NCERT Solutions for Class 10 Maths Chapter 13: Surface Areas and Volumes
The solutions for NCERT Class 10 Chapter 13 for the topic of Surface Areas and Volumes will take you through a variety of basic ideations of geometry regarding three-dimensional shapes. This chapter is of great significance as it deals with the calculation of the surface areas and volumes of different geometric figures, such as cubes, cuboids, spheres, hemispheres, cones, and cylinders. These solutions are so designed that the students comprehend and solve real-world application problems of geometry, which assures them a good understanding of the subject.
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The NCERT Solutions for Class 10 Maths Chapter 13: Surface Areas and Volumes are tailored to help the students master the concepts that are key to success in their classrooms. The solutions given in the PDF are developed by experts and correlate with the CBSE syllabus of 2023-2024. These solutions provide thorough explanations with a step-by-step approach to solving problems. Students can easily get a hold of the subject and learn the basics with a deeper understanding. Additionally, they can practice better, be confident, and perform well in their examinations with the support of this PDF.
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Access Answers to NCERT Solutions for Class 10 Maths Chapter 13: Surface Areas and Volumes
Students can access the NCERT Solutions for Class 10 Maths Chapter 13: Surface Areas and Volumes. Curated by experts according to the CBSE syllabus for 2023–2024, these step-by-step solutions make Maths much easier to understand and learn for the students. These solutions can be used in practice by students to attain skills in solving problems, reinforce important learning objectives, and be well-prepared for tests.
Exercise 13.1
A hemispherical depression is cut out from one face of a cubical wooden block such that the diameter I of the hemisphere is equal to the edge of the cube. Determine the surface area of the remaining solid.
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.
Surface area of capsule = 2πrh + 2 × 2πr2
Diameter of capsule = 5mm
Radius of cylinder = Radius of hemisphere = 5/2 = 2.5mm
Height (h) =14 − 2.5 − 2.5 = 9mm
TSA = 2πr (h + 2r)
= 2 × 22/7 × 2.5 (9 + 5)
= 220 mm2
2 cubes each of volume 64 cm3 are joined end to end. Find the surface area of the resulting cuboid.
Let l be length of each cube
Volume of each cube = l3 = 64 cm3
⇒ l3 = 64 cm3
⇒ l = 4 cm
If, we join 2 cubes each having side equal to 4 cm, we get a cuboid
Length of cuboid = L = 8 cm
Height of cuboid = H = 4 cm
Breadth of cuboid = B = 4 cm
We know that Surface Area of Cuboid = 2(L.B + B.H + H.L)
= 2((8 x 4) + (4 x 4) + (4 x 8))
= 2(32 + 16 + 32)
= 2(80)
= 160 cm2
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.
Height of the cylindrical portion 13 − 7 = 6 cm.
Area of a Curved surface of cylindrical portion is,
= 2πrh
= 2 × 22/7 × 7 × 6
= 264 cm2
Area of a curved of hemispherical portion is
=2πr2
=2 × 22/7 × 7 × 7
= 308 cm2
∴ total surface area is = 308 + 264 = 572 cm2.
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.
Total Surface area of toy = Surface area of hemisphere without base + Surface area of cone without base
Total height = 15.5 = r + h
h = 15.5 − 3.5 = 12 cm
A = 2πr2 + πr√(r2 + (15.5 − r)2)
A = 2π × 3.52 + π × 3.5√(3.52 + 122)
A = 214.5 cm2
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.
(i) A cubical block of side 7 cm is surrounded by a hemisphere.
∴ Diameter of hemisphere is = 7 cm.
Radius of hemisphere, r = 3.5 cm
∴ Surface area of hemisphere = Curved surface area - Area of base of hemisphere
= 2πr2 − πr2
= 2 × 22/7 × (7/2)2 − 22/7 × (7/2)2
= 38.5
(ii) The curved surface area of square,
= 6 × I2
= 6 × (7)2
= 6 × 49
= 294 sq.cm.
∴ Total surface area of cubical block = Curved surface area of cubical block + Surface area of hemisphere
= 294 + 38.5
= 332.5 sq.cm.
A tent is in the shape of a cylinder surmounted buy 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.)
Radius of the cylindrical base = 2m
Height of the cylindrical is = 2.1m
Slant Height of conical top is = 2.8
Curved Surface Area of cylindrical portion is,
= 2πrh
= 2π × 2 × 2.1
= 8.4πm2
Curved Surface Area of conical portion is,
= πrl
= π × 2 × 2.8
= 5.6πm2
Total Curved Surface Area
= 8.4π + 5.6π
= 14 × 22/7
= 44m2
Therefore,
Cost of canvas = Rate\times Surface Area
= 500 × 44
= 22000
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.
