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Applied Maths Chapter 4 Solutions
Mensuration
EXERCISE- 4.2
Q.1 Dimensions of the cuboidal hall are 10 m x 12 m x 15 m. Find the length of the longest rod that can be fitted into the hall.
Ans. Diagonal of a cuboid
d^2 = l^2 + b^2 + h^2
d^2 = (10)^2 + (12)^2 + (15)^2
d^2 = 100 + 144 + 225
d^2 = 469
d = √469
d = 21.65 m
Q.2 Nishant owns a house of dimensions 10 feet x 8 feet x 9 feet. His house has two entry doors on the opposite sides, of dimensions 7 feet by 3 feet each, and two windows on opposite walls of dimensions 3 feet x 2 feet each. He wants wallpaper to be done in his house. If the wallpaper that he selects is 2 feet wide and the cost is ₹25 per foot. Find the total cost paid by Nishant. If the labour charges Rs.10 per sq. feet.
Ans. Area of wallpaper = Curved surface area – (area of door) x 2 – (area of window) x 2
= 2h (l + b) – (l x b) x 2 – (l x b) x 2
= 2 x 9 (10 + 8) – (7 x 3) x 2 – (3 x 2) x 2
= 18 x 18 – 21 x 2 – 6 x 2
= 325 – 42 – 12
= 324 – 54
= 270 sq. feet
Width of wallpaper = 2 feet
Length of wallpaper = 270/2 = 135 feet
Cost of wallpaper = 135 x 25 = Rs.3,375
Cost of Labour = 270 x 10 = Rs.2,700
Therefore,
Total cost = 3375 + 2700 = Rs.6075
Q.3 The volume of a cuboidal block of silver is 10368 cm^3. If its dimensions are in the ratio 3:2:1, find
(i) the dimensions of the block.
(ii) the cost of gold polishing its entire surface at Rs.0.50 per
Cm^2.
Ans.(i) Volume = 10368
l x b x h = 10368
(3x) x (2x) x (x) = 10368
x^3 = 10368/6
= 1728
x^3 = 12^3
x = 12
Therefore, length = 3 x 12 = 36 cm
breath = 2 x 12 = 24 cm
height = 12cm
Total surface area = 2 (lb + bh + hl)
= 2 (36 x 24 + 24 x 12 + 12 x 36)
= 2 x 12 x 12 (3 x 2 + 2 x 1 + 1 x 3)
= 2 x 12 x 12(6 + 2 + 3)
= 288 x 11
= 3168 cm^2
(ii) Cost of polishing = Rs.050/cm^2
Total cost = 3168 x 0.50
= Rs.1584
Q.4 A rectangular plot is 24 m long and 20 m wide. A cubical pit of edge 4 m is dug at each of the four corners of the field and the soil removed is evenly spread over the remaining part of the plot. By what height does the remaining plot get raised?
Ans. The volume of earth dug from the pit = 4 x a^3
= 4 x (4)^3
= 4 x 64 = 256 m^3
Area of remaining part = 24 x 20 – (4 x 4 x 4)
= 480 – 64
= 416 m^2
The height of the pit increased
46 x h = 256
h = 256/46
= 8/13 m
Q.5 A rectangular container whose base is a square of side 6 cm, stands on a horizontal table and holds water up to 1 cm from the top. When a cube is placed in the water and is completely submerged, the water rises to the top, and 2 cm^3 of water overflows. Calculate the volume of the cube.
Ans. Volume of cube = Volume of water + Volume of water overflow
= 6 x 6 x 1 + 2
= 36 + 2
= 38cm^3
Q.6 A hollow square-shaped tube open at both ends, is made of iron. The internal square is of 5 cm side and the length of the tube is 8 cm. There are 192 cm^3 of iron in this tube. Find its thickness.
