Rs Aggarwal 2017 Solutions for Class 9 Math Chapter 1 Real Numbers are provided here with simple step-by-step explanations. These solutions for Real Numbers are extremely popular among Class 9 students for Math Real Numbers Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Rs Aggarwal 2017 Book of Class 9 Math Chapter 1 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Rs Aggarwal 2017 Solutions. All Rs Aggarwal 2017 Solutions for class Class 9 Math are prepared by experts and are 100% accurate.

#### 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

(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:

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.

Here, l = 8 m; b = 6 m; h = 2.5 m
Capacity of the closed rectangular cistern = volume of the cistern = $l×b×h$
∴ Capacity of the cistern

Also, area of the iron sheet required to make the cistern = total surface area of the cistern
= 2(lb+lh +bh)

#### Question 3:

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 6.40 per square metre.

We need to white wash the four walls of the room
Here, l = 9 m; b = 8 m; h = 6.5 m
Area of the four walls =

Now, the door and the windows are not to be white washed.
Area of the door =
Area of a window =

Area to be white washed = area of four walls − (area of door + area of two windows)
= 221 − (3 + 2 $×$ 1.5)
= 221 − 6 = 215 m2
Cost of whitewashing = Rs 6.40/m2
∴ Cost of whitewashing 215 m2 = Rs 6.40 $×$ 215 = Rs 1376

#### Question 4:

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?

Number of planks =

Volume of one plank =

∴ Number of planks =

#### Question 5:

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)?

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 6:

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?

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 7:

An open rectangular cistern when measured from outside is 1.35 m long, 1.08 m broad and 90 cm deep. It is made up of iron, which is 2.5 cm thick. Find the capacity of the cistern and the volume of the iron used.

External length of the cistern = 1.35 m = 135 cm
External breadth of the cistern = 1.08 m = 108 m
External height of the cistern = 90 cm
∴ External volume of  the cistern

In order to find the internal volume of the cistern, we need to determine the internal dimensions of the cistern.
Internal length of the cistern = 135 − (2.5 + 2.5) cm  = 130 cm
Internal breadth of the cistern = 108 − (2.5 + 2.5) = 103 cm                    [ Subtracting the thickness of iron from both sides ]
Internal height of the cistern = 90 − 2.5 cm = 87.5 cm
∴ Internal volume of the cistern =

i.e., capacity of the cistern = 1171625 cm3
∴ Volume of iron used  = external volume of the cistern − internal volume of the cistern
= (1312200 − 1171625) cm3 = 140575 cm3

#### Question 8:

A river 2 m deep and 45 m wide is flowing at the rate of 3 km per hour. Find the volume of water that runs into the sea per minute.

Breadth of the river = 45 m
Depth of the river = 2 m
Rate of flow of the river = 3 km/h =
∴ Volume of the water that runs into the sea per minute =

#### Question 9:

A box made of sheet metal costs Rs 1620 at Rs 30 per square metre. If the box is 5 m long and 3 m wide, find its height.

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 10:

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

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

#### Question 11:

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?

Volume of the dining hall =

Volume of air required by each person = 5 m3

∴ Capacity of the dining hall =

#### Question 12:

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?

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 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.

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:

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

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 15:

Find the volume, the lateral surface area, the total surface area and the diagonal of a cube, each of whose edges measures (a) 9 m, (b) 6.5 cm.

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

(b)  Here, a = 6.5 cm

Volume of the cube =
Lateral surface area of the cube =
Total surface area of the cube =
∴ Diagonal of the cube =

#### Question 16:

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

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 17:

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

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 18:

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

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

∴ Surface area of the cube =

#### Question 19:

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.

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 20:

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

Volume of the water that falls on the ground =

#### Question 1:

Find the volume and curved surface area of a right circular cylinder of height 21 cm and base radius 5 cm.

Here, r = 5 cm; h = 21 cm

#### Question 2:

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.

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

#### Question 3:

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.

Here, r = 10.5 cm; h = 60 cm

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

#### Question 4:

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

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

#### Question 5:

The curved surface area of a cylinder is 4400 cm2 and the circumference of its base is 10 cm. Find the height and the volume of the cylinder.

Suppose that the radius is r cm and the height is h cm.
Here, curved surface area = 4400 cm2; circumference = 10 cm

#### Question 6:

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.

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 7:

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.

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 8:

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.

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

Now, total surface area − curved surface area =

#### Question 9:

The sum of the height and radius of the base of a solid cylinder is 37 m. If the total surface area of the cylinder be 1628 m2, find its volume.

