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#### Question 1:

Find the volume, the lateral surface are and the total surface area of the cuboid whose dimensions are:
(i) length = 12 cm, breadth = 8 cm and height = 4.5 cm
(ii) length = 26 m, breadth = 14 m and height = 6.5 m
(iii) length = 15 m, breadth = 6 m and height = 5 dm
(iv) length = 24 m, breadth = 25 cm and height = 6 m

#### Answer:

(i)
Here, l = 12 cm, b = 8 cm, h = 4.5 cm
Volume of the cuboid = $l×b×h\phantom{\rule{0ex}{0ex}}$

Total Surface area = 2(lb + lh+ bh)

Lateral surface area = $2\left(l+b\right)×h$

(ii)
Here, l = 26 m; b = 14 m; h =6.5 m
Volume of the cuboid =  $l×b×h\phantom{\rule{0ex}{0ex}}$

Total surface area = 2(lb + lh+ bh)

Lateral surface area =

(iii)
Here, l = 15 m; b = 6 m; h = 5 dm = 0.5 m
Volume of the cuboid =  $l×b×h\phantom{\rule{0ex}{0ex}}$

Total surface area = 2(lb + lh+ bh

Lateral surface area =

(iv)
Here, l = 24 m; b = 25 cm = 0.25 m; h =6 m
Volume of the cuboid =

Total Surface area = 2(lb + lh+ bh)

Lateral surface area =

#### Question 2:

A matchbox measures 4 cm × 2.5 cm × 1.5 cm. What is the volume of a packet containing 12 such matchboxes?

#### Answer:

Volume of each matchbox = 4 × 2.5 × 1.5 = 15 cm3

∴ Volume of 12 matchboxes = 12 × 15 = 180 cm3

Thus, the volume of a packet containing 12 such matchboxes is 180 cm3.

#### Question 3:

A cuboidal water tank is 6 m long, 5 m wide and 4.5 m deep. How many litres of water can it hold? (Given, 1 m3 = 1000 litres.)

#### Answer:

Volume of water in the tank = Length × Breadth × Height = 6 × 5 × 4.5 = 135 m3

∴ Volume of water in litres = 135 × 1000 = 135000 L        (1 m3 = 1000 L)

Thus, the water tank can hold 135000 L of water.

#### Question 4:

The capacity of a cuboidal tank is 50000 litres of water. Find the breadth of the tank if its length and depth are respectively 10 m and 2.5 m. (Given, 1000 litres = 1 m3.)

#### Answer:

Capacity of the tank = 50000 L = $\frac{50000}{1000}$ = 50 m3           (1000 L = 1 m3)

Length of the tank = 10 m

Height (or depth) of the tank = 2.5 m

Now,

Volume of the cuboidal tank = Length × Breadth × Height

∴ Breadth of the tank =  = 2 m

Thus, the breadth of the tank is 2 m.

#### Question 5:

A godown measures 40 m × 25 m × 15 m. Find the maximum number of wooden crates, each measuring 1.5 m × 1.25 m × 0.5 m, that can be stored in the godown.

#### Answer:

Volume of the godown = 40 m × 25 m × 15 m = 15000 m3

Volume of each wooden crate = 1.5 m × 1.25 m × 0.5 m = 0.9375 m3

∴ Maximum number of wooden crates that can be stored in the godown

#### Question 6:

How many planks of dimensions (5 m × 25 cm × 10 cm) can be stored in a pit which is 20 m long, 6 m wide and 80 cm deep?

#### Answer:

Number of planks =

Volume of one plank =

∴ Number of planks =

#### Question 7:

How many bricks will be required to construct a wall 8 m long, 6 m high and 22.5 cm thick if each brick measures (25 cm × 11.25 cm × 6 cm)?

#### Answer:

Length of the wall = 8 m = 800 cm
Breadth of the wall = 22.5 cm
Height of the wall = 6 m = 600 cm
i.e., volume of wall

Length of the brick = 25 cm
Breadth of the brick = 11.25 cm
Height of the brick = 6 cm
i.e., volume of one brick

∴ Number of bricks = = $\frac{10800000}{1687.5}=6400$

#### Question 8:

Find the capacity of a closed rectangular cistern whose length is 8 m, breadth 6 m and depth 2.5 m. Also, find the area of the iron sheet required to make the cistern.

#### Answer:

Length of the cistern, l = 8 m

Breadth of the cistern, b = 6 m

Height (or depth) of the cistern, h = 2.5 m

∴ Capacity of the cistern

= Volume of the cistern

= l × b × h

= 8 × 6 × 2.5

= 120 m3

Also,

Area of the iron sheet required to make the cistern

= Total surface area of the cistern

= 2(lbbhhl)

= 2(8 × 6 + 6 × 2.5 + 2.5 × 8)

= 2 × 83

= 166 m2

#### Question 9:

The dimensions of a room are (9 m × 8 m × 6.5 m). It has one door of dimensions (2 m × 1.5 m) and two windows, each of dimensions (1.5 m × 1 m). Find the cost of whitewashing the walls at Rs 25 per square metre.

#### Answer:

Length of the room, l = 9 m

Breadth of the room, b = 8 m

Height of the room, h = 6.5 m

Now,

Area of the walls to be whitewashed

= Curved surface area of the room − Area of the door − 2 × Area of each window

= 2h(lb) − 2 m × 1.5 m − 2 × 1.5 m × 1 m

= 2 × 6.5 × (9 + 8) − 3 − 3

= 221 − 6

= 215 m2

∴ Cost of whitewashing the walls at Rs 25 per square metre

= Area of the walls to be whitewashed × Rs 25 per square metre

= 215 × 25

= Rs 5,375

#### Question 10:

A wall 15 m long, 30 cm wide and 4 m high is made of bricks, each measuring (22 cm × 12.5 cm × 7.5 cm). If $\frac{1}{12}$ of the total volume of the wall consists of mortar, how many bricks are there in the wall?

#### Answer:

Length of the wall = 15 m = 1500 cm
Breadth of the wall = 30 cm
Height of the wall = 4 m = 400 cm
Volume of wall =

Also, volume of the mortar =

Total volume of the bricks in the wall = volume of the wall − volume of the mortar
= (18000000 − 1500000) cm3= 16500000 cm3

∴ Number of bricks =

#### Question 11:

How many cubic centimetres of iron are there in an open box whose external dimensions are 36 cm, 25 cm and 16.5 cm, the iron being 1.5 cm thick throughout? If 1 cm3 of iron weighs 15 g, find the weight of the empty box in kilograms.

#### Answer:

The external dimensions of the box are 36 cm, 25 cm and 16.5 cm.

Thickness of the iron = 1.5 cm

∴ Inner length of the box = 36 − 1.5 −1.5 = 33 cm

Inner breadth of the box = 25 − 1.5 −1.5 = 22 cm

Inner height of the box = 16.5 − 1.5 = 15 cm

Now,

Volume of iron in the open box

= Volume of the outer box − Volume of the inner box

= 36 × 25 × 16.5 − 33 × 22 × 15

= 14850 − 10890

= 3960 cm3

It is given that 1 cm3 of iron weighs 15 g.

∴ Weight of the empty box = 3960 × 15 = 59400 g = $\frac{59400}{1000}$ = 59.4 kg         (1 kg = 1000 g)

#### Question 12:

A box made of sheet metal costs ₹ 6480 at ₹ 120 per square metre. If the box is 5 m long and 3 m wide, find its height.

#### Answer:

Length of the box =5 m
Breadth of the box =3 m

Area of the sheet required =

Let h m be the height of the box.
Then area of the sheet = total surface area of the box

∴ The height of the box is 1.5 m.

#### Question 13:

The volume of a cuboid is 1536 m3. Its length is 16 m, and its breadth and height are in the ratio 3 : 2. Find the breadth and height of the cuboid.

#### Answer:

Length of the cuboid = 16 m
Suppose that the breadth and height of the cuboid are 3x m and 2x m, respectively.

∴ The breadth and height of the cuboid are 12 m and 8 m, respectively.

#### Question 14:

How many persons can be accommodated in a dining hall of dimensions (20 m × 16 m × 4.5 m), assuming that each person requires 5 cubic metres of air?

#### Answer:

Volume of the dining hall =

Volume of air required by each person = 5 m3

∴ Capacity of the dining hall =

#### Question 15:

A classroom is 10 m long, 6.4 m wide and 5 m high. If each student be given 1.6 m2 of the floor area, how many students can be accommodated in the room? How many cubic metres of air would each student get?

#### Answer:

Length of the classroom = 10 m
Breadth of the classroom = 6.4 m
Height of the classroom = 5 m
Area of the floor = length $×$ breadth = 10 $×$ 6.4 m2

Now, volume of the classroom=

#### Question 16:

The surface area of a cuboid is 758 cm2. Its length and breadth are 14 cm and 11 cm respectively. Find its height.

#### Answer:

Length of the cuboid = 14 cm
Breadth of the cuboid = 11 cm
Let the height of the cuboid be x cm.
Surface area of the cuboid = 758 cm2

∴ The height of the cuboid is 9 cm.

#### Question 17:

In a shower, 5 cm of rain falls. Find the volume of water that falls on 2 hectares of ground.

#### Answer:

Volume of the water that falls on the ground =

#### Question 18:

Find the volume, the lateral surface area, the total surface area and the diagonal of a cube, each of whose edges measures 9 m. (Take $\sqrt{3}=1.73$.)

#### Answer:

Here, a = 9 m
Volume of the cube =
Lateral surface area of the cube =
Total surface area of the cube =
∴ Diagonal of the cube =

#### Question 19:

The total surface area of a cube is 1176 cm2. Find its volume.

#### Answer:

Suppose that the side of cube is x cm.
Total surface area of cube = 1176 sq cm

i.e., the side of the cube is 14 cm.

∴ Volume of the cube =

#### Question 20:

The lateral surface area of a cube is 900 cm2. Find its volume.

#### Answer:

Suppose that the side of cube is x cm.
Lateral surface area of the cube = 900 cm2

i.e., the side of the cube is 15 cm.
∴ Volume of the given cube =

#### Question 21:

The volume of a cube is 512 cm3. Find its surface area.

#### Answer:

Suppose that the side of the given cube is x cm.
Volume of the cube = 512 cm3

∴ Surface area of the cube =

#### Question 22:

Three cubes of metal with edges 3 cm, 4 cm and 5 cm respectively are melted to form a single cube. Find the lateral surface area of the new cube formed.

#### Answer:

Three cubes of metal with edges 3cm, 4cm and 5 cm are melted to form a single cube.

∴ Volume of the new cube = sum of the volumes the old cubes

Suppose the edge of the new cube = x cm
Then we have:

∴ Lateral surface area of the new cube =

#### Question 23:

Find the length of the longest pole that can be put in a room of dimensions (10 m × 10 m × 5 m).

#### Answer:

Length of the longest pole = length of the diagonal of the room

#### Question 24:

The sum of length, breadth and depth of a cuboid is 19 cm and the length of its diagonal is 11 cm. Find the surface area of the cuboid.

#### Answer:

Let the length, breadth and height (or depth) of the cuboid be l cm, b cm and h cm, respectively.

∴ lbh = 19           .....(1)

Also,

Length of the diagonal = 11 cm

Squaring (1), we get

(l + b + h)2 = 192

⇒ l2b2 + h+ 2(lbbhhl) = 361

⇒ 121 + 2(lb + bh + hl) = 361                      [Using (2)]

⇒ 2(lb + bh + hl) = 361 − 121 = 240 cm2

Thus, the surface area of the cuboid is 240 cm2.

