Let ABC is cone, which is converted into a frustum by cutting the top. Let r₁ and r₂ be the radius of two ends and h be its height. In ΔABG and ΔADF, DF∥BG ∴ ΔABG ∼ ΔADF DF/BG = AF/AG = AD/AB ⇒ r²/r₁ = h₁ - h/h₁ ⇒ r²/r₁ = 1 - h/h₁ = 1 - 1/l₁ ⇒ r₂/r₁ = 1 - h/h₁ ⇒ h/h₁ = 1 - r₂/r₁ = (r₁ - r₂/r₁) ⇒ h₁/Read more
Let ABC is cone, which is converted into a frustum by cutting the top. Let r₁ and r₂ be the radius of two ends and h be its height.
In ΔABG and ΔADF, DF∥BG ∴ ΔABG ∼ ΔADF
DF/BG = AF/AG = AD/AB
⇒ r²/r₁ = h₁ – h/h₁ ⇒ r²/r₁ = 1 – h/h₁ = 1 – 1/l₁
⇒ r₂/r₁ = 1 – h/h₁ ⇒ h/h₁ = 1 – r₂/r₁ = (r₁ – r₂/r₁)
⇒ h₁/h = (r₁/r₁ – r₂) ⇒ h₁ = r₁h/r₁ – r₂
Volume of frustum = Volume of cone ABC- Volume of cone ADE
= (1/3)πr₁²h₁ – (1/3)πr₂²(h₁ – h) = 1/3π[r₁²h₁ – r₂²(h₁ – h)]
= (1/3)π[r₁²(r₁h/r₁ – r₂) – r₂²(r₁h/r₁ – r₂) – h)] = (1/3)π[r₁³h/r₁ – r₂) – r₂²(r₁h – r₁h + r₂h/r₁ – r₂)]
= (1/3)π [(r₁³h/r₁ – r₂) – (r₂³h/r₁ -r₂)] = (1/3)πh[r₁³ – r₂³/r₁ – r₂] = (1/3)πh [(r₁ – r₂)(r₁² + r₁r₂ + r₂²/r₁ – r₂] = (1/3)πh(r₁² + r₁r₂ + r₂².
Two formed cones (which are formed by revolving the right angle triangle) are given. Hypotenuse AC = √(3² + 4²) = √25 = 5 cm Area of right angled triangle ABC = 1/2 × AB × AC ⇒ 1/2 × AC × OB = 1/2 × 4 × 3 ⇒ 1/2 × 5 × OB = 6 ⇒ OB = 12/5 = 2.4 cm Volume of double cone = Volume of cone 1 + Volume of coRead more
Two formed cones (which are formed by revolving the right angle triangle) are given.
Hypotenuse AC = √(3² + 4²) = √25 = 5 cm
Area of right angled triangle ABC = 1/2 × AB × AC
⇒ 1/2 × AC × OB = 1/2 × 4 × 3
⇒ 1/2 × 5 × OB = 6
⇒ OB = 12/5 = 2.4 cm
Volume of double cone = Volume of cone 1 + Volume of cone 2
= 1/3πr²h₁ + 1/3πr²h₂
= 1/3πr²(h₁ + h₂) = 1/3πr²(OA + OC)
1/3 × 3.14 × (2.4)² (5)
= 30.14 cm³
Curved surface area of double cone = Curved surface area of cone 1 + curved surface area of cone 2
= πrl₁ + πrl₂ = πr[4 + 3]
= 3.14 × 2.4 × 7 = 52.75 cm².
Volume of cistern = 150 × 120 × 110 = 1980000 cm³ Volume of water in cistern = 129600 cm³ Volume of bricks to be filled in cistern = 1980000 - 129600 = 1850400 cm³ Let, the number of bricks = n Therefore, Volume of n bricks = n × 22.5 × 7.5 × 6.5 = 1096.875 n As each brick absorbs 1/17 of its volumeRead more
Volume of cistern = 150 × 120 × 110 = 1980000 cm³
Volume of water in cistern = 129600 cm³
Volume of bricks to be filled in cistern = 1980000 – 129600 = 1850400 cm³
Let, the number of bricks = n
Therefore, Volume of n bricks = n × 22.5 × 7.5 × 6.5 = 1096.875 n
As each brick absorbs 1/17 of its volume, therefore, the volume of water obsorbed by n bricks = n/17 × 1096.875
According to question,
18504000 + n/17 × 1096.875 = 1096.875n
⇒ 18504000 = 16n/17 × 1096.875n ⇒ n = 1792.41
Hence, 1792 bricks can be put into the cistern without overflow the water.
