**ML Aggarwal ICSE Solutions Class 10 Math 17th Chapter Mensuration**

ML Aggarwal ICSE Solutions Class 10 Math Chapter 17 Mensuration

**Exercise 17.1**

**(1) Find the total surface area of a cylinder of radius 5 cm and height 10 cm. Leave your answer in terms of ****π.**

**Solution:**

Given that,

Radius of a cylinder = l = 5 cm

Height of a cylinder = h = 10 cm

We know that, TSA of a cylinder is given by

Total surface area of a cylinder} = 2 π r (h + r)

= 2 π 5 (10 + 5)

= 2 π 5 × 15

= 10 × 15 π

TSA = 150 π cm^{2} is the required total surface area of a given cylinder.

**(2) An electric geyser is cylindrical in shape, having a diameter of 35 cm and height 1.2 m. neglecting the thickness of its walls, calculate (i) its outer lateral surface area**

**(ii) its capacity in liters.**

**Solution:**

Given that, diameter of cylindrical geyser = d = 35 cm

Radius of cylindrical geyser = d/2 = r = 17.5 cm

Height of cylindrical geyser = h = 1.2 m = 120 cm

(i) Then outer lateral surface area of cylindrical geyser is given by LSA = 2 π rh

= 2 × 22/7 × 17.5 × 120

LSA = 13200 cm^{2}

This is the required outer lateral surface area of cylinder.

(ii) Capacity of a cylinder is given by

Capacity Volume of cylindrical geyser} = π r^{2}h

= 22/7 × (17.5)^{2} × 120

= 115500 cm^{3}

Volume = 115. 5 liters

This is the required capacity of given cylindrical geyser.

**(3) A school provides milk to the students daily in cylindrical glasses of diameter 7 cm. If the glass is filled with milk up to a height of 12 cm, find how many liters of milk is needed to serve 1600 students.**

**Solution:**

Given that, diameter of glass (cylindrical) = d = 7 cm

Then, Radius of cylindrical glass = r = 7/2 = 3.5 cm

Height of cylindrical glass = h = 12 cm

Total no. of students in a school = 1600

Then, volume of a cylindrical glass is given by

Volume = π r^{2}h

= 22/7 × (3.5)^{2} × 12

V = 462 cm^{3}

This is the volume of a each glass. But, there are total 1600 students in a school. Then, amount of milk required for 1600 students is given by

V = 1600 × 462

V = 739200 cm^{3}

V = 739.2 liters

Thus, 739.2 liters of milk is needed to serve 1600 students daily.

**(4) In the given fig, a rectangular tin foil of size 22 cm by 16 cm is wrapped around to form a cylinder of height 16 cm. Find the volume of the cylinder.**

**Solution:**

Given that, Length of rectangular foil (l) = 22 cm

Breath of rectangular foil (b) = 16 cm

When we are folding foil length wise, then cylinder is formed whose radius

2r = 22

R = 22/2 = 11 cm

R = 11 cm

Then, volume of the cylinder formed is given by

V = π r^{2}h

= (22/7) × (11)^{2} × 16

V = 6084 cm^{3} is the required volume of a given cylinder.

**(6) A road roller (in the shape of a cylinder) has a diameter 0.7m and its width is 1.2 m. Find the least number of revolutions that the roller must make in order to level a playground of size 120m by 44m.**

**Solution:**

Given that, diameter of a road roller = d = 0.7 m

Radius of a road roller = r = 0.35m

Width (h) = 1.2m

Then, Curved surface area of the road roller is given by

CSA = 2 π rh

= 2 × 22/7 × 0.35 × 1.2

CSA = 2.64m^{2}

Now, Area of the playground is given by,

Area = l × 6 = 120 × 44 = 5280 m^{2}

Thus, No. of revolutions = Area of Playground/Curved S. Area = 5280/2.64

Thus, the revolutions that the roller must make in order to level a playground are 2000.

**(7) (i) If the volume of a cylinder of height 7 cm is 448 ****π ****cm ^{3}. Find the lateral surface area and total surface area.**

**(ii) A wooden pole is 7m high and 20cm in diameter. Find its weight, if the wood weighs 225 kg per m ^{3}.**

**Solution:**

(i) Given that,

Volume of a cylinder = V = 448 π cm^{3}

Height of a cylinder = h = 7 cm

We have, Volume of a cylinder is given by,

V = π r^{2}h

448 π = π r^{2}(7)

448/7 = r^{2}

^{ }r^{2} = 64

r = 8 cm is the required of the given cylinder.

- Lateral surface area of a cylinder is given by

LSA = 2 π rh

= 2 × π × 8 × 7

LSA = || 2 π cm^{2}

is the required Lateral surface area of a cylinder.

- Total surface area of a cylinder is given by

TSA = 2 π r (r + h)

= 2 × π × 8 × (8 + 7)

= 2 × π × 8 × 15

TSA = 240 π cm^{2}

This is the required total surface area of a cylinder.

(ii) Given that,

Height of wooden pole (h) = 7m

Diameter of pole (d) = 20cm

Radius of pole = r = 10 cm

Then, Volume = π r^{2}h

= 22/7 × 10 × 10/100 × 100 × 7 m^{3} = 22/100 m^{3}

Here, weight of wood used = 225 kg/m^{3}

Hence, total weight = 22/100 × 225 kg = 99/2 = 49.5 kg

**(8) The circumference of the cylindrical vessel is 132 cm and its height is 25 cm. **

**Find (i) radius of cylinder**

**(ii) Volume of cylinder**

**Solution:**

Given that, height of cylindrical vessel (h) = 25 cm

Circumference of base = 132 cm

(i) We have, Circumference of circular base of a cylinder} = 2 π r

132 = 2 × 22/7 × r

r = 21 cm is the required radius of the cylinder.

(ii) Volume of a cylinder is given by,

V = π r^{2}h

= 22/7 × 21 × 21 × 25

V = 34650 cm^{3} is the required of given cylinder.

**(9) The area of the curved surface of a cylinder is 4400 cm ^{2}. and the circumference of its base is 110 cm. **

**Find (i) The height of the cylinder**

**(ii) The volume of the cylinder**

**Solution:**

Given that, Area of Curved surface of a cylinder} = CSA = 4400 cm^{2}

Circumference of base of a cylinder = 110 cm

(i) Curved surface area of a cylinder is given by

CSA = 2 π rh

4400 = 2 × 22/7 × 17.5 × h

h = 4400 × 7/ 2 × 22 × 17.5 = 40 cm

h = 40 cm is the required height of the cylinder.

Given that, Circumference of base = 2 π r

110 = 2 × 22/7 × r

r = 110 × 7/ 2 × 22 = 17.5 cm

r = 17.5 is the required radius of given cylinder.

(ii) Volume of a cylinder is given by,

V = π r^{2}h

= 22/7 × (17.5)^{2} × 40

V = 38500 cm^{3} is the required volume of a given cylinder.

**(10) A cylinder has a diameter of 20 cm. The area of curved surface is 1000 cm ^{2}. **

**Find (i) The height of the cylinder correct to one decimal place.**

**(ii) The volume of the cylinder correct to one decimal place.**

**Solution:**

Given that, diameter of a cylinder (d) = 20 cm

Then, Radius of a cylinder (r) = 10 cm

Curved Surface Area = 1000 cm^{2}

(i) The curved surface area of a cylinder is given by

CSA = 2 πrh

1000 = 2 × 3.14 × 10 × h

h = 1000/62.8 = 15.9cm

h = 15.9 cm is the required height of the given cylinder.

(ii) Volume of the cylinder is given by,

V = π r^{2}h

= 3.14 × 10 × 10 × 15.9

V = 4992.6 cm^{2}

This is the required volume of the given cylinder.

