Bihar Board Class 10th Maths Chapter 13 Surface Areas and Volumes Solution

 

BSEB Bihar Board Class 10th Maths Solutions Chapter 13 Surface Areas and Volumes Ex 13.1

Question 1.
2 cubes each of volume 64 cm³ are joined end to end. Find the surface area of the resulting cuboid.
Solution:
Let the length of each edge of the cube be a cm.
Then, Volume = 64 cm³ So, a³ = 64 or a = 4
When two cubes of equal volumes (i.e equal edges) are joined end to end, we get a cuboid such that its
l = Length = 4 cm + 4 cm = 8 cm
b = Breadth = 4 cm
and, h = Height = 4 cm
∴Surface area of the cuboid = 2(lb + bh + hl)
= 2(8 x 4 + 4 x 4 + 4 x 8) cm²
= 2(32 + 16 + 32) cm²
= (2 x 80) cm² = 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:
Here, the radius r of hemi – sphere 142 = 7
the height h of cylinder = (13 – 7) cm = 6 cm.
Clearly, radius of the base of cylindrical part is also r = 7cm.

Surface area of the vessel
= Curved surface area of the cylindrical part + Curved surface area of hemispherical part
= (2πh + 2π²) = 2π(h + r)
= 2 x 227 x 7 x 13 cm²
= 572 cm².

Question 3.
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 of the toy.
Solution:
We have : VO’ = 15.5 cm and OA = 00′ = 3.5 cm.
Let r be the radius of the base of cone and h be the height of conical part of the toy. Then, r = OA = 3.5 cm,

h = VO = VO’ – OO’
= (15.5 – 3.5) cm = 12 cm
Also, radius of the hemisphere = OA = r = 3.5 cm.
Total surface area of the toy = Curved surface area of cone + Curved surface area of hemisphere
= πrl + 2πr² = πr(l + 2 r), where l – Slant height of the cone.
Now, l = OA2+OV2−−−−−−−−−−√ = (3.5)2+122−−−−−−−−−−√
= 12.25+144−−−−−−−−−√ = 156.25−−−−−√ = 12.5 cm
Therefore, surface area of the toy = πr(l + 2r)
= 227 x 3.5 x (12.5 + 2 x 3.5) cm²
= 227 x 3.5 x 19.5 cm² = 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:
The greatest diameter that a hemisphere can have = 7 cm.
Surface area of the solid after surmounting hemisphere = 6l² – TCR² + 2πR², where l = side of cubical block and R = radius of hemisphere

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:
Edge of the cube = l
Diameter of the hemsiphere = l
∴ Radius of the hemisphere = l2
∴ Area of the remaining solid after cutting out the hemispherical depression
= 6l² – π(l2)² + 2π(l2)
= 6l² + π(l2)²
= 6l² + π x l24
= l24(24 + π)

Question 6.
A medicine capsule is in the shape of a cylinder with two hemispheres stuck to each of its ends (See figure). The length of the entire capsule is 14 mm and the diameter of the capsule is 5 mm.

Find its surface area.
Solution:
Let r mm be the radius and h mm be the height of the cylinder. Then,

r = 52 mm = 2.5 mm
and, h = (14 – 2 x 52) mm
= (14 – 5) mm = 9 mm
Also, the radius of hemisphere r = 52 cm
Now, surface area of the capsule = Curved surface area of cylinder + Surface area of two hemispheres
= (2πrh + 2 x 2πr²)
= 2πr(h + 2 r)
= 2 x 227 x 52 x (9 + 2 x 52) mm²
= 227 x 5 x 14 mm² = 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 Rs 500 per m². (Note that the base of the tent will not be covered with canvas.)
Solution:
We have : Total canvas used = Curved surface area of cylinder + Curved surface area of cone

= (2πrh + πrl)
= πr(2h + l)
= 227 x 2 x ( 2 x 2.1 + 2.8) m²
= 227 x 2 x 7 m²
= 44 m²
Now, cost of 1 m² canvas for the tent = Rs 500
So, cost of 44 m² canvas for the tent = Rs 44 x 500 = Rs 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:
Radius of the cylinder = 142 cm – 0.7 cm
Height of the cylinder = 2.4 cm
Radius of the cone = 0.7 cm
Height of the cone = 2.4 cm
Slant height of the cone