Ans. The volume of iron = Volume of the outer cuboid – Volume of the inner cuboid
192 = x * x * x – 5 * 5 * 8
192 = 8(x^2 – 25)
192/8 = x^2 – 25
24 + 24 = x^2
x^ 2 = 49
x = 7
Thickness = 7
y + y + 5 = 7
2y = 7 – 5
y = 1
Therefore,
thickness = 1 cm
Q.7 A rectangular pit 1.4 m long, 90 cm broad, and 70 cm deep was dug and 1000 bricks of base 21 cm by 10.5 cm were made from the earth dug out. Find the height of each brick.
Ans. The volume of earth dug from the pit = Volume of 1000 bricks
l x b x h = (l x b x h)1000
1.4 x 100 x 90 x 70 = (21 x 10.5 x h)1000
(1.4 x 100 x 90 x 70)/(21 x 10.5 x 1000) = h
h = 4 cm
Q.8 The inner dimensions of a closed wooden box are 2 m, 1.2 m, and 0.75 m. The thickness of the wood is 2.5 cm. Find the cost of wood required to make the box if 1 m^3 of wood costs Rs.5400.
Ans. Thickness = 2.5 cm = 0.025 m
length = 2m, breadth = 1.2m, height = 0.75m [inner]
Length = 2 + 0.025 + 0.025 = 2.05m [outer]
Breadth = 1.2 + 0.025 + 0.025 = 1.25m [outer]
Height = 0.75 + 0.025 + 0.025 = 0.8 [outer]
Volume of wood = (2.05 x 1.25 x 0.8) – (2 x 1.2 x 0.75)
= 2.05 – 1.8
= 0.25m^3
Cost of wood = 0.25 x 5400
= Rs.1350
Q.9 The ratio between the curved surface area and the total surface area of a right circular cylinder is 1:2. If the total surface area of the cylinder is 616 cm^2, then find the volume.
Ans. The ratio of the curved surface area & and total surface area
2лrh : 2лr (r + h) = 1 : 2
Total surface area = 616 cm^2
2лrh/2лr(r + h) = 1/2
r + h = 2h
h = r
Putting value of r in 2лr(r + h) = 161
2лr (r + r) = 616
2 x 22/7 x 2r^2 = 616
r^2 = (7 x 616)/(2 x 22 x 2)
r^2 = 49
r = 7
Therefore, h = 7
Volume = лr^2h
= 22/7 x 7 x 7 x 7
= 1078 cm^3
Q.10 Rainwater which falls on a flat rectangular surface of length 6 m and breadth 4 m is transferred into a cylindrical vessel of internal radius 20 cm. What will be the height of water in the cylindrical vessel if the rainfall is 1 cm? Give your answer to the nearest integer. (Тakе л= 3.14)
Ans. The volume of water stored on a rectangular surface = Volume of water in a cylinder
600 x 400 x 1 = 3.14 x 20 x 20 x h
(600 x 400)/(3.14 x 20 x 20) = h
h = 191.08 cm
h = 191 cm
Q.11 The curved surface of the cylindrical reservoir 12 m deep is plastered from the inside with a concrete mixture at the rate of Rs.15 per m^2. If the total payment made is Rs.5652, then find the capacity of this reservoir in litres. (Use л = 3.14)
Ans. Curved surface area = 2лrh
Cost = Rs.15/m^2
C.S.A x cost price per m^2 = Total cost
2лrh x 15 = 5652
2 x 3.14 x r x 12 x 15 = 5652
r= (5652 x 100)/(2 x 3.14 x 12 x 15)
r = 5 m
Capacity = лr^2h
= 3.14 x 5 x 5 x 12
= 942 m^3
= 942000 liters [1 m^3 = 1000 L ]
Q.12 If the height and radius of a cone are tripled, then find the ratio of the volume of the new cone and that of the original cone.
Ans. Let, the height & and radius of the original cone be h & r respectively
height & and radius of the new cone be 3h & 3r
V2/V1 = [1/3л(3r)^2(3h)]/[1/3лr^2h]
V2/V1 = (3r)^2(3h)/r^h
= 27r^2h/r^5
= 27: 1
Q.13 A cone is surmounted on a hemisphere as shown in the adjoining figure. The radius of the solid formed is 6 cm and its height is 21 cm, find
(i) curved surface area of the solid.