**(11) The barrel of a fountain pen, cylindrical in shape, 7 cm long and 5 mm in diameter. A full barrel of ink in the pen will be used up when writing 310 words on an average. How many words use up a bottle of ink containing one – fifth of a liter?**

**Solution:**

Given that, Height of cylindrical barrel of pen (h) = 7 cm

Diameter (d) = 5 mm

Radius (r) = 2.5 mm = 0.25 cm

Then, Volume of a barrel is given by,

V = π r^{2}h

= 22/7 × 0.25 × 0.25 × 7

V = 1.375 cm^{3}

Given that, Ink in a bottle = one fifth of a litre

= 1/5 × 1000

= 200 ml

And, No. of words written using full barrel of ink = 310

No. of words written using 200 ml of ink = 200/1.375 × 310

= 45090.90 words

Thus, the no. of words written using the ink are found to be 45100 words.

**(12) Find the ratio between the total surface area of a cylinder to its curved surface area given that its height and radius are 7.5 cm and 3.5 cm.**

**Solution:**

Given that, Radius of cylinder (r) = 3.5 cm

Height of a cylinder (h) = 7.5 cm

Total Surface Area of a cylinder = 2 π (r + h) r

Curved Surface area of a cylinder = 2 π r + h

Then, TSA/CSA = 2 π r (r + h)/ 2 π r h = r + h/h

= (3.5 + 7.5)/ 7.5

TSA/CSA = 11/7.5 = 22/15

Thus, the ratio of total surface area of cylinder and curved surface area of a cylinder is found to be 22:15.

**(13) The radius of the base of a right circular cylinder is halved and the height is doubled. What is the ratio of the volume of a new cylinder to that of the original cylinder?**

**Solution:**

Let us consider, r is the radius of circular base of right circular cylinder.

Height be ‘h’

Then, Volume of a cylinder = V = π r^{2}h

Given that, Radius of circular base of cylinder is halved & height is doubled then

r → r/2 and h → 2h

Then, Volume of new cylinder = V^{1} = π r^{2}h

V^{1} = π (r/2)^{2} (2h)

V^{1} = 1/2 π r^{2}h

The ratio of new cylinder’s volume to the volume of original cylinder is given by

V^{1}/v = ^{1/}_{2} π r^{2}h/ π r^{2}h = 1/2

= V^{1}: V = 1:2 is the required ratio.

**(15) The ratio between the curved surface & the total surface of a cylinder is 1:2. Find the volume of the cylinder, given that its total surface area is 616 cm ^{2}.**

**Solution:**

Given that,

The ratio between the curved surface area & the total surface area of a cylinder is 1:2.

And, total surface area = 616 cm^{2}

CSA/TSA = 1/2 = CSA = 1/2 (TSA) = 1/2 × 616

CSA = 308 cm^{2}

We have, Curved Surface Area = 2 π r h

308 = 2 π r h

π r h = 308/2

r h = 49 —– (1)

Total Surface area = 2 π r h + 2 π r^{2}

616 = 308 + 2 π r^{2}

154 = π r^{2}

r^{2} = 154 × 7/22 = 49

= r = 7 cm put in (1) = h = 49/7 = 7 h = 7 cm

Now, Volume of a cylinder is given by,

V = π r^{2}h = 22/7 × 7^{2} × 7 = 1078 cm^{3}

V = 1078 cm^{3} this is the required volume of the given cylinder.

**(16) Two cylindrical jars contain the same amount of milk. If their diameters are in the ratio 3:4, find the ratio of their heights.**

**Solution:**

Given that,

The two cylindrical jars contains the same amount of milk. diameter of both cylinders are in the ratio d_{1}:d_{2} = 3:4

And also, ratio of their radius r_{1}:r_{2} = 3:4

From given condition,

V_{1} = V_{2}

π r2_{1} h_{1} = π r^{2}_{2} h_{2}

h_{1}/h_{2} = r^{2}_{2}/r_{1}^{2} = (r_{2}/r_{1})^{2} = (4/3)^{2} = 16/9

h_{1}/h_{2} = 16/9 is the required ratio of heights two given cylinders.

**(18) A cylindrical tube open at both ends is mode of metal. The internal diameter of the tube is 11.2 cm and its length is 21 cm. The metal thickness is 0.4 cm. Calculate the Volume of the metal.**

**Solution:**

Given that,

Internal diameter of cylindrical tube = 11.2 cm

Then, Internal radius = r = 5.6 cm

Length of the tube (h) = 21 cm

Thickness of metal = 0.4 cm

Thus, outer radius of cylindrical tube = 5.6 + 0.4 = 6 cm = R

Then, Volume of the metal is given by

V = π r^{2} h – π r^{2} h

= π h (R^{2} – r^{2})

= 22/7 × 21 × (6^{2} – 5.6^{2})

= 66 × (6 + 5.6) (6 – 5.6)

= 66 × 11.6 × 0.4

V = 306.24 cm^{3} this is the required volume of the metal.

**(19) A lead pencil consists of a cylinder of wood with a shod cylinder of graphite filled in the interior. The diameter of the pencil is 7 mm and the diameter of the graphite is 1 mm. If the length of the pencil is 14 cm, find the volume of the wood and that of the graphite.**

**Solution:**

Given that, diameter of the pencil = 7 mm

Radius of the pencil = 7/2 mm = 7/20 cm = R

And, Diameter of graphite = 1 mm

Then Radius of graphite = r = 1/2 mm = 1/20 cm

Thus, Volume of a graphite is given by

V = π r^{2} h

= 22/7 × (1/20)^{2} × 14 = 11/100

V = 0.11 cm^{3}

Now, Volume of the wood is given by,

V = π r^{2} h – π r^{2} h

= π h (R^{2} – r^{2})

= 22/7 × 14 [(7/20)^{2} – (1/20)^{2}]

= 44 × 48/400

V = 5.28 cm^{3} This is the required volume of the wood.

**(20) A cylindrical roller made of iron is 2 m long. Its inner diameter is 35 cm and the thickness is 7 cm all round. Find the weight of the roller in kg, if 1 cm ^{3} of iron weight 8 g.**

**Solution:**

Given that, Length of cylindrical roller (h) = 2 m = 200 cm

Diameter = 35 cm

Inner radius = 35/2 = 17.5 cm = r

Thickness = 7 cm

Then, outer radius = R = 17.5 + 7 = 24.5 cm

Thus, Volume of iron used in roller is given by

V = π r^{2} h – π r^{2} h

= π h (R^{2} – r^{2})

= 22/7 × 200 [(49/_{2})^{2} – (35/_{2})^{2}]

= 22/7 × 50 (49^{2} – 35^{2})

= 22/7 × 50 × 1176

V = 184800 cm^{3}

But given that, 1 cm^{3} of iron weighs 8 g.

Then, weight of the roller = 184800 × 8

= 1478400 g

= 1478.4 kg

Thus, the weight of the roller is found to be 1478.4 kg.

**Exercise 17.2**

**(1) Find the curved surface area of a right circular cone whose slant height is 10 cm and base radius is 7 cm.**

**Solution:**

Given that, Slant height of RCC = 10 cm

Base Radius of RCC = r = 7 cm

Then, Curved surface area of RCC is given by,

CSA of RCC = π r l

= 22/7 × 7 × 10

CSA = 220 cm^{2}

This is the required curved Surface Area of Right Circular Cone.

**(2) Diameter of the base of a cone is 10.5 cm and slant height 10 cm. Find its curved surface area.**

**Solution:**

Given that, diameter of base of a cone (d) = 10.5 cm

= Radius of base of a cone (r) = 5.25 cm

Slant height of a cone = 10cm

Then, curved surface area of a cone is given by,

CSA = π r l

= 22/7 × 5.25 × 10

CSA = 165 cm^{2} this is the required curved surface area of a cone.

