.4 × 1.4 × 1.4
= 13.552 + 11.498
= 25.05 cm³
Volume of 45 Gulab Jamun
= 45 × 25.5
= 1,127.25 cm³
Volume of sugar syrup = 30% of volume of 45 Gulab Jamun
= 30/100 × 1127.25
= 338.17 cm³
Height of cylindrical part (h) = 2.1 m Diameter of cylindrical part = 4 m Radius of cylindrical part (r) = 2 m Height of conical part (l) = 2.8 m Area of convas used = CSA conical part + CSA of cylindrical part = πrl + 2πrh = π × 2 × 2.8 + 2π × 2 × 2.1 = 2π(2.8 + 4.2) = 2 × (22/7) × 7 = 44m² The CosRead more
Height of cylindrical part (h) = 2.1 m
Diameter of cylindrical part = 4 m
Radius of cylindrical part (r) = 2 m
Height of conical part (l) = 2.8 m
Area of convas used = CSA conical part + CSA of cylindrical part
= πrl + 2πrh
= π × 2 × 2.8 + 2π × 2 × 2.1
= 2π(2.8 + 4.2) = 2 × (22/7) × 7 = 44m²
The Cost of 1 m² canvas = Rs500
Therefore, the cost of 44m² canvas = 44 × Rs.500 = Rs.22000.
A wooden article was made by scooping out a hemisphere from each end of a solid cylinder, If the height of the cylinder is 10 cm, and its base is of radius 3.5 cm, find the total surface area of the article.
Radius of hemispherical part = radius of cylindrical part (r) = 3.5 cm Height of cylindrical part (h) = 10 m The total surface area of article = CSA of cylindrical part + CSA of two hemispherical part = 2πrh + 2 × 2πr² = 2π × 3.5 × 10 + 2 × 2π × 3.5 ×3.5 = 70π + 49π = 119π = 119 × (22/7) = 17 × 22 =Read more
Radius of hemispherical part = radius of cylindrical part (r) = 3.5 cm
See lessHeight of cylindrical part (h) = 10 m
The total surface area of article
= CSA of cylindrical part + CSA of two hemispherical part
= 2πrh + 2 × 2πr²
= 2π × 3.5 × 10 + 2 × 2π × 3.5 ×3.5
= 70π + 49π = 119π
= 119 × (22/7) = 17 × 22 = 374 cm²
Hence, the total surface area of the article is 374 cm²
A solid is in the shape of a cone standing on a hemisphere with both their radii being equal to 1 cm and the height of the cone is equal to its radius. Find the volume of the solid in terms of p.
Height of conical part (h) = dadius of conical part (r) = 1 cm Radius of conical part (r) = radius of hemispherical part (r) = 1 cm Volume of Solid = Volume of conical part + volume of hemispherical part = 1/3πr²h + 2/3πr³ = 1/3π. 1² . 1 + 2/3π. 1³ = πcm³
Height of conical part (h) = dadius of conical part (r) = 1 cm
See lessRadius of conical part (r) = radius of hemispherical part (r) = 1 cm
Volume of Solid = Volume of conical part + volume of hemispherical part
= 1/3πr²h + 2/3πr³
= 1/3π. 1² . 1 + 2/3π. 1³ = πcm³
Rachel, an engineering student, was asked to make a model shaped like a cylinder with two cones attached at its two ends by using a thin aluminium sheet. The diameter of the model is 3 cm and its length is 12 cm. If each cone has a height of 2 cm, find the volume of air contained in the model that Rachel made. (Assume the outer and inner dimensions of the model to be nearly the same.)
Height of conical part (h₁) = 2 cm Radius of conical part (r) = Radius of cylindrical part (r) = 3/2 cm Height of cylindrical part (h₂) = 12 - 2 × Height of conical part = 12 - 2 × 2 = 8 cm The volume of air contained in the modal = Volume of cylindrical part + 2 × Volume of conical part = πr²h₂ + 2Read more
Height of conical part (h₁) = 2 cm
See lessRadius of conical part (r) = Radius of cylindrical part (r) = 3/2 cm
Height of cylindrical part (h₂) = 12 – 2 × Height of conical part
= 12 – 2 × 2 = 8 cm
The volume of air contained in the modal
= Volume of cylindrical part + 2 × Volume of conical part
= πr²h₂ + 2 × 1/3πr²h₁
= π(3/2)² . 8 + 2 × 1/3π(3/2)².2
= 18π + 3π = 21π = 21 × 22/7 = 66 cm³.
A gulab jamun, contains sugar syrup up to about 30% of its volume. Find approximately how much syrup would be found in 45 gulab jamuns, each shaped like a cylinder with two hemispherical ends with length 5 cm and diameter 2.8 cm
Radium of hemispherical part (r) = Radius of cylindrical part (r) = 2.8/2 = 1.4 cm Height of Hemispherical part = Radius of hemispherical part = 1.4 cm Height of cylindrical part (h2) = 5 -2 × height of hemispherical part = 5 - 2 × 1.4 = 2.2 cm Volume of 1 Gulab jamun = Volume of cylindrical part +Read more
Radium of hemispherical part (r) = Radius of cylindrical part (r) = 2.8/2 = 1.4 cm
Height of Hemispherical part = Radius of hemispherical part = 1.4 cm
Height of cylindrical part (h2) = 5 -2 × height of hemispherical part = 5 – 2 × 1.4 = 2.2 cm
Volume of 1 Gulab jamun = Volume of cylindrical part + 2 × Volume of hemispherical part.
= πr²h + 2 × 2/3πr³ = πr²h + 4/3πr³
= π(1.4)² (2.2) + 4/3π(1.4)³
= 22/7 × 1.4 × 1.4 × 2.2 + 4/3 × 22/7 × 1.4 × 1.4 × 1.4
= 13.552 + 11.498
= 25.05 cm³
Volume of 45 Gulab Jamun
See less= 45 × 25.5
= 1,127.25 cm³
Volume of sugar syrup = 30% of volume of 45 Gulab Jamun
= 30/100 × 1127.25
= 338.17 cm³
A tent is in the shape of a cylinder surmounted by a conical top. If the height and diameter of the cylindrical part are 2.1 m and 4 m respectively, and the slant height of the top is 2.8 m, find the area of the canvas used for making the tent. Also, find the cost of the canvas of the tent at the rate of ` 500 per m2. (Note that the base of the tent will not be covered with canvas.)
Height of cylindrical part (h) = 2.1 m Diameter of cylindrical part = 4 m Radius of cylindrical part (r) = 2 m Height of conical part (l) = 2.8 m Area of convas used = CSA conical part + CSA of cylindrical part = πrl + 2πrh = π × 2 × 2.8 + 2π × 2 × 2.1 = 2π(2.8 + 4.2) = 2 × (22/7) × 7 = 44m² The CosRead more
Height of cylindrical part (h) = 2.1 m
See lessDiameter of cylindrical part = 4 m
Radius of cylindrical part (r) = 2 m
Height of conical part (l) = 2.8 m
Area of convas used = CSA conical part + CSA of cylindrical part
= πrl + 2πrh
= π × 2 × 2.8 + 2π × 2 × 2.1
= 2π(2.8 + 4.2) = 2 × (22/7) × 7 = 44m²
The Cost of 1 m² canvas = Rs500
Therefore, the cost of 44m² canvas = 44 × Rs.500 = Rs.22000.