(ii) volume of the solid.
Ans. Height of cone (h) = 21 – 6 = 15 cm
(i) L^2 = r^2 + h^2
= (6)^2 + (15)^2
= 36 + 225
= 261
l = √261 = 16.15cm
C.S.A = лrl + 2лr^
= лr(l + 2r)
= 22/7 x 6 (16.15 + 2 x 6)
= 132/7 x 28.15
= 531 cm^2
(ii) Volume of cone + Volume of hemisphere = Volume of solid
1/3лr^2h + 2/3лr^3
1/3лr^2 (h + 2r)
1/3 x 22/7 x 36 (15 + 2 x 6)
264/7 x 27
= 1018.29 cm^3
Q.14 A cuboidal gold block of dimensions 9 cm x 11 cm x 12 cm is melted and recast into spherical balls of radius 3 mm. Find the number of spherical balls formed.
Ans. The volume of the cuboid block = (Volume of the spherical ball) n
l x b x h = 4/3лr^3
9 x 11 x 12 = 4/3 x 22/7 x 3/10 x 3/10 x 3/10 x n
n = (9 x 11 x 12 x 7 x 5 x 10 x 10)/(2 x 22 x 3 x 3)
n = 10500 balls
Q.15 Monica has a piece of canvas whose area is 552 m^2. She uses it to make a conical tent with a base radius of 7 m. Assuming that all the stitching margins and the wastage incurred while cutting amount to approximately 2 m^2, find the volume of the tent that can be made with it.
Ans. Area of canvas used in tent = 552 – 2 = 550m^2
Curved surface area = 550
лrl = 550
22/7 x 7 x l = 550
l = 550/22
l = 25
l^2 = r^2 + h^2
(25)^2 = (7)^2 + h^2
625 = 49 + h^2
625 – 49 = h^2
h^2 = 576
h = 24
Volume of tent = 1/3лr^2h
= 1/3 x 22/7 x 7 x 7 x 24
= 22 x 7 x 8
= 1232 m^3
Q.16 The volume of a conical tent is 1232 m3 and the area of the base floor is 154 m2. Calculate the:
(i) radius of the floor.
(ii) height of the tent.
(iii) The length of the canvas required to cover this conical tent if its width is 2 m. 17. Find the radius and the curved surface area of the cone made from a quadrant of a circle of radius 28 cm.
Ans.
Volume = 1232 m^3, лr^2 = 154
1/3лr^2h = 1232
1/3 x 154 x h = 1232
h = (1232 x 3)/154
= 24 m
Therefore,
(ii) height of tent = 24 m.
(i) лr^2 = 154
22/7 x r^2 = 154
r^2 = (154 x 7)/22
r^2 = 49
r = 7
(iii) l^2 = r^2 + h^2
= (7)^2 + (24)^2
= 49 + 576
= 625
l = 25
Area of canvas = лrl
= 22/7 x 7 x 25
= 550 m^2
l x b = 550
l x 2 = 550
l = 550/2
= 275 m
Hence, the length of the cloth = 275 m
Q.18 A hemispherical dome of a building needs to be painted from the outside. If the circumference of the base of the dome is 17.6 m, find the cost of painting it at the rate of 8 per 100 cm^2.
Ans. Circumference of base = 17.6
2лr = 17.6
2 x 22/7 x r = 176/10
r = 28/10 m
Curved surface area = 2лr^2
= 2 x 22/7 x 28/10 x 28/10
= (44 x 112)/100
= 49.28 m^2
As we know, 1 m = 100 cm
1 m^2 = 10000 cm^2
Therefore, Curved surface area = 492800 cm^2
Rate = Rs. 8/100 cm^2
Cost = C.S.A x rate
= 492800 x 8/100
= 4928 x 8
= Rs.39424
Q.19 Two solid spheres made of the same metal have weights 5920 g and 740 g respectively. Determine the radius of the larger sphere, if the diameter of the smaller one is 5 cm.
