Rs Aggarwal 2015 Solutions for Class 10 Math Chapter 19 Volume And Surface Area Of Solids are provided here with simple step-by-step explanations. These solutions for Volume And Surface Area Of Solids are extremely popular among Class 10 students for Math Volume And Surface Area Of Solids Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Rs Aggarwal 2015 Book of Class 10 Math Chapter 19 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Rs Aggarwal 2015 Solutions. All Rs Aggarwal 2015 Solutions for class Class 10 Math are prepared by experts and are 100% accurate.

#### Question 1:

A tent is in the shape of a right circular cylinder up to a height of 3 m and conical above it. The total height of the tent is 13.5 m and the radius of its base is 14 m. Find the cost of cloth required to make the tent at the rate of Rs 80 per square metre.

Radius of the cylinder = 14 m
Radius of the base of the cone = 14 m
Height of the cylinder (h) = 3 m
Total height of the tent = 13.5 m
Surface area of the cylinder

Height of the cone

Surface area of the cone = $\pi r\sqrt{{r}^{2}+{h}^{2}}$

Total surface area

∴ Cost of cloth

#### Question 2:

A circus tent is cylindrical to a height of 3 m and conical above it. If its base radius is 52.5 m and the slant height of the conical portion is 53 m, find the area of canvas needed to make the tent.

Radius of the cylinder = 52.5 m
Radius of the base of the cone = 52.5 m
Slant height (l) of the cone = 53 m
Height of the cylinder (h) = 3 m
Curved surface area of the cylindrical portion
Curved surface area of the conical portion

Thus, the total area of canvas required for making the tent

#### Question 3:

A wooden article was made by scooping out a hemisphere from each end of a cylinder, as shown in the figure. If the height of the cylinder is 20 cm and its base diameter is 7 cm, find the total surface area of the article when it is ready.

Base diameter = 7 cm
Radius of the base =
Height of the cylinder = 20 cm

Curved/lateral surface area of the cylinder

Curved surface area of hemisphere$=2{\mathrm{\pi r}}^{2}$
Therefore, curved surface area of two hemispheres

Total surface area of the article

#### Question 4:

A wooden article was made by scooping out a hemisphere from each end of a solid cylinder, as shown in the figure. If the height of the cylinder is 12 cm and its base radius is 4.2 cm, find the total surface area of the article. Also, find the volume of the wood left in the article.

Height of the cylinder = 12 cm
Radius of base = 4.2 cm

Total surface area of the object

Volume of wood left in the article

#### Question 5:

A rocket is in the form of a circular cylinder closed at the lower end and a cone of the same radius is attached to the top. The radius of the cylinder is 2.5 m, its height is 21 m and the slant height of the cone is 8 m. Calculate the total surface area of the rocket.

Radius of the cylinder = 2.5 m
Height of the cylinder = 21 m

Curved surface area of the cylinder
Radius of the cone = 2.5 m
Slant height of the cone = 8 m
Curved surface area of the cone

Area of circular base
∴ Total surface area of rocket

#### Question 6:

A toy is in the form of a cone mounted on a hemisphere of common base radius 7 cm. The total height of the toy is 31 cm. Find the total surface area of the toy.

Radius of the common base of the hemisphere and cone = 7 cm
Curved surface area of the hemisphere

Height of the cone

Curved surface area of the cone

Total surface area of the toy

#### Question 7:

A container, shaped like a right circular cylinder of diameter 12 cm and height 15 cm, is full of ice cream. The ice cream is to be filled in cones of height 12 cm and diameter 6 cm with a hemispherical shape on the top. Find the number of such cones which can be filled with ice cream.

Diameter of the container = 12 cm
Thus, radius of the container = 6 cm
Height of the container = 15 cm

Volume of the container

Height of a cone = 12 cm
Diameter of a cone = 6 cm
Radius of a cone = 3 cm
Volume of a cone
Volume of hemisphere

Volume of ice-cream in each cone =

Number of cones of required dimension that can be filled with the given ice-cream$=\frac{1697.143}{169.714}=10$

#### Question 8:

A gulab jamun, when ready for consumption, contains sugar syrup that is about 30% of its volume. Find approximately, how much syrup would be found in 45 such gulab jamuns, each shaped like a cylinder with two hemispherical ends, if the complete length of each is 5 cm and diameter of each is 2.8 cm.

Diameter of the cylinder = 2.8 cm
So, radius of the cylinder = 1.4 cm

Length of cylinder

Volume of cylinder
Volume of two hemispheres
Total volume of each gulab jamun
​Volume of sugar syrup

Total sugar syrup in 45 gulab jamuns

#### Question 9:

A toy is in the form of a cone mounted on a hemisphere of diameter 7 cm. The total height of the toy is 14.5 cm. Find the volume and the total surface area of the toy.

A figure representing the toy is shown below:

Height of cone = 14.5 cm - 3.5 cm = 11 cm
The slant height of the conical part can be expressed using Pythagoras' theorem as shown:

Volume of the toy = Volume of the conical part + Volume of the hemisphere
Therefore,

Total area of the toy = Curved surface area of the cone + Curved surface area of the hemisphere.
Therefore,

#### Question 10:

A tent is in the form of a right circular cylinder surmounted by a cone. The diameter of the cylinder is 24 m. The height of the cylindrical portions is 11 m while the vertex of the cone is 16 m above the ground. Find the area of the canvas required for the tent.

The tent consists of a cylinder, topped by a cone.
Diameter of cylinder = 24 m
Radius of cylinder = 12 m
Total height of the tent = 16 m
Height of cylindrical portion = 11 m
​Height of cone

Curved surface area of cylinder
Curved surface area of cone
Area of canvas required for tent = Total surface area

#### Question 11:

A vessel is in the form of a hemispherical bowl surmounted by a hollow cylinder. The diameter of the hemisphere is 21 cm and the total height of the vessel is 14.5 cm. Find its capacity.

Diameter of the hemisphere = 21 cm
Therefore, radius of the hemisphere = 10.5 cm

Volume of the hemisphere

Height of the cylinder
Volume of the cylinder

Total volume

#### Question 12:

The adjoining figure represents a solid consisting of a cylinder surmounted by a cone at one end and a hemisphere at the other. Find the volume of the solid.

Diameter of the cylindrical part =7 cm
Therefore,radius of the cylindrical part = 3.5 cm

Volume of hemisphere

Volume of the cylinder

Height of cone

Volume of the cone

Total volume

#### Question 13:

A solid is composed of a cylinder with hemispherical ends. If the whole length of the solid is 98 cm and the diameter of each of its hemispherical ends is 28 cm, find the cost of polishing the surface of the solid at the rate of 15 paise per sq cm.

The arrangement is shown in the figure:

Diameter of the cylinder = 28 cm
Therefore, radius of the cylinder = Radius of the hemispheres= 14 cm

Surface area of two hemispheres

Length of cylinder

Curved surface area of cylinder

Total surface area

Cost of polishing the entire object

#### Question 14:

A solid is in the form of a right circular cone mounted on a hemisphere. The radius of the hemisphere is 2.1 cm and the height of the cone is 4 cm. The solid is placed in a cylindrical tub full of water in such a way that the whole solid is submerged in water. If the radius of the cylinder is 5 cm and its height is 9.8 cm, find the volume of the water left in the tub.

The object is shown in the figure below.

Radius of hemisphere = 2.1 cm
Volume of hemisphere

Height of cone = 4 cm
Volume of cone
Volume of the object

Volume of cylindrical tub

When the object is immersed in the tub, volume of water equal to the volume of the object is displaced from the tub.

Volume of water left in the tub =

#### Question 15:

From a solid cylinder whose height is 8 cm and radius 6 cm, a conical cavity of height 8 cm and base radius 6 cm is hollowed out. Find the volume of the remaining solid. Also, find the total surface area of the remaining solid.

Volume of the solid left = Volume of cylinder - Volume of cone

The slant length of the cone, $l=\sqrt{{r}^{2}+{h}^{2}}=\sqrt{36+64}=10cm$

Total surface area of final solid = Area of base circle + Curved surface area of cylinder + Curved surface area of cone

#### Question 16:

A spherical glass vessel has a cylindrical neck that is 7 cm long and 4 cm in diameter. The diameter of the spherical part is 21 cm. Find the quantity of water the vessel can hold.

Diameter of spherical part = 21 cm
Radius of the spherical part = 10.5 cm

Volume of spherical part of the vessel
Diameter of cylinder = 4 cm
Radius of cylinder = 2 cm
Height of cylinder = 7 cm

Volume of the cylindrical part of the vessel

Total volume of the vessel

#### Question 17:

A farmer connects a pipe of internal diameter 25 cm from a canal to a cylindrical tank in his field, which is 12 m in diameter and 2.5 m deep. If water flows through the pipe at the rate of 3.6 km/h, in how much time will the tank be filled? Also, find the cost of water if the canal department charges at the rate of Rs 0.07/m3.

Height of tank = 2.5 m
Diameter of tank = 12 m
Radius of tank = 6 m
Volume of the cylindrical tank

Rate of flow

Internal diameter of the pipe = 25 cm
Internal radius of the pipe = 12.5 cm = 0.125 m

Quantity of water flowing through the pipe in 1 sec
Time taken to fill the tank

Cost of water is Rs 0.07/m3.
Thus, cost of filling the tank

#### Question 18:

A juice seller serves his customers units a glass as shown in the figure. The inner diameter of the cylindrical glass is 5 cm but the bottom of the glass has a hemispherical portion raised, which reduces the capacity of the glass. If the height of the glass is 10 cm, find the apparent capacity of the glass and its actual capacity.

Inner diameter of a cylindrical glass = 5 cm
Inner radius of the glass = 2.5 cm
Height of the glass = 10 cm
Apparent capacity of the glass = Volume of cylinder

Volume of hemisphere at the bottom

Actual capacity of the glass

#### Question 1:

A solid metal cone with base radius of 12 cm and height 24 cm is melted to form solid spherical balls of diameter 6 cm each. Find the number of balls thus formed.

