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#### Question 1:

Two cubes each of volume 27 cm3 are joined end to end to form a solid. Find the surface area of the resulting cuboid.     [CBSE 2011, 14]

#### Question 2:

The volume of a hemisphere is $2425\frac{1}{2}$ cm3. Find its curved surface area.                                                   [CBSE 2012]

#### Question 3:

If the total surface area of a solid hemisphere is 462 cm2, then find its volume.                                                   [CBSE 2014]

= 718.67 cm3

#### Question 4:

A 5-m-wide cloth is used to make a conical tent of base diameter 14 m and height 24 m. Find the cost of cloth used at the rate of 25 per metre.                                                                                                                                                                                           [CBSE 2014]

#### Question 5:

If the volumes of two cones are in the ratio of 1:4 and their diameters are in the ratio of 4:5, then find the ratio of their heights.

So, the ratio of their heights is 25:64.

#### Question 6:

The slant height of a conical mountain is 2.5 km and the area of its base is 1.54 km2. Find the height of the mountain.

Let r, h and l be the base radius, the height and the slant height of the conical mountain, respectively.

So, the height of the mountain is 2.4 km.

#### Question 7:

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

Let r and h be the base radius and the height of the solid cylinder, respectively.

#### Question 8:

The surface area of a sphere is 2464 cm2. If its radius be doubled, then what will be the surface area of the new sphere?

Let the radii of the given sphere and the new sphere be r and R, respectively.

So, the surface area of the new sphere is 9856 cm2.

#### Question 9:

A military tent of height 8.25 m is in the form of a right circular cylinder of base diameter 30 m and height 5.5 m surmounted by a right circular cone of same base radius. Find the length of canvas used in making the tent, if the breadth of the canvas is 1.5 m.   [CBSE 2012]

We have,
the radii of bases of the cone and cylinder, r = 15 m,
the height of the cylinder, = 5.5 m,
the height of the tent = 8.25 m
Also, the height of the cone,

As, the width of the canvas = 1.5 m

Hence, the length of the tent used for making the tent is 825 m.

#### Question 10:

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 11:

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 12:

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 13:

A solid is in the shape of a cone surmounted on a hemisphere, the radius of each of them being 3.5 cm and the total height of the solid is
9.5 cm. Find the volume of the solid.                                                                                                                                         [CBSE 2012]

So, the volume of the solid is 166.83 cm3.

#### Question 14:

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 15:

A toy is in the shape of a cone mounted on a hemisphere of same base radius. If the volume of the toy is 231 cm3 and its diameter is 7 cm, then find the height of the toy.                                                                                                                                                       [CBSE 2012]

#### Question 16:

A cylindrical container of radius 6 cm and height 15 cm is filled with ice-cream. The whole ice-cream has to be distributed to 10 children in equal cones with hemispherical tops. If the height of the conical portion is 4 times the radius of its base, then find the radius of the ice-cream cone.

So, the radius of the ice-cream cone is 3 cm.

#### Question 17:

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 18:

A toy is in the form of a cylinder with hemispherical ends. If the whole length of the toy is 90 cm and its diameter is 42 cm, then find the cost of painting the toy at the rate of 70 paise per sq cm.                                                                                                                           [CBSE 2014]

#### Question 19:

A medicine capsule is in the shape of a cylinder with two hemispheres stuck to each of its ends. The length of the entire capsule is 14 mm and the diameter of the capsule is 5 mm. Find its surface area.

So, the surface area of the medicine capsule is 220 mm2.

#### Question 20:

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 21:

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 22:

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 23:

From a solid cylinder of height 2.8 cm and diameter 4.2 cm, a conical cavity of the same height and same diameter is hollowed out. Find the total surface area of the remaining solid.                                                                                                                               [CBSE 2014]

So, the total surface area of the remaining solid is 73.92 cm2.

#### Question 24:

From a solid cylinder of height 14 cm and base diameter 7 cm, two equal conical holes each of radius 2.1 cm and height 4 cm are cut off. Find the volume of the remaining solid.                                                                                                                                       [CBSE 2011]

So, the volume of the remaining solid is 502.04 cm3.

