# Question-1

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

Solution:
Volume of each cube = 64 cm3
If "a" is the side of cube, a3 = 64

â‡’ a =

â‡’ a = 4 cm.
âˆ´
Surface area of a cube = 6a2

= 6(4)2

= 6 (16)
= 96 cm2
âˆ´
Surface area of resulting cuboid = 2(Surface area of each cube) - (Surface area of one side)
= (2 Ã— 96) â€“ 2a2
= 192 â€“ 2(4)2
= 192 â€“ 32
= 160 cm2.

# Question-2

A vessel is in the form of a hollow hemisphere mounted by a hollow cylinder. The diameter of the hemisphere is 14 cm and the total height of the vessel is 13 cm. Find the inner surface area of the vessel.

Solution:

Diameter of a hemisphere = 14 cm âˆ´ Radius of a hemisphere (r) = 7 cm âˆ´ Surface area of hemisphere = 2Ï€ r2

= 2 Ã— Ã— 7 Ã— 7
= 308 cm2

Diameter of a cylinder = 14 cm âˆ´ Radius of a cylinder (r) = 7 cm
Total height of the vessel = Height of the cylinder + Radius of the hemisphere âˆ´ Height of the cylinder = Total height of the vessel - Radius of the hemisphere
= 13 - 7
= 6 cm âˆ´ Surface area of a cylinder = 2Ï€ rh
= 2 Ã— Ã— 7 Ã— 6
= 264 cm2

Inner Surface area of the vessel = Surface area of hemisphere + Surface area of a cylinder
= 308 + 264
= 572 cm2.

# Question-3

A toy is the form of a cone of radius 3.5 cm is 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.

Solution:

Total height of the toy = Height of the cone + Radius of the cone
Height of the cone = 15.5 - 3.5
= 12 cm
Radius of the cone (r) = 3.5 cm.
Height of the cone (h) = 12 cm
Slant height (l) =  =
=
= 12.5 cm
Surface area of the cone = Ï€ rl âˆ´ Surface area of a cone =
= 137.5 cm2

Radius of the hemisphere = 3.5 cm.
Surface area of a hemisphere = 2Ï€ r2

= 2 Ã— Ã— 3.5 Ã— 3.5
= 77 cm2

Total Surface area of the toy = Surface area of a cone + Surface area of a hemisphere
= 137.5 + 77
= 214.5 cm2.

# Question-4

A cubical block of side 7 cm is surmounted by a hemisphere. What is the greatest diameter the hemisphere can have? Find the surface area of the solid.

Solution:

Greatest diameter of the hemisphere = Length of an edge of the cube = 7 cm

Radius of the hemisphere (r) =
Total surface area of the solid = Surface area of the cube + Curved surface area of the hemisphere â€“ Base area of the hemisphere
= 6 Ã— (edge)2 + 2Ï€ r2 - Ï€ r2

= 6 Ã— (7)2 + Ï€ r2
=(6 Ã— 49) +
= 294 + 38.5
= 332.5 cm2

âˆ´ Total surface area of the solid = 332.5 cm2.

# Question-5

A hemispherical depression is cut out from one face of a cubical wooden block such that the diameter l of the hemisphere is equal to the edge of the cube. Determine the surface area of the remaining solid.

Solution:
Surface area of the cubical wooden box = 6(edge)2
Let the edge of a cube = l

âˆ´
Surface area of a cube = 6(l)2
Diameter of a hemisphere = l
=> Surface area of a hemisphere = 2
p r2

= 2
Ã— p Ã—  Ã—

= 2 Ã— p Ã—
=
p Ã—
Surface area of the remaining solid = Surface area of a cube - Surface area of a hemisphere
= 6(l)2 + (
p Ã—
= l2 [6 + p/2]

# Question-6

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.

Solution:

The length of the entire capsule = 14 mm
Diameter of the capsule = 5 mm

(Since the capsule is in the shape of cylinder with two hemispheres to each of its ends).

Radius of the hemisphere (r) =
= 2.5 mm

(Since the radius of both the hemispheres are included in the entire length of the capsule)
Height of the cylinder (h) = 14 - 5
= 9 mm

Surface area of the capsule = C.S.A. of two hemisphere + C.S.A of the cylinder
= 2(2Ï€ r2) + 2Ï€ rh
= 2Ï€ r[2r + h]
= 2 Ã—
= 2 Ã—
= 2 Ã—
= 220 mm2

âˆ´ The surface area of the capsule = 220 mm2.