#### Question 25:

Each edge of a cube is increased by 50%. Find the percentage increase in the surface area of the cube.

#### Answer:

Let the initial edge of the cube be a units.

∴ Initial surface area of the cube = 6a2 square units

New edge of the cube = a + 50% of a$a+\frac{50}{100}a$ = 1.5a units

∴ New surface of the cube = 6(1.5a)2 = 13.5a2 square units

Increase in surface area of the cube = 13.5a2 − 6a2 = 7.5a2 square units

∴ Percentage increase in the surface area of the cube

#### Question 26:

If V is the volume of a cuboid of dimensions a, b, c and S is its surface area then prove that $\frac{1}{V}=\frac{2}{S}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)$.

#### Answer:

Let the length, breadth and height of the cuboid be ab and c, respectively.

∴ Surface area of the cuboid, S = 2(abbcca

Volume of the cuboid, Vabc

Now,

$\frac{S}{V}=\frac{2\left(ab+bc+ca\right)}{abc}\phantom{\rule{0ex}{0ex}}⇒\frac{S}{V}=2\left(\frac{ab}{abc}+\frac{bc}{abc}+\frac{ca}{abc}\right)\phantom{\rule{0ex}{0ex}}⇒\frac{S}{V}=2\left(\frac{1}{c}+\frac{1}{a}+\frac{1}{b}\right)\phantom{\rule{0ex}{0ex}}⇒\frac{1}{V}=\frac{2}{S}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)$

#### Question 27:

Water in a canal, 30 dm wide and 12 dm deep, is flowing with a velocity of 20 km per hour. How much area will it irrigate, if 9 cm of standing water is densired?

#### Answer:

Width of the canal = 30 dm = 3 m                (1 m = 10 dm)

Depth of the canal = 12 dm = 1.2 m

Speed of the water flow = 20 km/h = 20000 m/h

∴ Volume of water flowing out of the canal in 1 h = 3 × 1.2 × 20000 = 72000 m3

Height of standing water on field = 9 cm = 0.09 m            (1 m = 100 cm)

Assume that water flows out of the canal for 1 h. Then,

Area of the field irrigated

Thus, the area of the field irrigated is 80 hectares.

Disclaimer: In this question time is not given, so the question is solved assuming that the water flows out of the canal for 1 hour.

#### Question 28:

A solid metallic cuboid of dimensions (9 m × 8 m × 2 m) is melted and recast into solid cubes of edge 2 m. Find the number of cubes so formed.

#### Answer:

Volume of the solid metallic cuboid = 9 m × 8 m × 2 m = 144 m3

Volume of each solid cube = (Edge)3 = (2)3 = 8 m3

∴ Number of cubes formed =  = 18

Thus, the number of cubes so formed is 18.

#### Question 1:

The diameter of a cylinder is 28 cm and its height is 40 cm. Find the curved surface area, total surface area and the volume of the cylinder.

#### Answer:

Here, r = 28/2 = 14 cm; h = 40 cm

#### Question 2:

A patient in a hospital is given soup daily in a cylindrical bowl of diameter 7 cm. If the bowl is filled with soup to a height of 4 cm, how much soup the hospital has to prepare daily to serve 250 patients?

#### Answer:

Radius of the bowl, r

Height of soup in the bowl, h = 4 cm

Volume of soup in one bowl =

∴ Amount of soup the hospital has to prepare daily to serve 250 patients

= Volume of soup in one bowl × 250

= 154 × 250

= 38500 cm3

#### Question 3:

The pillars of a temple are cylindrically shaped. Each pillar has a circular base of radius 20 cm and height 10 m. How much concrete mixture would be required to build 14 such pillars?

#### Answer:

Radius of each pillar, r = 20 cm = 0.2 m   (1 m = 100 cm)

Height of each pillar, h = 10 m

Volume of concrete mixture used in each pillar =

∴ Amount of concrete mixture required to build 14 such pillars

= Volume of concrete mixture used in each pillar × 14

=

= 17.6 m3

#### Question 4:

A soft drink is available in two packs: (i) a tin can with a rectangular base of length 5 cm, breadth 4 cm and height 15 cm, and (ii) a plastic cylinder with circular base of diameter 7 cm and height 10 cm. Which container has greater capacity and by how much?

#### Answer:

(i) Length of tin can, l = 5 cm

Breadth of tin can, b = 4 cm

Height of tin can, h = 15 cm

∴ Volume of soft drink in tin can = l × b × h = 5 × 4 × 15 = 300 cm3

(ii) Radius of plastic cylinder, r$\frac{7}{2}$ cm

Height of plastic cylinder, h = 10 cm

∴ Volume of soft drink in plastic cylinder = $\mathrm{\pi }{r}^{2}h=\frac{22}{7}×{\left(\frac{7}{2}\right)}^{2}×10$ = 385 cm3

So, the capacity of the plastic cylinder pack is greater than the capacity of the tin can pack.

Difference in the capacities of the two packs = 385 − 300 = 85 cm3

Thus, the capacity of the plastic cylinder pack is 85 cm3 more than the capacity of the tin can pack.

#### Question 5:

There are 20 cylindrical pillars in a building, each having a diameter of 50 cm and height 4 m. Find the cost of cleaning them at ₹ 14 per m2.

#### Answer:

Radius of each pillar, r = $\frac{50}{2}$ = 25 cm = 0.25 m        (1 m = 100 cm)

Height of each pillar, h = 4 m

∴ Surface area of each pillar =

Surface area of 20 pillars = Surface area of each pillar × 20 =

Rate of cleaning = ₹ 14 per m2

∴ Total cost of cleaning the 20 pillars

= Surface area of 20 pillars × Rate of cleaning

= ₹ 1760

#### Question 6:

The curved surface area of a right circular cylinder is 4.4 m2. If the radius of its base is 0.7 m, find its (i) height and (ii) volume.

#### Answer:

Radius of the cylinder, r = 0.7 m

Curved surface area of cylinder = 4.4 m2

(i) Let the height of the cylinder be h m.

Thus, the height of the cylinder is 1 m.

(ii) Volume of the cylinder = $\mathrm{\pi }{r}^{2}h=\frac{22}{7}×{\left(0.7\right)}^{2}×1$ = 1.54 m3

Thus, the volume of the cylinder is 1.54 m3.

#### Question 7:

The lateral surface area of a cylinder is 94.2 cm2 and its height is 5 cm. Find (i) the radius of its base and (ii) its volume. (Take π = 3.14.)

#### Answer:

Height of the cylinder, h = 5 cm

Lateral (or curved) surface area of cylinder = 94.2 cm2

(i) Let the radius of the cylinder be r cm.

Thus, the radius of the cylinder is 3 cm.

(ii) Volume of the cylinder = $\mathrm{\pi }{r}^{2}h=3.14×{\left(3\right)}^{2}×5$ = 141.3 cm3

Thus, the volume of the cylinder is 141.3 cm3.

#### Question 8:

The capacity of a closed cylindrical vessel of height 1 m is 15.4 litres. Find the area of the metal sheet needed to make it.

#### Answer:

Height of the cylinder, h = 1 m

Capacity of the cylinder = 15.4 L = 15.4 × 0.001 m3 = 0.0154 m3

∴ Area of the metal sheet needed to make the cylinder

= Total surface area of the cylinder

Thus, the area of the metal sheet needed to make the cylinder is 0.4708 m2.

#### Question 9:

The inner diameter of a cylindrical wooden pipe is 24 cm and its outer diameter is 28 cm. The length of the pipe is 35 cm. Find the mass of the pipe, if 1 cm3 of wood has a mass of 0.6 g.

#### Answer:

Inner radius of the wooden pipe, r$\frac{24}{2}$ = 12 cm

Outer radius of the wooden pipe, R$\frac{28}{2}$ = 14 cm

Length of the wooden pipe, h = 35 cm

∴ Volume of wood in the pipe =

It is given that 1 cmof wood has a mass of 0.6 g.

∴ Mass of the pipe = Volume of wood in the pipe × 0.6 = 5720 × 0.6 = 3432 g = $\frac{3432}{1000}$ = 3.432 kg

Thus, the mass of the pipe is 3.432 kg.

#### Question 10:

In a water heating system, there is a cylindrical pipe of length 28 m and diameter 5 cm. Find the total radiating surface in the system.

#### Answer:

Length of the cylindrical pipe, h = 28 m

Radius of the cylindrical pipe, r$\frac{5}{2}$ = 2.5 cm = 0.025 m         (1 m = 100 cm)

∴ Total radiating surface in the system

= Curved surface area of the cylindrical pipe

Thus, the total radiating surface in the system is 4.4 m2.

#### Question 11:

Find the weight of a solid cylinder of radius 10.5 cm and height 60 cm if the material of the cylinder weighs 5 g per cm2.

#### Answer:

Here, r = 10.5 cm; h = 60 cm

∴ Weight of cylinder = volume of cylinder $×$weight of cylinder per gram

#### Question 12:

The curved surface area of a cylinder is 1210 cm2 and its diameter is 20 cm. Find its height and volume.

#### Answer:

Curved surface area = 1210 cm2
Suppose that the height of cylinder is h cm.
We have r = 10 cm

#### Question 13:

The curved surface area of a cylinder is 4400 cm2 and the circumference of its base is 110 cm. Find the height and the volume of the cylinder.

#### Answer:

Let r be the radius and h be the height of the cylinder.
Circumference of its base(circle) = 110 cm.

Curved surface area of a cylinder = 4400 cm2.

Also, Volume of the cylinder =

#### Question 14:

The radius of the base and the height of a cylinder are in the ratio 2 : 3. If its volume is 1617 cm3, find the total surface area of the cylinder.

#### Answer:

Suppose that the radius of the base and the height of the cylinder are 2x cm and 3x cm, respectively.

Hence, radius = 7 cm; height of the cylinder = 10.5 cm

#### Question 15:

The total surface area of a cylinder is 462 cm2. Its curved surface area is one-third of its total surface area. Find the volume of the cylinder.

#### Answer:

Total surface area = 462 cm2

Given: Curved surface area =$\frac{1}{3}$$×$total surface area =

Now, total surface area − curved surface area =

Now, curved surface area = 154 cm2

#### Question 16:

The total surface area of a solid cylinder is 231 cm2 and its curved surface area is $\frac{2}{3}$ of the total surface area. Find the volume of the cylinder.

#### Answer:

Curved surface area = $\frac{2}{3}$$×$total surface area =

Now, total surface area − curved surface area =

#### Question 17:

The ratio between the curved surface area and the total surface area of a right circular cylinder is 1 : 2. Find the volume of the cylinder if its total surface area is 616 cm2.

#### Answer:

Suppose that the curved surface area and the total surface area of the right circular cylinder are x cm2 and 2x cm2.
Then we have:
2x = 616
x = 308 sq cm
Hence, the curved surface area of the cylinder is 308 sq cm.
Let  r cm and h cm be the radius and height of the cylinder, respectively.

#### Question 18:

A cylindrical bucket, 28 cm in diameter and 72 cm and high, is full of water. The water is emptied into a rectangular tank, 66 cm long and 28 cm wide. Find the height of the water level in the tank.

#### Answer:

Given: Diameter of the cylindrical bucket = 28 cm
i.e., radius = 14 cm
Height of the cylindrical bucket, h1 = 72 cm
Length of the rectangular tank, l = 66 cm
Breadth of the rectangular tank, b = 28 cm
Let the height of the rectangular tank be h cm.
The water from the cylindrical bucket is emptied into the rectangular tank.
i.e., volume of the bucket = volume of the tank

∴ Height of the rectangular tank = 24 cm

#### Question 19:

The barrel of a fountain pen, cylindrical in shape, is 7 cm long and 5 mm in diameter. A full barrel of ink in the pen will be used up on writing 330 words on an average. How many words would use up a bottle of ink containing one-fifth of a litre?