Ans. Dimension of the smaller sphere:
radius = 5/2
Volume = 4/3лr^3
= 4/3л(5/2)^3
Dimension of a larger sphere
Radius = x
Volume = 4/3лx^3
Comparing both the sphere
= [4/3л(5/2)^3]/[4/3лx^3] = 740/5920
= [125/8]/x^3 = 74/592
= 125/8x^3 = 74/592
= 125 = x^3
Therefore,
x = 5
Q.20 The water for a factory is stored in a hemispherical tank whose internal diameter is 14 m. The tank contains 50 kilolitres of water. Water is pumped into the tank to fill it. Find the volume of water pumped into the tank.
Ans. Dimension of tank:
radius = 7 m
Volume = 2/3лr^3
= 2/3 x 22/7 x 343
= (2 x 23 x 49)/3
= (44 x 49)/3
= 2156/3
= 718.6 m^3
Water already in tank = 50 KL
1 KL = 1000 L
1 m^3 = 1000 L
Therefore, 1 KL = 1 m^3
Water required in tank = 50 m^3
Water required = 718.6 – 50
= 668.6 m^3
Q.21 A cone of maximum size is carved out from a cube of edge 14 cm. Find the surface area of the cone and the remaining solid after the cone is carved out.
Ans. Dimension of Cone:
Height = 14 cm
Diameter = 14cm
Radius = 7 cm
l = √h^2 + r^2
= √(14)^2 + (7)^2
= √245
= 7√5
Surface area = лrl + лr^2
= 22/7 x 7 x 7√5 + 22/7 x 7^2
= 154√5 + 154
= 154 [√5 + 1] cm^2
Total surface area of remaining solid = 6a^2 – лr^2 + лrl
= 6 x (14)^2 – 22/7 x 49 + 22/7 x 7 x 7√5
= 6 x 196 – 154 + 154√5
= 1176 – 154 + 154√5
= 1022 + 154√5 cm^2
Q.22 The adjoining diagram shows a model of a rocket consisting of a cylinder surmounted by a cone at one end. The dimensions of the model are a common radius of 3 cm, height of cone = 4 cm and total height = 14 cm. If the model is drawn to a scale of 1:500, find
(i) the total surface area of the rocket in лm^2.
(ii) the total volume of the rocket in лm^3.
Ans. According to the model is drawn to a scale of 1:500,
Radius = 15 m
Height = 20 m
l = √h^2 + r^2
= √(20)^2 + (15)^2
= 625
= 25 m
(i) Total surface area = C.S.A of cone + C.S.A of cylinder + Area of base
T.S.A = лrl + 2лrh + лr^2
= л(15 x 25 + 2 x 15 x 50 + 15^2)
= л(375 x 1500 x 225)
= 2100л m^2
(ii) Volume = лr^2h + 1/3 лr^2h
= л(15^2 x 50 + 1/3 x 15^2 x 20)
= л(225 x 50 + 1/3 x 225 x 20)
= л(11250 + 75 x 20)
= л(11250 + 1500)
= л(12750)
Therefore, volume = 12750л m^3
Q.23 A toy is in the shape of a right circular cylinder with a hemisphere on one end and a cone on the other. The height and radius of the cylindrical part are 13 cm and 5 cm respectively. The radii of the hemispherical and conical parts are the same as that of the cylindrical part. Calculate the surface area of the toy if the height of the conical part is 12 cm.
Ans. Dimension of cylinder:
h = 13 cm
r = 5 cm
Dimension of the hemisphere:
r = 5 cm
h = 12 cm
l = √h^2 + r^2
= √12^2 + 5^2
= 13 cm
Surface Area = C.S.A (hemisphere) + C.S.A (cylinder) + C.S.A (Cone)
= 2лr^2 + 2лrh + лrl
= лr [2r + 2h + l]
= 22/7 x 5 [2 x 5 + 2 x 13 + 13]
= 110/7 [10 + 26 + 13]
= 110/7 x 49
= 110 x 7
= 770 cm^2
Q.24 A girl fills a cylindrical bucket 32 cm in height and 18 cm in radius with sand. She empties the bucket on the ground and makes a conical heap of sand. If the height of the conical heap is 24 cm, find
(i) the radius and
(ii) the slant height of the heap.