**(3) Curved surface area of a cone is 308 cm ^{2} and its slant height is 14 cm. **

**Find, (i) Radius of the base**

**(ii) total surface area of the cone**

**Solution:**

Given that, curved surface area of a cone = 308 cm^{2}

Slant height of a cone (l) = 14 cm

(i) We have, curved surface area of a cone} = π r l

308 = 22/7 × r × 14

r = 308 × 7/22 × 14

r = 7 cm this is the required radius of a cone.

(ii) Now, total surface area of a cone is given by

TSA of a cone = Base + Area + CSA

= π r^{2} + π r l

= (22/7 × 7^{2}) + 308

= 22/7 × 49 + 308

= 154 + 308

TSA = 462 cm^{2} this is the required total surface area of a cone.

**(4) Find the volume of the right circular cone with**

**(i) Radius 6 cm and height 7 cm**

**(ii) Radius 3.5 cm and height 12 cm**

**Solution:**

(i) Given that,

Radius of RCC (r) = 6 cm

Height of RCC (h) = 7 cm

We have, Volume of RCC is given by

V = 1/3 π r^{2} h

V = 1/3 × (22/7) (6)^{2} × 7

= 22 × 12

V = 264 cm^{3}

This is the required volume of a given RCC.

(ii) Given that,

Radius of RCC (r) = 3.5 cm

Height of RCC (h) = 12 cm

We have, volume of RCC is given by,

V = 1/3 π r^{2} h

= (1/3) (22/7) (3.5)^{2} (12)

= 22/7 × 12.25 × 4

V = 154 cm^{3}

This is the required volume of a given RCC.

**(6) A conical pit of top diameter 3.5 m is 12 m deep. What is its capacity in kiloliters?**

**Solution:**

Given that, diameter of conical pit = 3.5 m

= Radius of conical pit (r) = 1.75 m

depth (h) = 12 m

Then, capacity of conical pit (volume of a cone) is given by,

V = 1/3 π r^{2} h

= 1/3 × (22/7) × (1.75)^{2} × 12

V = 38.5 m^{3}

V = 38.5 Kilolitres

**∵ **1 m^{3} = 1 Kilolitres

This is the required capacity of conical pit in Kilolitres.

**(7) If the volume of a right circular cone of height 9 cm is 48 ****π ****cm ^{3}, find the diameter of its base.**

**Solution:**

Given that, Height of RCC = h = 9 cm

Volume of RCC = V = 48 π cm^{3}

We have, Volume of Right circular cone is given by

V = 1/3 π r^{2} h

48 (22/7) = (1/3) × (22/7) × r^{2} × 9

3r^{2} = 48

r^{2} = 16

r = 16 cm

This is the required radius of the given RCC.

**(8) The height of a cone is 15 cm. If its volume is 1570 cm ^{3}. Find the radius of the base.**

**Solution:**

Given that, Height of a cone (h) = 15 cm

Volume of a cone (v) = 1570 cm^{3}

We have, Volume of a cone is given by

V = 1/3 π r^{2} h

1570 = (1/3) (3.14) r^{2} × 15

5 × 3.14 × r^{2} = 1570

r^{2} = 1570/5 × 3.14 = 100

r = 10 cm

This is the required radius of given cone.

**(9) The slant height and base diameter of a conical tomb are 25 m and 14 m respectively. Find the cost of white washing its curved surface area at the rate of Rs. 210 per 100 m ^{2}.**

**Solution:**

Given that, Slant height of conical tomb (l) = 25 m

Base diameter (d) = 14 m

= Base Radius (r) = 7 m

We have, curved Surface area of a cone is given by

CSA = π r l

= 22/7 × 7 × 25

CSA = 550 m^{2} is the required curved surface area of given conical tomb.

Given that, Rate of washing its curved surface area per

100 m^{2} = Rs. 210

Then, total cost = 550/100 × 210 = Rs. 1155

Thus, the total cost of washing its curved surface area is found to be Rs. 1155.

**(10) A conical tent is 10 m high and the radius of its base is 24 m. Find (i) Slant height of the tent**

**(ii) Cost of the canvas required to make the tent, if the cost of 1m ^{2} canvas is Rs. 70**

**Solution:**

Given that, Height of conical tent (h) = 10 m

Radius of base (r) = 24 m

We have, l^{2} = h^{2} + r^{2}

I^{2} = 10^{2} + 24^{2} = 100 + 576 = 676

I^{2} = 676

I = 26 m is the required slant height of the tent.

(ii) Curved surface area of a cone is given by,

π r l = 22/7 × 24 × 26

= 13728/7 m^{2}

Given that, the cost of 1m^{2} canvas is Rs. 70.

Then, total cost of 13728/7 m^{2} canvas is found to be

Total cost = 13728/7 × 70

= Rs. 137280

**(11) A joker’s cap is in the form of a right circular cone of base radius 7 cm and height 24 cm. Find the area of the cloth required to make 10 such caps.**

**Solution:**

Given that, Base radius of conical cap = r = 7 cm

Height of conical cap (h) = 24 cm

We have, l^{2} = h^{2} + r^{2}

l^{2} = 24^{2} + 7^{2} = 576 + 49

l^{2} = 625

l = 25 cm is the required slant height.

We have, curved surface area of a cone is given by

CSA = π r l

= 22/7 × 7 × 25

= 22 × 25

CSA = 550 cm^{2}

Thus, for one cap the cloth required is 550 cm^{2}.

Then, the area of the cloth required to make 10 such caps is found to be

550 × 10 = 5500 cm^{2}.

**(13) Find what length of canvas 2m in width is required to make a conical tent 20m in diameter and 42 m in slant height allowing 10% for folds and the stitching. Also find the cost of the canvas at the rate of RS. 80 per meter.**

**Solution:**

Given that, diameter of base of conical tent (d) = 20 m

= Radius of base of conical tent (r) = 10 m

Slant height (l) = 42 m

Then, curved surface area of conical tent is given by,

CSA = π r l

= 22/7 × 10 × 42

CSA = 1320 m^{2}

This the area of the canvas required.

Now, 10% of this canvas is used for folds and stitches,

then actual cloth needed = 1320 + 10% of 1320

= 1320 + 10/100 × 1320

= 1320 + 132

= 1452 m^{2}

Width of cloth = 2m

Length of cloth = Area/width = 1452/2 = 726 m

Given that, cost of canvas is Rs. 80 per meter.

Then, total cost required is = 80 × 726

= 58080 Rs.

**(14) The perimeter of the base of a cone is 44 cm and the slant height is 25 cm. Find the volume and the curved surface area of the cone?**

**Solution:**

Given that, Perimeter of base of cone = 44 cm

= 2 π r = 44

r = 44/2 π = 44/2 × 7/22

r = 7 cm is the base radius.

Now, slant height (l) = 25

We have, l^{2} = r^{2} + h^{2}

h^{2} = l^{2} – r^{2} = 25^{2} – 7^{2}

h^{2} = 625 – 49

h^{2} = 576

h = 24 cm is the required of a cone.

Then, Volume of the cone is given by,

V = 1/3 π r^{2} h

= 1/3 × 22/7 × 49 × 8

V = 1232 cm^{3} is the required volume of a cone.

Now, curved surface area of a cone is given by,

CSA = π r l

= 22/7 × 7 × 25

= 22 × 25

CSA = 550 cm^{2}

This is the curved surface area of a cone.

**(15) The volume of a right circular cone is 9856 cm ^{3} and the area of its base is 616 cm^{2}. **

**Find (i) the slant height of cone.**

**(ii) total surface area of cone**

**Solution:**

Given that, Volume of RCC = 9856 cm^{3}

Base area of RCC = 616 cm^{2}

= π r^{2} = 616

r^{2} = 616 × 7/22 = 196

r = 14 cm is the required base radius.

Now, Volume of a cone = 1/3 π r^{2} h

9856 = 1/3 × 22/7 × 14 × 14 × h

h = (9856 × 3 × 7/ 14 × 14)

h = 48 cm is the required height of a cone.