#### Answer:

Height of the barrel = h = 7 cm
Radius of the barrel = r = 2.5 mm = 0.25 cm
Volume of the barrel =

i.e., 1.375 cm3 of ink is used for writing 330 words
Now, number of words that could be written with one-fifth of a litre, i.e., $\frac{1}{5}×1000$ cm3 =
∴ 48000 words would use up a bottle of ink containing one-fifith of a litre.

#### Question 20:

1 cm3 of gold is drawn into a wire 0.1 mm in diameter. Find the length of the wire.

#### Answer:

Let r cm be the radius of the wire and h cm be the height of the wire.
Volume of the gold = 1 cm3
1 cm3 of gold is drawn into a wire of diameter 0.1 mm.
Here, r

Hence, length of the wire = 127.27 m

#### Question 21:

If 1 cm3 of case iron weighs 21 g, find the weight of a cast iron pipe of length 1 m with a bore of 3 cm in which the thickness of the metal is 1 cm.

#### Answer:

Internal radius of the pipe = 1.5 cm
External radius of the pipe = (1.5 + 1) cm = 2.5 cm
Height of the pipe = 1 m = 100 cm
Volume of the cast iron = total volume of the pipe − internal volume of the pipe
$=\mathrm{\pi }×\left(2.5{\right)}^{2}×100-\mathrm{\pi }×{\left(1.5\right)}^{2}×100\phantom{\rule{0ex}{0ex}}=\frac{22}{7}×100×\left[6.25-2.25\right]\phantom{\rule{0ex}{0ex}}=\frac{2200}{7}×4=\frac{8800}{7}{\mathrm{cm}}^{3}$
1 cm3 of cast iron weighs 21 g.

∴ Weight of $\frac{8800}{7}{\mathrm{cm}}^{3}$cast iron =

#### Question 22:

A cylindrical tube, open at both ends, is made of metal. The internal diameter of the tube is 10.4 cm and its length is 25 cm. The thickness of the metal is 8 mm everywhere. Calculate the volume of the metal.

#### Answer:

Inner radius of the cylindrical tube, r = 5.2 cm
Height of the cylindrical tube, h = 25 cm
Outer radius of the cylindrical tube, R = (5.2 + 0.8) cm = 6 cm

#### Question 23:

It is required to make a closed cylindrical tank of height 1 m and base diameter 140 cm from a metal sheet. How many square metres of the sheet are required for the same?

#### Answer:

Height of the cylindrical tank, h = 1 m

Radius of the cylindrical tank, r$\frac{140}{2}$ = 70 cm = 0.7 m       (1 m = 100 cm)

∴ Area of the metal sheet required to make the cylindrical tank

= Total surface area of the cylindrical tank

Thus, the area of metal sheet required to make the cylindrial tank is 7.48 m2.

#### Question 24:

A juiceseller has a large cylindrical vessel of base radius 15 cm filled up to a height of 32 cm with orange juice. The juice is filled in small cylindrical glasses of radius 3 cm up to a height of 8 cm, and sold for ₹ 15 each. How much money does he received by selling the juice completely?

#### Answer:

Radius of the vassel, R = 15 cm

Height of orange juice in the vassel, H = 32 cm

∴ Volume of orange juice in the vassel =

Radius of the glass, r = 3 cm

Height of orange juice in the glass, h = 8 cm

∴ Volume of orange juice in each glass =

Number of glasses of orange juice sold by the juiceseller

Rate of each glass of orange juice = ₹ 15

∴ Total money received by the juiceseller

= Number of glasses of orange juice sold by the juiceseller × Rate of each glass of orange juice

= 100 × 15

= ₹ 1,500

Thus, the total money received by the juiceseller by selling the juice completely is ₹ 1,500.

#### Question 25:

A well with inside diameter 10 m is dug 8.4 m deep. Earth taken out of it is spread all around it to a width of 7.5 m to form an embankment. Find the height of the embankment.

#### Answer:

Inner radius of the well, r$\frac{10}{2}$ = 5 m

Depth of the well, h = 8.4 m

Suppose the outer radius of the embankment is R m.

Width of the embankment = 7.5 m

∴ R − r = 7.5 m

⇒ R = 7.5 + 5 = 12.5 m

Let the height of the embankment be H m.

Now,

Volume of earth used to form the embankment = Volume of earth dugged out of the well

Thus, the height of the embankment is 2.1 m.

#### Question 26:

How many litres of water flows out of a pipe having an area of cross section of 5 cm2 in 1 minute, if the speed of water in the pipe is 30 cm/sec?

#### Answer:

Speed of the water = 30 cm/s

Area of the cross section = 5 cm2

Volume of the water flowing out of the pipe in one second = Area of the cross section × 30 cm = 5 × 30 = 150 cm3

Now, 1 minute = 60 seconds

∴ Volume of the water flowing out of the pipe in 60 seconds

= Volume of the water flowing out of the pipe in one second × 60

= 150 × 60

= 9000 cm3

$\frac{9000}{1000}$              (1 L = 1000 cm3)

= 9 L

Thus, 9 L of water flows out of the given pipe in 1 minute.

#### Question 27:

A cylindrical water tank of diameter 1.4 m and height 2.1 m is being fed by a pipe of diameter 3.5 cm through which water flows at the rate of 2 m per second. In how much time will the tank be filled?

#### Answer:

Radius of the water tank, R$\frac{1.4}{2}$ = 0.7 m

Height of the water tank, H = 2.1 m

∴ Capacity of the water tank = $\mathrm{\pi }{R}^{2}H=\mathrm{\pi }{\left(0.7\right)}^{2}×2.1$ m3

Speed of the water flow = 2 m/s

Radius of the pipe, r$\frac{3.5}{2}$ = 1.75 cm = 0.0175 m

Area of the cross section of the pipe = $\mathrm{\pi }{r}^{2}=\mathrm{\pi }{\left(0.0175\right)}^{2}$ m2

Volume of the water flowing out of the pipe in one second = Area of the cross section of the pipe × 2 m = $\mathrm{\pi }{\left(0.0175\right)}^{2}×2$ m3

Let the time taken to fill the tank be t seconds.

∴ Volume of the water flowing out of the pipe in t seconds

= Volume of the water flowing out of the pipe in one second × t

$\mathrm{\pi }{\left(0.0175\right)}^{2}×2×t$ m3

Now,

Volume of the water flowing out of the pipe in t seconds = Capacity of the water tank

$⇒t=\frac{1680}{60}$

⇒ t = 28 minutes

Thus, the tank will be filled in 28 minutes.

#### Question 28:

A cylindrical container with diameter of base 56 cm contains sufficient water to submerge a rectangular solid of iron with dimensions (32 cm × 22 cm × 14 cm). Find the rise in the level of water when the solid is completely submerged.

#### Answer:

Volume of the rectangular solid of iron = 32 cm × 22 cm × 14 cm

Radius of the container, r = $\frac{56}{2}$ = 28 cm

Let the rise in the level of water in the container when rectangular solid of iron is submerged in it be h cm.

∴ Volume of the water displaced in the container = $\mathrm{\pi }{r}^{2}h=\mathrm{\pi }×{\left(28\right)}^{2}×h$

When the rectangular solid of iron is submerged in the container, then the volume of water displaced in the container is equal to the volume of the rectangular solid of iron.

Thus, the rise in the level of water in the container is 4 cm.

#### Question 29:

Find the cost of sinking a tube-well 280 m deep, having a diameter 3 m at the rate of ₹ 15 per cubic metre. Find also the cost of cementing its inner curved surface at ₹ 10 per square metre.

#### Answer:

Height of the tube-well, h = 280 m

Radius of the tube-well, r$\frac{3}{2}$ m

∴ Volume of the tube-well = $\mathrm{\pi }{r}^{2}h=\frac{22}{7}×{\left(\frac{3}{2}\right)}^{2}×280$ = 1980 m3

Rate of sinking the tube-well = ₹ 15 per cubic metre

∴ Cost of sinking the tube-well = Volume of the tube-well × Rate of sinking the tube-well = 1980 × 15 = ₹ 29,700

Thus, the cost of sinking the tube-well is ₹ 29,700.

Inner curved surface of the tube-well = $2\mathrm{\pi }rh=2×\frac{22}{7}×\frac{3}{2}×280$ = 2640 m2

Rate of cementing = ₹ 10 per square metre

∴ Cost of cementing the inner curved surface of the tube-well

= Inner curved surface of the tube-well × Rate of cementing

= 2640 × 10

= ₹ 26,400

Thus, the cost of cementing the inner curved surface of the tube-well is ₹ 26,400.

#### Question 30:

Find the length of 13.2 kg of copper wire of diameter 4 mm, when 1 cubic centimetre of copper weighs 8.4 g.

#### Answer:

Mass of copper wire = 13.2 kg = 13.2 × 1000 = 13200 g        (1 kg = 1000 g)

Volume of 8.4 g of copper wire = 1 cm3

∴ Volume of 13200 g (or 13.2 kg) of copper wire = $\frac{13200}{8.4}$ cm3

Let the length of the copper wire be l cm.

Radius of the copper wire, r$\frac{4}{2}$ = 2 mm = 0.2 cm            (1 cm = 10 mm)

∴ Volume of the copper wire = $\mathrm{\pi }{r}^{2}l=\frac{22}{7}×{\left(0.2\right)}^{2}×l$ cm3

Thus, the length of the copper wire is 125 m.

#### Question 31:

It costs ₹ 3300 to paint the inner curved surface of a cylindrical vessel 10 m deep at the rate of Rs 30 per m2. Find the
(i) inner curved surface area of the vessel,
(ii) inner radius of the base, and
(iii) capacity of the vessel.

#### Answer:

Total cost of paining the inner curved surface of the cylinderical vassel = ₹ 3,300

Rate of painting = ₹ 30 per m2

(i) Inner curved surface area of the vassel

Thus, the inner curved surface area of the vassel is 110 m2.

(ii) Depth of the vassel, h = 10 m

Let the inner radius of the base be r m.

∴ Inner curved surface area of the vasssel = $2\mathrm{\pi }rh=2×\frac{22}{7}×r×10$

Thus, the inner radius of the base is 1.75 m.

(iii) Capacity of the vassel = $\mathrm{\pi }{r}^{2}h=\frac{22}{7}×{\left(1.75\right)}^{2}×10=$96.25 m3

Thus, the capacity of the vassel is 96.25 m3.

#### Question 32:

The difference between inside and outside surfaces of a cylindrical tube 14 cm long, is 88 cm2. If the volume of the tube is 176 cm3, find the inner and outer radii of the tube.

#### Answer:

Let the inner and outer radii of the tube be r cm and R cm, respectively.

Length of the cylindrical tube, h = 14 cm

Outer curved surface of the cylinder − Inner curved surface of the cylinder = 88 cm2        (Given)

Volume of the tube = 176 cm3                (Given)

Adding (1) and (2), we get

2R = 5

⇒ R = 2.5 cm

Putting R = 2.5 cm in (2), we get

2.5 + r = 4

⇒ r = 4 − 2.5 = 1.5 cm

Thus, the inner and outer radii of the tube are 1.5 cm and 2.5 cm, respectively.