Radius of the cone = 12 cm
Height of the cone = 24 cm

Volume

Radius of each ball = 3 cm
Volume of each ball
Total number of balls formed by melting the cone

#### Question 2:

The internal and external diameters of a hollow hemispherical shell are 6 cm and 10 cm, respectively. It is melted and recast into a solid cone of base diameter 14 cm. Find the height of the cone so formed.

Internal diameter of the hemispherical shell = 6 cm
Therefore, internal radius of the hemispherical shell = 3 cm

External diameter of the hemispherical shell = 10 cm
External radius of the hemispherical shell = 5 cm

Volume of hemispherical shell

Radius of cone = 7 cm
Let the height of the cone be h cm.
Volume of cone

The volume of the hemispherical shell must be equal to the volume of the cone. Therefore,

#### Question 3:

A hemispherical bowl of internal diameter 30 cm contains some liquid. This liquid is to be poured into cylindrical bottles of diameter 5 cm and height 6 cm each. Find the number of bottles required.

Inner diameter of the bowl = 30 cm
Inner radius of the bowl
Inner volume of the bowl = Volume of liquid

Radius of each bottle = 2.5 cm
Height = 6 cm

Volume of each bottle$={\mathrm{\pi r}}^{2}\mathrm{h}=\mathrm{\pi }×\frac{5}{2}×\frac{5}{2}×6=\frac{75\mathrm{\pi }}{2}{\mathrm{cm}}^{3}$

Total number of bottles required

#### Question 4:

A solid metallic sphere of diameter 21 cm is melted and recast into a number of smaller cones, each of diameter 3.5 cm and height 3 cm. Find the number of cones so formed.

Diameter of sphere = 21 cm
Radius of sphere $=\frac{21}{2}cm$

Volume of sphere

Diameter of the cone = 3.5 cm
Radius of the cone $=\frac{3.5}{2}=\frac{7}{4}cm$
Height = 3 cm

Volume of each cone

Total number of cones

#### Question 5:

A spherical cannon ball, 28 cm in diameter, is melted and recast into a right circular conical mould with base diameter of 35 cm. Find the height of the cone.

Diameter of cannon ball = 28 cm
Radius of the cannon ball = 14 cm

Volume of ball

Diameter of base of cone = 35 cm

Radius of base of cone $=\frac{35}{2}cm$
Let the height of the cone be h cm.

Volume of cone

From the above results and from the given conditions,
Volume of ball = Volume of cone

#### Question 6:

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 sphere = 3 cm
Radius of first ball = 1.5 cm
Radius of second ball = 2 cm
Let radius of the third ball be r cm.
Volume of third ball = Volume of original sphere - Sum of volumes of other two balls

$=\frac{4}{3}\mathrm{\pi }×{3}^{3}-\left(\frac{4}{3}\mathrm{\pi }×{\frac{3}{2}}^{3}+\frac{4}{3}\mathrm{\pi }×{2}^{3}\right)\phantom{\rule{0ex}{0ex}}=\frac{4}{3}\mathrm{\pi }×3×3×3-\left(\frac{4}{3}\mathrm{\pi }×\frac{3}{2}×\frac{3}{2}×\frac{3}{2}+\frac{4}{3}\mathrm{\pi }×2×2×2\right)\phantom{\rule{0ex}{0ex}}=4\mathrm{\pi }×3×3-\left(\mathrm{\pi }×\frac{3×3}{2}+\frac{4}{3}\mathrm{\pi }×2×2×2\right)\phantom{\rule{0ex}{0ex}}=36\mathrm{\pi }-\left(\mathrm{\pi }\frac{9}{2}+\frac{32}{3}\mathrm{\pi }\right)\phantom{\rule{0ex}{0ex}}=\left(\frac{36×6-9×3-32×2}{6}\right)\mathrm{\pi }\phantom{\rule{0ex}{0ex}}=\left(\frac{216-27-64}{6}\right)\mathrm{\pi }=\frac{125\mathrm{\pi }}{6}$

Therefore,

#### Question 7:

A spherical shell of lead, whose external and internal diameters are 24 cm and 18 cm, is melted and recast into a right circular cylinder 37 cm high. Find the diameter of the base of the cylinder.

External diameter of the shell = 24 cm
External radius of the shell = 12 cm
Internal diameter of the shell = 18 cm
Internal radius of the shell = 9 cm

Volume of the shell

Height of cylinder = 37 cm
Let radius of cylinder be r cm.

Volume of cylinder

Volume of the shell = Volume of cylinder

So, diameter of the base of the cylinder = 2r = 12 cm.

#### Question 8:

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 hemisphere = 9 cm

Volume of hemisphere

Height of cone = 72 cm
Let the radius of the cone be r cm.

Volume of the cone

The volumes of the hemisphere and cone are equal.
Therefore,

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

#### Question 9:

A spherical ball of diameter 21 cm is melted and recast into cubes, each of side 1 cm. Find the number of cubes so formed.

Diameter of the spherical ball= 21 cm

Radius of the ball
Volume of spherical ball
Volume of each cube

Number of cubes =

#### Question 10:

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

Radius of the sphere = R = 8 cm
Volume of the sphere

Radius of each new ball = r = 1 cm
Volume of each ball

Total number of new balls that can be made

#### Question 11:

Marbles of diameter 1.4 cm are dropped into a cylindrical beaker of diameter 7 cm, containing some water. Find the number of marbles that should be dropped into the beaker so that the water level rises by 5.6 cm.

Diameter of each marble = 1.4 cm
Radius of each marble = 0.7 cm
Volume of each marble

The water rises as a cylindrical column.
Volume of cylindrical column filled with water

Total number of marbles

#### Question 12:

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

Radius of solid sphere = 3 cm
Volume of the sphere

Radius of each new ball = 0.3 cm
Volume of each new ball

Total number of balls $=\frac{\frac{4}{3}\mathrm{\pi }×3×3×3}{\frac{4}{3}\mathrm{\pi }×\frac{3}{10}×\frac{3}{10}×\frac{3}{10}}=1000$

#### Question 13:

The diameter of a sphere is 42 cm. It is melted and drawn into a cylindrical wire of diameter 2.8 cm. Find the length of the wire.

Diameter of sphere = 42 cm
Radius of sphere = 21 cm

Volume of sphere

Diameter of wire = 2.8 cm
Radius of wire = 1.4 cm
Let the length of the wire be cm.
Volume of the wire$={\mathrm{\pi r}}^{2}\mathrm{l}=\mathrm{\pi }×1.4×1.4×\mathrm{l}$

The volume of the sphere is equal to the volume of the wire.
​Therefore,

So, the wire is 63 m long.

#### Question 14:

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.

Diameter of sphere = 6 cm
Radius of sphere = 3 cm

Volume of sphere

Diameter of wire = 2 mm = 0.2 cm
Radius of wire = 0.1 cm
Let length of the wire becm.
Volume of the wire$=\pi {r}^{2}l=\mathrm{\pi }×0.1×0.1×l$

The volume of the sphere is equal to the volume of the wire.

Therefore,

Thus, the wire drawn by melting the sphere is 36 m long.

#### Question 15:

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.

Diameter of sphere = 18 cm
Radius of the sphere = 9 cm

Volume of sphere

Length of wire = 108 m = 10800 cm
Let radius of the wire be r cm.
Volume of the wire

The volume of the sphere and the wire are the same.
Therefore,

The diameter of the wire is 0.6 cm.

#### Question 16:

Three solid metallic spheres of radii 3 cm, 4 cm and 5 cm, respectively, are melted to form a single solid sphere. Find the diameter of the resulting sphere.

Let the radius of the new sphere formed be $r$.
The volume of this new sphere will be equal to the sum of the volumes of the smaller spheres that were melted.
Therefore,

The diameter of the resulting sphere is 12 cm.

#### Question 1:

If the radii of the ends of a 42-cm-high bucket are 16 cm and 11 cm, determine its capacity.

The bucket is shaped like a frustum.
Let the larger radius be R = 16 cm
Smaller radius = r = 11 cm
Height = h = 42 cm

Therefore capacity of the bucket can be expressed as:

#### Question 2:

The radii of the circular ends of a solid frustum of a cone are 33 cm and 27 cm, and its slant height is 10 cm. Find its capacity and total surface area.

Greater radius = R = 33 cm
Smaller radius = r = 27 cm
Slant height = l = 10 cm

Using the formula for height of a frustum:

Height =  h =

Capacity of the frustum

Surface area of the frustum

#### Question 3:

A bucket is in the form of a frustum of a cone. Its depth is 15 cm and the diameters of the top and the bottom are 56 cm and 42 cm, respectively. Find how many litres of water can the bucket hold.

Greater diameter of the frustum  = 56 cm
Greater radius of the frustum = R = 28 cm
Smaller diameter of the frustum = 42 cm
Radius of the smaller end of the frustum = r = 21 cm
Height of the frustum = h = 15 cm

Capacity of the frustum

#### Question 4:

A bucket made up of a metal sheet is in the form of a frustum of a cone of height 16 cm and radii of its lower and upper ends are 8 cm and 20 cm, respectively. Find the cost of the bucket if the cost of metal sheet used is Rs 15 per 100 cm2.

Greater radius of the frustum = R = 20 cm
Smaller radius of the frustum = r = 8 cm
Height of the frustum = h = 16 cm

Slant height, l, of the frustum

Surface area of the frustum

100 cm2 of metal sheet costs Rs 15.
So, total cost of the metal sheet used for fabrication

#### Question 5:

A bucket made up of a metal sheet is in the form of frustum of a cone. Its depth is 24 cm and the diameters of the top and bottom are 30 cm and 10 cm, respectively. Find the cost of completely filling the bucket with milk at the rate of Rs 20 per litre and the cost of metal sheet used if it costs Rs 10 per 100 cm2.