Surface area of the remaining solid = Curved surface area of cylinder + Curved surface area of the cone + Area of upper circular base of cylinder
= 2πrh + πrl + πr2 = πr(2h + l + r)
= 227 x 0.7 x (2 x 2.4 + 2.5 + 0.7) cm²
= 22 x 0.1 x (4.8 + 2.5 + 0.7) cm²
= 2.2 x 8.0 cm²
= 17.6 cm²
= 18 cm²

Question 9.
A wooden article was made by scooping out a hemisphere from each end of a solid cylinder, as shown in figure. 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:
Surface area of the article when it is ready = Curved surface area of cylinder + 2 x Curved surface area of hemisphere

= 2 πrh + 2 x 2πr²
= 2πr(h + 2r), where r = 3.5 cm and h = 10 cm
= 2 x 227 x 3.5 x (10 + 2 x 3.5) cm²
= 2 x 227 x 3.5 x 17 cm² = 374 cm².

BSEB Bihar Board Class 10th Maths Solutions Chapter 13 Surface Areas and Volumes Ex 13.2

Question 1.
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 n.
Solution:
Volume of the solid = Volume of the cone + Volume of the hemisphere

Question 2.
Rachel, an engineering student was asked to make a model shaped like a cylinder with two cones attached at its two ends by using 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 the Rachel made. (Assume the outer and inner dimensions of the model to be nearly the same).
Solution:
Volume of the air contained in the model = Volume of the cylindrical portion of the model + Volume of its two conical ends

Question 3.
A gulab jamun, contains sugar syrup up to about 30% of its volume. Find approximately how much syrup would be found in 45 gulab jamuns, each shaped like a cylinder with two hemispherical ends with length 5 cm and diameter is 2.8 cm (see figure).

[Annual Paper (Delhi) 2008]
Solution:
Volume of the gulab jamun = Volume of the cylindrical portion + Volume of the two hemispherical ends

Volume of 45 gulab jamuns
= 45 x \(\frac { 22 }{ 7 }\) x 1.96 x \(\frac { 12.2 }{ 3 }\) cm³

Quantity of syrup in gulab jamuns = 30% of their volume
= \(\frac { 30 }{ 100 }\) x 45 x \(\frac { 22 }{ 7 }\) x 1.96 x \(\frac { 12.2 }{ 3 }\) cm³
= \(\frac{9 \times 11 \times 1.96 \times 12.2}{7}\) cm³
= 338.184 cm³ = 338 cm³ (approx.)

Question 4.
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 (see figure).

Solution:
Volume of wood in the entire stand = Volume of the cuboid – 4 x Volume of a conical depression
= lbh – 4 x \(\frac { 1 }{ 3 }\) πr²H, where l, b, h are dimensions of cuboid, r is radius of the cone and H is its depth
= 15 x 10 x 3.5 cm³ – 4 x \(\frac { 1 }{ 3 }\) x \(\frac { 22 }{ 7 }\) x 0.5 x 0.5 x 1.4 cm³
= (525 – 1.47) cm³ = 523.53 cm³.

Question 5.
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 conical vessel h = 8 cm.
Its radius r = 5 cm.

Radius of the lead shot = 0.5 cm
Volume of one spherical lead shot

Question 6.
A solid iron pole consists of a cylindrical 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 π = 3.14)
Solution:
Volume of the solid iron pole = Volume of the base cylindrical portion + Volume of the other cylindrical portion
= πr1² + πr2² h2
= (3.14 x (12)² x 220 + 3.14 x (8)² x 60) cm³
= (3.14 x 144 x 220 + 3.14 x 64 x 60) cm³
= (99475.2 + 12057.6) cm³
= 111532.8 cm³
Hence, the mass of the pole

= (111532.8 x 8) grams
= \(\left(\frac{111532.8 \times 8}{1000}\right)\) kg
= 892.26 kg.