# Question-7

A tent is in the shape of a cylinder surmounted by a conical top. If the height and diameter of the cylindrical part are 2.1 m and 4 m respectively, and the slant height of the top is 2.8 m, find the area of the canvas used for making the tent. Also, find the cost of the canvas of the tent at the rate of  Rupees 500 per m2. (Note that the base of the tent will not be covered with canvas.)

Solution:

A tent is in the shape of cylinder surmounted by a conical top.
Height of the cylinder (h) = 2.1 m
Diameter of the cylinder = 4 m â‡’ Radius of the cylinder (r) = 2 m [since diameter = 2(radius)]
Slant height of the cone (l) = 2.8 m
Area of the canvas = Surface area of the cone + Surface area of the cylinder
= Ï€ rl + 2Ï€ rh
= Ï€ r (
l  + 2h)
=
=
=
= 44 m2

Cost of the canvas of the tent per m2  500
Cost of 44 m2 of the canvas = 500 Ã— 44
=
22000
âˆ´ The area of the canvas required to make the tent = 44 m2
Total cost of the canvas =
 22000.

# Question-8

From a solid cylinder whose height is 2.4 cm and diameter 1.4 cm, a conical cavity of the same height and same diameter is hollowed out. Find the total surface area of the remaining solid to the nearest cm2.

Solution:

Height of the hollow cylinder = 2.4 cm
Diameter of the cylinder = 1.4 cm
Radius of the cylinder (r) = 0.7 m
Surface area of a cylinder = 2
Ï€rh cm2
Radius of the cone = 0.7 cm
Height of the cone = 2.4 cm
Slant height of the cone (l) ==
= = 2.5 cm
Surface area of a cone = Ï€ r
l  cm2 âˆ´ The total surface area of the remaining solid = Surface area of a cylinder + Surface area of a cone + Area of the circular base
= 2
Ï€rh + Ï€ rl  + Ï€ r2

= Ï€ r(2h +
l + r)
=
= 17.6 cm2.

# Question-9

A wooden article was made by scooping out a hemisphere 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, find the total surface area of the article.

Solution:

Height of a cylinder = 10 cm.
Radius of a cylinder = 3.5 cm.
Surface area of the cylinder = 2rh cm2
Radius of a hemisphere = 3.5 cm.
Surface area of a hemisphere = 2r2 cm2 âˆ´ Total surface area of the article = Surface area of the cylinder + 2(Surface area of a hemisphere)
= 2rh + 4r2

= 2r(h + 2r)
= 2 Ã— Ã— 3.5(10 + 2(3.5))
= 2 Ã— Ã— 3.5 (10 + 7)
= 2 Ã— Ã— 3.5 Ã— 17
= 374 cm2.

# Question-10

A solid is in the shape of a cone standing on a hemisphere with both their radii being equal to 1 cm and the height of the cone is equal to its radius. Find the volume of the solid in terms of .

Solution:

Radius of a cone = 1 cm
Height of the cone = 1 cm
Volume of a cone = Ï€ r2h
Radius of a hemisphere = 1 cm
Volume of a hemisphere =

Volume of the solid = Volume of a cone + Volume of a hemisphere
= r2h +
= r2()
= Ã— 12 Ã— ()
= Ã— 1 Ã— ()
= Ã—
= cm3.

# Question-11

Rachel, an engineering student, was asked to make a model shaped like a cylinder with two cones attached at its two ends by using a thin aluminium sheet. The diameter of the model is 3 cm and its length is 12 cm. If each cone has a height of 2 cm, find the volume of air contained in the model that Rachel made. (Assume the outer and inner dimensions of the model to be nearly the same.).

Solution:

Total height of the model = 12 cm
Height of each cone (H) = 2 cm âˆ´ Height of the cylinder = Total height - 2(Height of each cone)
= 12 - 2(2)
= 12 - 4
= 8 cm
Diameter of the cylinder = 3 cm
Radius of the cylinder = = 1.5 cm
Height of the cylinder (h) = 8 cm
Volume of the cylinder = Ï€ r2h cu. units
Radius of the cone = 1.5 cm
Height of each cone (H) = 2 cm

Volume of a cone = Ï€ r2H cu. units âˆ´ Volume of air contained in the model = Volume of the cylinder + 2(Volume of a cone)
= Ï€ r2h + 2(Ï€ r2H)
= Ï€ r2(h + )
= Ã— 1.5 Ã— 1.5 (8 + )
= Ã— 1.5 Ã— 1.5 ()
= Ã— 1.5 Ã— 1.5 Ã—
= 66 cm3.