Greater diameter of the bucket = 30 cm
Radius of the bigger end of the bucket = R = 15 cm
Diameter of the smaller end of the bucket = 10 cm
Radius of the smaller end of the bucket = r = 5 cm
Height of the bucket = 24 cm

Slant height, l

Capacity of the frustum

A litre of milk cost Rs 20.
So, total cost of filling the bucket with milk

Surface area of the bucket

Cost of 100 cm2 of metal sheet is Rs 10.
So, cost of metal used for making the bucket

#### Question 6:

A tent is made in the form of a frustum of a cone surmounted by another cone. The diameters of the base and the top of the frustum are 20 m and 6 m, respectively, and the height is 24 m. If the height of the tent is 28 m and the radius of the conical part is equal to the radius of the top of the frustum, find the quantity of canvas required.

For the lower portion of the tent:
Diameter of the base= 20 m
Radius, R, of the base = 10 m
Diameter of the top end of the frustum = 6 m
Radius of the top end of the frustum = r = 3 m

Height of the frustum = h = 24 m

Slant height = l

For the conical part:
Radius of the cone's base = r = 3 m
Height of the cone = Total height - Height of the frustum = 28-24 = 4 m

Slant height, L, of the cone =

Total quantity of canvas = Curved surface area of the frustum + curved surface area of the conical top

#### Question 7:

A tent consists of a frustum of a cone, surmounted by a cone. If the diameter of the upper and lower circular ends of the frustum be 14 m and 26 m, respectively, the height of the frustum be 8 m and the slant height of the surmounted conical portion be 12 m, find the area of the canvas required to make the tent. (Assume that the radii of the upper circular end of the frustum and the base of the surmounted conical portion are equal.)

For the frustum:
Upper diameter = 14 m
Upper Radius, r = 7 m
Lower diameter = 26 m
Lower Radius, R = 13 m

Height of the frustum= h = 8 m
Slant height l =

For the conical part:
Radius of the base = r = 7 m
Slant height = L =12 m

Total surface area of the tent = Curved area of frustum + Curved area of the cone

#### Question 8:

A metallic right circular cone is 20 m high and has a vertical angle of 60°. It is cut into two parts at the middle of its height by a plane parallel to the base. If the frustum so obtained is drawn into a wire of diameter $\frac{1}{16}$ cm, find the length of the wire.

Consider the frustum created by cutting the original cone.
Consider the vertical axis.
Vertical angle of the cone = 60o
Angle of the central axis with slant height = 30o

Height of the cone = 20 cm
Height of the smaller cone =  Height of the frustum = 10 cm

In the smaller cone,

Height of the original cone = 20 m

Volume of the frustum

Let the length of the wire be x cm.
Diameter$=\frac{1}{16}cm$
Radius$=\frac{1}{32}cm$

Volume of the wire

Volume of the wire = Volume of the frustum
Therefore,

So, the length of the wire drawn is 7964.44 m.

#### Question 9:

A milk container is in the form of frustum of a cone of height 18 cm with radii of its upper and lower ends being 8 cm and 32 cm, respectively. Find the amount of milk that can be used to completely fill the container and the cost of doing so at the rate of Rs 20 per litre.

Greater radius of the frustum = R = 32 cm
Smaller radius of the frustum = r = 8 cm
Height of the frustum = h = 18 cm

Capacity of the frustum

Cost of 1 l is Rs 20.
So, total cost of filling the container with milk

#### Question 1:

A cylindrical pencil sharpened at one end is a combination of
(a) a cylinder and a cone
(b) a cylinder and frustum of a cone
(c) a cylinder and a hemisphere
(d) two cylinders

(a) a cylinder and a cone
A cylindrical pencil sharpened at one end is a combination of a cylinder and a cone.

#### Question 2:

A shuttlecock used for playing badminton is a combination of

(a) cylinder and a hemisphere
(b) frustum of a cone and a hemisphere
(c) a cone and a hemisphere
(d) a cylinder and a sphere

(b) frustum of a cone and a hemisphere
A shuttlecock used for playing badminton is a combination of frustum of a cone and a hemisphere.

#### Question 3:

A funnel is a combination of

(a) a cylinder and a cone
(b) a cylinder and a hemisphere
(c) a cylinder and frustum of a cone
(d) a cone and hemisphere

(c) a cylinder and frustum of a cone
A funnel is a combination of a cylinder and frustum of a cone.

#### Question 4:

A surahi is a combination of

(a) a sphere and a cylinder
(b) a hemisphere and a cylinder
(c) a cylinder and a cone
(d) two hemispheres

(a) a sphere and a cylinder
A surahi is a combination of a sphere and a cylinder.

#### Question 5:

The shape of a glass (tumbler) is usually in the form of

(a) a cylinder
(b) frustum of a cone
(c) a cone
(d) a sphere

(b) frustum of a cone
The shape of a glass (tumbler) is usually in the form of frustum of a cone.

#### Question 6:

The shape of the gilli used in a gilli-danda game is a combination of

(a) a cone and a cylinder
(b) two cylinders
(c) two cones and a cylinder
(d) two cylinders and a cone

(c) two cones and a cylinder
The shape of the gilli used in a gilli-danda game is a combination of two cones and a cylinder.

#### Question 7:

A plumbline (sahul) is a combination of

(a) a hemisphere and a cone
(b) a cylinder and a cone
(c) a cylinder and frustum of a cone
(d) a cylinder and a sphere

(a) a hemisphere and a cone
A plumbline (sahul) is a combination of a hemisphere and a cone.

#### Question 8:

A cone is cut by a plane parallel to its base and the upper part is removed. The part that is left  is called

(a) a cone
(b) a sphere
(c) a cylinder
(d) frustum of a cone

(d) frustum of a cone
A cone is cut by a plane parallel to its base and the upper part is removed. The part that is left is called frustum of a cone.

#### Question 9:

During conversion of a solid from one shape to another, the volume of the new shape will
(a) decrease
(b) increase
(c) remain unaltered
(d) be doubled

(c) remain unaltered
During conversion of a solid from one shape to another, the volume of the new shape will remain unaltered.

#### Question 10:

In a right circular cone, the cross-section made by a plane parallel to the base is a
(a) sphere
(b) hemisphere
(c) circle
(d) a semicircle

(c) circle
In a right circular cone, the cross-section made by a plane parallel to the base is a circle.

#### Question 11:

A solid piece of iron in the form a cuboid of dimensions (49 cm × 33 cm × 24 cm) is moulded into a solid sphere. The radius of the sphere is
(a) 19 cm
(b) 21 cm
(c) 23 cm
(d) 25 cm

(b) 21 cm
Volume of the cuboid $=\left(l×b×h\right)$ =
Let the radius of the sphere be r cm.
Volume of the sphere$=\frac{4}{3}\mathrm{\pi }{r}^{3}$

The volume of the sphere and the cuboid are the same.
Therefore,

Hence, the radius of the sphere is 21 cm.

#### Question 12:

A cubical ice-cream brick of edge 22 cm is to be distributed among some children by filling ice-cream cones of radius 2 cm and height 7 cm up to the brim. How many children will get the ice-cream cones?
(a) 163
(b) 263
(c) 363
(d) 463

(c) 363
The edge of the cubical ice-cream brick = a = 22 cm
Volume of the cubical ice-cream brick$={\left(a\right)}^{3}$

Radius of an ice-cream cone = 2 cm
Height of an ice-cream cone = 7 cm
Volume of each ice-cream cone$=\frac{1}{3}\mathrm{\pi }{r}^{2}h$

Number of ice-cream cones

$=\frac{22×22×22×3×7}{22×2×2×7}\phantom{\rule{0ex}{0ex}}=363$
Hence, the number of ice-cream cones is 363.

#### Question 13:

A mason constructs a wall of dimensions (270 cm × 300 cm × 350 cm) with bricks, each of size (22.5 cm × 11.25 cm × 8.75 cm) and it is assumed that $\frac{1}{8}$ space is covered by the mortar. Number of bricks used to construct the wall is
(a) 11000
(b) 11100
(c) 11200
(d) 11300

(c) 11200
Volume of wall =
$\frac{1}{8}\mathrm{th}$ of the wall is covered with mortar.
So,
Volume of the wall filled with bricks
Volume of each brick

Number of bricks used to construct the wall

$=\frac{7×270×300×350×32}{8×9×225×35}$
= 11200

Hence, the number of bricks used to construct the wall is 11200.

#### Question 14:

Twelve solid spheres of the same size are made by melting a solid metallic cylinder of base diameter 2 cm and height 16 cm. The diameter of each sphere is
(a) 2 cm
(b) 3 cm
(c) 4 cm
(d) 6 cm

(a) 2 cm
Let the diameter of each sphere be d cm.
Let r and be the radii of the sphere and the cylinder, respectively,
and h be the height of the cylinder.

Therefore,

Hence, the diameter of each sphere is 2 cm.

#### Question 15:

The diameters of two circular ends of a bucket are 44 cm and 24 cm, and the height of the bucket is 35 cm. The capacity of the bucket is
(a) 31.7 litres
(b) 32.7 litres
(c) 33.7 litres
(d) 34.7 litres

(b) 32.7 litres
Let R and r be the radii of the top and base of the bucket, respectively, and let h be its height.

Then,

Capacity of the bucket = Volume of the frustum of the cone

Hence, the capacity of the bucket is 32.7 litres.

#### Question 16:

The slant height of a bucket is 45 cm and the radii of its top and bottom are 28 cm and 7 cm, respectively. The curved surface area of the bucket is
(a) 4953 cm2
(b)  4952 cm2
(c) 4951 cm2
(d) 4950 cm2

(d) 4950 cm2
Let the radii of the top and bottom of the bucket be R and r and let its slant height be l.
Then,
Curved surface area of the bucket$=\mathrm{\pi }l\left(R+r\right)$

Hence, the curved surface area of the bucket is 4950 cm2.