Question 7.
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:
Volume of the cylinder = πr²h

Volume of water left in the cylinder = Volume of the cylinder – Volume of the solid

Question 8.
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 n = 3.14.
Solution:
Volume of spherical vessel = Volume of cylindrical part + Volume of spherical part

= π(1)²(8) + \(\frac { 4 }{ 3 }\) π(4.25)³
= 3.14[8 + \(\frac { 4 }{ 3 }\) x 76.765625] cm³
= 3.14[8 + 102.353] cm³
= 3.14 x 110.353 = 346.51 cm³
∴ Here answer is incorrect. Correct answer is 346.51 cm³.

BSEB Bihar Board Class 10th Maths Solutions Chapter 13 Surface Areas and Volumes Ex 13.3

Question 1.
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:
Volume of the sphere = 43 πr³
= 43 x π x (4.2)² cm³
If h is the height of a cylinder of radius 6 cm. Then, its volume
= n(6)²h cm³ = 36πh cm³
Since the volume of metal in the form of sphere and cylinder remains the same, we have
36πh = 43 x π x 4.2 x 4.2 x 4.2 3
or h = 136 x 43 x 4.2 x 4.2 x 42
or h = 2.744
Hence, the height of the cylinder is 2.744 cm.

Question 2.
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:
Sum of the volumes of 3 given spheres = 43π(r1³+r2³+r3³)
= 43π(6³ + 8³ + 10³) cm³
= 43π(216 + 512 + 1000) cm³ 3
= 43π(1728) cm³
Let R be the radius of the new spheres whose volume is the sum of the volumes of the three given spheres.

∴ 43πr³ = 43π(1728)
or R³ = 1728
or R³ = (12)³
or R = 12
Hence, the radius of the resulting sphere is 12 cm.

Question 3.
A 20 m deep well with diameter 7 m is dug and the earth from digging is evenly spread out to form a platform 22 m by 14 m. Find the height of the platform.
Solution:
Let h m be the required height of the platform.
The shape of the platform will be like the shape of a cuboid of dimensions 22 m x 14 m x h m.
The volume of the platform will be equal to the volume of the earth dug out from the well.
Now, the volume of the earth
= Volume of the cylindrical well
= πr²h = 227 x 12.25 x 20 m³ [∵ r = 72 m = 3.5 m]
= 770 m³
Also, the volume of the platform = 22 x 14 x h m³
But volume of the platform – Volume of the well
i.e., 22 x 14 x h = 770
h = 77022×14
Height of the platform = 2.5 m.

Question 4.
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:
Let h be the required height of the embankment.

The shape of the embankment will be like the shape of a cylindrical shell of internal radius 1.5 m and external radius (4 + 1.5) m = 5.5 m (see figure).
The volume of the embankment will be equal to the volume of earth dug out from the well.
Now, the volume of the earth
= Volume of the cylindrical well
= π x (1.5)² x 14 m³ = 31.5π m³
Also, the volume of the embankment
= π(5.5² – 1.52)h m³
= π(5.5 + 1.5 x 5.5 – 1.5)h m³
= π x 7 x 4h m³
= 28πh m³
Hence, we have :
28πh = 31.5π
So h = 31.528 = 1.125
Hence, the required height of the embankment is 1.125 m.

Question 5.
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 hemispherical shape on the top. Find the number of such cones which can be filled with ice cream. [Annual Paper (Delhi) 2008]
Solution:
Volume of the cylinder

Volume of a cone having hemispherical shape on the top
= 13πr²h + 23πr³ = 13 πr²(h + 2r)
= 13π(62)² (12 + 2 x 62) cm³
= 13π x 3² x 18 cm³
Let the number of cone that can be filled with ice cream
Then, 13π x 32 x 18 x n = π x 6² x 15
n = π×6×6×15π×3×3×18 x 3 = 10

Question 6.
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 x 10 cm x 3.5 cm?
Solution:
The shape of the coin will be like the shape of a cylinder of radius 1.752 cm
0.875 cm and of height 2 mm = 210 cm = 0.2 cm.
Its volume
= πr²h
= 227 x 0.875 x 0.875 x 0.2 cm³
= 0.48125 cm³
Volume of the cuboid
= 5.5 x 10 x 3.5 cm³ = 192.5 cm³
Number of coins required to form the cuboid
=  Volume of the cuboid  Volume of a coin 
= 192.50.48125
= 400
Volume of a coin 0.48125
Hence, 400 coins must be melted to form a cuboid of required dimensions.