#### Question 25:

A metallic cylinder has radius 3 cm and height 5 cm. To reduce its weight, a conical hole is drilled in the cylinder. The conical hole has a radius of $\frac{3}{2}$ cm and its depth is $\frac{8}{9}$ cm. Calculate the ratio of the volume of metal left in the cylinder to the volume of metal taken out in conical shape.                                                                                                                                                                                 [CBSE 2015]

So, the ratio of the volume of metal left in the cylinder to the volume of metal taken out in conical shape is 133 : 2.

#### Question 26:

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 27:

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 28:

From a cubical piece of wood of side 21 cm, a hemisphere is carved out in such a way that the diameter of the hemisphere is equal to the side of the cubical piece. Find the surface area and volume of the remaining piece.                                                                                 [CBSE 2014]

#### Question 29:

A cubical block of side 10 cm is surmounted by a hemisphere. What is the largest diameter that the hemisphere can have? Find the cost of painting the total surface area of the solid so formed, at the rate of 5 per 100 sq cm. [Use $\mathrm{\pi }$ = 3.14]                              [CBSE 2015]

hence, the cost of painting the total surface area of the solid is ₹33.92.

#### Question 30:

A toy is in the shape of a right circular cylinder with a hemisphere on one end and a cone on the other. The radius and height of the cylindrical part are 5 cm and 13 cm, respectively. The radii of the hemispherical and the conical parts are the same as that of the cylindrical part. Find the surface area of the toy, if the total height of the toy is 30 cm.

So, the surface area of the toy is 770 cm2.

#### Question 31:

The inner diameter of a glass is 7 cm and it has a raised portion in the bottom in the shape of a hemisphere, as shown in the
figure. If the height of the glass is 16 cm, find the apparent capacity and the actual capacity of the glass.

#### Question 32:

A wooden toy is in the shape of a cone mounted on a cylinder, as shown in the figure. The total height of the toy is 26 cm, while the height of the conical part is 6 cm. The diameter of the base of the conical part is 5 cm and that of the cylindrical part is 4 cm. The conical part and the cylindrical part are respectively painted red and white. Find the area to be painted by each of these colours. [Take $\mathrm{\pi }=\frac{22}{7}$]

#### Question 1:

The dimensions of a metallic cuboid are . It is melted and recast into a cube. Find the surface area of the cube.
[CBSE 2011]

So, the surface area of the cube is 38400 cm2.

#### Question 2:

A cone of height 20 cm and radius of base 5 cm is made up of modelling clay. A child reshapes it in the form of a sphere. Find the diameter of the sphere.                                                                                                                                                                                    [CBSE 2011]

So, the diameter of the sphere is 10 cm.

#### Question 3:

Metallic spheres of radii 6 cm, 8 cm and 10 cm, respectively are melted to form a single solid sphere. Find the radius of the resulting sphere.
[CBSE 2012]

So, the radius of the resulting sphere is 12 cm.

#### Question 4:

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 5:

The radii of internal and external surfaces of a hollow spherical shell are 3 cm and 5 cm, respectively. It is melted and recast into a solid cylinder of diameter 14 cm. Find the height of the cylinder.                                                                                                     [CBSE 2012]

So, the height of the cylinder is $\frac{8}{3}$ cm.

#### Question 6:

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 7:

A copper rod of diameter 2 cm and length 10 cm is drawn into a wire of uniform thickness and length 10 m. Find the thickness of the wire.
[CBSE 2012]

So, the thickness of the wire is 0.2 cm or 2 mm.

#### Question 8:

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 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 9:

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 10:

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 11:

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 12:

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 13:

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 14:

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

Volume of spherical ball
Volume of each cube

Number of cubes =

#### Question 15:

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 16:

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 17:

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 18:

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 19:

A hemispherical bowl of internal radius 9 cm is full of water. Its contents are emptied into a cylindrical vessel of internal radius 6 cm. Find the height of water in the cylindrical vessel.                                                                                                                                   [CBSE 2012]

So, the height of the water in the cylindrical vessel is 13.5 cm.

#### Question 20:

A hemispherical tank, full of water, is emptied by a pipe at the rate of $\frac{25}{7}$ litres per second. How much time will it take to empty half the tank if the diameter of the base of the tank is 3 m?                                                                                                                       [CBSE 2012]

So, the time taken to empty half the tank is 16 minutes and 30 seconds.

#### Question 21:

The rain water from a roof of 44 m $×$ 20 m drains into a cylindrical tank having diameter of base 4 m and height 3.5 m. If the tank is just full, then find the rainfall in cm.                                                                                                                                                                  [CBSE 2014]

So, the height of the rainfall is 5 cm.