Rainfall in the velley = 10 cm = 0.1 m. Area of valley = 97280 km² = 97280000000 m² Volume of rain water = 97280000000 × 0.1 = 9728000000 m³ Volume of one river = 1072000 × 75 × 3 = 241200000 m³ Volume of there rivers = 3 × 241200000 = 723600000 m³ Hence, 723600000 m³ ≈ 9728000000 m³
Rainfall in the velley = 10 cm = 0.1 m. Area of valley = 97280 km² = 97280000000 m²
Volume of rain water = 97280000000 × 0.1 = 9728000000 m³
Volume of one river = 1072000 × 75 × 3 = 241200000 m³
Volume of there rivers = 3 × 241200000 = 723600000 m³
Hence, 723600000 m³ ≈ 9728000000 m³
Upper radius of frustum (r₁) = 9 cm and lower radius of frustum (r₂) = 4 cm Therefore, the radius of cylindrical part = 4 cm Height of frustum (h₁) = 22 - 10 = 12 cm Height of cylindrical part (h₂) = 10 cm Slant height of frustum (l) = √(r₁ - r₂)² + h² = √(9 - 4)² + 12² = 13 cm Area of required tinRead more
Upper radius of frustum (r₁) = 9 cm and lower radius of frustum (r₂) = 4 cm
Therefore, the radius of cylindrical part = 4 cm
Height of frustum (h₁) = 22 – 10 = 12 cm
Height of cylindrical part (h₂) = 10 cm
Slant height of frustum (l) = √(r₁ – r₂)² + h² = √(9 – 4)² + 12² = 13 cm
Area of required tin = Curved surface area of frustum + curved surface area of cylindrical part = π(r₂ + r₂)l + 2πr₂h₂
= 22/7 × (9 + 4) × 13 + 2 × 22/7 × 4 × 10
= 22/7 [169 + 80] = (22 × 249/7) = 782(4/7) cm²
Let ABC is a cone, which is converted into a frustum by cutting the top. let r₁ and r₂ be the radius of two ends and h be its height. In ΔABG and ΔADF, DF ∥ BG ∴ ΔABG ∼ ΔADF DF/BG = AF/AG = AD/AB ⇒ r₂/r₁ = (h₁ - h/h₁) = (l₁ - l/l₁) ⇒ r₂/r₁ = 1 - h/h₁ = 1 - l/l₁ ⇒ r₂/r₁ = 1 - l/l₁ ⇒ l/l₁ = 1 - r₂/r₁Read more
Let ABC is a cone, which is converted into a frustum by cutting the top. let r₁ and r₂ be the radius of two ends and h be its height.
In ΔABG and ΔADF, DF ∥ BG
∴ ΔABG ∼ ΔADF
DF/BG = AF/AG = AD/AB
⇒ r₂/r₁ = (h₁ – h/h₁) = (l₁ – l/l₁) ⇒ r₂/r₁ = 1 – h/h₁ = 1 – l/l₁
⇒ r₂/r₁ = 1 – l/l₁ ⇒ l/l₁ = 1 – r₂/r₁ = r₁ – r₂/r₁
⇒ l₁/l = (r₁/r₁ – r₂) ⇒ l₁ = r₁l/r₁ – r₂
Curved surface area of frustum = CSA of cone ABC – CSA of cone ADE = πr₁l₁ – πr₂(l₁ – l)
= πr₁(r₁l/r₁ – r₂) – πr₂((r₁l/r₁ – r₂) – l) = (πr₁²l/r₁ – r₂) – πr₂ (r₁l – r₁l + r₂l/r₁ – r₂)
= (πr₁²l/r₁ -r₂) – (πr₂²l/r₁ – r₂) = πl (r₁² – r₂²/r₁ – r₂) = π(r₁ + r₂)l
total surface area of frustum = Curved surface area + Area of upper circle + Area of lower circle.