#### Question 17:

The volumes of two spheres are in the ratio 64 : 27. The ratio of their surface areas is
(a) 9 : 16
(b) 16 : 9
(c) 3 : 4
(d) 4 : 3

(b) 16 : 9
Let the radii of the spheres be R and r.
Then, ratio of their volumes
$=\frac{\frac{4}{3}\mathrm{\pi }{R}^{3}}{\frac{4}{3}{\mathrm{\pi r}}^{3}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$

Therefore,

$⇒{\left(\frac{R}{r}\right)}^{3}={\left(\frac{4}{3}\right)}^{3}\phantom{\rule{0ex}{0ex}}⇒\frac{R}{r}=\frac{4}{3}$
Hence, the ratio of their surface areas$=\frac{4\mathrm{\pi }{R}^{2}}{4{\mathrm{\pi r}}^{2}}$

$=\frac{{R}^{2}}{{r}^{2}}\phantom{\rule{0ex}{0ex}}={\left(\frac{R}{r}\right)}^{2}\phantom{\rule{0ex}{0ex}}={\left(\frac{4}{3}\right)}^{2}\phantom{\rule{0ex}{0ex}}=\frac{16}{9}\phantom{\rule{0ex}{0ex}}=16:9$

#### Question 18:

A hollow cube of internal edge 22 cm is filled with spherical marbles of diameter 0.5 cm and $\frac{1}{8}$ space of the cube remains unfilled. Number of marbles required is
(a) 142296
(b) 142396
(c) 142496
(d) 142596

(a) 142296
Since $\frac{1}{8}\mathrm{th}$ of the cube remains unfulfilled,
volume of the cube =

Space filled in the cube

Radius of each marble

Volume of each marble$=\frac{4}{3}\mathrm{\pi }{r}^{3}$

Therefore, number of marbles required$=\left(\frac{7×1331×24×7}{11}\right)$
= 142296

#### Question 19:

A metallic spherical shell of internal and external diameters 4 cm and 8 cm, respectively, is melted and recast in the form of a cone of base diameter 8 cm. The height of the cone is
(a) 12 cm
(b) 14 cm
(c) 15 cm
(d) 8 cm

(b) 14 cm
Let the internal and external radii of the spherical shell be r and R, respectively.
Volume of the spherical shell$=\frac{4}{3}\mathrm{\pi }\left({R}^{3}-{r}^{3}\right)$

Radius of the cone =
Volume of the cone$=\frac{1}{3}\mathrm{\pi }{r}^{2}h$

Therefore,

#### Question 20:

A medicine capsule is in the shape of a cylinder of diameter 0.5 cm with a hemisphere tucked at each end. The length of the entire capsule is 2 cm. The capacity of the capsule is
(a) 0.33 cm2
(b) 0.34 cm2
(c) 0.35 cm2
(d) 0.36 cm2

(d) 0.36 cm3
Radius of the capsule

= 0.25 cm

Let the length of the cylindrical part of the capsule be x cm.

Then,

$0.25+x+0.25=2\phantom{\rule{0ex}{0ex}}⇒0.5+x=2\phantom{\rule{0ex}{0ex}}⇒x=1.5$

Hence, the capacity of the capsule
$=\left(2×\frac{2}{3}\mathrm{\pi }r\right)$3+πr2h

3
3 = 0.36 cm3

#### Question 21:

The length of the longest pole that can be kept in a room (12 m × 9 m ×8 m) is
(a) 29 m
(b) 21 m
(c) 19 m
(d) 17 m

(d) 17 m
Length of the longest pole that can be kept in a room =Length of the diagonal of the room

#### Question 22:

The length of the diagonal of a cube is . Its total surface area is
(a) 144 cm2
(b) 216 cm2
(c) 180 cm2
(d) 108 cm2

(b) 216 cm2

Let the edge of the cube be a cm.
Then, length of the diagonal = $\sqrt{3}a$
Or,

Therefore, the total surface area of the cube = 6a2

#### Question 23:

The volume of a cube is 2744 cm2. Its surface area is
(a) 196 cm2
(b) 1176 cm2
(c) 784 cm2
(d) 588 cm2

(b) 1176 cm2
Let the edge of the cube be a cm.
Then, volume of the cube = a3
Or,

Therefore, surface area of the cube$=6{a}^{2}$

#### Question 24:

The total surface area of a cube is 864 cm2. Its volume is
(a) 3456 cm3
(b) 432 cm3
(c) 1728 cm3
(d) 3456 cm3

(c) 1728 cm3
Let the edge of the cube be a.
Total surface area of the cube = 6a2
Therefore,

Therefore, volume of the cube = a3

#### Question 25:

How many bricks, each measuring (25 cm × 11.25 cm × 6 cm), will be required to construct a wall (8 m × 6 m × 22.5 cm)?
(a) 8000
(b) 6400
(c) 4800
(d) 7200

(b) 6400

Volume of the wall
Volume of each brick =

Number of bricks

$=\frac{800×600×22.5}{25×11.25×6}\phantom{\rule{0ex}{0ex}}=6400$

#### Question 26:

The area of the base of a rectangular tank is 6500 cm2 and the volume of water contained in it is 2.6 m3. The depth of water in the tank is
(a) 3.5 m
(b) 4 m
(c) 5 m
(d) 8 m

(b) 4 m
Area of the base of a rectangular tank

Let the depth of the water be d metres.
Then,

#### Question 27:

The volume of a wall, 5 times as high as it is broad and 8 times as long as it is high, is 128 m3. The breadth of the wall is
(a) 30 cm
(b) 40 cm
(c) 22.5 cm
(d) 25 cm

Note : It should be 128 m3 instead of 12.8 m3

(b) 40 cm
Let the breadth of the wall be x cm.
Then, its height = 5x cm
and its length
= 40x cm
Hence, the volume of the wall
It is given that the volume of the wall = 128 m3.
Therefore,

#### Question 28:

If the areas of three adjacent faces of a cuboid are x, y and z, respectively, the volume of the cuboid is
(a) xyz
(b) 2xyz
(c) $\sqrt{xyz}$
(d) $3\sqrt{xyz}$

(c) $\sqrt{xyz}$
Let the length of the cuboid = l
breadth of the cuboid = b
and height of the cuboid = h
Since, the areas of the three adjacent faces are x, y and z, we have:

$lb=x\phantom{\rule{0ex}{0ex}}bh=y\phantom{\rule{0ex}{0ex}}lh=z$

Therefore,

Hence, the volume of the cuboid .

#### Question 29:

The sum of length, breadth and height of a cuboid is 19 cm and its diagonal is $5\sqrt{5}$ cm. Its surface area is
(a) 361 cm2
(b) 125 cm2
(c) 236 cm2
(d) 486 cm2

(c) 236 cm2
Let l, b and h be the length, breadth and height of the cuboid.
Then,

Therefore,

Hence, the surface area of the cuboid is .

#### Question 30:

If each edge of a cube is increased by 50%, the percentage increase in the surface area is
(a) 50%
(b) 75%
(c) 100%
(d) 125%

(d) 125%
Let the original edge of the cube be a units.
Then, the original surface area of the cube = 6a2 units

New edge of the cube = 150% of a
$=\frac{150a}{100}\phantom{\rule{0ex}{0ex}}=\frac{3a}{2}$
Hence, new surface area$=6×{\left(\frac{3a}{2}\right)}^{2}$
$=\frac{27{a}^{2}}{2}$
Increase in area$=\left(\frac{27{a}^{2}}{2}-6{a}^{2}\right)$
$=\frac{15{a}^{2}}{2}$
% increase in surface area$=\left(\frac{15{a}^{2}}{2}×\frac{1}{6{a}^{2}}×100\right)%$
= 125 %

#### Question 31:

How many bags of grain can be stored in a cuboidal granary (8 m × 6 m × 3 m), if each bag occupies a space of 0.64 m3?
(a) 8256
(b) 90
(c) 212
(d) 225

(d) 225
Volume of the cuboidal granary
Volume of each bag

Number of bags that can be stored in the cuboidal granary
$=\left(\frac{8×6×3}{0.64}\right)$
= 225

#### Question 32:

A cube of side 6 cm is cut into a number of cubes, each of side 2 cm. The number of cubes formed is
(a) 6
(b) 9
(c) 12
(d) 27

(d) 27
Volume of the given cube
Volume of each small cube

Number of cubes formed
$=\left(\frac{6×6×6}{2×2×2}\right)$
= 27

#### Question 33:

Rainfall in an area is 5 cm. The volume of the water that falls on 2 hectares of land is
(a) 100 m3
(b) 10 m3
(c) 1000 m3
(d) 10000 m3

(c) 1000 m3
Volume of water that falls on 2 hectares of land

#### Question 34:

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

(c) 1 : 9
Let the edges of the two cubes be a and b.
Then, ratio of their volumes$=\frac{{a}^{3}}{{b}^{3}}$
Therefore,

The ratio of their surface areas$=\frac{6{a}^{2}}{6{b}^{2}}$
Therefore,

Hence, the ratio of their surface areas is 1:9.

#### Question 35:

The diameter of the base of a cylinder is 4 cm and its height is 14 cm. The volume of the cylinder is
(a) 176 cm3
(b) 196 cm3
(c) 276 cm3
(d) 352 cm3

(a) 176 cm3
Volume of the cylinder$=\mathrm{\pi }{r}^{2}h$

#### Question 36:

The diameter of a cylinder is 28 cm and its height is 20 cm. The total surface area of the cylinder is
(a) 2993 cm2
(b) 2992 cm2
(c) 2292 cm2
(d) 2229 cm2

(b) 2992 cm2
The total surface area of the cylinder$=2\mathrm{\pi }r+2{r}^{2}$

#### Question 37:

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 of the cylinder $=2\mathrm{\pi }rh\phantom{\rule{0ex}{0ex}}$$=2×\frac{22}{7}×r×14$

Therefore,

Hence, the volume of the cylinder$=\mathrm{\pi }{r}^{2}h$

#### Question 38:

The curved surface are of a cylinder is 1760 cm2 and its base radius is 14 cm. The height of the cylinder is
(a) 10 cm
(b) 15 cm
(c) 20 cm
(d) 40 cm

(c) 20 cm
Curved surface area of the cylinder$=2\mathrm{\pi }rh$
$=2×\frac{22}{7}×14×h$
Therefore,

Hence, the height of the cylinder is 20 cm.