Radius = 0.7 cm and height = 2.4 cm
Total surface area of structure = Curved surface area of cylinder + Area of top of cylinder + Curved surface area of cone
Curved surface area of cylinder = 2πrh
= 2π × 0.7 × 2.4
= 3.36 π cm2
Area of top:
= πr2
= π × 0.72
= 0.49π cm2
Slant height of cone can be calculated as follows:
l = √(h2+r2)
= √(2.42 + 0.72)
= √(5.76 + 0.49)
= √6.25 = 2.5 cm
Curved surface area of cone:
= πrl
= π × 0.7 × 2.5
= 1.75π cm2
Hence, remaining surface area of structure
= 3.36π + 0.49π + 1.75π
= 5.6π = 17.6 cm2
= 18 cm2 (approx)
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.
Total surface area of the article = Lateral surface area of cylinder + 2 × Curved surface area of hemisphere
We know that the lateral surface are of a cylinder = 2πrh.
And the curved surface area of a hemisphere = 2πr2.
∴ TSA of article = 2πrh + 2 × 2πr2
⇒ 2 × 22/7 × 7/2 × 10 + 2 × 2 × 22/7 × (7/2)2
⇒ 22 × 10 + 22 × 7
⇒ 22 (10 + 7)
⇒ 22 × 17
∴ TSA = 374 cm2
Hence, the total surface area of the article is 374 cm2.
Exercise 13.2
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 π.
Total Volume = Volume of cone + Volume of Hemisphere
We know that
Volume of cone = 1/3 πr2h
Volume of hemisphere = 2/3πr3
= 1/3πr2h + 2/3πr3
= 1/3 πr2(h+2r)
= 1/3 × π × 1 × 1 × (1+2)
= π cm3
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.)
Height of cylinder = 12 – 4 = 8 cm, radius = 1.5 cm, height of cone = 2 cm
Volume of cylinder = πr2h
= π × 1.52 × 8
= 18π cm3
Volume of a cone = 1/3 π × 1.52 × 2
= 1.5π cm3
Total volume = Volume of cylinder + 2 Volume of a cone
= 1.5π + 1.5π + 18π
= 21π = 66 cm3
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 2.8 cm (see figure).
It is known that the gulab jamuns are similar to a cylinder with two hemispherical ends.
So, the total height of a gulab jamun = 5 cm.
Diameter = 2.8 cm
So, radius = 1.4 cm
∴ The height of the cylindrical part = 5 cm–(1.4+1.4) cm
=2.2 cm
Now, total volume of One Gulab Jamun = Volume of Cylinder + Volume of two hemispheres
= πr2h+(4/3)πr3
= 4.312π+(10.976/3) π
= 25.05 cm3
We know that the volume of sugar syrup = 30% of total volume
So, volume of sugar syrup in 45 gulab jamuns = 45×30%(25.05 cm3)
= 45×7.515 = 338.184 cm3
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 id 1.4 cm. Find the volume of wood in the entire stand (see figure).
Volume of cuboid = length x width x height
We know the cuboid’s dimensions as 15 cmx10 cmx3.5 cm
So, the volume of the cuboid = 15x10x3.5 = 525 cm3
Here, depressions are like cones and we know,
Volume of cone = (⅓)πr2h
Given, radius (r) = 0.5 cm and depth (h) = 1.4 cm
∴ Volume of 4 cones = 4x(⅓)πr2h
= 1.46 cm2
Now, volume of wood = Volume of cuboid – 4 x volume of cone
= 525-1.46 = 523.54 cm2
A vessel is in the form of an inverted cone. Its height is 8 cm and the radius of the 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.
It is known that,
Volume of cone = volume of water in the cone
= ⅓πr2h = (200/3)π cm3
Now,
Total volume of water overflown= (¼)×(200/3) π =(50/3)π
Volume of lead shot
= (4/3)πr3
= (1/6) π
Now,
The number of lead shots = Total Volume of Water over flown/ Volume of Lead shot
= (50/3)π/(⅙)π
= (50/3)×6 = 100
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. ( π = 3.14 ).