= π(r₁ + r₂)l + πr₂² + πr₁²
In ΔAEG, EG/AG = tan 30° ⇒ EG/10 = 1/√3 ⇒ EG = 10/√3 In ΔABD, BD/AD = tan 30° ⇒ BD/20 = 1/√3 ⇒ BD = 20√3 Radius of upper part of frustum (r₁) = 10/√3 Radius of lower part of frustum (r₂) = 20/√3 Height of frustum (h) = 10 cm Volume of frustum = 1/3πh(r₁² + r₂² + r₁r₂) = 1/3π × 10 × [(10/√3)² + (20/√Read more
In ΔAEG,
EG/AG = tan 30°
⇒ EG/10 = 1/√3
⇒ EG = 10/√3
In ΔABD,
BD/AD = tan 30°
⇒ BD/20 = 1/√3
⇒ BD = 20√3
Radius of upper part of frustum (r₁) = 10/√3
Radius of lower part of frustum (r₂) = 20/√3
Height of frustum (h) = 10 cm
Volume of frustum = 1/3πh(r₁² + r₂² + r₁r₂)
= 1/3π × 10 × [(10/√3)² + (20/√3)² + 10/√3 × 20/√3]
= 1/3 × 22/7 × 10 × (100/3 + 400/3 + 200/3)
= 1/3 × 22/7 × 10 × (700/3) = 22000/9 cm³
Radius of wire (r) = 1/2 × 1/16 = 1/32 cm
Let the lenght of wire = l
Volume of wire = area of cross- section of wire × length of wire
= (πr²) (l)
= π(1/32)²× l
Volume of frustum = Volume of wire
⇒ 22000/9 = π(1/32)² × l
⇒ 22000/9 = 22 /7 × 1/1024 × l
⇒ l = 22000/9 × 7/22 × 1024
= 796444.44 cm
7964. 44 m
Hence, the length of wire is 7964.44 m.
It can be observed that 1 round of wire will cover 3 mm height of cylinder. Therefore, the number of round = Height of cylinder/Diameter of wire = 12/0.3 = 40 Lenght of wire for one round = radius of cylinder = 2πr = 2π × 5 = 10π Length of wire to wound 40 times = 40 × 10π = (400 × 22/7) = 8800/7 =Read more
It can be observed that 1 round of wire will cover 3 mm height of cylinder.
Therefore, the number of round = Height of cylinder/Diameter of wire = 12/0.3 = 40
Lenght of wire for one round = radius of cylinder
= 2πr = 2π × 5 = 10π
Length of wire to wound 40 times = 40 × 10π
= (400 × 22/7) = 8800/7
= 1257.14 cm = 12.57 m
Radius of wire = 0.15 cm
Volume of wire = Area of cross- section of wire × length
= π(0.15)² × 1257.14 = 88.898 cm³
Mass = Volume × Density = 88.898 × 8.88 = 789.41 gm
Circumference of upper part of frustum = 18 cm ⇒ 2πr₁ = 18 ⇒r₁ = 9/π Circumference of lower part of frustum = 6 cm ⇒ 2πr₂ = 6 ⇒r₂ = 3/π Height of frustum = 4 cm Curved surface area of the frustum = π (r₁ + r₂)l = π (9/π+3/π)4 = 12 × 4 = 48 cm² Hence, the curved area of the frustum is 48 cm².
Circumference of upper part of frustum = 18 cm
⇒ 2πr₁ = 18 ⇒r₁ = 9/π
Circumference of lower part of frustum = 6 cm
⇒ 2πr₂ = 6 ⇒r₂ = 3/π
Height of frustum = 4 cm
Curved surface area of the frustum = π (r₁ + r₂)l
= π (9/π+3/π)4 = 12 × 4 = 48 cm²
Hence, the curved area of the frustum is 48 cm².