We have  r + h = 37, where r m is the radius and h m is the height of the given cylinder.
i.e., h = 37 − r
Total surface area = 1628 m2

Then h = 37 − 7 = 30 m

#### Question 10:

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.

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 11:

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

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 12:

The radii of two cylinder are in the ratio 2 : 3 and their heights are in the ratio 5 : 3. Calculate the ratio of their volumes and the ratio of their curved surfaces.

Suppose that the radii of the cylinders are 2r and 3r and the respective heights are 5h and 3h.
Now, ratio of their volumes =

Also, ratio of their curved surface areas = $\frac{\mathit{2}\pi \mathit{×}\mathit{2}r\mathit{×}\mathit{5}h}{\mathit{2}\pi \mathit{×}\mathit{3}r\mathit{×}\mathit{3}h}=\frac{10}{9}=10:9$

#### Question 13:

A powder tin has a square base with side 12 cm and height 17.5 cm. Another is cylindrical with diameter of its base 12 cm and height 17.5 cm. Which has more capacity and by how much?

Length of the powder tin =12 cm
Breadth of the powder tin =12 cm
Height of the powder tin =17.5 cm

Now, radius of the cylinder = 6 cm
Height of the cylinder = 17.5 cm

∴ The capacity of the powder tin is more than that of the cylinder by (2520 $-$1980) cm3 = 540 cm3

#### Question 14:

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.

Given: Diameter of the cylindrical bucket = 28 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 15:

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.

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 16:

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.

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 17:

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?

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 18:

A lead pencil consists of a cylinder of wood with a solid cylinder of graphite fitted into it. The diameter of the pencil is 7 mm, the diameter of the graphite is 1 mm and the length of the pencil is 10 cm. Calculate the weight of the whose pencil, if the specific gravity of the wood is 0.7 g/cm2 and that of the graphite is 2.1 g/cm2.

Here, r = ; h = 10 cm, where r and h are the radius and height of the pencil, respectively.

Similarly, weight of the wood = volume of the wood $×$specific gravity of wood

∴ Total weight of the pencil = weight of the wood + weight of graphite
= (0.165 + 2.64)  g =  2.805 g

#### Question 1:

Find the volume, curved surface area and the total surface area of a cone having base radius 35 cm and height 84 cm.

Here, h = 84 cm and r = 35 cm
Volume of the cone =

Also,  l =

Curved surface area = $\mathrm{\pi }rl$

∴ Total surface area =

#### Question 2:

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.

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

Volume of the cone =

Curved surface area of the cone =

∴ Total surface area =

#### Question 3:

The volume of a right circular cone is (100π) cm3 and its height is 12 cm. Find its slant height and its curved surface area.

Volume of the cone = (100π) cm3  and  h = 12 cm

#### Question 4:

The circumference of the base of a cone is 44 cm and its slant height is 25 cm. Find the volume and curved surface area of the cone.

Slant height, l = 25 cm
Circumference of the base of the cone = 44 cm

#### Question 5:

A cone of slant height 25 cm has a curved surface are 550 cm2. Find the height and volume of the cone.

Slant height of the cone = 25 cm
Curved surface area of the = 550 cm2

#### Question 6:

Find the volume of a cone having radius of the base 35 cm and slant height 37 cm.

Radius of the base of the cone, r = 35 cm
Slant height of the cone, l = 37 cm

∴ Volume of the cone =

#### Question 7:

The curved surface area of a cone is 4070 cm2 and its diameter is 70 cm. Find its slant height.

Diameter of the cone = 70 cm
i.e., radius, r = 35 cm
Curved surface area = 4070 cm2

Hence, slant height of the cone = 37 cm

#### Question 8:

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?

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 9:

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.

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 10:

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.

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 11:

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.

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 12:

A conical tent is to accommodate 11 persons. Each person must have 4 m2 of the space on the ground and 20 m3 of air to breath. Find the height of the cone.

Suppose that the radius of base is r m and the height is h m.
Area of base of the tent = number of people $×$ area required by 1 person
= 11$×$4 = 44 m2

Now, volume of the cone = number of people $×$volume of air to breadth for 1 person
= 11$×$20 = 220 m3

∴ Height of the tent = 15 m

#### Question 13:

A cylindrical bucket, 32 cm high and 18 cm of radius of the base, is filled with sand. This bucket is emptied on the ground and a conical heap of sand is formed. If the height of the conical heap is 24 cm, find the radius and the slant height of the heap.