#### Question 22:

The rain water from a 22 m $×$ 20 m roof drains into a cylindrical vessel of diameter 2 m and height 3.5 m. If the rain water collected from the roof fills $\frac{4}{5}$th of the cylindrical vessel, then find the rainfall in centimetre.                                                                                      [CBSE 2015]

So, the height of the rainfall is 2 cm.

#### Question 23:

A solid right circular cone of height 60 cm and radius 30 cm is dropped in a right circular cylinder full of water, of height 180 cm and radius 60 cm. Find the volume of water left in the cylinder, in cubic metres.                                                                        [CBSE 2015]

So, the volume of water left in the cylinder is 1.98 m3.

#### Question 24:

Water is flowing through a cylindrical pipe of internal diameter 2 cm, into a cylindrical tank of base radius 40 cm, at the rate of 0.4 m per second. Determine the rise in level of water in the tank in half an hour.                                                                         [CBSE 2013]

So, the rise in level of water in the tank in half an hour is 45 cm.

Disclaimer: The answer given in the textbook is incorrect. The same has been corrected above.

#### Question 25:

Water is flowing at the rate of 6 km/hr through a pipe of diameter 14 cm into a rectangular tank which is 60 m long and 22 m wide. Determine the time in which the level of water in the tank will rise by 7 cm.                                                                                   [CBSE 2011]

So, the time in which the level of water in the tank will rise by 7 cm is 1 hour.

#### Question 26:

Water in a canal, 6 m wide and 1.5 m deep, is flowing at a speed of 4 km/hr. How much area will it irrigate in 10 minutes if 8 cm of standing water is needed for irrigation?                                                                                                                                                           [CBSE 2014]

Hence, it will irrigate 7.5 hectare of area in 10 minutes.

#### Question 27:

A farmer connects a pipe of internal diameter 25 cm from a canal into 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/hr, then in how much time will the tank be filled? Also, find the cost of water if the canal department charges at the rate of 0.07 per m3.                                                                                                                      [CBSE 2009C]

#### Question 28:

Water running in a cylindrical pipe of inner diameter 7 cm, is collected in a container at the rate of 192.5 litres per minute. Find the rate of flow of water in the pipe in km/hr.                                                                                                                                                [CBSE 2013]

So, the rate of flow of water in the pipe is 3 km/hr.

#### Question 29:

150 spherical marbles, each of diameter 1.4 cm, are dropped in a cylindrical vessel of diameter 7 cm containing some water, which are completely immersed in water. Find the rise in the level of water in the vessel.                                                        [CBSE 2014]

So, the rise in the level of water in the vessel is 5.6 cm.

Disclaimer: The diameter of the spherical marbles should be 1.4 cm instead 14 cm. The has been corrected above.

#### Question 30:

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 31:

In a village, a well with 10 m inside diameter, is dug 14 m deep. Earth taken out of it is spread all around to a width 5 m to form an embankment. Find the height of the embankment. What value of the villagers is reflected here?                                      [CBSE 2014]

So, the height of the embankment is $\frac{14}{3}$ m.

Value: We must lanour hard to make maximum use of the available resources.

Disclaimer: The answer provided in the textbook is incorrect. It has been corrected above.

#### Question 32:

In a corner of a rectangular field with dimensions , a well with 14 m inside diameter is dug 8 m deep. The earth dug out is spread evenly over the remaining part of the field. Find the rise in the level of the field.                                                                                    [HOTS]

So, the rise in the level of the field is 2 m.

#### Question 33:

A copper wire of diameter 6 mm is evenly wrapped on a cylinder of length 18 cm and diameter 49 cm to cover its whole surface. Find the length and the volume of the wire. If the density of the copper be 8.8 g per cm3, then find the weight of the wire.                           [HOTS]

#### Question 34:

A right triangle whose sides are 15 cm and 20 cm (other than hypotenuse), is made to revolve about its hypotenuse. Find the volume and surface area of the double cone so formed. (Choose value of $\mathrm{\pi }$ as found appropriate)                                                                       [HOTS]

#### Question 1:

A drinking glass is in the shape of a frustum of a cone of height 14 cm. The diameters of its two circular ends are 16 cm and 12 cm. Find the capacity of the glass.                                                                                                                                                               [CBSE 2012)

So, the capacity of the glass is 2170.67 cm3.