Radius of lower part of cap (r₁) = 10 cm Radius of upper part of cap (r₂) = 4 cm Slant height of cap = 15 cm Area of material used for making it = CSA of frustum + Area of upper part = π(r₁ + r₂)l + πr₂ = π(10 + 4) × 15 + π × 4² = 210π + 16π = 226π = 226 × 22/7 = 710(2/7) cm² Hence, the area of mateRead more
Radius of lower part of cap (r₁) = 10 cm
Radius of upper part of cap (r₂) = 4 cm
Slant height of cap = 15 cm
Area of material used for making it = CSA of frustum + Area of upper part
= π(r₁ + r₂)l + πr₂
= π(10 + 4) × 15 + π × 4²
= 210π + 16π = 226π
= 226 × 22/7 = 710(2/7) cm²
Hence, the area of material used for making is 710(2/7) cm².
Derive the formula for the volume of the frustum of a cone, given to you in Section 13.5, using the symbols as explained.
Let ABC is cone, which is converted into a frustum by cutting the top. Let r₁ and r₂ be the radius of two ends and h be its height. In ΔABG and ΔADF, DF∥BG ∴ ΔABG ∼ ΔADF DF/BG = AF/AG = AD/AB ⇒ r²/r₁ = h₁ - h/h₁ ⇒ r²/r₁ = 1 - h/h₁ = 1 - 1/l₁ ⇒ r₂/r₁ = 1 - h/h₁ ⇒ h/h₁ = 1 - r₂/r₁ = (r₁ - r₂/r₁) ⇒ h₁/Read more
Let ABC is cone, which is converted into a frustum by cutting the top. Let r₁ and r₂ be the radius of two ends and h be its height.
See lessIn ΔABG and ΔADF, DF∥BG ∴ ΔABG ∼ ΔADF
DF/BG = AF/AG = AD/AB
⇒ r²/r₁ = h₁ – h/h₁ ⇒ r²/r₁ = 1 – h/h₁ = 1 – 1/l₁
⇒ r₂/r₁ = 1 – h/h₁ ⇒ h/h₁ = 1 – r₂/r₁ = (r₁ – r₂/r₁)
⇒ h₁/h = (r₁/r₁ – r₂) ⇒ h₁ = r₁h/r₁ – r₂
Volume of frustum = Volume of cone ABC- Volume of cone ADE
= (1/3)πr₁²h₁ – (1/3)πr₂²(h₁ – h) = 1/3π[r₁²h₁ – r₂²(h₁ – h)]
= (1/3)π[r₁²(r₁h/r₁ – r₂) – r₂²(r₁h/r₁ – r₂) – h)] = (1/3)π[r₁³h/r₁ – r₂) – r₂²(r₁h – r₁h + r₂h/r₁ – r₂)]
= (1/3)π [(r₁³h/r₁ – r₂) – (r₂³h/r₁ -r₂)] = (1/3)πh[r₁³ – r₂³/r₁ – r₂] = (1/3)πh [(r₁ – r₂)(r₁² + r₁r₂ + r₂²/r₁ – r₂] = (1/3)πh(r₁² + r₁r₂ + r₂².
A right triangle, whose sides are 3 cm and 4 cm (other than hypotenuse) is made to revolve about its hypotenuse. Find the volume and surface area of the double cone so formed. (Choose value of π as found appropriate.)
Two formed cones (which are formed by revolving the right angle triangle) are given. Hypotenuse AC = √(3² + 4²) = √25 = 5 cm Area of right angled triangle ABC = 1/2 × AB × AC ⇒ 1/2 × AC × OB = 1/2 × 4 × 3 ⇒ 1/2 × 5 × OB = 6 ⇒ OB = 12/5 = 2.4 cm Volume of double cone = Volume of cone 1 + Volume of coRead more
Two formed cones (which are formed by revolving the right angle triangle) are given.