(i) We have, l^{2} = h^{2} + r^{2}

l^{2} = 48^{2} + 14^{2}

l^{2} = 2304 + 196

l^{2} = 2500

l = 50 cm is the required slant height of a cone.

(ii) Total surface area of a cone is given by

TSA = π r (l + r)

= 22/7 × 14 × (50 + 14)

= 22 × 2 × 64

TSA = 2816 cm^{2}

This is the required total surface area of a cone.

**(16) A right triangle with sides 6 cm, 8 cm, and 10 cm is revolved about the side 8 cm. Find the volume and the curved surface of the cone so formed.**

**Solution:**

Given that, A right triangle with sides: 6 cm, 8 cm, 10 cm. It is revolved about the side 8 cm then cone is formed.

= Radius (r) = 6 cm

Height (h) = 8 cm

Slant height (l) = 10 cm

Then, Volume of cone formed is given by,

V = 1/3 π r^{2} h

V = 1/3 × 3.14 × 6^{2} × 8

V = 3.14 × 12 × 8

V = 301.44 cm^{3} is the required volume of the cone.

Now, Curved surface area of a cone is given by,

CSA = π r l

= 3.14 × 6 × 10

CSA = 188.4 cm^{2} is the required CSA of a cone.

**(17) The height of a cone is 30 cm. A small cone is cut off at the top by a plane parallel to its base. If its volume be 1/27 ^{th} of the volume of the given cone, at what height above the base is the section cut?**

**Solution:**

Given that, Height of cone (H) = 30 cm

A small cone is cut off from the top of the big cone given.

Volume of given big cone (V) = 1/3 π R^{2} H

Given that,

Volume of small cone = 1/27^{th} of the volume of big cone

1/3 π R^{2} h = 1/27 (1/3 π R^{2} h)

r^{2} h = 1/27 R^{2} H

r^{2} h = 30/27 R^{2}

r^{2} h/ R^{2} = 30/27 = 10/9 —- (1)

Also, from fig, r/R = h/H

r/R = h/30 ——(2)

From (1) & (2) (h/30)^{2} × h = 10/9

h^{3}/ 900 = 10/9

h^{3} = 1000

h = 10 cm is the height of small cone.

Now, H – h = 30 – 10 = 20 cm

Thus, the small cone is cut at height of 20 cm above the base of big cone.

**(18) A semi – circular lamina of radius 35 cm is folded so that the two bounding radii are joined together to form a cone. Find **

**(i) the radius of cone**

**(ii) the lateral surface area of cone.**

**Solution:**

(i) Given that, Radius of semi – circular lamina (r) = 35 cm

After folding a semi – circular lamina a cone is formed,

The slant height of cone formed (l) = 35 cm

Let us consider, r_{1} be the radius of a cone formed.

Now, perimeter of semi – circular lamina} = base of cone

π r = 2 π r,

r = 2 r,

35 = 2r_{1}

r_{1} = 17.5 cm is the radius of a cone.

(ii) Now, the curved surface area of the cone formed is

CSA = π r_{1} l

= 22/7 × 17.5 × 35

CSA = 1925 cm^{2} is the required curved surface area of a cone.

**Exercise 17.3**

**(1) Find the surface area of a sphere of radius 14 cm.**

**Solution:**

Given that, Radius of a sphere = r = 14 cm

Then, surface area of a sphere is given by

SAS = 4 π r^{2}

= 4 × 22/7 × 14 × 14

= 4 × 22 × 2 × 14

SAS = 2464 cm^{2} is the required surface area of a given sphere.

**(2) Find the surface area of a sphere of diameter 21 cm.**

**Solution:**

Given that, diameter of a sphere (d) = 21 cm

= Radius of a sphere (r) = 10.5 cm

Then, Surface area of a sphere is given by,

SAS = 4 π r^{2}

= 4 × 22/7 × 10.5 × 10.5

SAS = 1386 cm^{2} is the required surface area of a given sphere.

**(3) A shot – put is a metallic sphere of radius 4.9 cm. If the density of the metal is 7.8 g per cm ^{3}, find the mass of the shot – put.**

**Solution:**

Given that, Radius of metallic shot – put = r = 4.9 cm

We know that,

Volume of sphere = 4/3 π r^{3}

= 4/3 × 22/7 × (4.9)^{3}

V = 493. 005

V = 493 cm^{3} is the required volume.

Now, density = 7.8 g per cm^{3}

We have, density = mass/volume

Then, Mass = density × volume

= 7.8 × 493

Mass = 3845.4g is the required mass of shot – put.

**(4) Find the diameter of a sphere whose surface area is 154 cm.**

**Solution:**

Given that, Surface area of sphere = 154 cm^{2}

We have, Surface area of a sphere is given by,

SAS = 4 π r^{2}

154 = 4 × 22/7 × r^{2}

r^{2} = 49/4

r = √49/2 = 7/2

r = 7/2 cm

Then, diameter of sphere = d = 2r = 2 × 7/2 = 7 cm

**(5) Find (i) the curved surface area**

**(ii) the total surface area of a hemisphere of radius 21 cm.**

**Solution:**

Given that, Radius of hemisphere (r) = 21 cm

(i) The Curved surface area of hemisphere is given by

CSA of hemisphere = 2 π r^{2}

= 2 × 22/7 × 21^{2}

CSA = 2772 cm^{2} is the required curved surface area of given hemisphere.

(ii) Now, total surface area of hemisphere is given by

TSA of hemisphere = 3 π r^{2}

= 3 × 22/7 × 21^{2}

= 4158 cm^{2}

is the required total surface area of hemisphere.

**(6) A hemispherical brass bowl has inner – diameter 10.5 cm. Find the cost of tin – plating it on the inside at the rate of Rs 16 per 100 cm ^{2}.**

**Solution:**

Given that,

The inner diameter of hemispherical bowl (d) = 10.5 cm

= Radius (r) = 5.25 cm

Then, Curved surface area of hemispherical bowl is given by,

CSA of hemisphere = 2 π r^{2}

= 2 × 22/7 × (5.25)^{2}

= 173.25 cm^{2}

Given that, the rate of plating is Rs. 16 per 100 cm^{2}.

Then, total cost required for tin – plating the surface area of 173.25 cm^{2} is given by

Total cost = 173.25 × 16/100 = 27.72 Rs.

**(7) The radius of a spherical balloon ****increases**** from 7 cm to 14 cm as air is jumped into it. Find the ratio of the surface area of the balloon in two cases.**

**Solution:**

Given that, original radius of balloon (r) = 7 cm

Radius of balloon after air filling = 14 cm

Then, Surface area of original balloon = 4 π r^{2}

Surface area of new balloon = 4 π r^{2}

= SAB_{1}/ SAB_{2} = 4 π r^{2}/ 4 π r^{2} = r^{2}/R^{2} = 72/142

S_{1}/S_{2} = 1/4

S_{1}: S_{2} = 1:4

This is the required ratio of balloons when it is empty & after filled with air.

**(8) A sphere and a cube have the same surface. Show that the ratio of volume of the sphere to that of the cube is ****√****6: ****√****π.**

**Solution:**

Let us consider, the side of the cube = a

Then, Surface area of cube = 6a^{2}

Surface area of sphere of radius ‘r’ = 4 π r^{2}

= SA of sphere = SA of cube

4 π r^{2} = 6a^{2}

π r^{2}/ a^{2} = 6/4

∴ r/a = √6/2 π

Now, Volume of sphere (v_{1}) = 4/3 π r^{3}

and Volume of cube (v_{2}) = a^{3}

Then, V_{1}/V_{2} = 4 π r^{3}/ 3a^{3}

= 4 π √6^{3}/3 (2π)^{3}

= (4π × 6 √6)/ (3 × 8 π × √π)

V_{1}/V_{2} = √6/ √π

V_{1}:V_{2} = √6: √π

Hence Proved.