#### Question 33:

A rectangular sheet of paper 30 cm × 18 cm can be transformed into the curved surface of a right circular cylinder in two ways namely, either by rolling the paper along its length or by rolling it along its breadth. Find the ratio of the volumes of the two cylinders, thus formed.

#### Answer:

The dimensions of the rectangular sheet of paper are 30 cm × 18 cm.

Let V1 and Vbe the volumes of the cylinders formed by rolling the rectangular sheet of paper along its length (i.e. 30 cm) and breadth (i.e. 18 cm), respectively.

Suppose r1 and h1 be the radius and height of the cylinder formed by rolling the rectangular sheet of paper along its length, respectively.

h1 = 18 cm

Also, suppose r2 and h2 be the radius and height of the cylinder formed by rolling the rectangular sheet of paper along its breadth, respectively.

h2 = 30 cm

Now,

$\frac{{V}_{1}}{{V}_{2}}=\frac{\mathrm{\pi }{\left(\frac{30}{2\mathrm{\pi }}\right)}^{2}×18}{\mathrm{\pi }{\left(\frac{18}{2\mathrm{\pi }}\right)}^{2}×30}=\frac{5}{3}$

⇒ V1 : V2 = 5 : 3

Thus, the ratio of the volumes of the two cylinders thus formed is 5 : 3.

#### Question 1:

Note Use $\mathrm{\pi }=\frac{22}{7}$, unless stated otherwise.
Find the curved surface area of a cone with base radius 5.25 cm and slant height 10 cm.

#### Answer:

Radius of the base, r = 5.25 cm

Slant height, l = 10 cm

∴ Curved surface area of the cone = $\mathrm{\pi }rl=\frac{22}{7}×5.25×10$ = 165 cm2

Thus, the curved surface area of the cone is 165 cm2.

#### Question 2:

Note Use $\mathrm{\pi }=\frac{22}{7}$, unless stated otherwise.
Find the total surface area of a cone, if its slant height is 21 m and diameter of its base is 24 m.

#### Answer:

Slant height, l = 21 m

Radius of the base, r$\frac{24}{2}$ = 12 m

∴ Total surface area of the cone =

Thus, the total surface area of the cone is .

#### Question 3:

Note Use $\mathrm{\pi }=\frac{22}{7}$, unless stated otherwise.
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 sheet required to make 10 such caps.

#### Answer:

Radius of the cone, r = 7 cm

Height of the cone, h = 24 cm

∴ Slant height of the cone, $l=\sqrt{{r}^{2}+{h}^{2}}=\sqrt{{7}^{2}+{24}^{2}}=\sqrt{49+576}=\sqrt{625}$ = 25 cm

Area of the sheet required to make one cap = Curved surface area of the cone = $\mathrm{\pi }rl=\frac{22}{7}×7×25$ = 550 cm2

∴ Area of the sheet required to make 10 such caps = 550 × 10 = 5500 cm2

Thus, the area of the sheet required to make 10 such jocker caps is 5500 cm2.

#### Question 4:

Note Use $\mathrm{\pi }=\frac{22}{7}$, unless stated otherwise.
The curved surface area of a cone is 308 cm2 and its slant height is 14 cm. Find the radius of the base and total surface area of the cone.

#### Answer:

Slant height, l = 14 cm

Let the radius of the base be r cm.

Curved surface area of the cone = 308 cm        (Given)

∴ Total surface area of the cone = $\mathrm{\pi }r\left(r+l\right)=\frac{22}{7}×7×\left(7+14\right)=\frac{22}{7}×7×21$ = 462 cm2

Thus, the radius of the base is 7 cm and the total surface area of the cone is 462 cm2.

#### Question 5:

Note Use $\mathrm{\pi }=\frac{22}{7}$, unless stated otherwise.
The slant height and base diameter of a conical tomb are 25 m and 14 m respectively. Find the cost of whitewashing its curved surface at the rate of ₹ 12 per m2.

#### Answer:

Slant height of the conical tomb, l = 25 m

Radius of the conical tomb, r = $\frac{14}{2}$ = 7 m

∴ Curved surface area of the conical tomb = $\mathrm{\pi }rl=\frac{22}{7}×7×25$ = 550 m2

Rate of whitewashing = ₹ 12 per m2

∴ Cost of whitewashing the conical tomb

= Curved surface area of the conical tomb × Rate of whitewashing

= 550 × 12

= ₹ 6,600

Thus, the cost of whitewashing the conical tomb is ₹ 6,600.

#### Question 6:

Note Use $\mathrm{\pi }=\frac{22}{7}$, unless stated otherwise.
A conical tent is 10 m high and the radius of its base is 24 m. Find the slant height of the tent. If the cost of 1 m2 canvas is ₹ 70, find the cost of canvas required to make the tent.

#### Answer:

Radius of the conical tent, r = 24 m

Height of the conical tent, h = 10 m

∴ Slant height of the conical tent, $l=\sqrt{{r}^{2}+{h}^{2}}=\sqrt{{24}^{2}+{10}^{2}}=\sqrt{576+100}=\sqrt{676}$ = 26 m

Curved surface area of the conical tent =

The cost of 1 m2 canvas is ₹ 70.

∴ Cost of canvas required to make the tent

= Curved surface area of the conical tent × ₹ 70

$=\frac{22}{7}×24×26×70$

= ₹ 1,37,280

Thus, the cost of canvas required to make the tent is ₹ 1,37,280.

#### Question 7:

A bus stop is barricated from the remaining part of the road by using 50 hollow cones made of recycled cardboard. Each one has a base diameter of 40 cm and height 1 m. If the outer side of each of the cones is to be painted and the cost of painting is ₹ 25 per m2, what will be the cost of painting all these cones? (Use π = 3.14 and $\sqrt{1.04}$ = 1.02).

#### Answer:

Radius of each cone, r = $\frac{40}{2}$ = 20 cm = 0.2 m         (1 m = 100 cm)

Height of each cone, h = 1 m

∴ Slant height of each cone, $l=\sqrt{{r}^{2}+{h}^{2}}=\sqrt{{\left(0.2\right)}^{2}+{1}^{2}}=\sqrt{0.04+1}=\sqrt{1.04}$ = 1.02 m

Curved surface area of each cone = $\mathrm{\pi }rl=3.14×0.2×1.02$ = 0.64056 m2

∴ Curved surface area of 50 cones = 0.64056 × 50 = 32.028 m2

Cost of painting = ₹ 25 per m2

∴ Total cost of painting all the cones

= Curved surface area of 50 cones × ₹ 25

= 32.028 × 25

= ₹ 800.70

Thus, the cost of painting all the cones is ₹ 800.70.

#### Question 8:

Note Use $\mathrm{\pi }=\frac{22}{7}$, unless stated otherwise.
Find the volume, curved surface area and the total surface area of a cone having base radius 35 cm and height is 12 cm.

#### Answer:

Radius of the cone, r = 35 cm

Height of the cone, h = 12 cm

∴ Slant height of the cone, $l=\sqrt{{r}^{2}+{h}^{2}}=\sqrt{{35}^{2}+{12}^{2}}=\sqrt{1225+144}=\sqrt{1369}$ = 37 cm

(i) Volume of the cone = $\frac{1}{3}\mathrm{\pi }{r}^{2}h=\frac{1}{3}×\frac{22}{7}×{\left(35\right)}^{2}×12$ = 15400 cm3

(ii) Curved surface area of the cone = $\mathrm{\pi }rl=\frac{22}{7}×35×37$ = 4070 cm2

(iii) Total surface area of the cone = $\mathrm{\pi }r\left(r+l\right)=\frac{22}{7}×35×\left(35+37\right)=\frac{22}{7}×35×72$ = 7920 cm2

#### Question 9:

Find the volume, curved surface area and the total surface area of a cone whose height and slant height are 6 cm and 10 cm respectively.

#### Answer:

Height of the cone, h = 6 cm
Slant height of the cone, l = 10 cm
Radius, r =

Volume of the cone =

Curved surface area of the cone =

∴ Total surface area =

#### Question 10:

Note Use $\mathrm{\pi }=\frac{22}{7}$, unless stated otherwise.
A conical pit of diameter 3.5 m is 12 m deep. What is its capacity in kilolitres?
HINT 1 m3 = 1 kilolitre.

#### Answer:

Radius of the conical pit, r = $\frac{3.5}{2}$ m

Depth of the conical pit, h = 12 m

∴ Capacity of the conical pit = $\frac{1}{3}\mathrm{\pi }{r}^{2}h=\frac{1}{3}×\frac{22}{7}×{\left(\frac{3.5}{2}\right)}^{2}×12$ = 38.5 m3 = 38.5 kL         (1 m3 = 1 kilolitre)

Thus, the capacity of the conical pit is 38.5 kL.

#### Question 11:

A heap of wheat is in the form of a cone of diameter 9 m and height 3.5 m. Find its volume. How much canvas cloth is required to just cover the heap? (Use π = 3.14).

#### Answer:

Radius of the heap, r = $\frac{9}{2}$ m = 4.5 m

Height of the heap, h = 3.5 m

​∴ Volume of the heap of wheat = $\frac{1}{3}\mathrm{\pi }{r}^{2}h=\frac{1}{3}×3.14×{\left(4.5\right)}^{2}×3.5$ = 74.1825 m3

Now,

Slant height of the heap, $l=\sqrt{{r}^{2}+{h}^{2}}=\sqrt{{\left(4.5\right)}^{2}+{\left(3.5\right)}^{2}}=\sqrt{20.25+12.25}=\sqrt{32.5}$ ≈ 5.7 m

∴ Area of the canvas cloth required to just cover the heap of wheat

= Curved surface area of the heap of wheat

Thus, the area of canvas cloth required to just cover the heap is approximately 80.541 m2 .

#### Question 12:

Note Use $\mathrm{\pi }=\frac{22}{7}$, unless stated otherwise.
A man uses a piece of canvas having an area of 551 m2, to make a conical tent of base radius 7 m. Assuming that all the stitching margins and wastage incurred while cutting, amount to approximately 1 m2, find the volume of the tent that can be made with it.

#### Answer:

Area of the canvas = 551 m2

Area of the canvas used in stitching margins and wastage incurred while cutting = 1 m2

∴ Area of the canvas used in making the tent = 551 − 1 = 550 m2

Radius of the tent, r = 7 m

Let the slant height and height of the tent be l m and h m, respectively.

Area of the canvas used in making the tent = 550 m2

Now,

Height of the tent, $h=\sqrt{{l}^{2}-{r}^{2}}=\sqrt{{25}^{2}-{7}^{2}}=\sqrt{625-49}=\sqrt{576}$ = 24 m

∴ Volume of the tent = $\frac{1}{3}\mathrm{\pi }{r}^{2}h=\frac{1}{3}×\frac{22}{7}×{\left(7\right)}^{2}×24$ = 1232 m3

Thus, the volume of the tent that can be made with the given canvas is 1232 m3.

#### Question 13:

How many metres of cloth, 2.5 m wide, will be required to make a conical tent whose base radius is 7 m and height 24 metres?

#### Answer:

Radius of the conical tent, r = 7 m
Height of the conical tent, h = 24 m

Width of the cloth = 2.5 m
∴ Length of the cloth =

#### Question 14:

Two cones have their heights in the ratio 1 : 3 and the radii of their bases in the ratio 3 : 1. Show that their volumes are in the ratio 3 : 1.

#### Answer:

Let the heights of the first and second cones be h  and 3h, respectively .
Also, let the radius of the first and second cones be 3r and r,  respectively.