See lessHypotenuse AC = √(3² + 4²) = √25 = 5 cm
Area of right angled triangle ABC = 1/2 × AB × AC
⇒ 1/2 × AC × OB = 1/2 × 4 × 3
⇒ 1/2 × 5 × OB = 6
⇒ OB = 12/5 = 2.4 cm
Volume of double cone = Volume of cone 1 + Volume of cone 2
= 1/3πr²h₁ + 1/3πr²h₂
= 1/3πr²(h₁ + h₂) = 1/3πr²(OA + OC)
1/3 × 3.14 × (2.4)² (5)
= 30.14 cm³
Curved surface area of double cone = Curved surface area of cone 1 + curved surface area of cone 2
= πrl₁ + πrl₂ = πr[4 + 3]
= 3.14 × 2.4 × 7 = 52.75 cm².
A cistern, internally measuring 150 cm × 120 cm × 110 cm, has 129600 cm3 of water in it. Porous bricks are placed in the water until the cistern is full to the brim. Each brick absorbs one-seventeenth of its own volume of water. How many bricks can be put in without overflowing the water, each brick being 22.5 cm × 7.5 cm × 6.5 cm?
Volume of cistern = 150 × 120 × 110 = 1980000 cm³ Volume of water in cistern = 129600 cm³ Volume of bricks to be filled in cistern = 1980000 - 129600 = 1850400 cm³ Let, the number of bricks = n Therefore, Volume of n bricks = n × 22.5 × 7.5 × 6.5 = 1096.875 n As each brick absorbs 1/17 of its volumeRead more
Volume of cistern = 150 × 120 × 110 = 1980000 cm³
See lessVolume of water in cistern = 129600 cm³
Volume of bricks to be filled in cistern = 1980000 – 129600 = 1850400 cm³
Let, the number of bricks = n
Therefore, Volume of n bricks = n × 22.5 × 7.5 × 6.5 = 1096.875 n
As each brick absorbs 1/17 of its volume, therefore, the volume of water obsorbed by n bricks = n/17 × 1096.875
According to question,
18504000 + n/17 × 1096.875 = 1096.875n
⇒ 18504000 = 16n/17 × 1096.875n ⇒ n = 1792.41
Hence, 1792 bricks can be put into the cistern without overflow the water.
In one fortnight of a given month, there was a rainfall of 10 cm in a river valley. If the area of the valley is 7280 km2, show that the total rainfall was approximately equivalent to the addition to the normal water of three rivers each 1072 km long, 75 m wide and 3 m deep.
Rainfall in the velley = 10 cm = 0.1 m. Area of valley = 97280 km² = 97280000000 m² Volume of rain water = 97280000000 × 0.1 = 9728000000 m³ Volume of one river = 1072000 × 75 × 3 = 241200000 m³ Volume of there rivers = 3 × 241200000 = 723600000 m³ Hence, 723600000 m³ ≈ 9728000000 m³
Rainfall in the velley = 10 cm = 0.1 m. Area of valley = 97280 km² = 97280000000 m²
See lessVolume of rain water = 97280000000 × 0.1 = 9728000000 m³
Volume of one river = 1072000 × 75 × 3 = 241200000 m³
Volume of there rivers = 3 × 241200000 = 723600000 m³
Hence, 723600000 m³ ≈ 9728000000 m³
An oil funnel made of tin sheet consists of a 10 cm long cylindrical portion attached to a frustum of a cone. If the total height is 22 cm, diameter of the cylindrical portion is 8 cm and the diameter of the top of the funnel is 18 cm, find the area of the tin sheet required to make the funnel.
