**Exercise 20A**

**Question **1. The height of a circular cylinder is 20 cm and the radius of its base is 7 cm. Find:**(i) the volume****(ii) the total surface area. Solution:**

Given that, Height of the cylinder (h) = 20cm

Radius of the base of the cylinder (r) = 7cm

**Question **2. The inner radius of a pipe is 2.1 cm. How much water can 12m of this pipe hold?

Solution:

It is given that, radius of pipe = 2.1cm and length of the pipe = 12m = 1200cm.

Volume of the cylinder = πr^{2} h

Volume of the cylinder = 22/7× (2.1)^{2} 1200

Volume of the cylinder = 16632cm^{3}

Hence, the pipe can hold 16632cm^{3} of water.

**Question **3. A cylinder of circumference 8cm and length 21cm rolls without sliding for 4(1/2) second at the rate of 9 complete rounds per second. Find:**(i) distance travelled by the cylinder in 4(1/2) seconds and****(ii) the area covered by the cylinder in 4(1/2) seconds.** **Solution:**

(i) It is given that, the circumference of cylinder = 8cm, then distance covered in 9 revolutions = 9 × 8 = 72cm.

Distance covered in 1 second = 72cm

**Question **4. How many cubic meters of earth must be dug out to make a well 28m deep and 2.8m in diameter? Also, find the cost of plastering its inner surface at Rs. 4.50 per sq meter.**Solution:**

It is given that, the diameter of the wall is 2.8m

Radius of the wall = 2.8/2

Radius of the wall = 1.4m

Depth of the wall = 28m

Volume of earth dug out = πr^{2} h

**Question **5. What length of solid cylinder 2 cm in diameter must be taken to recast into a hollow cylinder of external diameter 20 cm, 0.25 cm thick and 15 cm long?

Solution:

It is given that, the external diameter of hollow cylinder = 20cm

Radius = 20/2

Radius = 10cm

Thickness = 0.25cm

Internal radius = 10 – 0.25

Internal radius = 9.75cm

Length of the cylinder (h) = 15cm

Volume of the cylinder = πh(R^{2} – r^{2})

Volume of the cylinder = π×15((10)^{2} – (9.75)^{2} )

Volume of the cylinder = 15π(100 – 95.0625)

Volume of the cylinder = 15π(4.9375)

Diameter = 2cm

Radius = 1cm

Let us assumed that, h be the length,

Volume = πhr^{2}

Volume = πh(1)^{2}

Volume = πh

According to question,

πh = 15π(4.9375)

h = 15(4.9375)

h = 74.0625cm

Hence, the length of the cylinder is 74.0625cm.

**Question **6. A cylinder has a diameter of 20 cm. The area of curved surface is 100 sq. cm. Find:**(i) the height of the cylinder correct to one decimal place.****(ii) the volume of the cylinder correct to one decimal place.Solution:**

(i) It is given that, the diameter of the cylinder = 20cm

Radius = 10cm

Height = hcm

Curve Surface Area = 2πrh

100cm

^{2}= 2πrh

**Question **7. A metal pipe has a bore (inner diameter) of 5 cm. The pipe is 5 mm thick all round. Find the weight, in kilogram, of 2 metres of the pipe if 1 cm3 of the metal weights 7.7g.

Solution:

It is given that, Inner radius of the pipe (r) = 5/2

Radius = 2.5 cm

External radius of the pipe (R) = Inner radius of the pipe + Thickness of the pipe

External radius of the pipe (R) = 2.5cm + 0.5cm

External radius of the pipe (R) = 3cm

Length of the pipe (h) = 2m

Length of the pipe = 200cm

Volume of the pipe = External Volume – Internal Volume

Volume of the pipe = πR^{2} h – πr^{2} h

Volume of the pipe = πh(R^{2} – r^{2} )

Volume of the pipe = π×200×((3)2 – (2.5)2 )

Volume of the pipe = 200π(9 – 6.25)

Volume of the pipe = 200π(2.75)

Volume of the pipe = 200×22/7×(2.75)

Volume of the pipe = 1728.6cm^{3}

1cm3of the metal weights 7.79,

Weight of the pipe = (1728.6 × 7.7)g

Weight of the pipe = (1728.6×7.7/1000)kg

Weight of the pipe = 13.31 kg

**Question **8. A cylindrical container with diameter of base 42 cm contains sufficient water to submerge a rectangular solid of iron with dimensions 22cm × 14cm × 10.5cm. Find the rise in level of the water when the solid is submerged.