#### Question 39:

The ratio of the total surface area to the lateral surface area of a cylinder with base radius 80 cm and height 20 cm is
(a) 2 : 1
(b) 3 : 1
(c) 4 : 1
(d) 5 : 1

(d) 5 : 1
Ratio of the total surface area to the lateral surface area
$=\frac{2\mathrm{\pi }r\left(h+r\right)}{2\mathrm{\pi }rh}\phantom{\rule{0ex}{0ex}}=\frac{h+r}{h}\phantom{\rule{0ex}{0ex}}=\frac{\left(20+80\right)}{20}\phantom{\rule{0ex}{0ex}}=\frac{100}{20}\phantom{\rule{0ex}{0ex}}=\frac{5}{1}\phantom{\rule{0ex}{0ex}}=5:1$
Hence, the required ratio is 5:1.

#### Question 40:

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
The curved surface area of a cylindrical pillar
$=2\mathrm{\pi }rh\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$
Therefore, $2\mathrm{\pi }rh=264$
Volume of a cylinder$=\mathrm{\pi }{r}^{2}h$
Therefore, $\mathrm{\pi }{r}^{2}h=924$

Hence,

Therefore,

Hence, the height of the pillar is 6 m.

#### Question 41:

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

(d) 770 cm2
Let the common multiple be x.
Let the radius of the cylinder be 2x cm and its height be 3x cm.
Then, volume of the cylinder$=\mathrm{\pi }{r}^{2}h$
$=\frac{22}{7}×{\left(2x\right)}^{2}×3x$
Therefore,

$⇒{x}^{3}=\left(\frac{7}{2}×\frac{7}{2}×\frac{7}{2}\right)\phantom{\rule{0ex}{0ex}}⇒{x}^{3}={\left(\frac{7}{2}\right)}^{3}\phantom{\rule{0ex}{0ex}}⇒x=\frac{7}{2}$
Now,
Hence, the total surface area of the cylinder$=\left(2\mathrm{\pi }rh+2\mathrm{\pi }{r}^{2}\right)\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$

#### Question 42:

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
Let the radii of the two cylinders be 2r and 3r and their heights be 5h and 3h, respectively.

Then, ratio of their volumes$=\frac{\mathrm{\pi }×{\left(2\mathrm{r}\right)}^{2}×5h}{\mathrm{\pi }×{\left(3r\right)}^{2}×3h}$
$=\frac{4{r}^{2}×5}{9{r}^{2}×3}\phantom{\rule{0ex}{0ex}}=\frac{20}{27}\phantom{\rule{0ex}{0ex}}=20:27$

#### Question 43:

The heights of two circular cylinders of equal volume are in the ratio 1 : 2. The ratio of their radii is
(a) $1:\sqrt{2}$
(b) $\sqrt{2}:1$
(c) 1 : 2
(d) 1 : 4

(b) $\sqrt{2}:1$
Let the radii of the two cylinders be r and R and their heights be h and 2h, respectively.
Since the volumes of the cylinders are equal, therefore:

$\mathrm{\pi }×{r}^{2}×h=\mathrm{\pi }×{R}^{2}×2h$

Hence, the ratio of their radii is $\sqrt{2}:1$.

#### Question 44:

The radius of the base of a cone is 5 cm and its height is 12 cm. Its curved surface area is
(a) 60π cm2
(b) 65π cm2
(c) 30π cm2
(d) None of these

(b) 65π cm2
Given: r = 5 cm, h = 12 cm

Slant height of the cone, $l=\sqrt{{r}^{2}+{h}^{2}}$

Hence, the curved surface area of the cone$=\mathrm{\pi }rl$

#### Question 45:

The diameter of the base of a cone is 42 cm and its volume is 12936 cm3. Its height is
(a) 28 cm
(b) 21 cm
(c) 35 cm
(d) 14 cm

(a) 28 cm
Let h be the height of the cone.
Diameter of the cone = 42 cm
Radius of the cone = 21 cm

Then, volume of the cone$=\frac{1}{3}\mathrm{\pi }{r}^{2}h$

Therefore,

Hence, the height of the cone is 28 cm.

#### Question 46:

The area of the base of a right circular cone is 154 cm2 and its height is 14 cm. Its curved surface area is
(a) $154\sqrt{5}{\mathrm{cm}}^{2}$
(b) $154\sqrt{7}{\mathrm{cm}}^{2}$
(c) $77\sqrt{7}{\mathrm{cm}}^{2}$
(d) $77\sqrt{5}{\mathrm{cm}}^{2}$

(a) $154\sqrt{5}{\mathrm{cm}}^{2}$
Area of the base of the of a right circular cone$=\mathrm{\pi }{r}^{2}$
Therefore,

Now, r = 7 cm and h = 14 cm
Then, slant height of the cone, $l=\sqrt{{r}^{2}+{h}^{2}}$

Hence, the curved surface area of the cone$=\mathrm{\pi }rl$

#### Question 47:

On increasing the radii of the base and the height of a cone by 20%, its volume will increase by
(a) 20%
(b) 40%
(c) 60%
(d) 72.8%

(d) 72.8%
Let the original radius of the cone be r and height be h.
Then, original volume$=\frac{1}{3}\mathrm{\pi }{r}^{2}h$

Let $\frac{1}{3}\mathrm{\pi }{r}^{2}h=V$
New radius = 120% of r
$=\frac{120r}{100}\phantom{\rule{0ex}{0ex}}=\frac{6r}{5}$

New height = 120% of h
$=\frac{120h}{100}\phantom{\rule{0ex}{0ex}}=\frac{6h}{5}$

Hence, the new volume = $\frac{1}{3}\mathrm{\pi }×{\left(\frac{6\mathrm{r}}{5}\right)}^{2}×\frac{6\mathrm{h}}{5}$
$=\frac{216}{125}\left(\frac{1}{3}\mathrm{\pi }{r}^{2}h\right)\phantom{\rule{0ex}{0ex}}=\frac{216}{125}\mathrm{V}$

Increase in volume$=\left(\frac{216}{125}\mathrm{V}-\mathrm{V}\right)$

$=\frac{91\mathrm{V}}{125}$

Increase in % of the volume$=\left(\frac{91\mathrm{V}}{125}×\frac{1}{\mathrm{V}}×100\right)%$

= 72.8%

#### Question 48:

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

(a) 9 : 8
Let the radii of the base of the cylinder and cone be 3r and 4r and their heights be 2h and 3h, respectively.
Then, ratio of their volumes$=\frac{\mathrm{\pi }{\left(3r\right)}^{2}×\left(2h\right)}{\frac{1}{3}\mathrm{\pi }{\left(4r\right)}^{2}×\left(3h\right)}$
$=\frac{9{r}^{2}×2×3}{16{r}^{2}×3}\phantom{\rule{0ex}{0ex}}=\frac{9}{8}\phantom{\rule{0ex}{0ex}}=9:8$

#### Question 49:

A metallic cylinder of radius 8 cm and height 2 cm is melted and converted into a right circular cone of height 6 cm. The radius of the base of this cone is
(a) 4 cm
(b) 5 cm
(c) 6 cm
(d) 8 cm

(d) 8 cm
Radius of the cylinder = 8 cm
Height of the cylinder = 2 cm
Height of the cone = 6 cm

Volume of the cylinder  = Volume of the cone
Therefore,

Hence, the radius of the base of the cone is 8 cm.

#### Question 50:

The height of a conical tent is 14 m and its floor area is 346.5 m2. How much canvas, 1.1 wide, will be required for it?
(a) 490 m
(b) 525 m
(c) 665 m
(d) 860 m

(b) 525 m
Area of the floor of a conical tent$=\mathrm{\pi }{r}^{2}$
Therefore,

Height of the cone = 14 m

Slant height of the cone, $l=\sqrt{{r}^{2}+{h}^{2}}$

Area of the canvas = Curved surface area of the conical tent

Length of the canvas

#### Question 51:

The diameter of a sphere is 14 cm. Its volume is
(a) 1428 cm3
(b) 1439 cm3
(c) $1437\frac{1}{3}{\mathrm{cm}}^{3}$
(d) 1440 cm3

(c) $1437\frac{1}{3}{\mathrm{cm}}^{3}$
Volume of the sphere$=\frac{4}{3}\mathrm{\pi }{r}^{3}$

#### Question 52:

The ratio between the volume of two spheres is 8 : 27. What is the ratio between their surface areas?
(a) 2 : 3
(b) 4 : 5
(c) 5 : 6
(d) 4 : 9

(d) 4 : 9
Let the radii of the spheres be R and r, respectively.
Then, ratio of their volumes$=\frac{\frac{4}{3}\mathrm{\pi }{R}^{3}}{\frac{4}{3}{\mathrm{\pi r}}^{3}}$
Therefore,
$\frac{\frac{4}{3}\mathrm{\pi }{R}^{3}}{\frac{4}{3}{\mathrm{\pi r}}^{3}}=\frac{8}{27}\phantom{\rule{0ex}{0ex}}⇒\frac{{R}^{3}}{{r}^{3}}=\frac{8}{27}\phantom{\rule{0ex}{0ex}}⇒{\left(\frac{R}{r}\right)}^{3}={\left(\frac{2}{3}\right)}^{3}\phantom{\rule{0ex}{0ex}}⇒\frac{R}{r}=\frac{2}{3}$
Hence, the ratio between their surface areas$=\frac{4\mathrm{\pi }{R}^{2}}{4\mathrm{\pi }{r}^{2}}$
$=\frac{{R}^{2}}{{r}^{2}}\phantom{\rule{0ex}{0ex}}={\left(\frac{R}{r}\right)}^{2}\phantom{\rule{0ex}{0ex}}={\left(\frac{2}{3}\right)}^{2}\phantom{\rule{0ex}{0ex}}=\frac{4}{9}\phantom{\rule{0ex}{0ex}}=4:9$

#### Question 53:

A hollow metallic sphere with external diameter 8 cm and internal diameter 4 cm is melted and moulded into a cone of base radius 8 cm. The height of the cone is
(a) 12 cm
(b) 14 cm
(c) 15 cm
(d) 18 cm

DISCLAIMER : The answer to the question does not match the options given.