Upper radius of frustum (r₁) = 9 cm and lower radius of frustum (r₂) = 4 cm Therefore, the radius of cylindrical part = 4 cm Height of frustum (h₁) = 22 - 10 = 12 cm Height of cylindrical part (h₂) = 10 cm Slant height of frustum (l) = √(r₁ - r₂)² + h² = √(9 - 4)² + 12² = 13 cm Area of required tinRead more
Upper radius of frustum (r₁) = 9 cm and lower radius of frustum (r₂) = 4 cm
See lessTherefore, the radius of cylindrical part = 4 cm
Height of frustum (h₁) = 22 – 10 = 12 cm
Height of cylindrical part (h₂) = 10 cm
Slant height of frustum (l) = √(r₁ – r₂)² + h² = √(9 – 4)² + 12² = 13 cm
Area of required tin = Curved surface area of frustum + curved surface area of cylindrical part = π(r₂ + r₂)l + 2πr₂h₂
= 22/7 × (9 + 4) × 13 + 2 × 22/7 × 4 × 10
= 22/7 [169 + 80] = (22 × 249/7) = 782(4/7) cm²
Derive the formula for the curved surface area and total surface area of the frustum of a cone, given to you in Section 13.5, using the symbols as explained.
Let ABC is a cone, which is converted into a frustum by cutting the top. let r₁ and r₂ be the radius of two ends and h be its height. In ΔABG and ΔADF, DF ∥ BG ∴ ΔABG ∼ ΔADF DF/BG = AF/AG = AD/AB ⇒ r₂/r₁ = (h₁ - h/h₁) = (l₁ - l/l₁) ⇒ r₂/r₁ = 1 - h/h₁ = 1 - l/l₁ ⇒ r₂/r₁ = 1 - l/l₁ ⇒ l/l₁ = 1 - r₂/r₁Read more
Let ABC is a cone, which is converted into a frustum by cutting the top. let r₁ and r₂ be the radius of two ends and h be its height.
See lessIn ΔABG and ΔADF, DF ∥ BG
∴ ΔABG ∼ ΔADF
DF/BG = AF/AG = AD/AB
⇒ r₂/r₁ = (h₁ – h/h₁) = (l₁ – l/l₁) ⇒ r₂/r₁ = 1 – h/h₁ = 1 – l/l₁
⇒ r₂/r₁ = 1 – l/l₁ ⇒ l/l₁ = 1 – r₂/r₁ = r₁ – r₂/r₁
⇒ l₁/l = (r₁/r₁ – r₂) ⇒ l₁ = r₁l/r₁ – r₂
Curved surface area of frustum = CSA of cone ABC – CSA of cone ADE = πr₁l₁ – πr₂(l₁ – l)
= πr₁(r₁l/r₁ – r₂) – πr₂((r₁l/r₁ – r₂) – l) = (πr₁²l/r₁ – r₂) – πr₂ (r₁l – r₁l + r₂l/r₁ – r₂)
= (πr₁²l/r₁ -r₂) – (πr₂²l/r₁ – r₂) = πl (r₁² – r₂²/r₁ – r₂) = π(r₁ + r₂)l
total surface area of frustum = Curved surface area + Area of upper circle + Area of lower circle.
= π(r₁ + r₂)l + πr₂² + πr₁²
A metallic right circular cone 20 cm high and whose vertical angle is 60° is cut into two parts at the middle of its height by a plane parallel to its base. If the frustum so obtained be drawn into a wire of diameter 1/6 cm find the length of the wire.
In ΔAEG, EG/AG = tan 30° ⇒ EG/10 = 1/√3 ⇒ EG = 10/√3 In ΔABD, BD/AD = tan 30° ⇒ BD/20 = 1/√3 ⇒ BD = 20√3 Radius of upper part of frustum (r₁) = 10/√3 Radius of lower part of frustum (r₂) = 20/√3 Height of frustum (h) = 10 cm Volume of frustum = 1/3πh(r₁² + r₂² + r₁r₂) = 1/3π × 10 × [(10/√3)² + (20/√Read more
In ΔAEG,
See lessEG/AG = tan 30°
⇒ EG/10 = 1/√3
⇒ EG = 10/√3
In ΔABD,
BD/AD = tan 30°
⇒ BD/20 = 1/√3
⇒ BD = 20√3
Radius of upper part of frustum (r₁) = 10/√3
Radius of lower part of frustum (r₂) = 20/√3
Height of frustum (h) = 10 cm
Volume of frustum = 1/3πh(r₁² + r₂² + r₁r₂)
= 1/3π × 10 × [(10/√3)² + (20/√3)² + 10/√3 × 20/√3]
= 1/3 × 22/7 × 10 × (100/3 + 400/3 + 200/3)
= 1/3 × 22/7 × 10 × (700/3) = 22000/9 cm³
Radius of wire (r) = 1/2 × 1/16 = 1/32 cm
Let the lenght of wire = l
Volume of wire = area of cross- section of wire × length of wire
= (πr²) (l)
= π(1/32)²× l
Volume of frustum = Volume of wire
⇒ 22000/9 = π(1/32)² × l
⇒ 22000/9 = 22 /7 × 1/1024 × l
⇒ l = 22000/9 × 7/22 × 1024
= 796444.44 cm
7964. 44 m
Hence, the length of wire is 7964.44 m.