**(10) Find the volume of a Sphere whose surface area is 154 cm ^{2}.**

**Solution:**

Given that, surface area of sphere = 154 cm^{2}

We have, Surface area of Sphere = 4πr^{2}

154 = 4 × 22/7 × r^{2}

r^{2} = 49/4

r = 7/2 cm is the radius of sphere.

Now, Volume of the sphere = 4/3πr^{3}

= 4/3 × 22/7 × (7/2)^{3}

= 179.67 cm^{3}

Thus, Volume of a sphere is found to be 179.67 cm^{3}.

**(11) If the volume of a sphere is 179.2/3. Find the radius & surface area of sphere.**

**Solution:**

Given that, Volume of a sphere = 179.2/3 cm^{3}

We have, Volume of sphere = 4/3 πr^{3}

539/3 = 4/3 × (22/7)r^{3}

r^{3} = 49 × 718 = (7 × 7 × 7)/ (2 × 2 × 2)

= r = 7/2 = 3.5 cm

r = 3.5 cm is the required radius of given sphere.

Now, Surface area of sphere = 4πr^{2}

= 4 × 22/7 × (7/2)^{2}

= 22 × 7

SAS = 154 cm^{2}

This is the required surface area of given sphere.

**(12) A hemispherical bowl has a radius of 3.5 cm. What would be the volume of water it would contain?**

**Solution:**

Given that, Radius of hemispherical bowl (r) = 3.5 cm

Volume of hemisphere = 2/3 πr^{3}

= 2/3 × 22/7 × (7/2)^{3}

= 11 × 4916

= 539/6

V = 89.5/6 cm^{3}

This is the required volume of a given hemispherical bowl.

**(13) The surface area of a solid sphere is 1256 cm ^{2}. It is cut into two hemisphere. Find the total surface area and the volume of a hemisphere.**

**Solution:**

Given that, Surface area of a solid sphere = 1256 cm^{2}

We have, Surface area of sphere = 1256

4πr^{2} = 1256

4 × 3.14 × r^{2} = 1256

r^{2} = 1256 × (4 × 3.14)

r^{2} = 100

r = 10 cm is the required radius.

Now, total surface area of hemisphere = 3πr^{2}

3πr^{2} = 3 × 3.14 × 10^{2}

= 3 × 3.14 × 100

TSA of H = 942 cm^{2} is the required total surface area of hemisphere.

Now, Volume of the hemisphere = 2/3πr^{3}

= 2/3 × 22/7 × 10^{3}

= 2/3 × 3.14 × 1000

= 6280/3

V = 2093. cm^{3}

This is the required volume of a given hemisphere.

**Exercise 17.4**

**(1) The following fig. shows a ****cyboidal**** block of wood through which a circular cylindrical hole of the biggest size is drilled. Find the volume of the wood left in the block.**

**Solution:**

Given that,

diameter of biggest hole = 30 cm

Then, Radius of hole (r) = 30/2 = 15 cm

Height of hole (h) = 70 cm

The volume of cubical block is given by,

V = l × b × h

V = 70 × 30 × 30

V = 63000 cm^{3} is the required volume of block.

Then, Volume of a cylindrical hole is given by,

V = πr^{2}h

= 22/7 × 15^{2} × 70

= 22 × 225 × 70

V = 49500 cm^{3} is the required volume of cylindrical hole.

Now, Volume of the wood left in the block} = 63000 – 49500

= 135000 cm^{3}

Hence, the volume of the wood left in the block is found to be 13500 cm^{2}

**(2) The given fig. shows a solid trophy made of shining glass. If one cubic centimeter of glass costs Rs. 0.75. find the cost of the glass for making the trophy.**

**Solution:**

Given that, Side of cubical part = 28 cm

Radius of cylindrical part (r) = 28/2 = 14 cm

Height of cylinder (h) = 28 cm

Then, volume of a cube is given by,

Volume of cube = a^{3 }= 28^{3}

V_{1} = 21952 cm^{3}

Now, volume of the cylindrical part is given by,

V_{2} = πr^{2}h

= 22/7 × 14 × 14 × 28

V_{2} = 17248 cm^{3}

Then, total volume of the solid = (Volume of the cube) + (Volume of the cylinder)

= 21952 + 17284

TV = 39200 cm^{3}

The cost of 1 cm^{3} glass is Rs. 0.75

Then, total cost of glass = 39200 × 0.75

= Rs. 294000.

**(3) From a cube of edge 14 cm, a cone of maximum size is carved out. Find the volume of the remaining material.**

**Solution:**

Given that, edge/ side of the cube = 14 cm

Volume of cube (V) = (14)^{3} = 2744 cm^{3}

diameter of cone cut from cube = 14 cm

Volume of cube = a^{3}

V = 14^{3}

V_{1} = 2744 cm^{3}

Volume of cone = 1/3 πr^{2}h

= 1/3 × 22/7 × 7^{2} × 14

V_{2} = 215613 cm^{3}

Volume of remaining material = (Volume of the cube) – (Volume of cone)

= (3 × 2744 – 2156)/3

= (8232 – 2156)/3

= 6076/3

V = ₹ 025.1/3 cm^{3}

This is the volume of remaining material.

**(4) A cone of maximum volume is curved out of a block of wood of size. 20 cm × 10 cm × 10 cm. find the volume of the remaining wood.**

**Solution:**

Given that,

The dimensions of wooden block = 20 cm × 10 cm × 10 cm

Then, Volume of the wooden block = 20 × 10 × 10 = 2000 cm^{3}.

diameter of cone (d) = 10 cm

= Radius of cone (r) = 5 cm

Height of cone (h) = 20 cm

Now, Volume of the cone is given by,

V = 1/3 πr^{2}h

V = 1/3 × 22/7 × 5^{2} × 20

V = 11000/21 cm^{3}

Now, (Volume of remaining) = (Volume of block of wood) – (Volume of cone)

= (2000 – 11000)/21

= (21 × 2000 – 11000)/21

= (42000 – 11000)/21

= 31000/21

V = 1476.19 cm^{3}

This is the required volume of remaining wood.

**(5) 16 glass spheres each of radius 2 cm are packed in a cuboidal box of internal dimensions 16 cm × 8 cm × 8 cm and then the box is filled with water. Find the volume of the water filled in the box.**

**Solution:**

Given that, Dimensions of cuboidal box = 16 cm × 8 cm × 8 cm

Then, Volume of cuboidal box = l b h = 16 × 8 × 8 = 1024 cm^{3}

Radius of sphere (r) = 2 cm

Then, volume of the sphere = 4/3πr^{3}

= 4/3 × 22/7 × 2^{3}

V = (4 × 22 × 8)/ (3 × 7)

V = 704121 cm^{3}

Volume of 16 sphere = 16 × 704/21

= 11264/21 cm^{3}

= 536.38 cm^{3}

Now, Volume of water filled in the box} = (Volume of the box) – (Volume of 16 spheres)

= (1024 – 536.38)

V = 487.62 cm^{3}

This is the required volume of water filled in the box.

**(6) A cuboidal block of side 7 cm is sup mounted by a hemisphere. What is the greatest diameter that the hemisphere can have? Also, find the surface area of the solid.**

**Solution:**

Given that, Side of cuboidal block = a = 7 cm

diameter of hemisphere = 7 cm

= Radius of hemisphere (r) = 7/2 = 3.5 cm

Now, Surface area of hemisphere = 2πr^{2}

2πr^{2} = 2 × 22/7 × 3.5 × 3.5

S = 44 × 12.25/7

= 539/7

S = 77 cm^{2}

Surface area of cube = 6a^{2} = 6 × 7^{2} = 294 cm^{2}

Now, Surface area of base of hemisphere = πr^{2}

= 22/7 × 3.5 × 3.5

= 38.5 cm^{2}

Thus, (the surface area of solid) = (Surface area of the cube) + (Surface area of hemisphere) – (Surface area of base of hemisphere)

= 294 + 77 – 38.5

= 332.5 cm^{2}

This is the required surface area of solid.