∴ Ratio of volumes of the cones =

#### Question 15:

A cylinder and a cone have equal radii of their bases and equal heights. If their curved surface areas are in the ratio 8 : 5., show that the radius and height of each has the ratio 3 : 4.

#### Answer:

Suppose that the respective radii and height of the cone and the cylinder are r and h.
Then ratio of curved surface areas = 8 : 5
Let the curved surfaces areas be 8x and 5x.

∴ The ratio of the radius and height of the cone is 3 : 4.

#### Question 16:

A right circular cone is 3.6 cm high and the radius of its base is 1.6 cm. It is melted and recast into a right circular cone having base radius 1.2 cm. Find its height.

#### Answer:

Let the cone which is being melted be denoted by cone 1 and let the cone into which cone 1 is being melted be denoted by cone 2.

Height of cone 1 = 3.6 cm
Radius of the base of cone 1 = 1.6 cm
Radius of the base of cone 2 = 1.2 cm
Let h cm be the height of cone 2.
The volumes of both the cones should be equal.

∴ Height of cone 2 = 6.4 cm

#### Question 17:

A circus tent is cylindrical to a height of 3 metres and conical above it. If its diameter is 105 m and the slant height of the conical portion is 53 m, calculate the length of the canvas 5 m wide to make the required tent.

#### Answer:

Radius of the tent, r = 52.5 m
Height of the cylindrical portion of the tent, H = 3 m
Slant height of the conical portion of the tent, l = 53 m
The tent is a combination of a cylindrical and a conical portion.
i.e., area of the canvas = curved surface area of the cone + curved surface area of the cylinder
=

But area of the canvas = length $×$ breadth

∴ Length of the canvas =

#### Question 18:

An iron pillar consists of a cylindrical portion 2.8 m high and 20 cm in diameter and a cone 42 cm high is surmounting it. Find the weight of the pillar, given that 1 cm3 of iron weights 7.5 g.

#### Answer:

Height of the cylindrical portion of the iron pillar, h = 2.8 m = 280 cm
Radius of the cylindrical portion of the iron pillar, r = 20 cm
Height of the cone which is surmounted on the cylindrical portion, H = 42 cm
Now, volume of the pillar = volume of the cylindrical portion + volume of the conical portion
=

∴ Weight of the pillar = volume of the pillar $×$ weight per cubic cm

#### Question 19:

From a solid right circular cylinder with height 10 cm and radius of the base 6 cm, a right circular cone of the same height and base is removed. Find the volume of the remaining solid.

#### Answer:

Height of the cylinder = 10 cm
Radius of the cylinder = 6 cm
The respective heights and radii of the cone and the cylinder are the same.
∴ Volume of the remaining solid = volume of the cylinder − volume of the cone

#### Question 20:

Water flows at the rate of 10 metres per minute through a cylindrical pipe 5 mm in diameter. How long would it take to fill a conical vessel whose diameter art the surface 40 cm and depth 24 cm?

#### Answer:

Radius of the cylindrical pipe = 2.5 mm = 0.25 cm
Water flowing per minute = 10 m = 1000 cm
Volume of water flowing per minute through the cylindrical pipe = = 196.4 cm3
Radius of the the conical vessel = 40 cm
Depth of the vessel = 24 cm
Volume of the vessel =
Let the time taken to fill the conical vessel be x min.
Volume of water flowing per minute through the cylindrical pipe $×$ x = volume of the conical vessel

∴ The cylindrical pipe would take 51 min 12 sec to fill the conical vessel.

#### Question 21:

Note Use $\mathrm{\pi }=\frac{22}{7}$, unless stated otherwise.
A cloth having an area of 165 m2 is shaped into the form of a conical tent of radius 5 m. (i) How many students can sit in the tent if a student, on an average, iccupies on the ground? (ii) Find the volume of the cone.

#### Answer:

Radius of the conical tent, r = 5 m

Area of the base of the conical tent =

Average area occupied by a student on the ground = $\frac{5}{7}$ m2

∴ Number of students who can sit in the tent

Thus, the number of students who can sit in the tent is 110.

Let the slant height of the tent be l m.

Curved surface area of the tent = 165 m2

Let the height of the tent be h m.

$h=\sqrt{{l}^{2}-{r}^{2}}=\sqrt{{\left(10.5\right)}^{2}-{5}^{2}}=\sqrt{110.25-25}=\sqrt{85.25}$ ≈ 9.23 m

∴Volume of the tent = $\frac{1}{3}\mathrm{\pi }{r}^{2}h=\frac{1}{3}×\frac{22}{7}×{\left(5\right)}^{2}×9.23$ ≈ 241.74 m3

Thus, the volume of the conical tent is approximately 241.74 m3.

#### Question 1:

Find the volume and surface area of a sphere whose radius is:
(i) 3.5 cm
(ii) 4.2 cm
(iii) 5 m

#### Answer:

(i) Radius of the sphere = 3.5 cm
Now, volume =

∴ Surface area = $4\mathrm{\pi }{r}^{2}$

(ii) Radius of the sphere=4.2 cm
Now, volume =

∴ Surface area =

(iii) Radius of sphere=5 m
Now, volume =
=

∴ Surface area =

#### Question 2:

The volume of a sphere is 38808 cm3. Find its radius and hence its surface area.

#### Answer:

Volume of the sphere = 38808 cm3
Suppose that r cm is the radius of the given sphere.

∴ Surface area of the sphere =

#### Question 3:

Find the surface area of a sphere whose volume is 606.375 m3.

#### Answer:

Volume of the sphere = 606.375 m3

∴ Surface area =

#### Question 4:

Note Take $\mathrm{\pi }=\frac{22}{7}$, unless stated otherwise.
Find the volume of a sphere whose surface area is 154 cm2.

#### Answer:

Let the radius of the sphere be r cm.

Surface area of the sphere = 154 cm2

∴ Volume of the sphere =

Thus, the volume of the sphere is approximately 179.67 m3.

#### Question 5:

The surface area of a sphere is (576π) cm2. Find its volume.

#### Answer:

Surface area of the sphere = (576π) cm2
Suppose that r cm is the radius of the sphere.

#### Question 6:

How many lead shots, each 3 mm in diameter, can be made from a cuboid with dimensions (12 cm × 11 cm × 9 cm)?

#### Answer:

Here, l = 12 cm, b = 11 cm and h = 9 cm

Radius of one lead shot = 3 mm= $\frac{0.3}{2}\mathrm{cm}$

#### Question 7:

How many lead balls, each of radius 1 cm, can be made from a sphere of radius 8 cm?

#### Answer:

Radius of the sphere = 8 cm

Volume of the sphere =
Radius of one lead ball = 1 cm
Volume of one lead ball =

∴ Number of lead balls =

#### Question 8:

A solid sphere of radius 3 cm is melted and then cast into smaller spherical balls, each of diameter 0.6 cm. Find the number of small balls thus obtained.

#### Answer:

Radius of the solid sphere = 3 cm

Volume of the solid sphere =
=

Diameter of the spherical ball = 0.6 cm
Radius of the spherical ball = 0.3 cm

Volume of the spherical ball =

Now, number of small spherical balls =

∴ The number of small balls thus obtained is 1000

#### Question 9:

A metallic sphere of radius 10.5 cm is melted and then recast into smaller cones, each of radius 3.5 cm and height 3 cm. How many cones are obtained?

#### Answer:

Radius of the metallic sphere, r = 10.5 cm
Volume of the sphere =
Radius of each smaller cone, r2 = 3.05 cm
Height of each smaller cone = 3 cm

Number of cones obtained =
$=\frac{\frac{4}{3}\mathrm{\pi }{r}^{\mathit{3}}}{\frac{1}{3}\mathrm{\pi }{{r}_{2}}^{\mathit{2}}h}\phantom{\rule{0ex}{0ex}}=\frac{4×10.5×10.5×10.5}{3.5×3.5×3}\phantom{\rule{0ex}{0ex}}=126.006\approx 126$

∴ 126 cones are obtained.

#### Question 10:

How many spheres 12 cm in diameter can be made from a metallic cylinder of diameter 8 cm and height 90 cm?

#### Answer:

Diameter of each sphere, d=12 cm
Radius of each sphere, r=6 cm

Diameter of base of  the cylinder, D=8 cm
Radius of base of cylinder, R=4 cm
Height of the cylinder, h = 90 cm

Number of spheres=

∴ Five spheres can be made.

#### Question 11:

The diameter of a sphere is 6 cm. It is melted and drawn into a wire of diameter 2 mm. Find the length of the wire.

#### Answer:

Let the length of the wire be h cm.
Radius of the wire, r = 1 mm = 0.1 cm
Radius of the sphere, R = 3 cm
Now, volume of the sphere = volume of the cylindrical wire

∴ Length of the wire = 36 m

#### Question 12:

The diameter of a copper sphere is 18 cm. It is melted and drawn into a long wire of uniform cross section. If the length of the wire is 108 m, find its diameter.

#### Answer:

Radius of the copper sphere, R = 9 cm
Length of the wire, h = 108 m = 10800 cm
Volume of the sphere = volume of the wire
Suppose that r cm is the radius of the wire.

∴ Diameter of the wire = 0.6 cm

#### Question 13:

A sphere of diameter 15.6 cm is melted and cast into a right circular cone of height 31.2 cm. Find the diameter of the base of the cone.

#### Answer:

Radius of the sphere, r = 7.8 cm
Height of the cone, h = 31.2 cm
Suppose that R cm is the radius of the cone.
Now, volume of the sphere = volume of the cone

∴ The diameter of the cone is 15.6 cm.

#### Question 14:

A spherical cannonball 28 cm in diameter is melted and cast into a right circular cone would, whose base is 35 cm in diameter. Find the height of the cone.

#### Answer:

Radius of the spherical cannonball, R = 14 cm
Radius of the base of the cone, r = 17.5 cm
Let h cm be the height of the cone.
Now, volume of the sphere = volume of the cone

∴ The height of the cone is 35.84 cm.

#### Question 15:

A spherical ball of radius 3 cm is melted and recast into three spherical balls. The radii of two of these balls are 1.5 cm and 2 cm. Find the radius of the third ball.

#### Answer:

Radius of the original spherical ball = 3 cm
Suppose that the radius of third ball is r cm.
Then volume of the original spherical ball = volume of the three spherical balls

∴ The radius of the third ball is 2.5 cm.

#### Question 16:

The radii of two spheres are in the ratio 1 : 2. Find the ratio of their surface areas.

#### Answer:

Suppose that the radii are r and 2r.
Now, ratio of the surface areas  = $\frac{4\mathrm{\pi }{r}^{2}}{4\mathrm{\pi }{\left(2r\right)}^{2}}=\frac{{r}^{2}}{4{r}^{2}}=\frac{1}{4}$
= 1:4

∴ The ratio of their surface areas is 1 : 4.

#### Question 17:

The surface areas of two spheres are in the ratio 1 : 4. Find the ratio of their volumes.

#### Answer:

Suppose that the radii of the spheres are r and R.
We have:
$\frac{4\mathrm{\pi }{r}^{2}}{4{\mathrm{\pi R}}^{2}}=\frac{1}{4}\phantom{\rule{0ex}{0ex}}⇒\frac{r}{R}=\sqrt{\frac{1}{4}}=\frac{1}{2}$

Now, ratio of the volumes =

∴ The ratio of the volumes of the spheres is 1 : 8.

#### Question 18:

A cylindrical tub of radius 12 cm contains water to a depth of 20 cm. A spherical iron ball is dropped into the tub and thus the level of water is raised by 6.75 cm. What is the radius of the ball?