# Question-12

A gulab jamuncontains sugar syrup up to about 30% of its volume. Find approximately how much syrub would be found in 45 gulab jamuns, each shaped like a cylinder with two hemispherical ends with length 5 cm and diameter 2.8 cm.

Solution:

Diameter of a cylinder = 2.8 cm
Radius of the cylinder = = 1.4 cm
Height of the cylinder = 5 - 2.8 = 2.2 cm
Volume of the cylinder = Ï€ r2h
Radius of the hemisphere = 1.4 cm
Volume of the hemisphere =
Volume of gulab jamun = Volume of the cylinder + 2(Volume of the hemisphere)
= Ï€ r2h + (2 Ã— )
= Ï€ r2(h + )
= Ã— 1.4 Ã— 1.4 Ã— (2.2 + )
= Ã— 1.4 Ã— 1.4 Ã—
= 25.05 cm3

Volume of 45 gulab jamuns = 45 Ã— 25.05 = 1127.25 cm3
Volume of syrub in the gulab jamun = 30% of volume of gulab jamun
= = 338.175 cm3
Hence the volume of syrub is 338.175 cm3.

# Question-13

A pen stand made of wood is in the shape of a cuboid with four conical depressions to hold pens.

The dimensions of the cuboid are 15 cm by 10 cm by 3.5 cm. The radius of each of the depressions is 0.5 cm and the depth is 1.4 cm. Find the volume of wood in the entire stand.

Solution:
Dimensions of the cuboid are 15 cm by 10 cm by 3.5 cm. âˆ´ Volume of the cuboid = l Ã— b Ã— h
= 15 Ã— 10 Ã— 3.5
= 525 cm3.

Radius of the conical depression (r) = 0.5 cm.
Height of the conical depression (h) = 1.4 cm.

As the pen stand made of wood is in the shape of a cuboid with four conical depressions.

Volume of wood = volume of cuboid - 4(volume of cone)
= 525 - (4 Ã— )
= 525 - 1.47
= 523.53 cm3. âˆ´ Volume of wood = 523.53 cm3.

# Question-14

A vessel is in the form of an inverted cone. Its height is 8 cm and the radius of its top, which is open, is 5 cm. It is filled with water up to the brim. When lead shots, each of which is a sphere of radius 0.5 cm are dropped into the vessel, one-fourth of the water flows out. Find the number of lead shots dropped in the vessel.

Solution:

Height of the cone (h) = 8 cm.
Radius of the cone (r) = 5 cm.
When lead shots, are dropped into the vessel, one-fourth of water flows out

According to Archimedes principle weight of the liquid displaced is equal to the weight of the object dropped
Volume of water displaced = Volume of lead shot.
=
Ãž =
=
= 100 âˆ´ The number of lead shots = 100.

# Question-15

A solid iron pole consists of a cylinder of height 220 cm and base diameter 24 cm, which is surmounted by another cylinder of height 60 cm and radius 8 cm. Find the mass of the pole, given that 1 cm3 of iron has approximately 8 g mass. (Use = 3.14)

Solution:
Height of the cylinder (h1) = 220 cm
Diameter of the cylinder = 24 cm
Radius of the cylinder (r1) = 12 cm
Height of another cylinder (h2) = 60 cm
Radius of the cylinder (r2) = 8 cm
Volume of the pole = volume of 1st cylinder + volume of 2nd cylinder
= Ï€ r12h1 + Ï€ r22h2

= Ï€ (r12h1 + r22h2) = 3.14(122
Â´ 220 + 82 Â´ 60)
= 3.14 Ã— (31680 + 3840)
= 3.14 Ã— 35520 = 111532.8 cm3
Mass of 1 cm3 of iron pole = 8 g

Mass of 111634.29 cm3 of iron pole = 111532.8 Ã— 8
= 892262.4 g
= 892.26 kg.

# Question-16

A solid consisting of a right circular cone of height 120 cm and radius 60 cm standing on a hemisphere of radius 60 cm is placed upright in a right circular cylinder full of water such that it touches the bottom. Find the volume of water left in the cylinder, if the radius of the cylinder is 60 cm and its height is 180 cm.

