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Surface Areas and Volumes - NCERT Questions
Q 1.

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

SOLUTION:

Soln. : Volume of each cube = 64 cm3
Let the edge of each cube = 2
x3 = 64 cm3
x = 4 cm
Now, Length of the resulting cuboid l = 2x cm = 8 cm
Breadth of the resulting cuboid b = x cm = 4 cm
Height of the resulting cuboid h = x cm = 4 cm
∴ Surface area of the cuboid = 2 (lb + bh + hl)
= 2 [(8 × 4) + (4 × 4) + (4 × 8)] cm2
= 2 [32 + 16 + 32] cm2 = 2 [80] cm2 = 160 cm2.

Q 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.
 images

SOLUTION:

Soln. : For hemispherical
part, radius (r) ncert
∴ Curved surface area = 2πr2
ncert
Total height of vessel = 13 cm
∴ Height of cylinder = (13 - 7)cm = 6 cm and radius(r) = 7 cm
∴ Curved surface area of cylinder = 2πrh ncert
∴ Inner surface area of vessel = (308 + 264)cm2
= 572 cm2

Q 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 the total surface area of the toy.
ncert

SOLUTION:

Soln. : Let r and h be the radius of cone, hemisphere and height of cone
and h = (15.5 ncert 3.5) cm
= 12.0 cm
Also l2 = h2 + r2 = 122 + (3.5)2
= 156.25
l = 12.5 cm
Curved surface area of the conical part = πrl
Curved surface area of the hemispherical part = 2πr2
Total surface area of the toy = πrl + 2πr2 = πr (l + 2r)
ncert
= 11 × (12.5 + 7) cm2 = 11 × 19.5 cm2 = 214.5 cm2

Q 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.
ncert

SOLUTION:

Soln. : Let side of the block, l = 7 cm
∴ The greatest diameter of the hemisphere = 7 cm
Surface area of the solid
= [Total surface area of the cubical block] + [C.S.A. of the hemisphere]
- [Base area of the hemisphere]
= (6 × l2) + 2πr2 ncert πr2
[where l = 7 cm and r = ncert cm]
ncert
ncert
= 332.5 cm2

Q 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.
ncert

SOLUTION:

Soln. : Let l be the side of the cube.
∴ The greatest diameter of the hemisphere = l
⇒ Radius of the hemi-sphere
∴ Surface area of hemisphere = 2πr2
ncert
Base area of the hemisphere
Surface area of the cube
∴ Surface area of the remaining solid
ncert
ncert

Q 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.
ncert

SOLUTION:

Soln. : Diameter of the hemispherical part
= 2.5 mm
Curved surface area of one hemisphere part = 2πR2
∴ Surface area of both hemispherical parts
ncert
ncert
Entire length of capsule = 14mm
∴ Length of cylindrical part
= (14 - 2 × 2.5)mm = 9mm
∴ Area of cylindrical part = 2πrh
ncert
∴ Total surface area
ncert
ncert
= 220 mm2

Q 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 ₹ 500 per m2. (Note that the base of the tent will not be covered with canvas.)
ncert

SOLUTION:

Soln. : For cylindrical part :
Radius (r) and height (h) = 2.1 m
∴ Curved surface area
For conical part :
Slant height (l) = 2.8 m and base radius (r) = 2 m
∴ Curved surface area
∴ Total surface area
= [Curved surface area of the cylindrical part] + [Curved surface area of conical part]
ncert
ncert
Cost of the canvas used :
Cost of 1 m2 of canvas = ₹ 500
∴ Cost of 44 m2 of canvas = ₹ 500 × 44 = ₹ 22000.

Q 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 cm3.
ncert

SOLUTION:

Soln. : For cylindrical part:
Height = 2.4 cm and diameter = 1.4 cm
⇒ Radius (r) = 0.7 cm
∴ Total surface area of the cylindrical part
= 2πrh + 2πr2 = 2πr [h + r]
ncert
ncert
For conical part :
Base radius (r) = 0.7 cm and height (h) = 2.4 cm
∴ Slant height (l) ncert
ncert
∴ Curved surface area of the conical part
ncert
Base area of the conical part
ncert
Total surface area of the remaining solid
= [(Total surface area of cylindrical part)
+ (Curved surface area of conical part)
- (Base area of the conical part)]
ncert
Hence, total surface area to the nearest cm2 is 18cm2.