#### Answer:

Suppose that the radius of the ball is r cm.
Radius of the cylindrical tub =12 cm
Depth of the tub= 20 cm
Now, volume of the ball = volume of water raised in the cylinder

∴ The radius of the ball is 9 cm.

#### Question 19:

A cylindrical bucket with base radius 15 cm is filled with water up to a height of 20 cm. A heavy iron spherical ball of radius 9 cm is dropped into the bucket to submerge completely in the water. Find the increase in the level of water.

#### Answer:

Let h cm be the increase in the level of water.
Radius of the cylindrical bucket = 15 cm
Height up to which water is being filled = 20 cm
Radius of the spherical ball = 9 cm
Now, volume of the sphere = increased in volume of the cylinder

∴ The increase in the level of water is 4.32 cm.

#### Question 20:

The outer diameter of a spherical shell is 12 cm and its inner diameter is 8 cm. Find the volume of metal contained in the shell. Also, find its outer surface area.

#### Answer:

Outer radius of the spherical shell = 6 cm
Inner radius of the spherical shell = 4 cm

#### Question 21:

A hollow spherical shell is made of a metal of density 4.5 g per cm3. If its internal and external radii are 8 cm and 9 cm respectively, find the weight of the shell.

#### Answer:

Internal radius of the hollow spherical shell, r= 8 cm
External radius of the hollow spherical shell, R= 9 cm

Volume of the shell =
$=\frac{4}{3}\mathrm{\pi }\left({9}^{3}-{8}^{3}\right)\phantom{\rule{0ex}{0ex}}=\frac{4}{3}×\frac{22}{7}×\left(729-512\right)\phantom{\rule{0ex}{0ex}}=\frac{4×22×217}{21}\phantom{\rule{0ex}{0ex}}=\frac{88×31}{3}\phantom{\rule{0ex}{0ex}}=\frac{2728}{3}{\mathrm{cm}}^{3}$

Weight of the shell = volume of the shell $×$ density per cubic cm
=

∴ Weight of the shell = 4.092 kg

#### Question 22:

A hemisphere of lead of radius 9 cm is cast into a right circular cone of height 72 cm. Find the radius of the base of the cone.

#### Answer:

Radius of the hemisphere = 9 cm
Height of the right circular cone = 72 cm
Suppose that the radius of the base of the cone is r cm.
Volume of the hemisphere = volume of the cone

∴ The radius of the base of the cone is 4.5 cm.

#### Question 23:

A hemispherical bowl of internal radius 9 cm contains a liquid. This liquid is to be filled into cylindrical shaped small bottles of diameter 3 cm and height 4 cm. How many bottles are required to empty the bowl?

#### Answer:

Internal radius of the hemispherical bowl = 9 cm
Radius of a cylindrical shaped bottle = 1.5 cm
Height of a bottle = 4 cm

Number of bottles required to empty the bowl  =
$=\frac{\frac{2}{3}\mathrm{\pi }×{9}^{3}}{\mathrm{\pi }×1.{5}^{2}×4}\phantom{\rule{0ex}{0ex}}=\frac{2×9×9×9}{3×1.5×1.5×4}\phantom{\rule{0ex}{0ex}}=54$

∴ 54 bottles are required to empty the bowl.

#### Question 24:

A hemispherical bowl is made of steel 0.5 cm thick. The inside radius of the bowl is 4 cm. Find the volume of steel used in making the bowl.

#### Answer:

Internal radius of the hemispherical bowl = 4 cm
Thickness of a the bowl = 0.5 cm
Now, external radius of the bowl = (4 + 0.5 ) cm = 4.5 cm

Now, volume of steel used in making the bowl =  volume of the shell

∴ 56.83 cm3 of steel is used in making the bowl .

#### Question 25:

Note Take $\mathrm{\pi }=\frac{22}{7}$, unless stated otherwise.
A hemispherical bowl is made of steel, 0.25 cm thick. The inner radius of the bowl is 5 cm. Find the outer curved surface area of the bowl.

#### Answer:

Inner radius of the bowl, r = 5 cm

Let the outer radius of the bowl be R cm.

Thickness of the bowl = 0.25 cm         (Given)

∴ R − r = 0.25 cm

⇒ R = 0.25 + r = 0.25 + 5 = 5.25 cm

∴ Outer curved surface area of the bowl = $2\mathrm{\pi }{r}^{2}=2×\frac{22}{7}×{\left(5.25\right)}^{2}$ = 173.25 cm2

Thus, the outer curved surface area of the bowl is 173.25 cm2.

#### Question 26:

Note Take $\mathrm{\pi }=\frac{22}{7}$, unless stated otherwise.
A hemispherical bowl made of brass has inner diameter 10.5 cm. Find the cost of tin-plating it on the inside at the rate of ₹ 32 per 100 cm2.

#### Answer:

Inner radius of the bowl, r = $\frac{10.5}{2}$ = 5.25 cm

∴ Inner curved surface area of the bowl = $2\mathrm{\pi }{r}^{2}=2×\frac{22}{7}×{\left(5.25\right)}^{2}$ = 173.25 cm2

Rate of tin-plating = ₹ 32 per 100 cm2

∴ Cost of tin-plating the bowl on the inside

= Inner curved surface area of the bowl × Rate of tin-plating

$=173.25×\frac{32}{100}$

= ₹ 55.44

Thus, the cost of tin-plating the bowl on the inside is ₹ 55.44.

#### Question 27:

Note Take $\mathrm{\pi }=\frac{22}{7}$, unless stated otherwise.
The diameter of the moon is approximately one fourth of the diameter of the earth. What fraction of the volume of the earth is the volume of the moon?

#### Answer:

Let the radius of the moon and earth be r units and R units, respectively.

∴ 2r$\frac{1}{4}$ × 2R        (Given)

⇒ $r=\frac{R}{4}$            .....(1)

[Using (1)]

Thus, the volume of the moon is $\frac{1}{64}$ of the volume of the earth.

#### Question 28:

Note Take $\mathrm{\pi }=\frac{22}{7}$, unless stated otherwise.
Volume and surface area of a solid hemisphere are numerically equal. What is the diameter of the hemisphere?

#### Answer:

Let the radius of the solid hemisphere be r units.

Numerical value of surface area of the solid hemisphere = $3\mathrm{\pi }{r}^{2}$

Numercial value of volume of the solid hemisphere = $\frac{2}{3}\mathrm{\pi }{r}^{3}$

It is given that the volume and surface area of the solid hemisphere are numerically equal.

Thus, the diameter of the hemisphere is 9 units.

#### Question 1:

The length, breadth and height of a cuboid are 15 cm, 12 cm and 4.5 cm respectively. Its volume is
(a) 243 cm3
(b) 405 cm3
(c) 810 cm3
(d) 603 cm3

(c) 810 cm3

#### Question 2:

A cuboid is 12 cm long, 9 cm broad and 8 cm high. Its total surface area is
(a) 864 cm2
(b) 552 cm2
(c) 432 cm2
(d) 276 cm2

(b) 552 cm2

#### Question 3:

The length, breadth and height of a cuboid are 15 m, 6 m and 5 dm respectively. The lateral surface area of the cuboid is
(a) 45 m2
(b) 21 m2
(c) 201 m2
(d) 90 m2

#### Answer:

Length of the cuboid, l = 15 m

Breadth of the cuboid, b = 6 m

Height of the cuboid, h = 5 dm = 0.5 m           (1 m = 10 dm)

∴ Lateral surface area of the cuboid = 2h(lb) = 2 × 0.5 × (15 + 6) = 2 × 0.5 × 21 = 21 cm2

Hence, the correct answer is option (b).

#### Question 4:

A beam 9 m long, 40 cm wide and 20 cm high is made up of iron which weighs 50 kg per cubic metre. The weight of the beam is
(a) 27 kg
(b) 48 kg
(c) 36 kg
(d) 56 kg

#### Answer:

(c) 36 kg
Length = 9 m
Breadth = 40 cm = 0.4 m
Height = 20 cm = 0.2 m
∴ Weight of the beam = volume of the beam $×$weight of iron per cubic metre

#### Question 5:

The length of the longest rod that can be placed in a room of dimensions (10 m × 10 m × 5 m) is
(a) 15 m
(b) 16 m
(c)
(d) 12 m

#### Answer:

(a) 15 m

Length of longest rod = diagonal of the room
= diagonal of a cuboid

#### Question 6:

What is the maximum length of a pencil that can be placed in a rectangular box of dimensions (8 cm × 6 cm × 5 cm)?
(a) 8 cm
(b) 9.5 cm
(c) 19 cm
(d) 11.2 cm

#### Answer:

(d) 11.2 cm
Maximum length of the pencil =  diagonal of the box

#### Question 7:

The number of planks of dimensions (4 m × 5 m × 2 m) that can be stored in a pit which is 40 m long, 12 m wide and 16 m deep, is
(a) 190
(b) 192
(c) 184
(d) 180

#### Answer:

(b)  192

Number of planks =

#### Question 8:

How many planks of dimensions (5 m × 25 cm × 10 cm) can be stored in a pit which is 20 m long, 6 m wide and 50 cm deep?
(a) 480
(b) 450
(c) 320
(d) 360

#### Answer:

(a)  480

Length of the pit = 20 m
Breadth of the pit = 6 m
Height of the pit = 50 cm = 0.5 m

Length of the plank = 5m
Breadth of the plank = 25 cm = 0.25 m
Height of the plank = 10 cm = 0.1 m

∴ Number of planks =

$=\frac{20×6×0.5}{5×0.25×0.1}\phantom{\rule{0ex}{0ex}}=\frac{60×1000}{125}\phantom{\rule{0ex}{0ex}}=480$

#### Question 9:

How many bricks will be required to construct a wall 8 m long, 6 m high and 22.5 cm thick if each brick measures (25 cm × 11.25 cm × 6 cm)?
(a) 4800
(b) 5600
(c) 6400
(d) 5200

#### Answer:

(c)  6400

Length of the wall = 8 m = 800 cm
Breadth of the wall = 6 m = 600 cm
Height of the wall = 22.5 cm

Length of the brick = 25 cm
Breadth of the brick = 11.25 cm
Height of the brick = 6 cm

∴ Number of bricks required =

#### Question 10:

How many persons can be accommodated in a dining hall of dimensions (20 m × 15 m × 4.5 m), assuming that each person requires 5 m3 of air?
(a) 250
(b) 270
(c) 320
(d) 300

#### Answer:

(b) 270

Number of persons  =

∴ 270 persons can be accommodated.