Question 7.
A cylindrical bucket, 32 cm high and with radius of base 18 cm, is filled with sand. The 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:
Volume of the sand = Volume of the cylindrical bucket
= πr²h = π x 18 x 18 x 32 cm³
Volume of the conical heap

= 13πr²h, where r = ?, h = 24 cm
= 13πr2 x 24 cm³ = 8πr²
The volume of the conical heap will be equal to that of sand.
∴ 8πr² = π x 18 x 18 x 32
or r² = 18 x 18 x 4 = 18² x 2²
or r = 18 x 2 = 36
Here, slant height l = r2+h2−−−−−−√
So, l = 362+242−−−−−−−−√
= 12×12(3×3+2×2)−−−−−−−−−−−−−−−−−−√
= 129+4−−−−√ = 1213−−√
Hence, the radius of the conical heap is 36 cm and its slant height is 1213−−√ cm.

Question 8.
Water in a canal 6 m wide and 1.5 m deep, is flowing with a speed of 10 km/h. How much area will it irrigate in 30 minutes, if 8 cm of standing water is needed ?
Solution:
Width of the canal = 6 m
Depth of the canal = 1.5 m
Length of water column per hour = 10 km
Length of water column in 30 minutes or 12 hours
= 12 x 10 km = 5000 m
Volume of water flown in 30 minutes
= 1.5 x 6 x 5000 m³ = 45000 m³
Now, 8 cm = 8100 m
i.e., 0.08 m standing water is desired.
∴ Area irrigated in 30 minutes
=  Volume  Height 
= 450000.08
= 562500 m²
or 56.25 hectares

Question 9.
A farmer connects a pipe of internal diameter 20 cm from a canal into a cylindrical tank in his 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 pipe = 20 cm
So, Radius of the pipe = 10 cm
Length of water column per hour = 3 km
= 3 x 1000 x 100 cm
= 300000 cm
Volume of water flown in one hour = π x 100 x 300000 cm³
Tank to be filled = Volume of cylinder (with r = 5 m = 500 cm and h = 2 m = 200 cm)
= π x 500 x 500 x 200 cm³
Time required to fill the tank

BSEB Bihar Board Class 10th Maths Solutions Chapter 13 Surface Areas and Volumes Ex 13.4

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 4 cm and 2 cm. Find the capacity of the glass. S61. Capacity of the glass.
Solution:
V = \(\frac{\pi \times h}{3}\) x (R² + r² + Rr)
Here, R = 2 cm, r = 1 cm and h = 14 cm. 22
∴ V = \(\frac{\frac{22}{7} \times 14}{3}\) x (2² + 1² + 2 x 1)cm³
= \(\frac { 44 }{ 3 }\) x (4 + 1 + 2) cm³
= (\(\frac { 44 }{ 3 }\) x 7 )cm³ = \(\frac { 308 }{ 3 }\) cm³ = 102\(\frac { 2 }{ 3 }\) cm³

Question 2.
The slant height of a frustum of a cone is 4 cm and the perimeters (circumferences) of its circular ends are 18 cm and 6 cm. Find the curved surface area of the frustum.
Solution:
Slant height l = 4 cm (Given)

2πr1 = 6 gives πr1 = 3
and, 2πr2 = 18 gives πr2 = 9
Curved surface area of the frustum
= (πr1 + πr2)l
= (3 + 9) x 4 cm²
= 12 x 4 cm²
= 48 cm².