#### Question 2:

The radii of the circular ends of a solid frustum of a cone are 18 cm and 12 cm and its height is 8 cm. Find its total surface area.
[Use $\mathrm{\pi }$ = 3.14]                                                                                                                                                               [CBSE 2011]

So, the total surface area of the solid frustum is 2411.52 cm2.

#### Question 3:

A metallic bucket, open at the top, of height 24 cm is in the form of the frustum of a cone, the radii of whose lower and upper circular ends are 7 cm and 14 cm, respectively. Find
(i) the volume of water which can completely fill the bucket;
(ii) the area of the metal sheet used to make the bucket.                                                                                                                    [CBSE 2014]

#### Question 4:

A container, open at the top, is in the form of a frustum of a cone of height 24 cm with radii of its lower and upper circular ends as 8 cm and 20 cm, respectively. Find the cost of milk which can completely fill the container at the rate of 21 per litre.                         [CBSE 2014]

Hence, the cost of milk which can completely fill the container is ₹329.47.

#### Question 5:

A container, open at the top and made up of metal sheet, is in the form of a frustum of a cone of height 16 cm with diameters of its lower and upper ends as 16 cm and 40 cm, respectively. Find the cost of metal sheet used to make the container, if it costs 10 per 100 cm2.
[CBSE 2013]

Hence, the cost of metal sheet used to make the container is ₹196.11.

Disclaimer: The answer given in the textbook is incorrect. The same has been corrected above.

#### Question 6:

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 7:

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 8:

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 9:

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 10:

A container in the shape of a frustum of a cone having diameters of its two circular faces as 35 cm and 30 cm and vertical height 14 cm,
is completely filled with oil. If each cm3 of oil has mass 1.2 g, then find the cost of oil in the container if it costs 40 per kg.   [CBSE 2009C]

#### Question 11:

A bucket is in the form of a frustum of a cone and it can hold 28.49 litres of water. If the radii of its circular ends are 28 cm and 21 cm, then find the height of the bucket.                                                                                                                                                              [CBSE 2012]

So, the height of the bucket is 15 cm.

#### Question 12:

The radii of the circular ends of a bucket of height 15 cm are 14 cm and r cm (r < 14). If the volume of bucket is 5390 cm3, then find the value of r.                                                                                                                                                                                            [CBSE 2011]

So, the value of r is 7 cm.

#### Question 13:

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 total surface area. [Use $\mathrm{\pi }=3.14.\right]$

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

Surface area of the frustum

#### Question 14:

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 15:

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 16:

The perimeters of the two circular ends of a frustum of a cone are 48 cm and 36 cm. If the height of the frustum is 11 cm, then find its volume and curved surface area.                                                                                                                                                                     [CBSE 2014]

#### Question 17:

A solid cone of base radius 10 cm is cut into two parts through the midpoint of its height, by a plane parallel to its base. Find the ratio of the volumes of the two parts of the cone.                                                                                                                                                [CBSE 2013]

So, the ratio of the volume of the two parts of the cone is 1 : 7.

#### Question 18:

The height of a right circular cone is 20 cm. A small cone is cut off at the top by a plane parallel to the base. If its volume be $\frac{1}{8}$ of the volume of the given cone, then at what height above the base is the section made?                                                           [HOTS] [CBSE 2014]

So, the section is made at the height of 10 cm above the base.

#### Question 19:

A solid metallic right circular cone 20 cm high and whose vertical angle is 60°, is cut into two parts at the middle of its height by a plane parallel to its base. If the frustum so obtained be drawn into a wire of diameter $\frac{1}{12}$ cm, then find the length of the wire.
[HOTS] [CBSE 2014]

So, the length of the wire is 4480 m.

#### Question 20:

A fez, the cap used by the Turks, is shaped like the frustum of a cone. If its radius on the open side is 10 cm, radius at the upper base is
4 cm and its slant height is 15 cm, then find the area of material used for making it.

So, the area of material used for making the fez is 710.28 cm2.

#### Question 21:

An oil funnel made of tin sheet consists of a 10 cm long cylindrical portion attached to a frustum of a cone. If the total height is 22 cm,
diameter of the cylindrical portion is 8 cm and the diameter of the top of the funnel is 18 cm, then find the area of the tin sheet required to make the funnel.

So, the area of the tin sheet required to make the funnel is 782.57 cm2.