A copper wire, 3 mm in diameter, is wound about a cylinder whose length is 12 cm, and diameter 10 cm, so as to cover the curved surface of the cylinder. Find the length and mass of the wire, assuming the density of copper to be 8.88 g per cm3.
It can be observed that 1 round of wire will cover 3 mm height of cylinder. Therefore, the number of round = Height of cylinder/Diameter of wire = 12/0.3 = 40 Lenght of wire for one round = radius of cylinder = 2πr = 2π × 5 = 10π Length of wire to wound 40 times = 40 × 10π = (400 × 22/7) = 8800/7 =Read more
It can be observed that 1 round of wire will cover 3 mm height of cylinder.
See lessTherefore, the number of round = Height of cylinder/Diameter of wire = 12/0.3 = 40
Lenght of wire for one round = radius of cylinder
= 2πr = 2π × 5 = 10π
Length of wire to wound 40 times = 40 × 10π
= (400 × 22/7) = 8800/7
= 1257.14 cm = 12.57 m
Radius of wire = 0.15 cm
Volume of wire = Area of cross- section of wire × length
= π(0.15)² × 1257.14 = 88.898 cm³
Mass = Volume × Density = 88.898 × 8.88 = 789.41 gm
The slant height of a frustum of a cone is 4 cm and the perimeters (circumference) of its circular ends are 18 cm and 6 cm. Find the curved surface area of the frustum.
Circumference of upper part of frustum = 18 cm ⇒ 2πr₁ = 18 ⇒r₁ = 9/π Circumference of lower part of frustum = 6 cm ⇒ 2πr₂ = 6 ⇒r₂ = 3/π Height of frustum = 4 cm Curved surface area of the frustum = π (r₁ + r₂)l = π (9/π+3/π)4 = 12 × 4 = 48 cm² Hence, the curved area of the frustum is 48 cm².
Circumference of upper part of frustum = 18 cm
See less⇒ 2πr₁ = 18 ⇒r₁ = 9/π
Circumference of lower part of frustum = 6 cm
⇒ 2πr₂ = 6 ⇒r₂ = 3/π
Height of frustum = 4 cm
Curved surface area of the frustum = π (r₁ + r₂)l
= π (9/π+3/π)4 = 12 × 4 = 48 cm²
Hence, the curved area of the frustum is 48 cm².
A fez, the cap used by the Turks, is shaped like the frustum of a cone If its radius on the open side is 10 cm, radius at the upper base is 4 cm and its slant height is 15 cm, find the area of material used for making it.
Radius of lower part of cap (r₁) = 10 cm Radius of upper part of cap (r₂) = 4 cm Slant height of cap = 15 cm Area of material used for making it = CSA of frustum + Area of upper part = π(r₁ + r₂)l + πr₂ = π(10 + 4) × 15 + π × 4² = 210π + 16π = 226π = 226 × 22/7 = 710(2/7) cm² Hence, the area of mateRead more
Radius of lower part of cap (r₁) = 10 cm
See lessRadius of upper part of cap (r₂) = 4 cm
Slant height of cap = 15 cm
Area of material used for making it = CSA of frustum + Area of upper part
= π(r₁ + r₂)l + πr₂
= π(10 + 4) × 15 + π × 4²
= 210π + 16π = 226π
= 226 × 22/7 = 710(2/7) cm²
Hence, the area of material used for making is 710(2/7) cm².