**(7) A wooden article was made by scooping out a hemisphere from each end of a solid cylinder (as shown in fig). 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.**

**Solution:**

Given that,

Height of cylinder (h) = 10 cm

Radius of the base (r) = 3.5 cm

Total surface area is given by,

TSA = (Curved surface area of a cylinder) + 2x (Curved surface area of hemisphere)

= 2πrh + 2 × 2πr^{2}

= 2 × 22/7 × 3.5 × (10 + 2 × 3.5)

= (15417) × (10 + 7)

= 22 × 17

TSA = 374 cm^{2} is the required total surface area of given article.

**(8) A toy is in the form of a cone of radius 3.5 cm mounted on a hemisphere of same radius. If the total height of the toy is 15.5 cm, find the total surface area of the toy.**

**Solution:**

Given that,

Total height of toy (h) = 15.5 cm

Radius of base of conical part (r) = 3.5 cm

Now, Height of cone (h) = 15.5 – 3.5 =12 cm

We have, l^{2} = h^{2} + r^{2}

l^{2} = 12^{2} + 3.5^{2}

l^{2} = 144 + 12.25 = 156.25

l = 12.5 cm

(Total surface area of the toy) = (curved surface area of cone) + (curved surface area of the hemisphere)

TSA = π r l + 2πr^{2}

= π r (l + 2r)

= 22/7 × 3.5 × (12.5 + 2 × 3.5)

= 11 × 19.5

TSA = 214.5 cm^{2}

This is the required surface area of the toy. ^{ }

**(10) A circus tent is in the shape of a cylinder sun mounted a cone. The diameter of the cylindrical portion is 24 m and its height is 11 m. If the vertex of the cone is 16 m. above the ground, find the area of the canvas used make the tent.**

**Solution:**

Given that, Radius of base of cylindrical part of the tent}

r = 24/2 = 12 m

Height of cylindrical part (H) = 11 m

As the vertex of a cone is 16 m above the ground, height of cone (h) = 16 – 11 = 5 m

Now, we have

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

l^{2} = 25 + 144

l^{2} = 169

l = 13 m

Radius of cone (r) = 12 m

(Area of canvas used to make tent) = 2πrH + π r l

= (2H + l) π r

= (264/7) × (22 + 13)

= 264 × 5

A = 1320 m^{2}

Thus, the area of the canvas used to make the tent is found to be 1320 m^{2}.

**(12) From a solid cylinder of height 30 cm and radius 7 cm, a conical cavity of height 24 cm and of base radius 7 cm is drilled out. Find the volume and total surface of the remaining solid.**

**Solution:**

Given that, Radius of solid cylinder (r_{1}) = 7 cm

Height of cylinder (H) = 30 cm

Height of cone (h) = 24 cm

Radius of cone (r) = 7 cm

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

l^{2} = 24^{2} + 7^{2}

l^{2} = 576 + 49

l^{2} = 625

l = 25 cm

(Now, Volume of remaining solid) = (Volume of cylinder) – (Volume of the cone)

= πr^{2}H – 1/3 πr^{2} h

V = πr^{2} (H – h/3)

V = πr^{2} (H – h/3)

= 22/7 × 7^{2} × (30 – 24/3)

= 22 × 7 × (30 – 8)

V = 3388 cm^{3} is the required volume of remaining solid.

Now, (Total surface area of remaining solid) = (Curved surface area of cylinder) + (Surface area of top of cylinder) + (Curved surface area of the wane)

= 2 π r H + π r^{2} + π r l

= π r (2H + r + l)

= 22/7 × 7 (2 × 30 + 7 + 25)

= 22 × (60 + 32)

= 22 × 92

TSA = 2024 cm^{2}

This is the required total surface area of the remaining solid.

**(13) The following fig. shows a hemisphere of radius 5 cm surmounted by a right circular cone of base radius 5 cm. find the volume of the solid if the height of the cone is 7 cm. Give your answer correct to two places of decimal.**

**Solution:**

Given that,

Height of conical portion (h) = 7cm

Radius of base of cone (r) = 5 cm

Now, Volume of solid = (Volume of hemisphere) + (Volume of the cone)

= (2/3 π r^{3}) + 1/3 π r^{2} h

= 1/3 π r^{2} (2r + h)

= 1/3 × 22/7 × 5^{2} × (10 + 7)

= 22/21 × 25 × 17

= 445.238

V = 445.24 cm^{3}

This is the required volume of solid.

**(15) A building is in the form of a cylinder surmounted by a hemisphere valted dome and contains 41 (19/21) m ^{3} of air. If the internal diameter of the dome is equal to its total height above the floor, find the height of the building.**

**Solution:**

Let us Consider, r be the radius of dome.

Then, internal diameter = 2r

Given that, internal diameter = total height

Then, total height of building = 2r

And height of hemispherical area = r

Then, height of cylindrical area = h = 2r – r = r

Now, (Volume of the building) = (Volume of cylindrical area) + (Volume of hemispherical area)

V = π r^{2} h + 2/3 π r^{3}

= π r^{3} + 2/3 π r^{3}

∵ h = r

V = π r^{3} (1 + 2/3)

= π r^{3} (3 + 2) / 3

V = (5/3) π r^{3}

Given that, Volume of building = 41 (19/21) = 880/21

(5/3) π r^{3} = 880/21

5/3 × 22/7 × r^{3} = 880/21

r^{3} = 880 × 3 × 7/ (5 × 22 × 11)

r^{3} = 880/110

r^{3} = 8

r = 2m is the required radius

Now, height of the building =2r = 2 × 2 = 4 m

**(16) A rocket is in the form of a right circular cylinder closed at the lower end and submitted by a Cone with the same radius as that of the cylinder. The diameter and the height of the cylinder are 6 cm and 12 cm respectively. If the slant height of the conical portion is 5 cm, find the total surface area and the volume of the rocket.**

**Solution:**

Given that,

Height of cylindrical portion = 12 cm (H)

diameter = 6 cm = Radius = r = 3 cm

Slant height of conical portion (l) = 5 cm

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

h^{2} = l^{2} – r^{2}

h = √25 – 9 = √16

= h = 4 cm

Now, (Total surface area of the rocket) = (Curved surface area of cylinder) + (base area of cylinder) + (Curved surface area of cone)

= 2πrH + πr^{2} + π r l

= π r (2H + r + l)

= 3.14 × 3 × (24 + 3 + 5)

= 3.14 × 3 × 32

= 301.44 cm^{2}

Thus, the total surface area of the rocket is found to be 301.44 cm^{2}

Now, Volume of rocket = (Volume of the cone) + (Volume of the cylinder)

= 1/3 π r^{2} h + π r^{2} H

= π r^{2} (h/3 + H)

= 3.14 × 3^{2} × (4/3 + 12)

= 3.14 × 9 × (4 + 36)/3

= 3.14 × 3 × 40

Volume of rocket = 376.8 cm^{3}

This is the required volume of the rocket.

**(17) The following fig. represents a solid consisting of a right circular cylinder with a hemisphere at one end and a cone at the other. Their common radius is 7 cm. The height of the cylinder and the cone are each of 4 cm. Find the volume of the solid.**

**Solution:**

Given that, Common radius RCC & hemisphere r = 7 cm

Height of the cone (h) = 4 cm

Height of the cylinder (H) = 4 cm

Then, Volume of solid = (Volume of the cone) + (Volume of the cylinder) + (Volume of the hemisphere)

= 1/3 π r^{2} h + π r^{2} H + 2/3 π r^{3}

= π r^{2} [h/3 + H + 2r/3]

= 22/7 × 7^{2} [4/3 + 4 + 14/3]

= 22 × 7 × (4 + 12 + 14) /3

Volume of solid = 1540 cm^{3}

This is the required volume of the given solid.