Q 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.
ncert

SOLUTION:

Soln. : Radius of the cylinder (r) = 3.5 cm
Height of the cylinder (h) = 10 cm
∴ Curved surface area = 2πrh
ncert
Curved surface area of a hemisphere = 2πr2
∴ Curved surface area of both hemispheres
ncert
Total surface area of the remaining solid
= (220 + 154) cm2 = 374 cm2.

Q 10.

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

SOLUTION:

Soln. : Here, r = l cm and h = l cm.
Volume of the conical part
and volume of the hemi-spherical part
∴ Volume of the solid shape

Q 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.)
ncert

SOLUTION:

Soln. : Here, diameter = 3 cm ⇒ Radius (r)
Total height = 12 cm
Height of a cone (h1) = 2 cm
∴ Height of both cones = 2 2 = 4 cm
⇒ Height of the cylinder (h2) = (12 - 4) cm = 8 cm.
Now, volume of the cylindrical part = πr2h2
Volume of both conical parts
∴ Volume of the whole model
ncert
ncert
ncert

Q 12.

A gulab jamun, contains sugar syrup up to about 30% of its volume. Find approximately how much syrup would be found in 45 gulab jamun, each shaped like a cylinder with two hemispherical ends with length 5 cm and diameter 2.5 cm (see figure).
ncert

SOLUTION:

Soln. : Since, a gulab jamun is like a cylinder with hemispherical ends.
Total height of the gulab jamun = 5 cm.
Diameter = 2.8 cm ⇒ Radius = 1.4 cm
∴ Length (height) of the cylindrical part
= 5 cm - (1.4 + 1.4) cm = 5 cm - 2.8 cm = 2.2 cm
ncert
Now, volume of the cylindrical part = πr2h
and volume of both the hemispherical ends
∴ Volume of a gulab jamun
ncert
ncert
ncert
ncert
Volume of 45 gulab jamuns
ncert
Since, the quantity of syrup in gulab jamuns = 30% of [volume] = 30% of ncert
ncert
= 338 cm2 (approx.)

Q 13.

A pen stand made up 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).
ncert

SOLUTION:

Soln. : Dimensions of the cuboid are 15 cm, 10 cm and 3.5 cm.
∴ Volume of the cuboid
= 525 cm3
Since each depression is conical in shape with base radius (r) = 0.5 cm and depth (h) = 1.4 cm,
∴ Volume of each depression
ncert
Since there are 4 depressions,
∴ Total volume of 4 depressions
Now, volume of the wood in entire stand
= [Volume of the wooden cuboid] - [Volume of 4 depressions]
ncert
= 523.53 cm2

Q 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-forth of the water flows out. Find the number of lead shots dropped in the vessel.
ncert

SOLUTION:

Soln. : Height of the conical vessel (h) = 8 cm
Base radius (r) = 5 cm
Volume of water in conical vessel
ncert
Now, Total volume of lead shots = ncert of [Volume of water in the cone]
ncert
Since, radius of spherical lead shot (r) = 0.5 cm
∴ Volume of 1 lead shot ncert
ncert
∴ Number of lead shots
=ncert
= 100
Thus, the required number of lead shots = 100.

Q 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 π = 3.14)
ncert

SOLUTION:

Soln. : Height of the big cylinder (h) = 220 cm
Base radius (r)
∴ Volume of the big cylinder = πr2h = π(12)2 × 220 cm3
Also, height of smaller cylinder (h1) = 60 cm
Base radius (r1) = 8 cm
∴ Volume of the smaller cylinder ncert
= π(8)2 × 60 cm3
∴ Volume of iron pole
= [Volume of big cylinder]
+ [Volume of the smaller cylinder]
= π × 220 × 122 + π × 60 × 82 cm3
= 3.14[220 × 12 × 12 + 60 × 8 × 8]cm3
ncert
ncert
Mass of iron
ncert = 892.2624 kg = 892.26 kg.

Q 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.
ncert

SOLUTION:

Soln. : Height of the conical part = 120 cm.
Base radius of the conical part = 60 cm.
∴ Volume of the conical part
Radius of the hemispherical part = 60 cm.
∴ Volume of the hemispherical part
∴ Volume of the solid = [Volume of conical part] + [Volume of hemispherical part]
ncert
ncert
ncert
Volume of the cylinder = πr2h
ncert ncert
⇒ Volume of water in the cylinder
ncert
∴ Volume of the water left in the cylinder
ncert
= 1131428.57142 cm3
= 1.13142857142 m3 = 1.131 m3 (approx).