#### Question 1:

A river 1.5 m deep and 36 m wide is flowing at the rate of 3.5 km/hr. Find the amount of water (in cubic metres) that runs into the sea per minute.

So, the amount of water that runs into the sea per minute is 3150 m3.

#### Question 2:

The volume of a cube is 729 cm3. Find its surface area.

So, the surface area of the cube is 486 cm2.

#### Question 3:

How many cubes of 10 cm edge can be put in a cubical box of 1 m edge?

So, the number of cubes that can be put in the cubical box is 1000.

#### Question 4:

Three cubes of iron whose edges are 6 cm, 8 cm and 10 cm, respectively are melted and formed into a single cube. Find the edge of the new cube formed.

So, the edge of the new cube so formed is 12 cm.

#### Question 5:

Five identical cubes, each of edge 5 cm, are placed adjacent to each other. Find the volume of the resulting cuboid.

So, the volume of the resulting cuboid is 625 cm3.

#### Question 6:

The volumes of two cubes are in the ratio 8 : 27. Find the ratio of their surface areas.

So, the ratio of the surface areas of the given cubes is 4 : 9.

#### Question 7:

The volume of a right circular cylinder with its height equal to the radius is $25\frac{1}{7}$ cm3. Find the height of the cylinder.

So, the height of the cylinder is 2 cm.

#### Question 8:

The ratio between the radius of the base and the height of a cylinder is 2 : 3. If the volume of the cylinder is 12936 cm3, then find the radius of the base of the cylinder.

So, the radius of the base of the cylinder is 14 cm.

#### Question 9:

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

So, the ratio of the volumes of the given cylinders is 20 : 27.

#### Question 10:

66 cubic cm of silver is drawn into a wire 1 mm in diameter. Calculate the length of the wire in metres.

So, the length of the wire is 84 m.

#### Question 11:

If the area of the base of a right circular cone is 3850 cm2 and its height is 84 cm, then find the slant height of the cone.

So, the slant height of the given cone is 91 cm.

#### Question 12:

A cylinder with base radius 8 cm and height 2 cm is melted to form a cone of height 6 cm. Calculate the radius of the base of the cone.

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

#### Question 13:

A right cylindrical vessel is full of water. How many right cones having the same radius and height as those of the right cylinder will be needed to store that water?

So, the number of cones that will be needed to store the water is 3.

#### Question 14:

The volume of a sphere is 4851 cm3. Find its curved surface area.

So, the curved surface area of the sphere is 1386 cm2.

#### Question 15:

The curved surface area of a sphere is 5544 cm2. Find its volume.

So, the volume of the sphere is 38808 cm3.

#### Question 16:

The surface areas of two spheres are in the ratio of 4 : 25. Find the ratio of their volumes.

So, the ratio of the volumes of the given spheres is 8 : 125.

#### Question 17:

A solid metallic sphere of radius 8 cm is melted and recast into spherical balls each of radius 2 cm. Find the number of spherical balls obtained.

So, the number of spherical balls obtained is 64.

#### Question 18:

How many lead shots each 3 mm in diameter can be made from a cuboid of dimensions 9 cm $×$ 11 cm $×$ 12 cm?

So, the number of lead shots that can be made from the cuboid is 84000.

#### Question 19:

A metallic cone of radius 12 cm and height 24 cm is melted and made into spheres of radius 2 cm each. How many spheres are formed?

So, the number of spheres so formed is 108.

#### Question 20:

A hemisphere of lead of radius 6 cm is cast into a right circular cone of height 75 cm. Find the radius of the base of the cone.

So, the radius of the base of the cone is 2.4 cm.

#### Question 21:

A copper sphere of diameter 18 cm is drawn into a wire of diameter 4 mm. Find the length of the wire.

So, the length of the wire is 243 m.

#### Question 22:

The radii of the circular ends of a frustum of height 6 cm are 14 cm and 6 cm, respectively. Find the slant height of the frustum.

So, the slant height of the frustum is 10 cm.

#### Question 23:

Find the ratio of the volume of a cube to that of a sphere which will fit inside it.

So, the ratio of the volume of the cube to that of the sphere is 6 : $\mathrm{\pi }$.

#### Question 24:

Find the ratio of the volumes of a cylinder, a cone and a sphere, if each has the same diameter and same height?