Question 3.
A fez, the cap used by the Turks, is shaped like the frustum of a cone (see figure). If its radius oh 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:
Here, R = 10 cm, r = 4 cm and l = 15 cm.
Area of the material used for making the fez = Curved surface area of frustum + The surface area of top circular section

Question 4.
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 milk which can completely fill the container, at the rate of Rs 20 per litre. Also, find the cost of metal sheet used to make the container, if it costs Rs 8 per 100 cm². (Take π = 3.14)
Solution:
Here, R – 20 cm, r = 8 cm and h = 16 cm.
Capacity of the container = Volume of the frustum

Cost of milk at the rate of Rs 20 per litre
= Rs (20 x \(\frac { 10449.92 }{ 1000 }\) = Rs 208.99 ≈ Rs 209
To find the slant height

l = \(\sqrt{16^{2}+12^{2}}\) cm
= \(\sqrt{256+144}\) cm
= \(\sqrt{400}\) cm = 20 cm
Curved surface area
= π (R + r)l
= \(\frac { 22 }{ 7 }\) x (20 + 8) x 20 cm² = \(\frac { 22 }{ 7 }\) x 28 x 20 cm²
= 1758.4 cm²
and area of the bottom
= πr² = \(\frac { 22 }{ 7 }\) x (8)² cm² = \(\frac { 22 }{ 7 }\) x 64 cm² = 200.96 cm²
∴ Total area of metal required
= 1758.4 cm² + 200.96 cm²
= 1959.36 cm²
Cost of metal sheet used to manufacture the container at the rate of Rs 8 per 100 cm²
= Rs \(\frac { 8 }{ 1000 }\) x 1959.36) = Rs 156.75

Question 5.
A 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 \(\frac { 1 }{ 16 }\) cm, find the length of the wire.
Solution:
Let ABC be the metallic cone and DECB be the equired frustum.
Let the two radii of the frustum be DO’ = r2 and BO = r1
From the ∆s ADO’ and ABO,
r2 = h1 tan 30°
= 10 x \(\frac{1}{\sqrt{3}}\)

r1 = (h1 + h2) tan 30°
= 20 x \(\frac{1}{\sqrt{3}}\)
Volume of the frustum DBCE

Volume of the wire of length l and diameter d
= π(\(\frac { d }{ 2 }\))² x l = \(\frac{\pi d^{2}}{4}\) x l [V = πr²h]
∴ Volume of the frustum = Volume of the wire drawn from it

BSEB Bihar Board Class 10th Maths Solutions Chapter 13 Surface Areas and Volumes Ex 13.5

Question 1.
A copper wire 3 mm in diameter is wound about a cylinder whose length is 12 cm, and diameter 10 cm, so as to cover the curved surface of the cylinder. Find the length and mass of the wire, assuming the density of copper to be 8.88 g per cm³.
Solution:
Curved surface area of the cylinder
= 2πrh = 2 x 3.14 x 5 x 12 cm²
= 120 x 3.14 cm²
Diameter i.e., width of the wire
= 3 mm = \(\frac { 3 }{ 10 }\) cm
Let length of the wire be l cm.
So, area of the wire = l x \(\frac { 3 }{ 10 }\) cm²
So, l x \(\frac { 3 }{ 10 }\) = 120 x 3.14
or l = \(\frac{120 \times 3.14 \times 10}{3}\) cm = 1256 cm
Now volume of wire = πr²h
= 3.14 x \(\frac { 3 }{ 20 }\) x \(\frac { 3 }{ 20 }\) x 1256 cm³
So, weight of wire = 3.14 x \(\frac { 9 }{ 400 }\) x 1256 x 8.88 g
= 789.98 g (approx.)

Question 2.
A right triangle, whose sides are 3 cm and 4 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 n as found appropriate.)
Solution:
In the right angled ∆ ABC,

So, the radius of the base for both the cones = \(\frac { 12 }{ 5 }\) cm.

∴ Volume of the cone ABB’
= \(\frac { 1 }{ 3 }\)π.BD².AD
= \(\frac { 1 }{ 3 }\) x \(\frac { 22 }{ 7 }\) x \(\frac { 144 }{ 25 }\) x \(\frac { 9 }{ 5 }\)cm³ = \(\frac { 9504 }{ 875 }\) cm³
Volume of the cone CBB’
= \(\frac { 1 }{ 3 }\)π.BD².CD
= \(\frac { 1 }{ 3 }\) x \(\frac { 22 }{ 7 }\) x \(\frac { 144 }{ 25 }\) x \(\frac { 16 }{ 5 }\)cm³ = \(\frac { 16896 }{ 875 }\) cm³
∴Volume of the double cone