**(18) A solid is in the form of a right circular cylinder with a hemisphere at one end a cone at the other end. Their common diameter is 3.5 cm and the height of the cylindrical and conical portions are 10 cm and 6 cm respectively. Find the volume of the solid.**

**Solution:**

Given that, Common diameter (d) = 3.5 cm

Then, Radius (r) = 1.75 cm

Height of cylindrical portion (h_{1}) = 10 cm

Height of conical portion (h_{2}) = 6 cm

Now, Volume of the solid is given by,

(Volume of solid) = (Volume of the cone) + (Volume of the cylinder) + (Volume of hemisphere)

= 1/3 π r^{2} h + π r^{2} H + 2/3 π r^{3}

= π r^{2} [h/3 + H +20/3]

= 3.14 × 1.75 × 1.75 × [6/3 + 10 + (2 × 1.75) /3]

= 3.14 × 3.0625 × 13.167

= 9.61625 × 13.167

= 126.617 cm^{3}

Volume of solid = 126.62 cm^{3}

This is the required volume of the solid.

**(20) The following fig. shows a model of solid consisting of a cylinder surmounted by a hemisphere at one end. If the model is drawn to a scale of 1:200, find**

**(i) The total surface area of the solid in ****π**** m ^{2}.**

**(ii) The volume of the solid in ****π**** liters.**

**Solution:**

(i) Given that,

Height of cylindrical portion (h) = 8 cm

Radius (r) = 3 cm

Scale = 1:200

Then, total surface are of solid is given by,

Total surface area = 2 π r^{2} + 2 π r h

= 2 π r (r + h)

= 2 × π × 3 (3 + 8)

= 6π × 11

Total surface area = 66π cm^{2} is the required total surface area.

Hence, total surface area of the solid is given by

Surface area of solid = 66π × (200)2/1

= (66 × 40,000)/ (100 × 100) π

Surface area of solid = 264π m^{2}

This is the required surface area of solid.

(ii) The volume of solid = 2/3 π r^{3} + π r^{2} h

= π r^{2} [2/3r + h]

= π × 3^{2} × [2/3 × 3 + 8]

= 9π (2 + 8)

Volume of solid = 90π

Given that, Scale = 1:200

Thus, Capacity of solid = 90π × (200)^{3}

= 90π × 8000000

= 720000000 cm^{3}

= (720000000/ (100 × 100 × 100)

= 720 π m^{3}

= 720 π × 1000 liters

Capacity of solid = 720000 π liters

This is the required volume of the solid.

**Exercise 17.5**

**(1) The diameter of a metallic sphere is 6 cm. The sphere is melted and drawn into a wire of uniform Cross – section. If the length of the wire is 36 m, find its radius.**

**Solution:**

Given that, diameter of metallic sphere =6 cm

Radius of sphere (r) = 3 cm

Volume of sphere = 4/3 π r^{3}

V = 4/3 π r^{3}

= 4/3 × π × 3^{3}

V = 36 π cm^{3}

Length of wire (h) = 36 m

Let ‘r’ be the radius of the wire,

π r^{2} h = 36 π

r^{2} × 36 × 100 = 36

r^{2} = 1/100 = (1/10)^{2} = 1/10 cm = r

r = 1 mm

**(2) A solid metallic sphere of radius 6 cm is melted and made into a solid cylinder of height 32 cm. find the**

**(i) Radius of the cylinder**

**(ii) Curved surface area of the cylinder.**

**Solution:**

Given that, Radius of metallic sphere (R) = 6 cm

Height of cylinder (h) = 32 cm

Then, Volume of metallic cylinder is given by,

(i) Volume of cylinder = Volume of metallic sphere

π r^{2} h = 4/3 π r^{3}

r^{2} (32) = 4/3 6^{3}

r^{2} = 4/3 × 6 × 6 ×/32 = 9

r = 3 cm is the required radius of the cylinder.

(ii) Curved Surface area of the cylinder is given by,

(Curved Surface area of cylinder) = 2 π r h

= 2 × 3.1 × 3 × 32

Curved surface area of cylinder = 595.2 cm^{2}

**(3) A Solid metallic hemisphere of radius 8cm is melted and recasted into right circular cone of base radius 6 cm. Determine the height of the cone.**

**Solution:**

Given that, Radius of hemisphere (r) = 8 cm

Volume of hemisphere = 2/3 π r^{3}

Radius of cone (R) = 6 cm

V = 2/3 π r^{3}

= 2/3 × 8^{3} × π

V = (1024/3) π cm^{3} is the required volume of hemisphere.

Given that, hemisphere is melted and converted into a cone then volume remains same.

= Volume of cone = 1/3 π R^{2} h

(1024)/3 π = 1/3 × π × 6^{2} × h

36 h = 1024

h = 1024/ 36

h = 28.44 cm is the required height of the cone.

** **

** (4) A rectangular water tank of base 11m × 6m contains water up to a height of 5 m. If the water in the tank is transferred to a cylindrical tank of radius 3.5, find the height of the water level in the tank.**

**Solution:**

Given that, Base of water tank = 11 m × 6 m

Height of water level in tank (h) = 5 m

And volume of water = 11 × 6 × 5 = 330 m^{3}

Now, volume of cylindrical tank is given by,

(Volume of cylindrical tank) = π r^{2} h

330 = 22/7 × 3.5 × 3.5 × h

h = 330 × 7/22 × 12.25

h = 8.57 m

This is the required height of water level in the tank.

**(5) The volume of a cone is the same as that of the cylinder whose height is 9 cm and diameter 40 cm. Find the radius of the base of the cone if its height is 108 cm.**

**Solution:**

Given that, diameter of cylinder = 40 cm

Then, radius of cylinder = (r) = 20 cm

Height of cylinder (h) = 9 cm

Now, volume of cylinder is given by

Height of cone = 108 cm

V = π r^{2} h

V = π × 20 × 20 × 9

V = 3600 π cm^{3}

And volume of cone = 1/3 π r^{2} h

3600 π = 1/3 × π × r^{2} × 108

r^{2} = 3 × 3600 π/ π × 108

= 3 × 3600/ 108

r = √100 = 10 cm

r = 10 cm

This is the required radius of the wane.

**(7) A hemispherical bowl of diameter 7.2 cm is filled completely with chocolate sauce. This source is poured into an invested cone of radius 4.8 cm. Find the height of the cone.**

**Solution:**

Given that, diameter of hemispherical bowl = 7.2 cm

Radius of hemispherical bowl (r) = 3.6 cm

Volume of hemispherical bowl = 2/3 π r^{3}

= 2/3 × 22/7 × 3.6 × 3.6 × 3.6

V = 97.76 cm^{3}

And, volume of cone = 1/3 π R^{2} h = 1/3 × 22/7 × 4.8 × 4.8 × h

V = 24.14 × h

But, Volume of cone = Volume of hemispherical bowl

24.14 × h = 97.76

h = 97.76/24.14

= 4.05 cm

This is the required height of the cone.

**(9) A hollow copper pipe of inner diameter 6cm and outer diameter 10 cm is melted and changed into a solid circular. cylinder of the same height as that of the pipe. Find the diameter of the solid cylinder.**

**Solution:**

Given that, (Inner diameter of hollow pipe) = 6 cm

Outer diameter of hollow pipe = 10 cm

Then, Inner radius (r) = 3 cm & outer radius (R) = 5 cm

Let us consider, ‘h’ be the height of the hollow pipe.

Volume of pipe = π R^{2} h – π r^{2} h

= π (5^{2} – 3^{2}) h

= π h (25 – 9)

= π R^{2} h

16 = R^{2}

R = 4 cm

Then, diameter of solid cylinder is found to be 8 cm.