So, the ratio of the volumes of the cylinder, the cone and the sphere is 3 : 1 : 2.

#### Question 25:

Two cubes each of volume 125 cm3 are joined end to end to form a solid. Find the surface area of the resulting cuboid.            [CBSE 2013C]

So, the surface area of the resulting cuboid is 250 cm2.

#### Question 26:

Three metallic cubes whose edges are 3 cm, 4 cm and 5 cm, are melted and recast into a single large cube. Find the edge of the new cube formed.                                                                                                                                                                                       [CBSE 2011]

So, the edge of the new cube so formed is 6 cm.

#### Question 27:

A solid metallic sphere of diameter 8 cm is melted and drawn into a cylindrical wire of uniform width. If the length of the wire is 12 m, then find its width.                                                                                                                                                                                      [CBSE 2013]

So, the width of the wire is $\frac{8}{15}$ cm.

#### Question 28:

A 5-m-wide cloth is used to make a conical tent of base diameter 14 m and height 24 m. Find the cost of cloth used, at the rate of 25 per metre.                                                                                                                                                                                                  [CBSE 2014]

So, the cost of the cloth used for making the tent is ₹2750.

#### Question 29:

A wooden toy was made by scooping out a hemisphere of same radius from each end of a solid cylinder. If the height of the cylinder is 10 cm
and its base is of radius 3.5 cm, then find the volume of wood in the toy.                                                                                         [CBSE 2013]

So, the volume of wood in the toy is $\frac{616}{3}$ cm3 or 205.33 cm3.

#### Question 30:

Three cubes of a metal whose edges are in the ratio 3 : 4 : 5 are melted and converted into a single cube whose diagonal is $12\sqrt{3}$ cm. Find the
edges of the three cubes.                                                                                                                                                                  [CBSE 2013C]

Hence, the edges of the given three metallic cubes are 6 cm, 8 cm and 10 cm.

#### Question 31:

A hollow sphere of external and internal diameters 8 cm and 4 cm, respectively is melted into a solid cone of base diameter 8 cm. Find the height of the cone.                                                                                                                                                                      [CBSE 2013C]

So, the height of the cone is 14 cm.

#### Question 32:

A bucket of height 24 cm is in the form of frustum of a cone whose circular ends are of diameter 28 cm and 42 cm. Find the cost of milk at the rate of 30 per litre, which the bucket can hold.                                                                                                                    [CBSE 2013C]

So, the cost of the milk which the bucket can hold is ₹702.24.

#### Question 33:

The interior of a building is in the form of a right circular cylinder of diameter 4.2 m and height 4 m surmounted by a cone of same diameter.
The height of the cone is 2.8 m. Find the outer surface area of the building.                                                                                    [CBSE 2014]

So, the outer surface area of the building is 75.9 m2.

#### Question 34:

A metallic solid right circular cone is of height 84 cm and the radius of its base is 21 cm. It is melted and recast into a solid sphere. Find the
diameter of the sphere.                                                                                                                                                                        [CBSE 2014]

So, the diameter of the solid sphere is 42 cm.

#### Question 35:

A toy is in the form of a cone of radius 3.5 cm mounted on a hemisphere of same radius. The total height of the toy is 15.5 cm. Find the total
surface area of the toy.                                                                                                                                                                        [CBSE 2012]

So, the total surface area of the toy is 214.5 cm2.

Disclaimer: The answer given in the textbook is incorrect. The same has been corrected above.

#### Question 36:

If the radii of the circular ends of a bucket 28 cm high, are 28 cm and 7 cm, then find its capacity and total surface area.             [CBSE 2011]

Disclaimer: The answer of the total surface area given in the textbook is incorrect. The same has been corrected above.

#### Question 37:

A bucket is in the form of a frustum of a cone with a capacity of 12308.8 cm3 of water. The radii of the top and bottom circular ends are
20 cm and 12 cm, respectively. Find the height of the bucket. [Use $\mathrm{\pi }$ = 3.14]                                                                              [CBSE 2006C]

So, the height of the bucket is 15 cm.

#### Question 38:

A milk container is made of metal sheet in the shape of frustum of a cone whose volume is $10459\frac{3}{7}$ cm3. The radii of its lower and upper circular ends are 8 cm and 20 cm, respectively. Find the cost of metal sheet used in making the container at the rate of 1.40 per cm2.
[CBSE 2010]

Hence, the cost of the metal sheet used for making the milk container is ₹2745.60.