Given that, the height of the big cylinder (H) = 220 cm
Radius of the base (R) = 24/2 = 12 cm
So, the volume of the big cylinder = πR2H
= π(12)2 × 220 cm3
= 99565.8 cm3
Now, the height of smaller cylinder (h) = 60 cm
Radius of the base (r) = 8 cm
So, the volume of the smaller cylinder = πr2h
= π(8)2×60 cm3
= 12068.5 cm3
∴ Volume of iron = Volume of the big cylinder + Volume of the small cylinder
= 99565.8 + 12068.5
=111634.5 cm3
We know,
Mass = Density x volume
So, mass of the pole = 8×111634.5
= 893 Kg (approx.)
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.
Here, the volume of water left will be = Volume of cylinder – Volume of solid
Given,
Radius of cone = 60 cm,
Height of cone = 120 cm
Radius of cylinder = 60 cm
Height of cylinder = 180 cm
Radius of hemisphere = 60 cm
Now,
Total volume of solid = Volume of Cone + Volume of hemisphere
Volume of cone = 1/3πr2h = 1/3 × π×602×120cm3 = 144×103π cm3
Volume of hemisphere = (⅔)×π×603 cm3 = 144×103π cm3
So, total volume of solid = 144×103π cm3 + 144×103π cm3 = 288 ×103π cm3
Volume of cylinder = π×602×180 = 648000 = 648×103 π cm3
Now, volume of water left will be = Volume of cylinder – Volume of solid
= (648-288) × 103×π = 1.131 m3
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 cm2 . Check whether she is correct, taking the above as the inside measurements and π = 3.14.
Given,
For the cylinder part, Height (h) = 8 cm and Radius (R) = (2/2) cm = 1 cm
For the spherical part, Radius (r) = (8.5/2) = 4.25 cm
Now, volume of this vessel = Volume of cylinder + Volume of sphere
= π×(1)2×8+(4/3)π(4.25)3
= 346.51 cm3
Exercise 13.3
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.
It is given that radius of the sphere (R) = 4.2 cm
Also, Radius of cylinder (r) = 6 cm
Now, let height of cylinder = h
It is given that the sphere is melted into a cylinder.
So, Volume of Sphere = Volume of Cylinder
∴ (4/3)×π×R3 = π×r2×h.
h = 2.74 cm
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.
For Sphere 1:
Radius (r1) = 6 cm
∴ Volume (V1) = (4/3)×π×r13
For Sphere 2:
Radius (r2) = 8 cm
∴ Volume (V2) = (4/3)×π×r23
For Sphere 3:
Radius (r3) = 10 cm
∴ Volume (V3) = (4/3)× π× r33
Also, let the radius of the resulting sphere be “r”
Now,
Volume of resulting sphere = V1+V2+V3
(4/3)×π×r3 = (4/3)×π×r13+(4/3)×π×r23 +(4/3)×π×r33
r3 = 63+83+103
r3 = 1728
r = 12 cm
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.
Answer:
Diameter of well = 7 m
It is given that the shape of the well is in the shape of a cylinder with a diameter of 7 m
So, radius = 7/2 m
Also, Depth (h) = 20 m
Volume of the earth dug out will be equal to the volume of the cylinder
Let the height of the platform = H
Volume of soil from well (cylinder) = Volume of soil used to make such platform
π×r2×h = Area of platform × Height of the platform
We know that the dimension of the platform is = 22×14
So, Area of platform = 22×14 m2
∴ π×r2×h = 22×14×H
⇒ H = 2.5 m
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.
Given, Depth (h1) of well = 14 m
Diameter of the circular end of the well =3 m
So, Radius (r1) = 3/2 m
Width of the embankment = 4 m
From the figure, it can be said that the embankment will be a cylinder having an outer radius (r2) as 4+(3/2) = 11/2 m and inner radius (r1) as 3/2m
Now, let the height of embankment be h2
∴ Volume of soil dug from well = Volume of earth used to form embankment
π×r12×h1 = π×(r22-r12) × h2
Solving this, we get,
The height of the embankment (h2) as 1.125 m.
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.
Number of cones will be = Volume of cylinder / Volume of ice cream cone
For the cylinder part,
Radius = 12/2 = 6 cm
Height = 15 cm
∴ Volume of cylinder = π×r2×h = 540π
For the ice cone part,
Radius of conical part = 6/2 = 3 cm
Height = 12 cm
Radius of hemispherical part = 6/2 = 3 cm
Now,
Volume of ice cream cone = Volume of conical part + Volume of hemispherical part
= (⅓)×π×r2×h+(⅔)×π×r3
= 36π +18π
= 54π
∴ Number of cones = (540π/54π)
= 10
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 10cm x 3.5?