Q 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 π = 3.14.

SOLUTION:

Soln. : Volume of the cylindrical part = πr2h
= 3.14 × 12 × 8 cm3

ncert
Radius of spherical part ncert
Volume of the spherical part ncert
ncert
Total volume of the glass-vessel
ncert
ncert
ncert
ncert
= 346.51 cm3 (approx.)
⇒ Volume of water in the vessel = 346.51 cm3
Since, the child finds the volume as 345 cm3
∴ The child's answer is not correct
The correct answer is 346.51 cm3.

Q 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:

Soln. : Radius of the sphere (r1) = 4.2 cm
∴ Volume of the sphere
ncert
Radius of the cylinder (r2) = 6 cm
Let h be the height of the cylinder
∴ Volume of the cylinder = πr2h
ncert
Since, Volume of the metallic sphere = Volume of the cylinder.
ncert
ncert
ncert
Hence, height of the cylinder = 2.744 cm

Q 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:

Soln. : Radii of the given spheres are:
r1 = 6 cm, r2 = 8 cm, r3 = 10 cm
⇒ Volume of the given spheres are:
ncert
∴ Total volume of the given spheres = V1 + V2 + V3
ncert
ncert
ncert
ncert
Let the radius of the new big sphere be R.
∴ Volume of the new sphere
ncert
Since, the two volume must be equal.
ncert
R3 = 1728 R = 12 cm
Thus, the required radius of the resulting sphere = 12 cm.

Q 20.

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:

Soln. : Diameter of the cylindrical well = 7 m
⇒ Radius of the cylinder (r)
Depth of the well (h) = 20 m
∴ Volume
= 22 × 7 × 5m3
⇒ Volume of the earth taken out = 22 × 7 × 5 m3
Now this earth is spread out to form a cuboidal platform having length = 22 m, breadth = 14 m
Let h be the height of the platform.
∴ Volume of the platform = 22 × 14 × h m3
∴ 22 × 14 × h = 22 × 7 × 5

Thus, the required height of the platform is 2.5 m.

Q 21.

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 shape of a circular ring of width 4 m to form an embankment. Find the height of the embankment.

SOLUTION:

Soln. : Diameter of cylindrical well (d) = 3 m
⇒ Radius of the cylindrical well
Depth of the well (h) = 14 m
∴ Volume

Let the height of the embankment = H metre.
Internal radius of the embankment (r) = 1.5 m.
External radius of the embankment R = (4 + 1.5) m
= 5.5 m.
∴ Volume of the embankment
= πR2H - πr2H = πH [R2 - r2] = πH (R + r) (R - r)
ncert
Since, Volume of the embankment = Volume of the cylindrical well
ncert

Thus, the required height of the embankment = 1.125 m.

Q 22.

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.
ncert

SOLUTION:

Soln. : For the circular cylinder:
Diameter = 12 cm
⇒ Radius (r) and height (h) = 15 cm
∴ Volume of total ice cream
ncert
For conical and hemispherical part of ice-cream:
Diameter = 6 cm ⇒ radius (R) = 3 cm
Height of conical part (H) = 12 cm
ncert
Volume = (Volume of the conical part) + (Volume of the hemispherical part)
ncert
ncert
Let number of ice-cream cones required to fill the total ice cream = n.
ncert
ncert
Thus, the required number of cones is 10.

Q 23.

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 × 10 cm × 3.5 cm?
ncert

SOLUTION:

Soln. : For a circular coin:
Diameter = 1.75 cm
⇒ Radius (r)
Thickness (h)
∴ Volume = πr2h

For a cuboid:
Length (l) = 10 cm,
Breadth (b) = 5.5 cm
and Height (h) = 3.5 cm
∴ Volume =
Number of coins

Thus, the required number of coins = 400.

Q 24.