Solution:

Height of cylinder (h1)= 180 cm
Radius of cylinder (r1) = 60 cm
Height of cone (h2) = 180 - 60 = 120 cm
Radius of cone (r2) = 60 cm
Radius of hemisphere (r3) = 60 cm
But r1= r2 = r3 = r = 60 cm
Volume of water left in the cylinder = Volume of cylinder - (Volume of cone + volume of hemisphere)
= Ï€ r2h1 -(Ï€ r2h2 + Ï€ r3)
= Ï€ r2 (h1 - h2 - r)
= Ã— 60 Ã— 60 (180 - (120) - (60))
= Ã— 3600 Ã— (100) = 1131428.572 cm3

= 1.131 m3.

# Question-17

A spherical glass vessel has a cylindrical neck 8 cm long, 2 cm in diameter; the diameter of the spherical part is 8.5 cm. By measuring the amount of water it holds, a child finds its volume to be 345 cm3. Check whether she is correct, taking the above as the inside measurements, and Ï€ = 3.14.

Solution:
Height of cylinder (h) = 8 m
Diameter of cylinder = 2 m
Radius of cylinder (r1) = 1 m
Radius of sphere (r2) = = 4.25 cm
Volume of water = Volume of cylinder + Volume of sphere
= Ï€ r12h + Ï€ r23

= Ï€ (r12h + r23)
= 3.14 Ã— (1 Ã— 8 + (4.25)3)
= 3.14 Ã— (8 +(76.77))
= 346.51 cm3.
No, the amount of water calculated by the child is wrong.

# Question-18

A metallic sphere of radius 4.2 cm is melted and recast into the shape of a cylinder of radius 6 cm. Find the height of the cylinder.

Solution:
Radius of the sphere (r1) = 4.2 cm
Height of cylinder = h cm
Radius of cylinder (r2) = 6 cm
Volume of a cylinder = Volume of sphere                     Ï€ r22h = Ï€ r13

6 Ã— 6 Ã— h = Ã— (4.2)3

h = Ã— (4.2)3 Ã—
= 2.74 cm.

# Question-19

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.

Solution:
Let the radius of the metallic spheres be of radii 6 cm, 8 cm, and 10 cm be r1, r2, r3 respectively.
Let "R" be the radius of resulting sphere.
Volume of three spheres = Volume of single solid sphere
Ï€ r13 + Ï€ r23 + Ï€ r33 = Ï€ R3
Ï€ [r13 + r23 + r33 ] = Ï€ R3
[(6)3 + (8)3 + (10)3] = R3
216 + 512 + 1000 = R3

1728 = R3

12 cm = R

âˆ´ Radius of the resulting sphere = 12 cm.

# Question-20

A well of diameter 3 m is dug 14 m deep. The earth taken out of it has been spread evenly all around it in the shape of a circular ring of width 4 m to form an embankment. Find the height of the embankment.

Solution:
align="center">
The radius of the well is 2/2 = 1 m and height = 14 m.
The height of the embankment =

# Question-21

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

Solution:-
Volume of Cylinder =
Volume of Conical Portion of Ice-cream =
=
volume of hemispherical portion of ice-cream=
Volume of Ice-cream =

So number of Ice-cream= = 12

# Question-22

How many silver coins, 1.75 cm in diameter and of thickness 2 mm, must be melted to form a cuboid of dimensions 5.5 cm Ã— 10 cm Ã— 3.5 cm?

Solution:
Volume of a coin Ã— Number of coins = Volume of a cuboid
Diameter of a coin = 1.75 cm
Radius of a coin = = 0.375 cm
Thickness of a coin (h) = 2 mm = 0.2 cm
Dimensions of a cuboid = 5.5 cm Ã— 10 cm Ã— 3.5 cm
Volume of a coin Ã— Number of coins = Volume of a cuboid Ï€ r12h Ã— Number of coins = 5.5 cm Ã— 10 cm Ã— 3.5 cm Ï€ Ã— (0.375)2 Ã— 0.2 Ã— Number of coins = 192.5
Number of coins = 192.5 Ã—
= 400 coins.

# Question-23

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

Solution:
Heap of the bucket (h) = 32 cm
Radius of the bucket (R) = 18 cm
Height of the cone (h) = 24 cm
Volume of the cylindrical bucket = volume of the cone Ï€ R2H = â‡’ Ï€ (18)(18) 32 = (24) â‡’ = r2 â‡’ 1296 = r2 â‡’ r = 36 cm
Slant height of the cone =
=
=
=
= 43.72 cm

The radius of the cone = 36 cm
Slant height of the cone = 43.27 cm.