For silver coin, Diameter = 1.75 cm
Radius (r) of circular end of coins = 1.75/2 = 0.875 cm
Now, the number of coins to be melted to form the required cuboids be “n”
So, Volume of n coins = Volume of cuboids
n × π × r2 × h1 = l × b × h
n×π×(0.875)2×0.2 = 5.5×10×3.5
Or, n = 400
A cylindrical bucket, 32 cm and 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.
let “r2” be the radius of the circular end of the conical heap.
We know that volume of the sand in the cylindrical bucket will be equal to the volume of sand in the conical heap.
∴ Volume of sand in the cylindrical bucket = Volume of sand in conical heap
π×r12×h1 = (⅓)×π×r22×h2
π×182×32 = (⅓)×π ×r22×24
Or, r2= 36 cm
And,
Slant height (l) = √(362+242) = 12√13 cm.
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?
Breadth (b) = 6 m and Height (h) = 1.5 m
It is also given that
The speed of canal = 10 km/hr
Length of canal covered in 1 hour = 10 km
Length of canal covered in 60 minutes = 10 km
Length of canal covered in 1 min = (1/60)x10 km
Length of canal covered in 30 min (l) = (30/60)x10 = 5km = 5000 m
We know that the canal is cuboidal in shape. So,
Volume of canal = lxbxh
= 5000x6x1.5 m3
= 45000 m3
Now,
Volume of water in canal = Volume of area irrigated
= Area irrigated x Height
So, Area irrigated = 56.25 hectares
∴ Volume of canal = lxbxh
45000 = Area irrigatedx8 cm
45000 = Area irrigated x (8/100)m
Or, Area irrigated = 562500 m2 = 56.25 hectares.
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?
Volume of water that flows in t minutes from pipe = t×0.5π m3
Radius (r2) of circular end of cylindrical tank =10/2 = 5 m
Depth (h2) of cylindrical tank = 2 m
Let the tank be filled completely in t minutes.
Volume of water filled in the tank in t minutes is equal to the volume of water flowing in t minutes from the pipe.
Volume of water that flows in t minutes from pipe = Volume of water in tank
t×0.5π = π×r22×h2
Or, t = 100 minutes
Exercise 13.4
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.
Here,
Radius (r1) of the upper base = 4/2 = 2 cm
Radius (r2) of lower the base = 2/2 = 1 cm
Height = 14 cm
Now, Capacity of glass = Volume of frustum of cone
So, Capacity of glass = (⅓)×π×h(r12+r22+r1r2)
= (⅓)×π×(14)(22+12+ (2)(1))
∴ The capacity of the glass = 102×(⅔) cm3
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.
Slant height (l) = 4 cm
Circumference of upper circular end of the frustum = 18 cm
∴ 2πr1 = 18
Or, r1 = 9/π
Similarly, circumference of lower end of the frustum = 6 cm
∴ 2πr2 = 6
Or, r2 = 3/π
Now, CSA of frustum = π(r1+r2) × l
= π(9/π+3/π) × 4
= 12×4 = 48 cm2
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.
For the lower circular end, radius (r1) = 10 cm
For the upper circular end, radius (r2) = 4 cm
Slant height (l) of frustum = 15 cm
Now,
The area of material to be used for making the fez = CSA of frustum + Area of upper circular end
CSA of frustum = π(r1+r2)×l
= 210π
And, Area of upper circular end = πr22
= 16π
The area of material to be used for making the fez = 210π + 16π = (226 x 22)/7 = 710 2/7
∴ The area of material used = 710 2/7 cm2
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 total cost of milk which can completely fill the container at the rate of Rs. 20 per liter. Also find the cost of metal sheet used to make the container, if it costs Rs. 8 per 100 cm2. ( take π = 3.14)
Answer:
Given,
r1 = 20 cm,
r2 = 8 cm and
h = 16 cm
∴ Volume of the frustum = (⅓)×π×h(r12+r22+r1r2)
= 1/3 ×3.14 ×16((20)2+(8)2+(20)(8))
= 1/3 ×3.14 ×16(400 + 64 + 160) = 10449.92 cm3 = 10.45 lit
It is given that the rate of milk = Rs. 20/litre
So, Cost of milk = 20×volume of the frustum
= 20 × 10.45
= Rs. 209
Now, slant height will be
l = 20 cm
So, CSA of the container = π(r1+r2)×l
= 1758.4 cm2
Hence, the total metal that would be required to make container will be = 1758.4 + (Area of bottom circle)
= 1758.4+πr2 = 1758.4+π(8)2
= 1758.4+201 = 1959.4 cm2
∴ Total cost of metal = Rs. (8/100) × 1959.4 = Rs. 157
A metallic right circular cone 20 cm high and whose vertical angle is 600 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 1/16 cm, find the length of the wire.