External diameter = 8 cm
Internal diameter = 4 cm
Let the external and internal radii of the hollow metallic sphere be R and r, respectively.
Then,
Then, volume of the hollow sphere:
$\frac{4}{3}\mathrm{\pi }\left[{\left(R\right)}^{3}-{\left(r\right)}^{3}\right]\phantom{\rule{0ex}{0ex}}$

$=\frac{4}{3}\mathrm{\pi }\left[{\left(4\right)}^{3}-{\left(2\right)}^{3}\right]$

Therefore,
Volume of the hollow sphere =  Volume of the cone formed

#### Question 54:

A metallic cone of base radius 2.1 cm and height 8.4 cm is melted and moulded into a sphere. The radius of the sphere is
(a) 2.1 cm
(b) 1.05 cm
(c) 1.5 cm
(d) 2 cm

(a) 2.1 cm
Radius of cone = 2.1 cm
Height of cone = 8.4 cm

Volume of cone =

Volume of the sphere$=\frac{4}{3}\mathrm{\pi }{r}^{3}$
Therefore,

Volume of cone = Volume of sphere

Hence, the radius of the sphere is 2.1 cm.

#### Question 55:

The volume of a hemisphere is 19404 cm3. The total surface area of the hemisphere is
(a) 4158 cm2
(b) 16632 cm2
(c) 8316 cm2
(d) 3696 cm2

(a) 4158 cm2
Volume of hemisphere$=\frac{2}{3}\mathrm{\pi }{r}^{3}$
Therefore,

Hence, the total surface area of the hemisphere$=3\mathrm{\pi }{r}^{2}$

#### Question 56:

The surface area of a sphere is 154 cm2. The volume of the sphere is
(a) $179\frac{2}{3}{\mathrm{cm}}^{3}$
(b) $359\frac{1}{3}{\mathrm{cm}}^{3}$
(c) $1437\frac{1}{3}{\mathrm{cm}}^{3}$
(d) None of these

(a) $179\frac{2}{3}{\mathrm{cm}}^{3}$
Surface area of a sphere$=4\mathrm{\pi }{r}^{2}$
Therefore,

Volume of the sphere$=\frac{4}{3}\mathrm{\pi }{r}^{3}$

#### Question 57:

The total surface area of a hemisphere of radius 7 cm is
(a) (588π) cm2
(b) (392π) cm2
(c) (147π) cm2
(d) (98π) cm2

(c) (147π) cm2

Radius of the hemisphere = 7 cm
Total surface area of the hemisphere = Curved surface area of hemisphere + Area of the circle$=3\mathrm{\pi }{r}^{2}$

#### Question 58:

The circular ends of a bucket are of radii 35 cm and 14 cm and the height of the bucket is 40 cm. Its volume is
(a) 60060 cm3
(b) 80080 cm3
(c) 70040 cm3
(d) 80160 cm3

(b) 80080 cm3
Let R and r be the radii of the top and base of the bucket, respectively, and let h be its height.

Then,

Volume of the bucket = Volume of the frustum of the cone

Hence, the volume of the bucket is .

#### Question 59:

If the radii of the ends of a bucket are 5 cm and 15 cm and it is 24 cm high, then its surface area is
(a) 1815.3 cm2
(b) 1711.3 cm2
(c) 2025.3cm2
(d) 2360 cm2

(b) 1711.3 cm2
Let R and r be the radii of the top and base of the bucket, respectively, and let h and l be its height and slant height.

Then,

$l=\sqrt{{h}^{2}+{\left(R-r\right)}^{2}}$

Surface area of the bucket$=\mathrm{\pi }\left[{r}^{2}+l\left(R+r\right)\right]$

#### Question 60:

A circus tent is cylindrical to a height of 4 m and conical above it. If its diameter is 105 m and its slant height is 40 m, the total area of canvas required is
(a) 1760 m2
(b) 2640 m2
(c) 3960 m2
(d) 7920 m2

(d) 7920 m2
Total area of the canvas required = (Curved surface area of the cylinder) + (Curved surface area of the cone)

#### Question 61:

Match the following columns:

 Column I Column II (a) A solid metallic sphere of radius 8 cm is melted and the material is used to make solid right cones with height 4 cm and base radius of 8 cm. How many cones are formed? (p) 18 (b) A 20-m-deep well with diameter 14 m is dug up and the earth from digging is evenly spread out to form a platform 44 m by 14 m. The height of the platform is ...........m. (q) 8 (c) A sphere of radius 6 cm is melted and recast in the shape of a cylinder of radius 4 cm. Then, the height of the cylinder is ......... cm. (r) 16 : 9 (d) The volumes of two spheres are in the ratio 64 : 27. The ratio of their surface areas is ....... . (s) 5

(a)
Volume of the sphere$=\frac{4}{3}\mathrm{\pi }{r}^{3}$
$=\left(\frac{4}{3}\mathrm{\pi }×{\left(8\right)}^{3}\right){\text{cm}}^{3}$

Volume of each cone$=\frac{1}{3}\mathrm{\pi }{r}^{2}h$

Number of cones formed

$=\frac{4\mathrm{\pi }×8×8×8×3}{3×\mathrm{\pi }×8×8×4}\phantom{\rule{0ex}{0ex}}=8$
Hence, $\left(a\right)⇒\left(q\right)$

(b)
Volume of the earth dug out = Volume of the cylinder
$=\mathrm{\pi }{r}^{2}h\phantom{\rule{0ex}{0ex}}=\frac{22}{7}×7×7×20$
Let the height of the platform be h.
Then, volume of the platform = volume of the cuboid

Therefore,

Hence, $\left(b\right)⇒\left(s\right)$

(c)
Volume of the sphere
$=\frac{4}{3}\mathrm{\pi }{r}^{3}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$

$=\frac{4}{3}\mathrm{\pi }×6×6×6$

Let h be the height of the cylinder.
Then, volume of the cylinder$={\mathrm{\pi r}}^{2}\mathrm{h}\phantom{\rule{0ex}{0ex}}$
$=\mathrm{\pi }×4×4×\mathrm{h}$
Therefore,

Hence, $\left(c\right)⇒\left(p\right)$

(d)
Let the radii of the spheres be R and r respectively.
Then, ratio of their volumes$=\frac{\frac{4}{3}\mathrm{\pi }{R}^{3}}{\frac{4}{3}{\mathrm{\pi r}}^{3}}$
Therefore,
$\frac{\frac{4}{3}\mathrm{\pi }{R}^{3}}{\frac{4}{3}{\mathrm{\pi r}}^{3}}=\frac{64}{27}\phantom{\rule{0ex}{0ex}}⇒\frac{{R}^{3}}{{r}^{3}}=\frac{64}{27}\phantom{\rule{0ex}{0ex}}⇒{\left(\frac{R}{r}\right)}^{3}={\left(\frac{4}{3}\right)}^{3}\phantom{\rule{0ex}{0ex}}⇒\frac{R}{r}=\frac{4}{3}$
Hence, the ratio of their surface areas$=\frac{4\mathrm{\pi }{R}^{2}}{4\mathrm{\pi }{r}^{2}}$
$=\frac{{R}^{2}}{{r}^{2}}\phantom{\rule{0ex}{0ex}}={\left(\frac{R}{r}\right)}^{2}\phantom{\rule{0ex}{0ex}}={\left(\frac{4}{3}\right)}^{2}\phantom{\rule{0ex}{0ex}}=\frac{16}{9}\phantom{\rule{0ex}{0ex}}=16:9$
Hence, $\left(d\right)⇒\left(r\right)$

#### Question 62:

Match the following columns:

 Column I Column II (a) The radii of the circular ends of a bucket, in the form of the frustum of a cone of height 30 cm, are 20 cm and 10 cm respectively. The capacity of the bucket is ........cm3. (p) 2418π (b) The radii of the circular ends  of a conical bucket of height 15 cm are 20 and 12 cm respectively. The slant height of the bucket is ........ cm. (q) 22000 (c) The radii of the circular ends of a solid frustum of a cone are 33 cm and 27 cm and its slant height is 10 cm. The total surface area of the bucket is .........cm2. (r) 12 (d) Three solid metallic spheres of radii 3 cm, 4 cm and 5 cm are melted to form a single solid sphere. The diameter of the resulting sphere is ........ cm. (s) 17

(a)
Let R and r be the top and base of the bucket and let h be its height.
Then, R = 20 cm, r = 10 cm and h = 30 cm.
Capacity of the bucket = Volume of the frustum of the cone
$=\frac{\mathrm{\pi }h}{3}\left({R}^{2}+{r}^{2}+Rr\right)\phantom{\rule{0ex}{0ex}}$

Hence, $\left(a\right)⇒\left(q\right)$

(b)
Let R and r be the top and base of the bucket and let h be its height.
Then, R = 20 cm, r = 12 cm and h = 15 cm.
Slant height of the bucket, $l=\sqrt{{h}^{2}+{\left(R-r\right)}^{2}}$