Solution:

It is given that, diameter of cylindrical container = 42cm

Radius (r) = 21cm

The dimensions of rectangle solid = 22cm × 14cm × 10.5cm

Volume of the solid = 22cm × 14cm × 10.5cm

Volume of water in the container = πr^{2} h

22/7×(21)^{2}×h = 22cm × 14cm × 10.5cm

22×3×21×h = 22cm×14cm×10.5cm

63h = 14cm×10.5cm

h = 14cm×10.5cm/63

h = 7/3

h = 2.33cm

**Question **9. A cylindrical container with internal radius of its base 10cm, contains water up to a height of 7 cm. Find the area of wetted surface of the cylinder.**Solution:**

It is given that, the internal radius of the cylindrical container = 10cm

Height of water = 7cm

Surface area of the wetted surface = 2πh+πr^{2}

Surface area of the wetted surface = πr(2h+r)

**Question **10. Find the total surface area of an open pipe of length 50 cm, external diameter 20 cm and internal diameter 6cm.

Solution:

It is given that, Length of the open pipe = 50cm

External diameter (d) = 20cm

External radius (r) = 20/2 = 10cm

Internal diameter (d) = 6cm

Internal radius (r) = 6/2 = 3cm

Surface area of pipe open from both sides = 2πRh + 2πrh

Surface area of pipe open from both sides = 2πh(R+r)

**Question **11. The height and radius of base of cylinder are in the ratio 3:1. If its volume is 1029π cm^{3}, find its total surface area.

Solution:

It is given that, the ratio between height and radius of a cylinder = 3:1

Volume = 1029π cm^{3}________(i)

Let us assumed that, radius of the base = r

Height = 3r

Volume of the cylinder = πr^{2} h

Volume of the cylinder = π×r^{2}×3r

Volume of the cylinder = 3πr^{3}_______(ii)

From equation (i) and (ii)

3πr^{3} = 1029π

r^{3} = 1029/3

r^{3} = 343

r = ^{3}√343

r = 7cm

Height of the cylinder = 3×7

Height of the cylinder = 21cm

Total surface area = 2πr(h+r)

Total surface area = 2×22/7×7(21+7)

Total surface area = 2×22×(28)

Total surface area = 1232cm^{2}

**Question **12. Find the minimum length in cm and correct to nearest whole number of the thin metal sheet required to make a hollow and closed cylindrical box of diameter 20cm and height 35cm. Given that the width of the metal sheet is 1 m. Also, find the cost of the sheet at the rate of Rs. 56 per m.**Find the area of metal sheet required, if 10% of it is wasted in cutting, overlapping, etc.** **Solution:**

It is given that, the height of the cylindrical box (h) = 35cm

Radius of the cylindrical box (r) = 10cm

Width of the sheet = 1m = 100cm

Area of sheet = Total surface area of the box

Length × Width = 2πr(r+h)

Area of the metal sheet = Length × Width

Area of the metal sheet = 28 × 100

Area of the metal sheet = 2800cm^{2}

Area of the metal sheet = 00.28m^{2}

It is given that, cost of the sheet at the rate of Rs. 56 per m^{2}

Cost of the sheet = Rs. 56 × 0.28

Cost of the sheet = Rs. 15.68

**Question **13. 3080cm^{3} of water is required to fill a cylindrical vessel completely and 2310 cm^{3} of water is required to fill it upto 5 cm below the top. Find:**(i) radius of the vessel.****(ii) height of the vessel.****(iii) wetted surface area of the vessel when it is half-filled with water. Solution:**

Let us assumed that, the radius of the cylindrical vessel be r and height be h.

Volume of cylindrical vessel = Volume of water filled

πr

^{2}h=3080

**Question **14. A metal cube of side 11 cm is completely submerged in water contained in a cylindrical vessel with diameter 28 cm. Find the rise in the level of water.

Solution:

Here,

Volume of the cube = Volume of water displaced in the cylinder

(11)^{3 }= πr^{2} h

**Question **15. A circular tank of diameter 2m is dug and the earth removed is spread uniformly all around the tank to form an embankment 2m in width and 1.6m in height. Find the depth of the circular tank.

Solution:

Let us assumed that, the depth of the circular tan be hm.