#### Question 11:

A river 1.5 m deep and 30 m wide is flowing at the rate of 3 km per hour. The volume of water that runs into the sea per minute is
(a) 2000 m3
(b) 2250 m3
(c) 2500 m3
(d) 2750 m3

#### Answer:

(b) 2250 m3

Length of the river = 1.5 m
Breadth of the river = 30 m
Depth of the river = 3 km = 3000 m

∴ Volume of water that runs into the sea per minute =

#### Question 12:

The lateral surface area of a cube is 256 m2. The volume of the cube is
(a) 64 m3
(b) 216 m3
(c) 256 m3
(d) 512 m3

#### Answer:

(d) 512  m3

Suppose that a m be the edge of the cube.
We have:

∴ Volume of the cube =

#### Question 13:

The total surface area of a cube is 96 cm2. The volume of the cube is
(a) 8 cm3
(b) 27 cm3
(c) 64 cm3
(d) 512 cm3

#### Answer:

(c)  64 cm3

Let a cm be the edge of the cube.
We have:

∴ Volume of the cube =

#### Question 14:

The volume of a cube is 512 cm3. Its total surface area is
(a) 256 cm2
(b) 384 cm2
(c) 512 cm2
(d) 64 cm2

#### Answer:

(b) 384 cm2

Suppose that a cm is the edge of the cube.
We have:

#### Question 15:

The length of the longest rod that can fit in a cubical vessel of side 10 cm, is
(a) 10 cm
(b) 20 cm
(c)
(d)

#### Answer:

(d)  $10\sqrt{3}$ cm103 cm
Length of the longest rod = body diagonal of the vessel

#### Question 16:

If the length of diagonal of a cube is , then its surface area is
(a) 192 cm2
(b) 384 cm2
(c) 512 cm2
(d) 768 cm2

#### Answer:

(b)  384 cm2
We have:

∴ Surface area of the cube =

#### Question 17:

If each edge of a cube is increased by 50%, then the percentage increase in its surface area is
(a) 50%
(b) 75%
(c) 100%
(d) 12%

#### Answer:

Let a be the edge of the cube.
Then the surface area is

Now, increased edge = $\left(a+\frac{50}{100}a\right)$ =

#### Question 18:

Three cubes of metal with edges 3 cm, 4 cm and 5 cm respectively are melted to form a single cube. The lateral surface area of the new cube formed is
(a) 72 cm2
(b) 144 cm2
(c) 128 cm2
(d) 256 cm2

#### Answer:

(b)  144  cm2
Volume of the new cube formed = total volume of the three cubes
Suppose that a cm is the edge of the new cube, then

∴ Lateral surface area of the new cube =

#### Question 19:

In a shower, 5 cm of rain falls. What is the volume of water that falls on 2 hectares of ground?
(a) 500 m3
(b) 750 m3
(c) 800 m3
(d) 1000 m3

(d) 1000 m3

#### Question 20:

Two cubes have their volumes in the ratio 1 : 27. The ratio of their surface areas is
(a) 1 : 3
(b) 1 : 8
(c) 1 : 9
(d) 1 : 18

#### Answer:

(c) 1 : 9
Suppose that the edges of the cubes are a and b.
We have:
$\frac{{a}^{3}}{{b}^{3}}=\frac{1}{27}\phantom{\rule{0ex}{0ex}}⇒{\left(\frac{a}{b}\right)}^{3}=\frac{1}{27}\phantom{\rule{0ex}{0ex}}⇒\frac{a}{b}=\frac{1}{3}$

∴ Ratio of the surface areas =

#### Question 21:

If each side of a cube is doubled, then its volume
(a) is doubled
(b) becomes 4 times
(c) becomes 6 times
(d) becomes 8 times

#### Answer:

(d) becomes 8 times

Suppose that the side of the cube is a.
When it is doubled, it becomes 2a.
New volume of the cube =

Hence, the volume becomes 8 times the original volume.

#### Question 22:

The diameter of the base of a cylinder is 6 cm and its height is 14 cm. The volume of the cylinder is
(a) 198 cm3
(b) 396 cm3
(c) 495 cm3
(d) 297 cm3

(b)  396 cm3

#### Question 23:

If the diameter of a cylinder is 28 cm and its height is 20 cm, then its curved surface area is
(a) 880 cm2
(b) 1760 cm2
(c) 3520 cm2
(d) 2640 cm2

(b)  1760 cm2

#### Question 24:

If the curved surface area of a cylinder is 1760 cm2 and its base radius is 14 cm, then its height is
(a) 10 cm
(b) 15 cm
(c) 20 cm
(d) 40 cm

#### Answer:

(c) 20 cm

Curved surface area = 1760 cm2
Suppose that h cm is the height of the cylinder.
Then we have:

#### Question 25:

The height of a cylinder is 14 cm and its curved surface area is 264 cm2. The volume of the cylinder is
(a) 308 cm3
(b) 396 cm3
(c) 1232 cm3
(d) 1848 cm3

#### Answer:

(b)  396 cm3

Curved surface area = 264 cm2.
Let r cm be the radius of the cylinder.
Then we have:

#### Question 26:

The curved surface area of a cylindrical pillar is 264 m2 and its volume is 924 m3. The height of the pillar is
(a) 4 m
(b) 5 m
(c) 6 m
(d) 7 m

#### Answer:

(c)  6 m

Curved surface area = 264 m2
Volume = 924 m3
Let r m be the radius and h m be the height of the cylinder.
Then we have:

#### Question 27:

The radii of two cylinders are in the ratio 2 : 3 and their heights are in the ratio 5 : 3. The ratio of their curved surface area is
(a) 2 : 5
(b) 8 : 7
(c) 10 : 9
(d) 16 : 9

#### Answer:

(c)  10 : 9

Suppose that the radii of the cylinders are 2r and 3r and their respective heights are 5h and 3h.
Then, ratio of the curved surface areas = $\frac{2\mathrm{\pi }\left(2r\right)\left(5h\right)}{2\mathrm{\pi }\left(3r\right)\left(3h\right)}$=10 : 9

#### Question 28:

The radii of two cylinders are in the ratio 2 : 3 and their heights are in the ratio 5 : 3. The ratio of their volumes is
(a) 27 : 20
(b) 20 : 27
(c) 4 : 9
(d) 9 : 4

#### Answer:

(b)  20 : 27

Suppose that the radii of the cylinders are 2r and 3r and their respective heights are 5h and 3h..

#### Question 29:

The ratio between the radius of the base and the height of a cylinder is 2 : 3. If its volume is 1617 cm3, then its total surface area is
(a) 308 cm2
(b) 462 cm2
(c) 540 cm2
(d) 770 cm2

#### Answer:

(d) 770 cm2
We have:
r: h = 2 : 3

Now, volume = 1617 cm3

h = 10.5 cm

#### Question 30:

Two circular cylinders of equal volume have their heights in the ratio 1 : 2. The ratio of their radii is
(a)
(b)
(c) 1 : 2
(d) 1 : 4

#### Answer:

(b)

Suppose that the heights of two cylinders are h and 2h whose radii are r and R, respectively.
Since the volumes of the cylinders are equal, we have:

#### Question 31:

The ratio between the curved surface area and the total surface area of a right circular cylinder is 1 : 2. If the total surface area is 616 cm2, then the volume of the cylinder is
(a) 1078 cm3
(b) 1232 cm3
(c) 1848 cm3
(d) 924 cm3

(a) 1078 cm3

We have:

#### Question 32:

In a cylinder, if the radius is halved and the height is doubled, then the volume will be
(a) the same
(b) doubled
(c) halved
(d) four times

#### Answer:

(c) halved

Suppose that the new radius is $\frac{1}{2}$r and the height is 2h.

#### Question 33:

The number of coins 1.5 cm in diameter and 0.2 cm thick to be melted to form a right circular cylinder of height 10 cm and diameter 4.5 cm is
(a) 540
(b) 450
(c) 380
(d) 472

#### Answer:

(b) 450

Number of coins =

#### Question 34:

The radius of a wire is decreased to one-third. If volume remains the same, the length will become
(a) 2 times
(b) 3 times
(c) 6 times
(d) 9 times

#### Answer:

(d) 9 times

Let the new radius be $\frac{1}{3}$r.
Suppose that the new height is H.
The volume remains the same.

#### Question 35:

The diameter of a roller, 1 m long, is 84 cm. If it takes 500 complete revolutions to level a playground, the area of the playground is
(a) 1440 m2
(b) 1320 m2
(c) 1260 m2
(d) 1550 m2

#### Answer:

(b)  1320  m2

Area covered by the roller in 1 revolution = 2$\mathrm{\pi }$rh

#### Question 36:

2.2 dm3 of lead is to be drawn into a cylindrical wire 0.50 cm in diameter. The length of the wire is
(a) 110 m
(b) 112 m
(c) 98 m
(d) 124 m

#### Answer:

(b) 112 m

Volume of the lead = volume of the cylindrical wire
Suppose that h m is the length of the wire.
As 2.2 dm3 of lead is to be drawn into a cylindrical wire of diameter 0.50 cm, we have:

#### Question 37:

The lateral surface area of a cylinder is
(a) πr2h
(b) πrh
(c) 2πrh
(d) 2πr2

#### Answer:

(c) 2πrh
The lateral surface area of a cylinder is equal to its curved surface area.
∴ Lateral surface area of a cylinder =

#### Question 38:

The height of a cone is 24 cm and the diameter of its base is 14 cm. The curved surface area of the cone is
(a) 528 cm2
(b) 550 cm2
(c) 616 cm2
(d) 704 cm2

(b) 550 cm2

#### Question 39:

The volume of a right circular cone of height 12 cm and base radius 6 cm, is
(a) (12π) cm3
(b) (36π) cm3
(c) (72π) cm3
(d) (144π) cm3

#### Answer:

(d) (144π) cm3

Volume of the cone =

#### Question 40:

How much cloth 2.5 m wide will be required to make a conical tent having base radius 7 m and height 24 m?
(a) 120 m
(b) 180 m
(c) 220 m
(d) 550 m

#### Answer:

(c)  220 m
Let the length of the required cloth be L m and its breadth be B m.
Here, B = 2.5 m
Radius of the conical tent =7 m
Height of the tent = 24 m
Area of cloth required = curved surface area of the conical tent

#### Question 41:

The volume of a cone is 1570 cm3 and its height is 15 cm. What is the radius of the cone?
(a) 10 cm
(b) 9 cm
(c) 12 cm
(d) 8.5 cm

#### Answer:

(a) 10 cm
Let r cm be the radius of the cone.
Volume = 1570  cm3

#### Question 42:

The height of a cone is 21 cm and its slant height is 28 cm. The volume of the cone is
(a)  7356 cm3
(b) 7546 cm3
(c) 7506 cm3
(d) 7564 cm3

(b)  7546 cm3

#### Question 43:

The volume of a right circular cone of height 24 cm is 1232 cm3. Its curved surface area is
(a)  1254 cm2
(b) 704 cm2
(c) 550 cm2
(d) 462 cm2

#### Answer:

(c) 550 cm2
Let r cm be the radius of the cone.
Volume of the right circular cone  = 1232 cm3
Then we have:

#### Question 44:

If the volumes of two cones be in the ratio 1 : 4 and the radii of their bases be in the ratio 4 : 5, then the ratio of their heights is
(a) 1 : 5
(b) 5 : 4
(c) 25 : 16
(d) 25 : 64

#### Answer:

(d) 25 : 64

Suppose that the radii of the cones are 4r and 5r and their heights are h and H, respectively .
It is given that the ratio of the volumes of the two cones is 1:4
Then we have:

#### Question 45:

If the height of a cone is doubled, then its volume is increased by
(a) 100%
(b) 200%
(c) 300%
(d) 400%

#### Answer:

(a) 100 %

Suppose that height of the cone becomes 2h and let its radius be r.
Then new volume of the cone =
Increase in volume = $\frac{2}{3}\mathrm{\pi }{r}^{\mathit{2}}h-\frac{1}{3}\mathrm{\pi }{r}^{\mathit{2}}h=\frac{1}{3}\mathrm{\pi }{r}^{\mathit{2}}h$
=$\frac{\frac{1}{3}\mathrm{\pi }{r}^{\mathit{2}}h}{\frac{1}{3}\mathrm{\pi }{r}^{\mathit{2}}h}×100%$=100%

Hence, the volume increases by 100%.