Hence, $\left(b\right)⇒\left(s\right)$

(c)
Let R and r be the top and base of the bucket and let l be its slant height.
Then, R = 33 cm, r = 27 cm and h = 10 cm
Total surface area of the bucket$=\mathrm{\pi }\left[{R}^{2}+{r}^{2}+l\left(R+r\right)\right]$

Hence, $\left(c\right)⇒\left(p\right)$

(d)
Let the diameter of the required sphere be d.
Then, volume of the sphere$=\frac{4}{3}\mathrm{\pi }{r}^{3}$
$=\frac{4}{3}\mathrm{\pi }{\left(\frac{\mathrm{d}}{2}\right)}^{3}$
Therefore,
$\frac{4}{3}\mathrm{\pi }{\left(\frac{\mathrm{d}}{2}\right)}^{3}=\frac{4}{3}\mathrm{\pi }{\left(3\right)}^{3}+\frac{4}{3}\mathrm{\pi }{\left(4\right)}^{3}+\frac{4}{3}\mathrm{\pi }{\left(5\right)}^{3}\phantom{\rule{0ex}{0ex}}⇒\frac{4}{3}\mathrm{\pi }\frac{{\mathrm{d}}^{3}}{8}=\frac{4}{3}\mathrm{\pi }×\left[{\left(3\right)}^{3}+{\left(4\right)}^{3}+{\left(5\right)}^{3}\right]\phantom{\rule{0ex}{0ex}}⇒\frac{{\mathrm{d}}^{3}}{8}=216$
d3
= 1728
d3
= 123
d  = 12 cm

Hence, $\left(d\right)⇒\left(r\right)$

#### Question 63:

Assertion (A)
If the radii of the circular ends of a bucket 24 cm high are 15 cm and 5 cm, respectively, then the surface area of the bucket is 545π cm2.

Reason(R)
If the radii of the circular ends of the frustum of a cone are R and r, respectively, and its height is h, then its surface area is

(a) Both Assertion (A) and Reason (R) are true and Reason (R) is a correct explanation of Assertion (A).
(b) Both Assertion (A) and Reason (R) are true but Reason (R) is not a correct explanation of Assertion (A).
(c) Assertion (A) is true and Reason (R) is false.
(d) Assertion (A) is false and Reason (R) is true.

Assertion (A):
Let R and r be the top and base of the bucket and let h be its height.
Then, R = 15 cm, r = 5 cm and h = 24 cm
Slant height, $l=\sqrt{{h}^{2}+{\left(R-r\right)}^{2}}$

Surface area of the bucket$=\mathrm{\pi }\left[{R}^{2}+{r}^{2}+l\left(R+r\right)\right]$

Thus, the area and the formula are wrong.

Note:
Question seems to be incorrect.

#### Question 64:

Assertion (A)
The outer surface of a hemisphere of radius 7 cm is to be painted. The total cost of painting at Rs 5 per cm2 is Rs 2300.

Reason (R)
The total surface area of a hemisphere is 3π r2.

(a) Both Assertion (A) and Reason (R) are true and Reason (R) is a correct explanation of Assertion (A).
(b) Both Assertion (A) and Reason (R) are true but Reason (R) is not a correct explanation of Assertion (A).
(c) Assertion (A) is true and Reason (R) is false.
(d) Assertion (A) is false and Reason (R) is true.

(d) Assertion (A) is false and Reason (R) is true.
Assertion (A):
Total surface area of the hemisphere$=3\mathrm{\pi }{r}^{2}$

Cost of painting at Rs 5 per cm2

= Rs 2310
Hence, Assertion (A) is false.

Reason (R): The given statement is true.

#### Question 65:

Assertion (A)
The number of coins of diameter 1.75 cm and 2 mm thickness that can be formed from a melted cuboid (10 cm × 5.5 cm × 3.5 cm) is 400.

Reason (R)
Volume of a cylinder of base radius r and height h $=\left(\mathrm{\pi }{r}^{2}h\right)$ cubic units. And area of a cuboid = (l × b × h) cubic units.

(a) Both Assertion (A) and Reason (R) are true and Reason (R) is a correct explanation of Assertion (A).
(b) Both Assertion (A) and Reason (R) are true but Reason (R) is not a correct explanation of Assertion (A).
(c) Assertion (A) is true and Reason (R) is false.
(d) Assertion (A) is false and Reason (R) is true.

(a) Both Assertion (A) and Reason (R) are true and Reason (R) is a correct explanation of Assertion (A).
Assertion (A):
Volume of the cuboid $=\left(l×b×h\right)$

Volume of each coin$=\mathrm{\pi }{r}^{2}h$

Number of coins

$=\frac{10×55×35×7×200×200×5}{10×10×22×175×175}\phantom{\rule{0ex}{0ex}}=400$
Hence, Assertion (A) is true.

Reason (R): The given statement is true.

#### Question 66:

Assertion (A)
If the volumes of two spheres are in the ratio 27 : 8, then their surface areas are in the ratio 3 : 2.

Reason (R)
Volume of a sphere = $\frac{4}{3}\mathrm{\pi }{R}^{3}$
Surface area of a sphere = $4\mathrm{\pi }{R}^{2}$

(a) Both Assertion (A) and Reason (R) are true and Reason (R) is a correct explanation of Assertion (A).
(b) Both Assertion (A) and Reason (R) are true but Reason (R) is not a correct explanation of Assertion (A).
(c) Assertion (A) is true and Reason (R) is false.
(d) Assertion (A) is false and Reason (R) is true.

(d) Assertion (A) is false and Reason (R) is true.
Assertion (A):
Let R and r be the radii of the two spheres.
Then, ratio of their volumes$=\frac{\frac{4}{3}\mathrm{\pi }{R}^{3}}{\frac{4}{3}{\mathrm{\pi r}}^{3}}$
Therefore,
$\frac{\frac{4}{3}\mathrm{\pi }{R}^{3}}{\frac{4}{3}{\mathrm{\pi r}}^{3}}=\frac{27}{8}\phantom{\rule{0ex}{0ex}}⇒\frac{{R}^{3}}{{r}^{3}}=\frac{27}{8}\phantom{\rule{0ex}{0ex}}⇒{\left(\frac{R}{r}\right)}^{3}={\left(\frac{3}{2}\right)}^{3}\phantom{\rule{0ex}{0ex}}⇒\frac{R}{r}=\frac{3}{2}$
Hence, the ratio of their surface areas$=\frac{4\mathrm{\pi }{R}^{2}}{4{\mathrm{\pi r}}^{2}}$
$=\frac{{R}^{2}}{{r}^{2}}\phantom{\rule{0ex}{0ex}}={\left(\frac{R}{r}\right)}^{2}\phantom{\rule{0ex}{0ex}}={\left(\frac{3}{2}\right)}^{2}\phantom{\rule{0ex}{0ex}}=\frac{9}{4}\phantom{\rule{0ex}{0ex}}=9:4$
Hence, Assertion (A) is false.

Reason (R): The given statement is true.

#### Question 67:

Assertion (A)
The curved surface area of a cone of base radius 3 cm and height 4 cm is 15π cm2.

Reason (R)
Volume of a cone = πr2h

(a) Both Assertion (A) and Reason (R) are true and Reason (R) is a correct explanation of Assertion (A).
(b) Both Assertion (A) and Reason (R) are true but Reason (R) is not a correct explanation of Assertion (A).
(c) Assertion (A) is true and Reason (R) is false.
(d) Assertion (A) is false and Reason (R) is true.

(c) Assertion (A) is true and Reason (R) is false.
Assertion (A):
Curved surface area of a cone$=\mathrm{\pi }r\sqrt{{r}^{2}+{h}^{2}}$

Hence, Assertion (A) is true.

Reason (R): The given statement is false.

#### Question 1:

A solid is hemispherical at the bottom and conical above it. If the surface areas of the two parts are equal, then the ratio of its radius and the height of its conical part is
(a) 1 : 3
(b) $1:\sqrt{3}$
(c) $\sqrt{3}:1$
(d) 1 : 1

(b) $1:\sqrt{3}$
Surface area of the hemisphere$=2\mathrm{\pi }{r}^{2}$
Surface area of the cone$=\mathrm{\pi }rl$
$=\mathrm{\pi }r\sqrt{{r}^{2}+{h}^{2}}$
Therefore,

#### Question 2:

The number of solid spheres, of diameter 6 cm each, that could be moulded to form a solid metallic cylinder of height 45 cm and diameter 4 cm is
(a) 3
(b) 4
(c) 5
(d) 6

(c) 5

Radius of sphere =
Volume of sphere$=\frac{4}{3}\mathrm{\pi }{r}^{3}$

Radius of cylinder =

Height of cylinder  = 45 cm

Volume of each cylinder$=\mathrm{\pi }{r}^{2}h$
$=\left(\mathrm{\pi }×2×2×45\right){\text{cm}}^{3}$

Number of solid spheres

$=\frac{\mathrm{\pi }×2×2×45×3}{4×\mathrm{\pi }×3×3×3}\phantom{\rule{0ex}{0ex}}=5$

#### Question 3:

A metallic spherical shell, of internal and external diameters 4 cm and 8 cm, is melted and recast as a cone of base diameter 8 cm. The height of the cone is
(a) 12 cm
(b) 14 cm
(c) 15 cm
(d) 18 cm

(b) 14 cm
Let the internal and external radii of the spherical shell be r and R, respectively.
Then,
Volume of the spherical shell$=\frac{4}{3}\mathrm{\pi }\left[{R}^{3}-{r}^{3}\right]$
$=\frac{4}{3}\mathrm{\pi }\left[{\left(4\right)}^{3}-{\left(2\right)}^{3}\right]$
Volume of the cone$=\frac{1}{3}\mathrm{\pi }{r}^{2}h$
$=\frac{1}{3}\mathrm{\pi }×{\left(4\right)}^{2}×h$
Therefore,

#### Question 4:

The radii of the top and bottom of a bucket of slant height 45 cm are 28 cm and 7 cm, respectively. The curved surface area of the bucket is
(a) 4900 cm2
(b) 4950 cm2
(c) 4960 cm2
(d) 4980 cm2

(b) 4950 cm2
Let R and r be the radii of the top and base of the bucket, respectively, and let l be its slant height.
Then, curved surface area of the bucket = Curved surface area of the frustum of the cone

#### Question 5:

A solid metal cone with base radius 12 cm and height 24 cm is melted to form solid spherical balls of diameter 6 cm each. Find the number of balls formed.