Diameter of the tank = 2m

Radius of the tank = 2/2 = 1m

Volume of the tank = πr^{2} h

Volume of the tank = π×(1)^{2} h

Volume of the tank = πh m^{3}

Volume of the embankment = Volume of hollow cylinder having height 1.6m

Volume of the embankment = πh(R^{2} – r^{2})

Volume of the embankment = π(1.6)(32 – 12)

Volume of the embankment = π(1.6)(9 – 1)

Volume of the embankment = π(1.6)(8)

Volume of the embankment = 12.8πm^{3}

Volume of tank = Volume of embankment

πh = 12.8π

h = 12.8m

**Question **16. The sum of the heights and the radius of a solid cylinder is 35 cm and its total surface area is 3080 cm^{2}, find the volume of the cylinder.

Solution:

Let us assumed that, r is radius and h be the height of a solid cylinder.

r + h = 35cm

Total surface area of a cylinder = 3080cm^{2}

2πr(h + r) = 3080

2×22/7×r(35) = 3080

2×22×r(5) = 3080

r = 3080/(2×22×5)

r = 14cm

It is given that,

14 + h = 35cm

h = 35-14

h = 21cm

Volume of cylinder = πr^{2} h

Volume of cylinder = 22/7×(14)^{2}×21

Volume of cylinder = 22/7×14×14×21

Volume of cylinder = 22×2×14×21

Volume of cylinder = 12936cm^{2}

**Question **17. The given figure shows a solid formed of a solid cube of side 40cm and a solid cylinder of radius 20cm and height 50cm attached to the cube as shown. Find the volume and the total surface area of the whole solid (Take π = 3.14)

**Solution:**

It is given that, Edge of a cube (l) = 40cm

Volume of a cube l^{3}= (40)^{3}

Volume of a cube l^{3}= 64000cm^{3}

Radius of a solid cylinder (r) = 20cm

Height of a solid cylinder (h) = 50cm

Volume of cylinder = πr^{2} h

Volume of cylinder = 22/7×(20)^{2}×50

Volume of cylinder = 22/7×20×20×50

Volume of cylinder = 62800cm^{3}

Volume of the solid = Volume of cube + Volume of cylinder

Volume of the solid = 64000 + 62800

Volume of the solid = 126800cm^{3}

Total surface area of the whole solid = Total surface area of a cube + Curve surface area of cylinder

Total surface area of the whole solid = 6l^{2} + 2πrh

Total surface area of the whole solid

Total surface area of the whole solid = 6(1600) + 6280

Total surface area of the whole solid = 9600 + 6280

Total surface area of the whole solid = 15880cm^{2}

**Question **18. A dosed cylindrical tank, made of thin iron sheet, has diameter = 8.4m and height 5.4m. How much metal sheet, to the nearest m^{2}, is used in making this tank, if 1/15 of the sheet actually used was wasted in making the tank?

Solution:

It is given that, the diameter of the cylindrical tank = 8.4/2 = 4.2m

Height of the cylindrical tank (h) = 5.4m

Total surface area of the cylindrical tank = 2πr(h + r)

Total surface area of the cylindrical tank = 2×22/7×4.2(5.4 + 4.2)

Total surface area of the cylindrical tank = 2×22×0.6(9.6)

Total surface area of the cylindrical tank = 253.44m^{2}

Area of sheet wasted in making the tank = 1/15×253.44

Area of sheet wasted in making the tank = 16.896m^{2}

Total sheet required = 253.44 + 16.896 = 270.34m^{2}

**Exercise 20 B**

**Question **1. Find the volume of a cone whose slant height is 17 cm and radius of base is 8 cm.

Solution:

It is given that, Slant height (l) = 17cm

Radius (r) = 8cm

l^{2} = r^{2} + h^{2}

(17)^{2} = (8)^{2} + h^{2}

289 = 64 + h^{2}

h^{2} = 289-64

h^{2} = 225

h = √225

h = 15

**Question **2. The curved surface area of a cone is 12320cm^{2}. If the radius of its base is 56cm, find its height.

Solution:

It is given that, the Curve surface area = 12320cm^{2} and the radius of base (r) = 56cm

Let us assumed that, the slant height = l

Curve surface area = πrl

12320 = πrl

**Question **3. The circumference of the base of a 12 m high conical tent is 66 m. Find the volume of the air contained in it.