Disclaimer: The answer given in the textbook is incorrect. The same has been corrected above.

#### Question 39:

A solid metallic sphere of diameter 28 cm is melted and recast into a number of smaller cones, each of diameter $4\frac{2}{3}$ cm and height 3 cm. Find the number of cones so formed.

So, the number of cones so formed is 672.

Disclaimer: The answer given in the textbook is incorrect. The same has been corrected above.

#### Question 40:

A cylindrical vessel with internal diameter 10 cm and height 10.5 cm is full of water. A solid cone of base diameter 7 cm and height 6 cm is completely immersed in water. Find the volume of water
(i) displaced out of the cylinder
(ii) left in the cylinder.                                                                                                                                                                       [CBSE 2009]

#### 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:

Choose the correct answer of the following:

The radius (in cm) of the largest right circular cone that can be cut out from a cube of edge 4.2 cm is

(a) 2.1                         (b) 4.2                         (c) 8.4                         (d) 1.05                                                                                     [CBSE 2011]

Since, the diameter of the base of the largest cone that can be cut out from the cube = Edge of the cube = 4.2 cm

So, the radius of the base of the largest cone = $\frac{4.2}{2}$ = 2.1 cm

Hence, the correct answer is option (a).

#### Question 13:

Choose the correct answer of the following question:

A metallic solid sphere of radius 9 cm is melted to form a solid cylinder of radius 9 cm. The height of the cylinder is

(a) 12 cm                         (b) 18 cm                         (c) 36 cm                        (d) 96 cm                                                                    [CBSE 2014]

Hence, the correct answer is option (a).

#### Question 14:

Choose the correct answer of the following question:

A rectangular sheet of paper 40 cm $×$ 22 cm, is rolled to form a hollow cylinder of height 40 cm. The radius of the cylinder (in cm) is

(a) 3.5                        (b) 7                        (c) 80                        (d) 5                                                                                                [CBSE 2014]

Hence, the correct answer is option (a).

#### Question 15:

Choose the correct answer of the following question:

The number of solid spheres, each of diameter 6 cm, that can be made by melting a solid metal cylinder of height 45 cm and diameter 4 cm, is

(a) 2                       (b) 4                       (c) 5                       (d) 6                                                                                                         [CBSE 2014]

Hence, the correct answer is option (c).

#### Question 16:

Choose the correct answer of the following question:

The surface areas of two spheres are in the ratio 16 : 9. The ratio of their volumes is

(a) 64 : 27                          (b) 16 : 9                          (c) 4 : 3                          (d) 163 : 93                                                             [CBSE 2013C]

Hence, the correct answer is option (a).

#### Question 17:

Choose the correct answer of the following question:

If the surface area of a sphere is 616 cm2, its diameter (in cm) is

(a) 7                         (b) 14                         (c) 28                         (d) 56                                                                                           [CBSE 2012]

Hence, the correct answer is option (b).

#### Question 18:

Choose the correct answer of the following question:

If the radius of a sphere becomes 3 times, then its volume will become

(a) 3 times                         (b) 6 times                         (c) 9 times                         (d) 27 times                                                         [CBSE 2011]

Hence, the correct answer is option (d).

#### Question 19:

Choose the correct answer of the following question:

If the height of a bucket in the shape of frustum of a cone is 16 cm and the diameters of its two circular ends are 40 cm and 16 cm, then its slant height is

(a) 20 cm                         (b) $12\sqrt{5}$ cm                         (c) $8\sqrt{13}$ cm                         (d) 16 cm                                                     [CBSE 2013C]

Hence, the correct answer is option (a).

#### Question 20:

Choose the correct answer of the following question:

A sphere of diameter 18 cm is dropped into a cylindrical vessel of diameter 36 cm, partly filled with water. If the sphere is completely submerged, then the water level rises by

(a) 3 cm                         (b) 4 cm                         (c) 5 cm                         (d) 6 cm                                                                          [CBSE 2011]

Hence, the correct answer is option (a).

#### Question 21:

Choose the correct answer of the following question:

A solid right circular cone is cut into two parts at the middle of its height by a plane parallel to its base. The ratio of the volume of the smaller cone to the whole cone is

(a) 1 : 2                            (b) 1 : 4                            (c) 1 : 6                            (d) 1 : 8                                                                     [CBSE 2012]

Hence, the correct answer is option (d).