Let the radius of the heap and slant height be R cm and l cm, respectively.
Radius of the cylindrical bucket = 18 cm
Height of the cylindrical bucket=32 cm
Height of the conical heap =24 cm
The volume of sand remains the same in the bucket and in the heap.

#### Question 14:

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.

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 15:

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.

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 16:

The height of a cone is 30 cm. A small cone  is cut off at the top by a plane parallel to the base. If its volume be $\frac{1}{27}$ of the volume of the given cone, at what height above the base, the section has been made? Suppose that the smaller cone has radius r cm, height h cm and the original cone has radius R cm.
Height of the original cone = 30 cm
Considering $△$ABC & $△$ADE, we have:

Therefore, we can say that
$\frac{r}{R}=\frac{h}{30}\phantom{\rule{0ex}{0ex}}r=\frac{hR}{30}$
Also, volume of a small cone = $\frac{1}{27}$ volume of the original cone

i.e., height of the small cone = 10 cm.

∴ The section has been made at a height of (30 − 10) cm = 20 cm above the base of original cone.

#### Question 17:

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.

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 18:

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?

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 1:

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

(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 =

Now, volume =
=

∴ Surface area =

#### Question 2:

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

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.

Volume of the sphere = 606.375 m3

∴ Surface area =

#### Question 4:

The surface area of a sphere is 394.24 cm2. Find its radius and volume.

Surface area = 394.24 m2
Suppose that r cm is the radius of the sphere.

#### Question 5:

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

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

#### Question 6:

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.

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

#### Question 7:

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

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 8:

Radius of the sphere = 8 cm

Volume of the sphere =
Volume of one lead ball =

∴ Number of lead balls =

#### Question 9:

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.

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 10:

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?

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 11:

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

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 12:

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.

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 13:

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.

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 14:

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.

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 15:

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.

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 16:

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.

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 17:

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

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 18:

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

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 19:

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?

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 20:

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.

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 21:

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.

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 22:

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?

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 23:

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.

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 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.

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 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 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

(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

(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

(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

(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

(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

(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

(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

(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

(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

(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

(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)

(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

(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%

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

(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

(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

(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

(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

(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

(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

(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

(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

(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

(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

(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

(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

(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

(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

(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

(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

(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

(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

(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

(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

(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%

(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

(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

(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

(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

(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

(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

(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

(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) $\frac{64{\mathrm{\pi r}}^{3}}{3}$

(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

(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

(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

(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

(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

(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

(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

(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

(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

(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

(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

(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

(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

(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

(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

(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

(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)

(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

(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

(c) 3 units

We have:

#### Question 74:

Which is false in case of a hollow cylinder?
(a) Curved surface area of a hollow cylinder = 2πh(R + r).
(b) Total surface area of a hollow cylinder = 2π(R + r)(h + Rr).
(c) Inner curved surface area of a hollow cylinder = 2 πh(Rr).
(d) Area of each end of a hollow cylinder = π(R2r2).

(c) Inner curved surface area of a hollow cylinder = 2 πh(Rr)

Inner curved surface area of a hollow cylinder is $2\mathrm{\pi }rh$.

#### Question 75:

Which of false?
(a) Volume of a hollow
(b) Volume of a $\mathrm{hemisphere}=\frac{2}{3}\pi {R}^{3}$
(c) Total surface area of a hemisphere = 3 πR2
(d) Curved surface area of a hemisphere = πR2

(d) Curved surface area of a hemisphere = πR2

Curved surface area of hemisphere is =

#### Question 76:

For a right circular cylinder of base radius = 7 cm and height = 14 cm, which is false?
(a) Curved surface area = 616 cm2.
(b) Total surface area = 924 cm2.
(c) Volume = 2156 cm3.
(d) Total area of the end faces = 154 cm2.

(d) Total area of the end faces = 154 cm2

Hence, (d) is false.

#### Question 77:

Which is false?
A metal pipe is 63 cm long. Its inner diameter is 4 cm and the outer diameter is 4.4 cm. Then,
(a) its inner curved surface area = 792 cm2
(b) its outer curved surface area = 871.2 cm2
(c) surface area of each end = 2.64 cm2
(d) its total surface area = 1665.84 cm2

(d) its total surface area = 1665.84 cm2

Height of the metal pipe = 63 cm
Its inner radius = 2 cm
Its outer radius = 2.2 cm

(a)
Hence, (a) is true.

(b)
Hence, (b) is true.

(c)
Hence, (c) is true.
(d)

Hence, (d) is false.