**(10) A hollow sphere of internal & external diameter 4 cm & 8 cm respectively, is melted into a cone of base diameter 8 cm. find the height of the cone.**

**Solution:**

Given that,

Internal diameter of hollow Sphere = 4 cm

Then, inner radius of hollow sphere = 2 cm

and, outer radius of hollow sphere = 4 cm

Thus, Volume of the hollow sphere is given by

V = 4/3 π (4^{3} – 2^{3}) —– (1)

Again, diameter of a cone = 8 cm

Then, Radius of a cone (r) = 4 cm

Then, volume of a cone = 1/3 π R^{2} h = 1/3 π × 4^{2} × h —— (2)

But, given that the volume of hollow sphere & cone are equal.

From (1) & (2) = 4/3 π (4^{3} – 2^{3}) = 1/3 π × 4^{2} × h

4 × (64 – 8) = 16 × h

= h = 14 cm

This is the required height of the cone.

**(12) A cylindrical can of internal diameter 21 cm contains water. A solid sphere whose diameter is 10.5 cm is lowered into the cylindrical can. The sphere is completely immersed in water. Calculate the rise in water level, assuming that no water overflows.**

**Solution:**

Given that, inner diameter of cylindrical can = 21 cm

Then, inner radius (R) = 21/2 cm

diameter of sphere = 10.5 cm

Then, Radius of sphere (r) = 10.5/2 = 21/4 cm

Let us consider, h be the height of water level raised.

Rise in (water level) volume = Volume of sphere immersed

π R^{2} h = 4/3 × π r^{3}

(21/2)^{2} π h = 4/3 × π × (21/4)^{3}

21/2 × 21/2 × h = 4/3 × 21/4 × 21/4 × 21/4

= h = 21/12

= 7/14

h = 1.75 cm

This is the required rise in water level.

**(13) There is water to a height of 14 cm in a cylindrical glass jar of radius 8 cm. Inside the water there is a sphere of diameter 12 cm completely immersed. By what height will the water 90 down when the sphere is removed?**

**Solution:**

Given that, Radius of glass jar (R) = 8 cm

diameter of sphere = 12 cm

Then, Radius of sphere (r) = 6 cm

When, the sphere immersed in the water is removed from the jar then water level decreases.

Let us consider, h is the height of water level decreased.

Volume of water decreased = Volume of the sphere

π R^{2} h = 4/3 π r^{3}

8^{2} π h = 4/3 π 6^{3}

h = 4/3 × 6 × 6 × 6/ 8 × 8

h = 4.5 cm

This is the required height by which the water level decreased.

**(15) A solid metallic circular cylinder of radius 14 cm and height 12 cm is melted and recast into small cubes of edge 2 cm. How many such cubes can be made from the solid cylinder.**

**Solution:**

Given that, Radius of solid metallic cylinder (r) = 14 cm

Height of solid cylinder (h) = 12 cm

Now, Volume of cylinder = π r^{2} h

= 14^{2} × 12 × π

= 196 × 12 × π

= 2352 × π

V = 7392 cm^{3} is the required volume of a cylinder.

Then, Volume of a cube = a^{3} = 2^{3} = 8 cm^{3}

V = 8 cm^{3} is the required volume of a cube.

Then, no. of cubes made from solid cylinder = 7392/8 = 924

**(16) How many shots each hawing diameter 3 cm can be made from a cuboidal lead solid of dimensions 9 cm × 11 cm × 12 cm?**

**Solution:**

Given that, diameter of shot = 3 cm

dimensions of cuboidal solid = 9cm × 11cm × 12cm

Then, Volume of cuboidal solid = 9 × 11 × 12 = 1188 cm^{3}

Radius of shot (r) = 3/2 cm

Volume of one shot = 4/3 π r^{3}

V = 4/3 π r^{3}

= 4/3 × π × 1.5^{3}

= 4/3 × 22/7 × 1.5^{3}

V = 297/21 cm^{3} is the required volume of one shot.

Then, no. of shots made from cuboidal lead of solid are

1188 × 21/297

= 84

Thus, the no. of shots made from cuboidal lead of solid is found to be 84.

**(17) A solid metal cylinder of radius 4 cm and height 21 cm is melted down and recast into sphere of radius 3.5 cm. Calculate the no. of spheres that can be made.**

**Solution:**

Given that, Radius of a solid metallic cylinder} (r) = 14 cm

Height of cylinder (h) = 21 cm

Then, Volume of cylinder = π r^{2} h

V = 22/7 × 14 × 14 × 21

V = 12936 cm^{3}

Given that, Radius of sphere (R) = 3.5 cm

Then, Volume of sphere = 4/3 π R^{3}

= 4/3 × 22/7 × 3.5 × 3.5 × 3.5

= 11 × 49/3

V = 539/ 3 cm^{3} is the required volume of sphere.

Then, no. of sphere made = Volume of metal cylinder/ Volume of Sphere

= 12936 × 3/539

No. of spheres made = 72

**(18) A metallic sphere of radius 10.5 cm is melted and then recast into small cense, each of radius 3.5 cm and height 3 cm. Find the no. of cones thus obtained.**

**Solution:**

Given that, Radius of metallic sphere (r) = 10.5 cm

Radius of cone (r) = 3.5 cm

Height of cone (h) = 3 cm

Then, Volume of sphere is given by

V = 4/3 π R^{3}

= 4/3 × π × 10.5^{3}

V = 1543.5 π cm^{3} is the required volume of sphere.

And Volume of cone is given by

V = 1/3 π r^{2} h = 1/3 × π × 3.5^{2} × h

V = 12.25 π cm^{3} is the required volume of cone.

Now, (No. of cones made from sphere) = Volume of sphere/ Volume of cone

= 1543.5 π/12.25 π

(no. of cones made from sphere) = 126

**(19) A certain no. of metallic cones each of radius 2 cm and height 3 cm are melted and recast in a solid sphere of radius 6 cm. Find the no. of cones.**

**Solution:**

Given that, radius of cone (r) = 2 cm

Height of cone (h) = 3 cm

Radius of sphere (R) = 6 cm

Volume of cone = 1/3 π r^{2} h

V = 1/3 × π × 2^{2} × 3

V = 4 π cm^{3} is the required volume of cone

And Volume of sphere = 4/3 π R^{3}

V = 4/3 × π × 6^{3}

V = 288 π cm^{3} is the required volume of sphere.

Then, no. of cones made from sphere = Volume of sphere/ Volume of cone

= 288 π / 4 π

No. of cones made from sphere = 72

**(20) A cylindrical can whose base is horizontal and of radius 3.5 cm contains sufficient water so that when a sphere is placed in the can, the water just covers the sphere. Given that, the sphere just fits into can. Calculate (i) the total surface area of the can in contact with water when sphere is in it.**

**(ii) The depth of water in the can before the sphere was put into the can. Give your answer as proper fractions.**

**Solution:**

Given that, Radius of cylindrical can = 3.5 cm

Radius of the Sphere = 3.5 cm

Water level height in the can = 3.5 × 2 = 7 cm

Height of cylinder (h) = 7 cm

(i) Then, (Total surface area of can in contact with water) = (Curved surface area of cylinder) + (base area of cylinder)

= 2 π r h + π r^{2}

= π r (2h + r)

= 22/7 × 3.5 × (2 × 7 + 3.5)

= 11 × 17.5

TSA = 192.5 cm^{2}

This is the required surface area of can in contact with water.

(ii) Let ‘d’ be the depth of the water before putting sphere in it.

Then, Volume of cylindrical can = (Volume of sphere) + (Volume of water)

π r^{2} h = 4/3 π r^{3} + π r^{2} d

π r^{2} h = π r^{2} [4/3^{r} + d]

h = 4/3^{r} + d

d = h – 4/3 r = 7 – 4/3 × 3.5

d = (21 – 14)/3

d = 7/3 is the required depth of water.