#### Question 22:

Choose the correct answer of the following question:

The radii of the circular ends of a bucket of height 40 cm are 24 cm and 15 cm. The slant height (in cm) of the bucket is

(a) 41                      (b) 43                      (c) 49                      (d) 51                                                                                                  [CBSE 2012]

Hence, the correct answer is option (a).

#### Question 23:

Choose the correct answer of the following question:

A solid is hemispherical at the bottom and conical (of same radius) above it. If the surface areas of the two parts are equal, then the ratio of its radius and the slant height of the conical part is

(a) 1 : 2                        (b) 2 : 1                        (c) 1 : 4                        (d) 4 : 1                                                                                  [CBSE 2011]

Hence, the correct answer is option (a).

#### Question 24:

Choose the correct answer of the following question:

If the radius of the base of a right circular cylinder is halved, keeping the height the same, then the ratio of the volume of the cylinder thus obtained to the volume of original cylinder is

(a) 1 : 2                         (b) 2 : 1                         (c) 1 : 4                         (d) 4 : 1                                                                             [CBSE 2012]

Hence, the correct answer is option (d).

#### Question 25:

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 26:

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 27:

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 28:

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 29:

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 30:

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 31:

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

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 32:

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$=\frac{4}{3}\mathrm{\pi }\left({R}^{3}-{r}^{3}\right)$

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

Therefore,

#### Question 33:

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

= 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 34:

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 35:

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 36:

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 37:

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 38:

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 39:

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 40:

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 41:

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 42:

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 43:

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 44:

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 45:

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 46:

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 47:

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 48:

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 49:

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 50:

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 51:

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 52:

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 53:

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 54:

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 55:

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 56:

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 57:

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 58:

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 59:

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 60:

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 61:

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 62:

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 63:

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 64:

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 65:

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 66:

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 67:

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 68:

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 69:

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 70:

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 71:

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 72:

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 73:

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 74:

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 75:

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 76:

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 77:

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 78:

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 79:

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 80:

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:

Find the number of solid spheres, each of diameter 6 cm, that could be moulded to form a solid metallic cylinder of height 45 cm and diameter 4 cm.

So, the number of solid spheres so moulded is 5.

#### Question 2:

Two right circular cylinders of equal volumes have their heights in the ratio 1 : 2. What is the ratio of their radii?

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

#### Question 3:

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, then find the total area of the canvas required.

So, the area of the canvas required to make the tent is 7920 m2.

#### Question 4:

The radii of the top and bottom of a bucket of slant height 45 cm are 28 cm and 7 cm, respectively. Find the curved surface area of the bucket.

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$=\frac{1}{3}\mathrm{\pi }{r}^{2}h$
$=\frac{1}{3}\mathrm{\pi }×{\left(12\right)}^{2}×24$

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?

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.

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

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.

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

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$=\frac{2}{3}\mathrm{\pi }{\mathrm{r}}^{3}$
$=\left(\frac{2}{3}\mathrm{\pi }×9×9×9\right){\text{cm}}^{3}$
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:

From a solid cylinder whose height is 15 cm and diameter 16 cm, a conical cavity of the same height and same diameter is hollowed out. Find the total surface area of the remaining solid. (Use $\mathrm{\pi }$ = 3.14)

So, the total surface area of the remaining solid is 1381.6 cm2.

Disclaimer: The answer given in the textbook is incorrect. The same has been corrected above.

#### Question 18:

A solid rectangular block of dimensions 4.4 m, 2.6 m and 1 m is cast into a hollow cylindrical pipe of internal radius 30 cm and thickness
​5 cm. Find the length of the pipe.

So, the length of the pipe is 112 m.

#### Question 19:

An open metal bucket is in the shape of a frustum of a cone, mounted on a hollow cylindrical base made of the same metallic sheet. The diameters of the two circular ends of the bucket are 45 cm and 25 cm, the total vertical height of the bucket is 40 cm and that of the cylindrical base is 6 cm. Find the area of the metallic sheet used to make the bucket. Also, find the volume of water the bucket can hold, in litres.

#### Question 20:

A farmer connects a pipe of internal diameter 20 cm from a canal into a cylindrical tank which is 10 m in diameter and 2 m deep. If the water flows through the pipe at the rate of 4 km/hr, then in how much time will the tank be filled completely?