Exercise 13.5
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 cm2
Given that,
Diameter of cylinder = 10 cm
So, radius of the cylinder (r) = 10/2 cm = 5 cm
∴ Length of wire in completely one round = 2πr = 3.14×5 cm = 31.4 cm
It is given that diameter of wire = 3 mm = 3/10 cm
∴ The thickness of cylinder covered in one round = 3/10 m
Hence, the number of turns (rounds) of the wire to cover 12 cm will be-
Now, the length of wire required to cover the whole surface = length of wire required to complete 40 rounds
40 x 31.4 cm = 1256 cm
Radius of the wire = 0.3/2 = 0.15 cm
Volume of wire = Area of cross-section of wire × Length of wire
= π(0.15)2×1257.14
= 88.898 cm3
We know,
Mass = Volume × Density
= 88.898×8.88
= 789.41 gm
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)
Let us consider the ABA
Here,
AS = 3 cm, AC = 4 cm
So, Hypotenuse BC = 5 cm
We have got 2 cones on the same base AA’ where the radius = DA or DA’
Now, AD/CA = AB/CB
By putting the value of CA, AB and CB we get,
AD = 2/5 cm
We also know,
DB/AB = AB/CB
So, DB = 9/5 cm
As, CD = BC-DB,
CD = 16/5 cm
Now, volume of double cone will be
Solving this we get,
V = 30.14 cm3
The surface area of the double cone will be
= 52.75 cm2
A cistern, internally measuring 150 cm x 120 cm x 110 cm has 129600 cm2 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 brick 22.5 cm x 7.5 cm x 6.5 cm?
Given that the dimension of the cistern = 150 × 120 × 110
So, volume = 1980000 cm3
Volume to be filled in cistern = 1980000 – 129600
= 1850400 cm3
Now, let the number of bricks placed be “n”
So, volume of n bricks will be = n×22.5×7.5×6.5
Now as each brick absorbs one-seventeenth of its volume, the volume will be
= n/(17)×(22.5×7.5×6.5)
For the condition given in the question,
The volume of n bricks has to be equal to volume absorbed by n bricks + Volume to be filled in cistern
Or, n×22.5×7.5×6.5 = 1850400+n/(17)×(22.5×7.5×6.5)
Solving this we get,
n = 1792.41
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.
From the question, it is clear that
Total volume of 3 rivers = 3×[(Surface area of a river)×Depth]
Given,
Surface area of a river = [1072×(75/1000)] km
And,
Depth = (3/1000) km
Now, volume of 3 rivers = 3×[1072×(75/1000)]×(3/1000)
= 0.7236 km3
Now, volume of rainfall = total surface area × total height of rain
= 0.7280 km3
For the total rainfall was approximately equivalent to the addition to the normal water of three rivers, the volume of rainfall has to be equal to volume of 3 rivers.
But, 0.7280 km3 = 0.7236 km3
So, the question statement is true.
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).
Given,
Diameter of upper circular end of frustum part = 18 cm
So, radius (r1) = 9 cm
Now, the radius of the lower circular end of frustum (r2) will be equal to the radius of the circular end of the cylinder
So, r2 = 8/2 = 4 cm
Now, height (h1) of the frustum section = 22 – 10 = 12 cm
And,
Height (h2) of cylindrical section = 10 cm (given)
Now, the slant height will be-
Or, l = 13 cm
Area of tin sheet required = CSA of frustum part + CSA of cylindrical part
= π(r1+r2)l+2πr2h2
Solving this we get,
Area of tin sheet required = 782 4/7 cm2
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.
Let ABC be a cone. From the cone the frustum DECB is cut by a plane parallel to its base. Here, r1 and r2 are the radii of the frustum ends of the cone and h be the frustum height.
Now, consider the ΔABG and ΔADF,
Here, DF||BG
So, ΔABG ~ ΔADF
The total surface area of frustum will be equal to the total CSA of frustum + the area of upper circular end + area of the lower circular end
= π(r1+r2)l+πr22+πr12
∴ Surface area of frustum = π[(r1+r2)l+r12+r22]
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