#### Question 78:

Assertion: The base radius of a cone is 7 cm and its slant height is 25 cm. The volume of the cone is 1232 cm3.
Reason: The volume of a right circular cone of base radius r and height h is $\frac{1}{3}\mathrm{\pi }{r}^{2}h$.
(a) Both Assertion and Reason are true and Reason is a correct explanation of Assertion.
(b) Both Assertion and Reason are true and Reason is not a correct explanation of Assertion.
(c) Assertion is true and Reason is false.
(d) Assertion is false and Reason is true.

(a) Both Assertion and Reason are true and Reason is a correct explanation of Assertion.

Reason (R) is clearly true.
Let r cm be the radius of base of the cone .

Reason (R) and assertion (A) are both true and R is the correct explanation of A.
Hence, the correct option is (a).

#### Question 79:

Assertion: The surface are of a sphere is 2464 cm2. Its volume is $11498\frac{2}{3}{\mathrm{cm}}^{3}\left(\mathrm{\pi }=\frac{22}{7}\right).$
Reason: The volume of a sphere of radius r is $\frac{4}{3}{\mathrm{\pi r}}^{3}$.
(a) Both Assertion and Reason are true and Reason is a correct explanation of Assertion.
(b) Both Assertion and Reason are true and Reason is not a correct explanation of Assertion.
(c) Assertion is true and Reason is false.
(d) Assertion is false and Reason is true.

(b) Both Assertion and Reason are true and Reason is not a correct explanation of Assertion.

Reason (R) is clearly true.
Assertion (A) is true but  reason (R)  is not the correct explanation of assertion (A).

#### Question 80:

Assertion: The outer and inner radii of a hollow cylinder 2 m 10 cm long are 5 cm and 3 cm respectively. If it is made up of copper, then volume of copper in it is 10560 cm3.
Reason: The volume of a hollow cylinder of length h, external radius R and internal radius r is given by .
(a) Both Assertion and Reason are true and Reason is a correct explanation of Assertion.
(b) Both Assertion and Reason are true and Reason is not a correct explanation of Assertion.
(c) Assertion is true and Reason is false.
(d) Assertion is false and Reason is true.

(a) Both Assertion and Reason are true and Reason is a correct explanation of Assertion.
Outer radius of the hollow cylinder, R = 5 cm
Inner radius of the hollow cylinder, r = 3 cm
Height of the hollow cylinder, h = 2 m 10 cm = 210 cm

∴ Volume of the hollow cylinder =
=

Reason (R) is clearly true.
Assertion (A) is true and reason (R)  is a correct explanation of assertion (A).

#### Question 81:

Assertion: If the radius of a sphere is doubled then the ratio of the volume of the first sphere to that of the second is 1 : 8.
Reason: A cone and a hemisphere have equal bases and equal volumes. The ratio of their heights is 1 : 2.
(a) Both Assertion and Reason are true and Reason is a correct explanation of Assertion.
(b) Both Assertion and Reason are true and Reason is not a correct explanation of Assertion.
(c) Assertion is true and Reason is false.
(d) Assertion is false and Reason is true.

(c)  Assertion is true and Reason is false.
Assertion (A):
Let r be the radius of the sphere. On doubling it, it becomes 2r.

Reason (R):
It is given that the cone and the hemisphere have equal bases and equal volumes.

2h = 23πr3 h = 2r h:r  = 2:1

Hence, reason (R) is false.

Assertion (A) is true but reason (R) is clearly false and (R) is not the explanation for (A).

#### Question 82:

Assertion: The curved surface area of a cone is 550 cm2 and its diameter is 14 cm. Then, its slant height is 25 cm.
Reason: The curved surface area of a cone having base radius r and slant height l is πrl.
(a) Both Assertion and Reason are true and Reason is a correct explanation of Assertion.
(b) Both Assertion and Reason are true and Reason is not a correct explanation of Assertion.
(c) Assertion is true and Reason is false.
(d) Assertion is false and Reason is true.

(a) Both Assertion and Reason are true and Reason is a correct explanation of Assertion.

Reason (R) is clearly true.

Reason (R) is clearly true.
Assertion (A) is true and reason(R)  is a correct explanation of assertion (A) .

#### Question 83:

A right circular cylinder just encloses a sphere of radius r (as shown in the figure). Then the surface area of the sphere is equal to the curved surface area of the cylinder. True.

Height of the cylinder = diameter of the sphere = 2r

∴ Surface area of the cylinder =

#### Question 84:

The largest possible right circular cone is cut out of a cube of edge r cm. The volume of the cone is $\frac{1}{12}\pi {r}^{3}$.