# Question-24

A farmer connects a pipe of internal diameter 20 cm from a canal into a cylindrical tank in her field, which is 10 m in diameter and 2 m deep. If water flows through the pipe at the rate of 3 km/h, in how much time will the tank be filled?

Solution:
Diameter of the cylindrical tank = 10 m
Radius of the cylindrical tank (R) = 5 m
Height of the cylindrical tank (H) = 2 m
Volume of cylindrical tank = Ï€ r2H
= Ï€ Ã— 52 Ã— 2
= 50 Ï€ m3

Internal Diameter of the pipe (d) = 20 cm
Internal Radius of the pipe (r) = 10 cm = m

Rate of flow of water = 3 km /hr = 3000 m/ hr

Let d pipe take t hours to fill up the tank

So, the volume of water that flows in t hours from the pipe = Area of cross section of pipe Ã— speed Ã— time
= Ï€  = 30 Ï€ t m3

âˆ´ Volume of water that flows from the pipe into the tank in "t" hours = volume of the tank â‡’ 30 Ï€ t = 50 Ï€ â‡’ 30 t = 50
t =
=
= 1 hr 40 min
= 100 minutes [âˆ´ 1 hour = 60 minutes].

# Question-25

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 4 cm and 2 cm. Find the capacity of the glass.

Solution:

Height of the frustum (h) = 14 cm
Radius of the circular ends R = 2 cm, r = 1 cm
Volume of the frustum = h (R2 + r2+ R
Ã— r)
= Ã— 14 (22 + 12 + 2(1))
= 102.67 cm3.

# Question-26

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

Solution:
Perimeter with R as radius = 2R = 18

Ãž R = 2.86 cm
Perimeter with r as radius = 2r = 6
r = 0.95 cm
Slant height of the frustum (l) = 4 cm
Curved surface area of the frustum = l(R + r)
=
= 47.90 cm2.

# Question-27

A fez, the cap used by the Turks, is shaped like the frustum of a cone (see Fig.). 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, find the area of material used for making it.

Solution:

In the problem, r1 = 4 cm, r2 = 10 cm
Slant height (l) = 15 cm
Curved surface area of the frustum = l(r1 + r2)
= (15)(4 +10)
= (15)(14)
= 660 cm2
Area of the top of the cap = r12 = (4)2 = cm2

Thus total area of the material used for making the cap = Curved surface area of the frustum + Area of the top of the cap
= 660 cm2 + cm2
= 710.29 cm2.

# Question-28

A container, opened from the top and made up of a metal sheet, is in the form of a frustum of a cone of height 16 cm with radii of its lower and upper ends as 8 cm and 20 cm, respectively. Find the cost of the milk which can completely fill the container, at the rate of 20 per litre. Also find the cost of metal sheet used to make the container, if it costs 8 per 100 cm2. (Take = 3.14)

Solution:

OO' = h = 16 cm
OM = r1 = 20 cm
O'N = r2 = 8 cm = OK
KN = h = 16 cm
KM = OM - OK = 20 - 8
= 12 cm
l2 = KN2 + KM2

= 256 + 144
= 400
l = 20 cm

Curved surface area of frustum =l(r1 + r2)
= 3.14 Ã— 20 (20 +8)
= 1758.4 cm2
Area of the base =
Ï€r2 = 3.14 Ã— (8)2 = 200.96
Total tin required = C.S.A + Area of the base
= 1758.4 cm2 + 200.96
= 1959.36
Cost of tin =
156.75.
Volume of the frustum = (r12 + r1r2 + r22)
= 16 Ã— ((20)2 + 20(8) + (8)2)
= 10459.43 cm3

Cost of the milk which can completely fill the container = 1046 Ã— 20
=
`209.20.

# Question-29

A metallic right circular cone 20 cm high and whose vertical angle is 60Â° 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 cm, find the length of the wire.

Solution:

Let ABC be the metallic cone, DECB is the required frustum
Let the two radii of the frustum be
DO' = r2 and BO = r1
From Î” ADO' and Î” ABO
r2 = h1(tan 30
Â°) = 10 Ã— =
r1 = (h1 + h2) tan 30Â°
= 20 tan 30Â° =
Volume of the frustum DECB = (r12 + r1r2 + r22)
=Ã— ()
= Ã— =

Let l be the length of the wire,
Diameter of the wire d =   cm,
âˆ´ Radius of the wire (R) =  cm

âˆ´ Volume of the frustum = Volume of the wire drawn from it
R2l

= 796444.44 cm or 7964.44 m

âˆ´ The length of the wire = 7964.44 m.