Solution:

It is given that the, Circumference of conical tent = 66cm

Height (h) = 12m

Circumference of circle = 2πr

2πr = 66

**Question **4. The radius and height of a right circular cone are in the ratio 5:12 and its volume is 2512 cubic cm. Find the radius and slant height of the cone. (Take π = 3.14)**Solution:**

It is given that, the ratio between radius and height = 5:12 and Volume = 2512 cubic cm

Let us assumed that, radius (r) = 5x and height (h) = 12x and slant height = l

l^{2} = r2 + h^{2}

l^{2} = (5x)^{2} + (12x)^{2}

l^{2} = 25x^{2} + 144x^{2}

l^{2} = 169x^{2}

l = √169x^{2}

l = 13x

x^{3} = 8

x = 2

Radius = 5x

Radius = 5×2

Radius = 10cm Height = 12x

Height = 12×2

Height = 24cm Slant Height = 13x

Slant Height = 13×2

Slant Height = 26cm

**Question **5. Two right circular cones x and y are made, x having three times the radius of y and y having half the volume of x. Calculate the ratio between the heights of x and y.**Solution:**

Let us assumed that, radius of cone y be r so, radius of cone x = 3r

Also, volume of cone y be v so, radius of cone x = 2v

Here, Height of the cone x = h1 and Height of the cone y = h_{2}

**Question **6. The diameters of two cones are equal. If their slant heights are in the ratio 5:4, find the ratio of their curved surface areas.**Solution:**

It is given that, the ratio of slant height = 5:4.

Let us assumed that, radius of each cone = r, Height of first cone = 5x and Height of second cone = 4x

Curved surface area = πrl

Curved surface area of first cone = πr(5x)

Curved surface area of first cone = 5πrx

Curved surface area of second cone = πr(4x)

Curved surface area of second cone = 4πrx

Ratio between Curved surface area = 5πrx/4πrx

Ratio between Curved surface area = 5:4.

**Question **7. There are two cones. The curved surface area of one is twice that of the other. The slant height of the latter is twice that of the former. Find the ratio of their radii.**Solution:**

Let us assumed that, the height of the first cone = l

Slant height of the second cone = 2l

Radius of the first cone = r_{1}

Radius of the first cone = r_{2}

Curve surface area of first cone = πr_{1} l

Curve surface area of first cone = πr^{2} (2l) = 2πr^{2} l

According to question,

**Question **8. A heap of wheat is in the form of a cone of diameter 16.8 m and height 3.5 m. Find its volume. How much cloth is required to just cover the heap?**Solution:**

It is given that, the diameter of the cone = 16.8m and radius the circle (r) = 16.8/2 = 8.4m

Height (h) = 3.5m

Cloth required or curved surface area = πrl

Cloth required or curved surface area = 22/7×8.4×9.1

Cloth required or curved surface area = 22×1.2×9.1

Cloth required or curved surface area = 240.24m^{2}

**Question **9. Find what length of canvas, 1.5m in width, is required to make a conical tent 48m in diameter and 7m in height. Given that 10% of the canvas is used in folds and stitching. Also, find the cost of the canvas at the rate of Rs. 24 per meter.

Solution:

It is given that, Diameter of the tent = 48m

Radius (r) = 24m and Height (h) = 7m

Here,

**Question **10. A solid cone of height 8cm and base radius 6cm is melted and re-casted into identical cones, each of height 2cm and diameter 1cm. Find the number of cones formed.

Solution:

It is given that, the height of the cone (h) = 8cm

Radius (r) = 6cm

**Question **11. The total surface area of a right circular cone of slant height 13 cm is 90π cm^{2}. Calculate:**(i) its radius in cm****(ii) its volume in cm3 take (π = 3.14)Solution:**

(i) It is given that, total surface area of cone = 90π cm

^{2}and slant height (l) = 13cm

Let us assumed that the radius be r.

Total surface area = πrl + πr

^{2}

Total surface area = πr(l + r)

90π = πr(l + r)

90 = r(13 + r)

r

^{2}+ 13r – 90 = 0

r

^{2}+ 18r – 5r – 90 = 0

r(r + 18) – 5(r + 18) = 0

(r – 5)(r + 18) = 0 r – 5 = 0

r = 5 r + 18 = 0

r = – 18

Radius can’t be negative.

Hence, the radius is 5

**Question **12. The area of the base of a conical solid is 38.5cm^{2} and its volume is 154cm^{3}. Find the curved surface area of the solid.

Solution:

It is given that, the area of the base (πr^{2}) = 38.5cm^{2} and the volume is 154cm^{3}.

We know that,

πr^{2} = 38.5

3.14×r^{2} = 38.5

r^{2} = 38.5/3.14

r^{2} = 12.26

r = √12.26

r = 3.5

Volume of the cone = 1/3 πr^{2} hcm^{3}

154×3 = 38.5×h

Curved surface area of the solid = πrl

Curved surface area of the solid = 3.14×3.5×12.5

Curved surface area of the solid = 137.44cm^{2}