Ans. Volume of bucket = лr^2h
= л x (18)^2 x 32 cm^3
Volume of heap = 1/3лr^2h
= 1/3 лr^2 x 24
= 8лr^2
Now,
Volume of bucket = Volume of heap
л x (18)^2 x 32 = 8лr^2
18^2 x 4 = r^2
r^2 = 18^2 x 2^2
r = 18 x 2 = 36 cm
l = √h^2 + r^2
= √24^2 + 36^2
= √(12×2)^2 + (12×3)^2
= √12^2 x 2^2 + 12^2 x 3^2
= √12^2(2^2 + 3^2)
= 12√4 + 9
= 12√13 cm
Q.25 A hollow sphere of internal and external radii 6 cm and 8 cm respectively is melted and recast into small cones of base radius 2 cm and height 8 cm. Find the number of cones formed.
Ans. Volume of hollow sphere = 4/3л [R^3 – r^3]
= 4/3л [8^3 – 6^3]
= 4/3л [512 – 216]
= 4/3л x 296
Volume of cone = 1/3лr^2h
= 1/3 л x 4 x 8
=4/3л x 8 cm^3
Volume of hollow sphere = Volume of 1 cone x no. of cones
4/3л x 296 = 296 4/3л x 8 x h
296/8 = h
h = 37
Q.26 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 into the vessel.
Ans. Dimension of cone:
height = 8 cm
radius = 5 cm
Volume of vessel = 1/3лr^2h
= 1/3л x 25 x 8
= 200/3л cm^3
Quantity of water falling out
= 1/4 x 200/3л cm^3
= 50/3л cm^3
Total volume of lead shots = 50/31л cm^3
Total volume = no. of shots x Volume of 1 shot
No. of shots = Total volume/Volume of 1 shot
= (50/3л)/4/3л(0.5)^3
= 50/(4 x 0.5 x 0.5 x 0.5)
= 50/0.5
= 50/(5/10)
= 500/5
= 100
Q.27 A container shaped like a right circular cylinder having a diameter of 12 cm and height of 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.
Ans. Diameter of cylinder = 12cm
Radius = 12/2 = 6 cm
Height = 15 cm
Volume of container = лr^2h
= л x 36 x 15 cm^3
The volume of 1 ice cream = Volume of cone + Volume of the hemisphere
= 1/3лr^2h + 2/3лr^3
= л/3 x r^2 [h + 2r]
= л/3 x 9 [12 + 6]
= л x 3 x 18
Volume of container = Volume of 1 ice cream + no. of cone
No. of cones = Volume of container/Volume of 1 ice cream
= (л x 36 x 15)/(л x 3 x 18)
= 10
FAQ’s related to Applied Maths Chapter 4 on Mensuration:
Q.1 What is mensuration?
Ans. Mensuration is the branch of mathematics that deals with the measurement of geometric figures, such as length, area, volume, and other related quantities.
Q.2 What are the basic formulas in mensuration?
Ans. Basic formulas in mensuration include formulas for calculating the perimeter and area of geometric shapes like squares, rectangles, triangles, circles, and the volume of solids like cubes, cylinders, and spheres.
Q.3 How is mensuration used in real life?
Ans. Mensuration is used in various real-life applications such as calculating the area of fields, determining the amount of material needed for construction projects, estimating the volume of containers, and designing objects in engineering and architecture.
These are few Frequently Asked Questions relating to Applied Maths Chapter 4
In Applied Maths chapter 4, you will explore fascinating topics that form the backbone of practical problem-solving techniques. Through clear explanations, illustrative examples, and step-by-step solutions, you’ll grasp complex concepts effortlessly. Whether you’re preparing for exams or simply eager to deepen your mathematical understanding, Applied Maths Chapter 4 promises an enriching learning experience that will set you on the path to success. Applied Maths Chapter 4, we delve deep into advanced mathematical concepts that are crucial for understanding.
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