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:

Soln. : For the cylindrical bucket:
Radius (r) = 18 cm and height (h) = 32 cm
Volume = πr2h
⇒ Volume of the sand
For the conical heap:
Height (H) = 24 cm & let radius of the base be (R).
∴ Volume of conical heap

∵ Volume of the conical heap sand = Volume of the sand



Let 'l' be the slant height of the conical heap of the sand.

l2 = R2 + H2
l2 = 242 + 362
l2 = (12 × 2)2 + (12 × 3)2
l2 = 122[22 + 32]
l2 = 122 × 13

Thus, the required radius = 36 cm and slant height

Q 25.

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:

Soln. : Width of the canal = 6 m, Depth of the canal = 1.5 m
Length of the water column in 1 hr = 10 km
∴ Length of the water column in 30 minutes
∴ Volume of water flown in

Since the above amount (volume) of water is spread in the form of a cuboid of height as 8 cm Let the area of the cuboid = a
∴ Volume of the cuboid = Area × Height

Thus,

= 562500m2 = 56.25 hectares
Thus, the required area = 56.25 hectares.

Q 26.

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:

Soln. : Diameter of the pipe = 20 cm
⇒ Radius of the pipe (r) = cm = 10 cm
Since, the water flows through the pipe at 3 km/hr.
∴ Length of water column per hour(h) = 3 km
= 3 × 1000 m = 3000 × 100 cm = 300000 cm.
∴ Volume of water = πr2h = π × 102 × 300000 cm3
= π × 30000000 cm3
Now, for the cylindrical tank,
Diameter = 10 m ⇒ Radius (R) = = 5 × 100 cm = 500 cm
Height (H) = 2 m = 2 × 100 cm = 200 cm
∴ Volume of the cylindrical tank = πr2H = π × (500)2 × 200 cm3
Now, time required to fill the tank
ncert
ncert
ncert

Q 27.

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:

Soln. : We have: r1 = 2 cm, r2 = 1 cm and h = 14 cm
Volume of the glass
ncert
= ncert πh (r12 + r22 + r1r2)
ncert
ncert
ncert

Q 28.

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:

Soln. : We have
ncert
Slant height (l) = 4 cm
r1 = 18 cm and 2πr2 = 6 cm
ncert and ncert
∴ Curved surface area of the frustum of the cone
= π(r1 + r2) l = (πr1 + πr2) l = (9 + 3 ) × 4 cm2
= 12 × 4 cm2 = 48 cm2.

Q 29.

A fez, the cap used by the turks, is shaped like the frustum of a cone (see figure). 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.
ncert

SOLUTION:

Soln. : Here, the radius of the open side (r1) = 10 cm
The radius of the upper base (r2) = 4 cm
Slant height (l) = 15 cm
∴ Area of the material required
= [Curved surface area of the frustum] + [Area of the top end]
= π(r1 + r2) l + πr22
ncert
ncert
ncert

Q 30.

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 π = 3.14)

SOLUTION:

Soln. : We have : r1 = 20 cm, r2 = 8 cm and h = 16 cm
ncert
∴ Volume of the frustum
ncert
ncert
ncert
ncert
ncert
∴ Cost of milk = ₹ 20 ncert
= ₹ 208.998 ≈ ₹ 209
Now, slant height of the given frustum
ncert
ncert
∴ Curved surface area = π(r1 + r2)l
ncert
ncert
Area of the bottom = πr2
ncert
∴ Total area of metal required
= 1758.4 cm2 + 200.96 cm2 = 1959.36 cm2
Cost of metal required = ₹ ncert
= ₹ 156.75.

Q 31.

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 ncert find the length of the wire.

SOLUTION:

Soln. : Let us consider the frustum DECB of the metallic cone ABC
ncert
Here, r1 = BO and r2 = DO
In AOB, ncert
ncert
In ADO, ncert
ncert
Now, the volume of the frustum DBCE
ncert
ncert
ncert
Let l be the length and D be diameter of the wire drawn from the frustum. Since the wire is in the form of a cylinder,
∴ Volume of the wire = πr2l ncert
ncert ncert
∵ [Volume of the frustum] = [Volume of the wire]
ncert
ncert
= 7964.44m
Thus, the required length of the wire = 7964.44 m

Q 32.