True.

Radius of the cone = $\frac{r}{2}$
Height of the cone = r
∴ Volume of the cone  =

#### Question 85:

If a sphere is inscribed in a cube, then the ratio of the volume of the cube to the volume of the sphere will be 6 : π.

True.
Let the side of the cube be a, then the radius of the sphere will be $\frac{a}{2}$.
Ratio of the volume of the cube and the sphere:

= 6 : $\mathrm{\pi }$

#### Question 86:

If the length of diagonal of a cube is , then the length of each edge of the cube is 3 cm.

False.
Let a cm be the side of the cube.
Then we have:

#### Question 1:

The radii of two cylinders are in the ratio of 2 : 3 and their heights are in the ratio of 5 : 3. Then the ratio of their volumes is
(a) 10 : 17
(b) 20 : 27
(c) 17 : 27
(d) 20 : 37

(b)  20 : 27

Volume of a cylinder = $\mathrm{\pi }{R}^{\mathit{2}}H$, where R = radius and H = height
Suppose that the radii of the cylinders are 2r and 3r and their respective heights are 5h and 3h.
∴ Ratio of volumes = $\frac{\mathrm{\pi }{\left(2\mathrm{r}\right)}^{2}×5\mathrm{h}}{\mathrm{\pi }{\left(3\mathrm{r}\right)}^{2}×3\mathrm{h}}=\frac{20{r}^{2}h}{27{r}^{2}h}=\frac{20}{27}$

#### Question 2:

The total surface area of a cone whose radius is $\frac{r}{2}$ and slant height 2l is
(a)
(b) $\pi r\left(\mathrm{l}+\frac{\mathrm{r}}{4}\right)$
(c)
(d) $2\pi rl$

(b) $\pi r\left(\mathrm{l}+\frac{\mathrm{r}}{4}\right)$

#### Question 3:

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) 1.6 cm
(b) 2.1 cm
(c) 2.4 cm
(d) 4.2 cm

(b) 2.1 cm

Let r cm be the radius of the sphere.
Radius of the base of the cone = 2.1 cm
Height of the cone = 8.4 cm
The cone is melted and recast into a sphere.
Then volume of sphere = volume of cone

#### Question 4:

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 balloon in the two cases is
(a) 1 : 4
(b) 1 : 3
(c) 2 : 3
(d) 2 : 1

(a)  1: 4

Ratio of the surface area of the balloon with radii 6 cm and 12 cm =    $\frac{4{{r}_{1}}^{2}}{4{{r}_{2}}^{2}}$

$=\frac{4{\left(6\right)}^{2}}{4{\left(12\right)}^{2}}\phantom{\rule{0ex}{0ex}}=\frac{36}{144}\phantom{\rule{0ex}{0ex}}=\frac{1}{4}$

#### Question 5:

A copper sphere of diameter 6 cm is melted and drawn into 36 cm long wire of uniform circular cross-section. Then, its radius is
(a) 2 cm
(b) 1.5 cm
(c) 1.2 cm
(d) 1 cm

(d)  1 cm

Ler r cm be the radius of the wire.
Diameter of the copper sphere = 6 cm
i.e.,  radius of the copper sphere = 3 cm
Length of the wire= 36 cm
The sphere is melted and drawn into a wire.
Then volume of wire = volume of copper sphere

Hence, the radius of the wire is 1 cm.

#### Question 6:

Find the lateral surface area and the total surface area of a cube of side 8 cm.

Side of the cube = 8 cm
Lateral surface area =

∴ Total surface area of the cube =

#### Question 7:

Find the lateral surface area and the total surface area of a cuboid of dimensions 40 cm × 30 cm × 20 cm.

Here, l = 40 cm, b = 30 cm, h = 20 cm

Lateral surface area of the cuboid =

∴ Total surface area of the cuboid =

#### Question 8:

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

Let r cm be the radii and h cm be the height of the cylinder.

#### Question 9:

The length and breadth of a room are in a ratio 3 : 2. The cost of carpeting the room at Rs 25 per m2 is Rs 1350 and the cost of papering the four walls at Rs 15 per m2 is Rs 2580. If one door and two windows occupy 8 m2, find the dimensions of the room.

Let the length, breadth and height be 3x m, 2x m and h m

Now, area of the floor =

#### Question 10:

If the radius of a sphere is increased by 10%, prove that its volume will be increased by 33.1%.