Radius of cone = 12 cm
Height of cone = 24 cm
Volume of the metallic cone$=\frac{1}{3}\mathrm{\pi }{r}^{2}h$
$=\frac{1}{3}\mathrm{\pi }×{\left(12\right)}^{2}×24$

Radius of spherical ball =

Volume of each spherical ball$=\frac{4}{3}\mathrm{\pi }{\mathrm{r}}^{3}$
$=\frac{4}{3}\mathrm{\pi }×{\left(3\right)}^{3}$

Number of balls formed

$=\frac{\mathrm{\pi }×12×12×24×3}{3×4×\mathrm{\pi }×3×3×3}\phantom{\rule{0ex}{0ex}}=32$

#### Question 6:

A hemispherical bowl of internal diameter 30 cm is full of a liquid. This liquid is poured into cylindrical bottles of diameter 5 cm and height 6 cm each. How many bottles are required?

Radius of hemispherical ball =
Volume of hemispherical bowl$=\frac{2}{3}\mathrm{\pi }{\mathrm{r}}^{3}$

Radius of each bottle = $\frac{5}{2}\mathrm{cm}$

Height of each bottle = 6 cm

Volume of each bottle$=\mathrm{\pi }{r}^{2}h$

Number of bottles required

$=\frac{2×\mathrm{\pi }×15×15×15×2×2}{3×\mathrm{\pi }×5×5×6}\phantom{\rule{0ex}{0ex}}=60$

#### Question 7:

A solid metallic sphere of diameter 21 cm is melted and recast into small cones of diameter 3.5 cm and height 3 cm each. Find the number of cones so formed.

Radius of sphere =
Volume of the metallic sphere$=\frac{4}{3}\mathrm{\pi }{\mathrm{r}}^{3}$

Radius of cone =
Height of cone =  3 cm

Volume of each small cone$=\frac{1}{3}\mathrm{\pi }{r}^{2}h$

Number of cones

$=\frac{4×\mathrm{\pi }×21×21×3×20×20}{3×2×2×2×\mathrm{\pi }×35×3×3×3}\phantom{\rule{0ex}{0ex}}=504$

#### Question 8:

The diameter of a sphere is 42 cm. It is melted and drawn into a cylindrical wire of diameter 2.8 cm. Find the length of the wire.

Radius of the sphere =
Volume of the sphere$=\frac{4}{3}\mathrm{\pi }{\mathrm{r}}^{3}$

Radius of the wire =
Let the length of the wire be h cm. Then,
Volume of the wire$=\mathrm{\pi }{r}^{2}h$

Therefore,

Hence, the length of the wire is 63 m.

#### Question 9:

A drinking glass is in the shape of the frustum of a cone of height 21 cm with 6 cm and 4 cm as the diameters of its two circular ends. Find the capacity of the glass.

Let R and r be the radii of the top and base, respectively, of the drinking glass and let its height be h.

Then,

Capacity of the glass = Capacity of the frustum of the cone

#### Question 10:

Two cubes, each of volume 64 cm3, are joined end to end. Find the total surface area of the resulting cuboid.

Volume of the cube = a3

Therefore,

Each side of the cube = 4 cm

Then,
Length of the cuboid
Breadth of the cuboid = 4 cm
Height of the cuboid = 4 cm

Total surface area of the cuboid $=2\left(lb+bh+lh\right)$

#### Question 11:

The radius of the base and the height of a solid right circular cylinder are in the ratio 2 : 3 and its volume is 1617 cm3. Find the total surface area of the cylinder.

Let the radius of the cylinder be 2x cm and its height be 3x cm.

Then, volume of the cylinder$=\mathrm{\pi }{r}^{2}h$
$=\frac{22}{7}×{\left(2x\right)}^{2}×3x$
Therefore,

$⇒{x}^{3}=\left(\frac{7}{2}×\frac{7}{2}×\frac{7}{2}\right)\phantom{\rule{0ex}{0ex}}⇒{x}^{3}={\left(\frac{7}{2}\right)}^{3}\phantom{\rule{0ex}{0ex}}⇒x=\frac{7}{2}$
Now,
Hence, the total surface area of the cylinder:

$\left(2\mathrm{\pi }rh+2\mathrm{\pi }{r}^{2}\right)\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$
​

#### Question 12:

A toy is in the form of a cone mounted on a hemisphere of common base radius 7 cm. The total height of the toy is 31 cm. Find the total surface area of the toy.

Radius of the hemisphere = Radius of the cone = 7 cm

Height of the cone

Slant height of the cone, $l=\sqrt{{r}^{2}+{h}^{2}}$

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

#### Question 13:

A hemispherical bowl of internal radius 9 cm is full of water. This water is to be filled in cylindrical bottles of diameter 3 cm and height 4 cm. Find the number of bottles needed in which the water can be filled.

Radius of hemisphere = 9 cm
Volume of hemisphere$=\frac{2}{3}\mathrm{\pi }{\mathrm{r}}^{3}$
$=\left(\frac{2}{3}\mathrm{\pi }×9×9×9\right){\text{cm}}^{3}$
Radius of each bottle =
Height of each bottle = 4 cm
Volume of each bottle$=\mathrm{\pi }{r}^{2}h$

Number of bottles
$=\frac{2\mathrm{\pi }×9×9×9×2×2}{3×\mathrm{\pi }×3×3×4}\phantom{\rule{0ex}{0ex}}=54$

#### Question 14:

The surface areas of a sphere and a cube are equal. Find the ratio of their volumes.

Surface area of the sphere$=4\mathrm{\pi }{r}^{2}$
Surface area of the cube$=6{a}^{2}$
Therefore,

Ratio of their volumes$=\frac{4}{3}\mathrm{\pi }{r}^{3}}{{a}^{3}}=\frac{4\mathrm{\pi }{r}^{3}}{3{a}^{3}}$

Thus, the ratio of their volumes is .

#### Question 15:

The slant height of the frustum of a cone is 4 cm and the perimeters (i.e. circumferences) of its circular ends are 18 cm and 6 cm. Find the curved surface area of the frustum.

Let R and r be the radii of the top and base of the frustum of the cone, respectively, and its slant height be l.

Then,

Curved surface area of the frustum$=\mathrm{\pi }l\left(R+r\right)$

#### Question 16:

A solid is composed of a cylinder with hemispherical ends. If the whole length of the solid is 104 cm and the radius of each hemispherical end is 7 cm, find the surface area of the solid.

Radius of the hemispherical end = 7 cm
Height of the hemispherical end = 7 cm
Height of the cylindrical part
Surface area of the solid = 2(curved surface area of the hemisphere) + (curved surface area of the cylinder)
$=\left[2\left(2\mathrm{\pi }{r}^{2}\right)+2\mathrm{\pi }rh\right]$

#### Question 17:

If the radii of the circular ends of a bucket that is 24 cm high are 5 cm and 15 cm, respectively, show that the surface area of the bucket is (545π) cm2.

Let R and r be the radius of the top and base, respectively, of the bucket and let be its height.

Then, R = 15 cm, r = 5 cm and h = 24 cm

Slant height, $l=\sqrt{{h}^{2}+{\left(R-r\right)}^{2}}$

Surface area of the bucket$=\mathrm{\pi }\left[{r}^{2}+l\left(R+r\right)\right]$

#### Question 18:

Water is flowing at the rate of 15 km per hour through a pipe of diameter 14 cm into a rectangular tank which is 50 m long and 44 m wide. Find the time in which the level of water in the tank rises by 21 cm.

Length of the tank = 50 m; width of the tank = 44 m
Required height to which water level has to rise$=\frac{21}{100}\text{m}$
Volume of water in the tank

Radius of the pipe$=\frac{7}{100}\text{m}$

Speed of water flowing through the pipe$=\left(15×1000\right)\text{m/hour}$
= 15000 m/hour

Volume of water flowing per hour$=\mathrm{\pi }{r}^{2}h$
$=\left(\frac{22}{7}×\frac{7}{100}×\frac{7}{100}×15000\right){\text{m}}^{3}$
= 231 m3

Time required to fill the tank$=\left(\frac{1}{231}×462\right)$
= 2 hours

#### Question 19:

A solid cylinder of diameter 12 cm and height 15 cm is melted and recast into toys in the shape of a right circular cone mounted on a hemisphere of radius 3 cm. If the height of the toy is 12 cm, find the number of toys so formed.

Radius of cylinder =
Height of cylinder = 15 cm
Volume of cylinder$=\mathrm{\pi }{r}^{2}h$

Radius of the hemisphere = 3 cm

Height of the cone = Total height of the toy - Radius of the hemisphere

Volume of each toy = (Volume of the hemisphere) + (Volume of the cone)

Number of toys

$=\left(\frac{11880×7}{7×990}\right)\phantom{\rule{0ex}{0ex}}=12$

#### Question 20:

The radii of two cylinders are in the ratio 2 : 3 and their heights are in the ratio 5 : 3. Find the ratio of their volumes.

Then, ratio of their volumes$=\frac{\mathrm{\pi }×4{r}^{2}×5h}{\mathrm{\pi }×9{r}^{2}×3h}$
$=\frac{20}{27}\phantom{\rule{0ex}{0ex}}=20:27$