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:

Soln. : Since diameter of the cylinder = 10 cm
∴ Radius of the cylinder (r)
⇒ Length of wire in one round = 2πr
= 2 × 3.14 × 5 cm = 31.4 cm
∵ Diameter of wire = 3 mm
∴ The thickness of cylinder covered in one round
⇒ Number of rounds (turns) of the wire to cover
∴ Length of wire required to cover the whole surface = Length of wire required to complete 40 rounds = 40 × 31.4 cm = 1256 cm
Now, radius of the wire
∴ Volume of wire = πr2l
∵ Density of wire = 8.88 gm/cm3
∴ Weight of the wire
= [Volume of the wire] × density


= 787.97 g = 788 g (approx.)

Q 33.

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 π as found appropriate.

SOLUTION:

Soln. : Let us consider the right BAC, right angled at A such that AB = 3 cm, AC = 4 cm
∴ Hypotenuse ncert
Obviously, we have obtained two cones on the same base AA' such that radius = DA or DA'
Now, ncert
[∴ ADB ~ ΔCAB]
ncert
Also ncert
Since, CD = BC - DB
ncert
Now, volume of the double cone
ncert
ncert
ncert
Surface area of the double cone = πrl1 + πrl2
ncert
ncert

Q 34.

A cistern, internally measuring 150 cm × 120 cm × 110 cm, has 129600 cm3 of water in it. Porous bricks are placed in the water until the cistern is full to 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 being 22.5 cm × 7.5 cm × 6.5 cm?

SOLUTION:

Soln. : ∵ Dimensions of the cistern are 150 cm, 120 cm and 100 cm.
∴ Volume of the cistern = 150 × 120 × 110 cm3 = 1980000 cm3
Volume of water contained in the cistern = 129600 cm3
∴ Free space (volume) which is not filled with water = (1980000 - 129600) cm3 = 1850400 cm3
Now, volume of one brick = (22.5 × 7.5 × 6.5) cm3 = 1096.875 cm3
∴ Volume of water absorbed by one brick ncert
Let n bricks can be put in the cistern.
∴ Volume of water absorbed by n bricks ncert
∴ [volume occupied by n bricks] = [(free space in the cistern + volume of water absorbed by n-bricks)]
ncert
ncert
ncert
ncert
= 1792.4102 ≈ 1792
Thus, 1792 bricks can be put in the cistern.

Q 35.

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:

Soln. : Volume of three rivers = 3 {(Surface area of a river) × Depth}
ncert
ncert
Volume of rainfall = (Surface area) × (Height of rainfall)
ncert
ncert
ncert
Since, 0.7236 km3 ≠ 9.728 km3
∴ The additional water in the three rivers is not equivalent to the rainfall.

Q 36.

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, find the area of the tin sheet required to make the funnel (see figure).
ncert

SOLUTION:

Soln. : We have,
For the cylindrical part
Diameter = 8 cm ⇒ Radius (r) = 4 cm
Height = 10 cm ⇒ Curved Surface area
ncert
For the frustum : ncert
Height (H) = 22 - 10 = 12 cm
∴ Slant height (l) ncert
ncert
ncert
∴ Surface area = π (r1 + r2)l
ncert
Area of tin required = [Area of the frustum] + [Area of cylindrical portion]
ncert
ncert

Q 37.

Derive the formula for the curved surface area and total surface area of the frustum of a cone

SOLUTION:

Soln. :
ncert
We have, Curved surface area of the frustum PQRS
ncert
= πr1l1 - πr2l2 ...(1)
Now,OC1Q OC2S [By AA similarity]
ncert
ncert
(∵ l1 = l + l2)
ncert ...(2)
Now, from (1), we get
Curved surface area of the frustum
ncert
ncert
= πl (r1 + r2) [From (2)]
Now, the total surface area of the frustum
= (curved surface area) + (base surface area) + (top surface area)
= πl (r1 + r2) + πr22 + πr12 = π (r1 + r2) l + π (r12 + r22)
= π [(r1 + r2) l + r12 + r22]

Q 38.

Derive the formula for the volume of the frustum of a cone.

SOLUTION:

Soln. : We have,
[Volume of the frustum RPQS]
= [Volume of right circular cone OPQ]
- [Volume of right circular cone ORS]
ncert
ncert ...(1)
ncert
Since OC1Q OC2S [By AA similarity]
ncert
ncert ...(2)
ncert
ncert
ncert ...(3)
From (1) and (2), we have Volume of the frustum RPQS
ncert
ncert
ncert
ncert [From (3)]