#### Question 46:

The curved surface area of one cone is twice that of the other while the slant height of the latter is twice that of the former. The ratio of their radii is
(a) 2 : 1
(b) 4 : 1
(c) 8 : 1
(d) 1 : 1

#### Answer:

(b) 4 : 1
If the slant height of the first cone is l, then the slant height of the second cone will be 2l.
Let the radii of the first and second cones be r and R, respectively.
Then we have:

$\mathrm{\pi }rl=2×\left(\mathrm{\pi }R×2l\right)\phantom{\rule{0ex}{0ex}}⇒r=4R\phantom{\rule{0ex}{0ex}}⇒\frac{r}{R}=\frac{4}{1}$

#### Question 47:

The ratio of the volumes of a right circular cylinder and a right circular cone of the same base and the same height will be
(a) 1 : 3
(b) 3 : 1
(c) 4 : 3
(d) 3 : 4

#### Answer:

(b)  3 : 1

It is given that the right circular cylinder and the right circular cone have the same base and height.
Suppose that the respective radii of bases and heights are equal to r and h.
Then ratio of their volumes = $\frac{\mathrm{\pi }{r}^{\mathit{2}}h}{\frac{1}{3}\mathrm{\pi }{r}^{\mathit{2}}h}=\frac{3}{1}$

#### Question 48:

A right circular cylinder and a right circular cone have the same radius and the same volume. The ratio of the height of the cylinder to that of the cone is
(a) 3 : 5
(b) 2 : 5
(c) 3 : 1
(d) 1 : 3

#### Answer:

(d)  1 : 3
It is given that the right circular cylinder and the right circular cone have the same radius and volume.
Suppose that the radii of their bases are equal to r and the respective heights of the cylinder and the cone are h and H.
As the volumes of the cylinder and the cone are the same, we have:
$\mathrm{\pi }{r}^{\mathit{2}}h=\frac{1}{3}\mathrm{\pi }{r}^{\mathit{2}}H\phantom{\rule{0ex}{0ex}}⇒\frac{\mathit{h}}{\mathit{H}}=\frac{1}{3}$

#### Question 49:

The radii of the bases of a cylinder and a cone are in the ratio 3 : 4 and their heights are in the ratio 2 : 3. Then, their volumes are in the ratio
(a) 9 : 8
(b) 8 : 9
(c) 3 : 4
(d) 4 : 3

#### Answer:

(a)  9 : 8

Suppose that the respective radii of the cylinder and the cone are 3r and 4r and their respective heights are 2h and 3h.

#### Question 50:

If the height and the radius of a cone are doubled, the volume of the cone becomes
(a) 3 times
(b) 4 times
(c) 6 times
(d) 8 times

#### Answer:

(d) 8 times

Let the new height and radius be 2h and 2r, respectively.

#### Question 51:

A solid metallic cylinder of base radius 3 cm and height 5 cm is melted to make n solid cones of height 1 cm and base radius 1 mm. The value of n is
(a) 450
(b) 1350
(c) 4500
(d) 13500

#### Answer:

(d) 13500
Radius of the cylinder = 3 cm
Height of the cylinder = 5 cm
Radius of the cone = 1 mm = 0.1 cm
Height of the cone = 1 cm

∴ Number of cones, n =
$=\frac{\mathrm{\pi }×{3}^{2}×5}{\frac{1}{3}\mathrm{\pi }×0.{1}^{2}×1}\phantom{\rule{0ex}{0ex}}=\frac{3×9×5}{0.01}\phantom{\rule{0ex}{0ex}}=13500$

#### Question 52:

A conical tent is to accommodate 11 persons such that each person occupies 4 m2 of space on the ground. They have 220 m3 of air to breathe. The height of the cone is
(a) 14 m
(b) 15 m
(c) 16 m
(d) 20 m

#### Answer:

(b) 15 m

Suppose that the height of the cone is h m.
Area of the ground =

Hence, the height of the cone is 15 m.

#### Question 53:

The volume of a sphere of radius 2r is
(a) $\frac{32{\mathrm{\pi r}}^{3}}{3}$
(b) $\frac{16{\mathrm{\pi r}}^{3}}{3}$
(c) $\frac{8{\mathrm{\pi r}}^{3}}{3}$
(d) $\frac{64{\mathrm{\pi r}}^{3}}{3}$

#### Answer:

(a) $\frac{32{\mathrm{\pi r}}^{3}}{3}$

#### Question 54:

The volume of a sphere of radius 10.5 cm is
(a) 9702 cm3
(b) 4851 cm3
(c) 19404 cm3
(d) 14553 cm3

#### Answer:

(b)  4851 cm3
Volume of the sphere =

#### Question 55:

The surface area of a sphere of radius 21 cm is
(a) 2772 cm2
(b) 1386 cm2
(c) 4158 cm2
(d) 5544 cm2

#### Answer:

(d)  5544 cm2

Surface area of sphere =
=

#### Question 56:

The surface area of a sphere is 1386 cm2. Its volume is
(a) 1617 cm3
(b) 3234 cm3
(c) 4851 cm3
(d) 9702 cm3

#### Answer:

(c)  4851 cm3

Surface area = 1386 cm2
Let r cm be the radius of the sphere.
Then we have:

#### Question 57:

If the surface area of a sphere is (144π) m2, then its volume is
(a) (288π) m3
(b) (188π) m3
(c) (300π) m3
(d) (316π) m3

#### Answer:

(a) (288π) m3

Surface area = (144π) m2
Ler r m be the radius of the sphere.
Then we have:

#### Question 58:

The volume of a sphere is 38808 cm3. Its curved surface area is
(a) 5544 cm2
(b) 8316 cm2
(c) 4158 cm2
(d) 1386 cm2

#### Answer:

(a) 5544 cm2
Let r cm be the radius of the sphere.
Then we have:

#### Question 59:

If the ratio of the volumes of two spheres is 1 : 8, then the ratio of their surface area is
(a) 1 : 2
(b) 1 : 4
(c) 1 : 8
(d) 1 : 16

#### Answer:

(b) 1 : 4
Suppose that r and R are the radii of the spheres.
Then we have:

$\frac{\frac{4}{3}\mathrm{\pi }{r}^{3}}{\frac{4}{3}\mathrm{\pi }{R}^{3}}=\frac{1}{8}\phantom{\rule{0ex}{0ex}}⇒{\left(\frac{r}{R}\right)}^{3}=\frac{1}{8}\phantom{\rule{0ex}{0ex}}⇒\frac{r}{R}=\frac{1}{2}$
∴ Ratio of surface area of spheres =$\frac{4\mathrm{\pi }{r}^{2}}{4\mathrm{\pi }{R}^{2}}={\left(\frac{r}{R}\right)}^{2}={\left(\frac{1}{2}\right)}^{2}=\frac{1}{4}$

#### Question 60:

A solid metal ball of radius 8 cm is melted and cast into smaller balls, each of radius 2 cm. The number of such balls is
(a) 8
(b) 16
(c) 32
(d) 64

#### Answer:

(d) 64

Number of balls =
= $\frac{\frac{4}{3}\mathrm{\pi }×{8}^{3}}{\frac{4}{3}\mathrm{\pi }×{2}^{3}}=\frac{512}{8}=64$

#### Question 61:

A cone is 8.4 cm high and the radius of its base is 2.1 cm. It is melted and recast into a sphere. The radius of the sphere is
(a) 4.2 cm
(b) 2.1 cm
(c) 2.4 cm
(d) 1.6 cm

#### Answer:

(b) 2.1 cm

Let r cm be the radius of the sphere.
Volume of the cone = volume of the sphere

#### Question 62:

A solid lead ball of radius 6 cm is melted and then drawn into a wire of diameter 0.2 cm. The length of wire is
(a) 272 m
(b) 288 m
(c) 292 m
(d) 296 m

#### Answer:

(b)  288 m

Let h m be the length of wire.
Volume of the lead ball = volume of the wire

#### Question 63:

A metallic sphere of radius 10.5 cm is melted and then recast into small cones, each of radius 3.5 cm and height 3 cm. Then number of such cones will be
(a) 21
(b) 63
(c) 126
(d) 130

#### Answer:

(c)  126

Number of cones =

#### Question 64:

How many lead shots, each 0.3 cm in diameter, can be made from a cuboid of dimensions 9 cm × 11 cm× 12 cm?
(a) 7200
(b) 8400
(c) 72000
(d) 84000

(d) 84000

#### Question 65:

The diameter of a sphere is 6 cm. It is melted and drawn into a wire of diameter 2 mm. The length of the wire is
(a) 12 m
(b) 18 m
(c) 36 m
(d) 66 m

#### Answer:

(c)  36 m
Let h m be the length of the wire.
Volume of the sphere = volume of the wire

#### Question 66:

A sphere of diameter 12.6 cm is melted and cast into a right circular cone of height 25.2 cm. The radius of the base of the cone is
(a) 6.3 cm
(b) 2.1 cm
(c) 6 cm
(d) 4 cm

#### Answer:

(a)  6.3 cm
Let r cm be the radius of base of the cone.
Volume of the sphere = volume of the cone

#### Question 67:

A spherical ball of radius 3 cm is melted and recast into three spherical balls. The radii of two of these balls are 1.5 cm and 2 cm. The radius of the third ball is
(a) 1 cm
(b) 1.5 cm
(c) 2.5 cm
(d) 0.5 cm

#### Answer:

(c) 2.5 cm

Let r cm be the radius of the third ball.
Volume of the original ball = volume of the three balls

#### Question 68:

The radius of a hemispherical balloon increases from 6 cm to 12 cm as air is being pumped into it. The ratio of the surface areas of the balloons in two cases is
(a) 1 : 4
(b) 1 : 3
(c) 2 : 3
(d) 1 : 2

#### Answer:

(a)  1 : 4
Ratio of the surface areas of balloon = $\frac{2\mathrm{\pi }×{6}^{2}}{2\mathrm{\pi }×{12}^{2}}=\frac{36}{144}=\frac{1}{4}$

#### Question 69:

The volumes of the two spheres are in the ratio 64 : 27 and the sum of their radii is 7 cm. The difference of their total surface areas is
(a) 38 cm2
(b) 58 cm2
(c) 78 cm2
(d) 88 cm2

#### Answer:

(d) 88 cm2

Suppose that the radii of the spheres are r cm and (7 − r) cm. Then we have:

Now, the radii of the two spheres are 3 cm and 4 cm.

#### Question 70:

A hemispherical bowl of radius 9 cm contains a liquid. This liquid is to be filled into cylindrical small bottles of diameter 3 cm and height 4 cm. How many bottles will be needed to empty thee bowl?
(a) 27
(b) 35
(c) 54
(d) 63

#### Answer:

(c) 54

Number of bottles  =

#### Question 71:

A cone and a hemisphere have equal bases and equal volumes. The ratio of their heights is
(a) 1 : 2
(b) 2 : 1
(c) 4 : 1
(d)

#### Answer:

(b) 2 : 1

Let the radii of the cone and the hemisphere be r and their respective heights be h and H.

#### Question 72:

A cone, a hemisphere and a cylinder stand on equal bases and have the same height. The ratio of their volumes is
(a) 1 : 2 : 3
(b) 2 : 1 : 3
(c) 2 : 3 : 1
(d) 3 : 2 : 1

#### Answer:

(a) 1 : 2 : 3
The cone, hemisphere and the cylinder stand on equal bases and have the same height.
We know that radius and height of a hemisphere are the same.
Hence, the height of the cone and the cylinder will be the common radius.
i.e., r = h
Ratio of the volumes of the cone, hemisphere and the cylinder:

#### Question 73:

If the volume and the surface area of a sphere are numerically the same, the nits radius is
(a) 1 unit
(b) 2 units
(c) 3 units
(d) 4 units

#### Answer:

(c) 3 units

We have:

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