Suppose that r is the radius of the sphere.
Volume of the original sphere =
Let us increase the radius of the sphere by 10%.
Then new radius = $\left(r+\frac{10}{100}r\right)$ = $\frac{110}{100}×r=\frac{11r}{10}$

Hence, proved

#### Question 11:

The surface area of a sphere of radius 5 cm is five times the area of the curved surface of a cone of radius 4 cm. Find the height and volume of the cone.

Let h cm be the height of cone and l cm be its slant height.
Radius of the cone = 4 cm

Now, surface area of the sphere = 5(Curved surface area of the cone)

#### Question 12:

A rectangular tank measuring 5 m × 4.5 m × 2.1 m is dug in the centre of the field measuring 13.5 m × 2.5 m. The earth dug out is spread over the remaining portion of the field. How much is the level of the field raised?

Length of the tank = 5 m
Breadth of the tank = 4.5 m
Height of the tank = 2.1 m
Volume of the earth dug out =

Now, area over which the earth is spread = area of the field − area of the earth dug out

∴ Height of the raised field =

#### Question 13:

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.

Area of sheet required for one cap = $\mathrm{\pi }rl$

∴ Area of sheet for 10 such caps = 5500 cm2

#### Question 14:

The volume of a right circular cone is 9856 cm3. If the diameter of its base is 28 cm, find the height of the cone.

Let h cm be the height of the cone.
Diameter of the base of the cone = 28 cm
∴ Radius of the base of the cone = 14 cm

We have:

Hence, the height of the given cone is 48 cm.

#### Question 15:

Into a circular drum of radius 4.2 m and height 3.5 m, how many full bags of wheat can be emptied if the space required for wheat in each bag is 2.1 m3?

Radius of the circular drum = 4.2 m
Height of the drum = 3.5 m
Number of bags =

The number of bags cannot be in decimals.
Hence, the number of  bags that can be emptied is 92.

#### Question 16:

A well with 10 m inside diameter is dug 14 m deep. Earth taken out of it is spread all around to a width of 5 m to form an embankment. Find the height of the embankment.

Let h m be the height of the embankment.
Internal radius of the well, r =5 m

Now, area of the embankment = $\mathrm{\pi }\left({10}^{2}-{5}^{2}\right)=\frac{22}{7}×75=\frac{1650}{7}{\mathrm{m}}^{2}$

∴ Height of the embankment =

#### Question 17:

How many metres of cloth 5 m wide will be required to make a conical tent, the radius of whose base is 7 m and whose height is 24 m?

Let l  m be the length of the cloth.
Width of the cloth =5 m
Radius of the base of the tent =7 m
Height of the tent =24 m
Now, area of the required cloth = curved surface area of the tent

∴ The length of the cloth is 110 m.

#### Question 18:

The volume of a solid cylinder is 1584 cm3 and its height is 14 cm. Find its total surface area.

Let r cm be the radius of the base of the cylinder.
Volume of  the solid cylinder = 1584 cm3
Height of the solid cylinder, h = 14 cm
We have:

#### Question 19:

The volume of two spheres are in the ratio 64 : 27. Find the difference of their surface areas if the sum of their radii is 7 cm.

Let the radii of the spheres be R cm and r cm.
Then we have:

#### Question 20:

The radius and height of a right circular cone are in the ratio 4 : 3 and its volume is 2156 cm3. Fin the curved surface area of the cone.

Let the radius and height of the cone be 4x cm and 3x cm, respectively.
Volume of the right circular cone =2156 cm3
We have:

Now, radius of the cone = 14 cm; height = 10.5 cm

#### Question 21:

The radius of the base of a cone is 14 cm and its height is 24 cm. Find the volume, curved surface area and the total surface area of the cone.

Here, r = 14 cm  and h = 24 cm

Volume of the cone =

Curved surface area of the cone = $\mathrm{\pi }rl\phantom{\rule{0ex}{0ex}}$

∴ Total surface area of the cone =

#### Question 22:

Two cylindrical vessels are filled with oil. Their radii are 15 cm and 10 cm respectively and their heights are 25 cm and 18 cm respectively. Find the radius of the cylindrical vessel. 33 cm in height which will just contain the oil of the two given vessels.

The radii of two cylindrical vessels that are filled with oil are 15 cm and 10 cm .
Let r cm be the radius of the new cylindrical vessel.
Height of the new vessel = 33 cm
According to the question, we have:
Volume of two cylindrical vessels = volume of the vessel that contains liquid of both the vessels

Hence, the radius of the new cylindrical vessel is 15 cm.

#### Question 23:

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