Curved surface area of the cone ABB’
= π.BD.CD
= \(\frac { 22 }{ 7 }\) x \(\frac { 12 }{ 5 }\) x 3 cm² = \(\frac { 792 }{ 35 }\) cm²
Curved surface area of the cone CBB’
= π.BD.CD
= \(\frac { 22 }{ 7 }\) x \(\frac { 12 }{ 5 }\) x 4 cm² = \(\frac { 1056 }{ 35 }\) cm²
∴ Surface area of the double cone =
= \(\left(\frac{792}{35}+\frac{1056}{35}\right)\) cm²
= \(\frac { 1848 }{ 35 }\) cm²
= 52.8 cm²

Question 3.
A cistern, internally measuring 150 cm x 120 cm x 110 cm, has 129600 cm³ of water in it. Porous bricks are placed in the water until the cistern is full of the brim. Each brick absorbs one-seventeenth of its own volume of water. How many bricks can be put in without overflowing the water, each brick being 22.5 cm x 7.5 cm x 6.5 cm?
Solution:
Volume of the cistern = 150 cm x 120 cm x 110 cm = 1980000 cm³
Let n bricks can be put in the cistern without over flowing water

Then, volume of n bricks = n x (22.5 x 7.5 x 6.5) cm³ = 1096.875 x n cm³
Volume of water absorbed by n bricks
= \(\frac { 1 }{ 17 }\) x 1096.875 x n cm³
= 64.52 x n cm³
∴1096.875 x n + 129600 – 64.52 x n

Question 4.
In one fortnight of a given month, there was a rainfall of 10 cm in a river valley. If the area of the valley is 97280 km2, show that the total rainfall was approximately equivalent to the addition to the normal water of three rivers each 1072 km long, 75 m wide and 3 m deep.
Solution:
Volume of water in the 3 rivers

So, the addition of water in the rivers is not equivalent to the rainfall.

Question 5.
An oil funnel 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, find the area of the tin sheet required to make the funnel (see figure).

Solution:
The diameter of the cylindrical portion is 8 cm.

Radius of top portion = 9 cm
Here, r = 4 cm, R = 9 cm, h = 12 cm
KM = (9-4) cm = 5 cm
∴ NM² = l² = KM² + KN2
= 5² + 12²
= 25 + 144 = 169
So, l = \(\sqrt{169}\) cm = 13 cm
Now, the tin required will be equal to the total surface area of the figure.
Area of the tin required = Curved surface area of the cylinder + Curved surface area of the frustum

Question 6.
Derive the formula for the curved surface area and total surface area of the frustum of a cone, given to you in Section 13.5, using the symbols as explained.
Solution:
Curved surface area of the frustum RPQS = Curved surface area of the right circular cpne OPQ – Curved surface area of the right circular cone ORS
= πr1l1 – πr2l2 … (1)

Now, since ∆ OC1Q and ∆ OC2S are similar, therefore

Putting the value of l1 in terms of r1, r2 and l2 in equation (1), we have :
Curved surface area of the frustum

∴Total surface area of the frustum
= Curved surface area of the frustum + πr1² + πr2²
= π(r1 + r2)l + πr1² + πr2²
= π[(r1 + r2)l + r1² + r2²]

Question 7.
Dev ive the formula for the volume of the frustum of a cone, given to you in Section 13.5, using the symbols as explained.
Solution:
Volume of the frustum RPQS = Volume of the right circular cone OPQ – Volume of the right circular cone ORS
= \(\frac { 1 }{ 3 }\)πr1²h1 – \(\frac { 1 }{ 3 }\)πr2²h2
= \(\frac { 1 }{ 3 }\)π(r1²h1 – r2²h2) … (1)
Now, since ∆ OC1Q and ∆ OC2S are similar, therefore

or h1 = \(\left(\frac{r_{1}}{r_{2}}-1\right)\) h2 = \(\left(\frac{r_{1}-r_{2}}{r_{2}}\right)\)h2 … (2)
Putting the value of in terms of rr r2 and h2 in eqn. (1), we have :
Volume of the frustum