Please refer to Surface areas and Volumes Class 9 Mathematics Exam Questions provided below. These questions and answers for Class 9 Mathematics have been designed based on the past trend of questions and important topics in your class 9 Mathematics books. You should go through all Class 9 Mathematics Important Questions provided by our teachers which will help you to get more marks in upcoming exams.
Class 9 Mathematics Exam Questions Surface areas and Volumes
Class 9 Mathematics students should read and understand the important questions and answers provided below for Surface areas and Volumes which will help them to understand all important and difficult topics.
Very Short Answer Type Questions:
Question. If the number of square centimetres in the surface area of a sphere is equal to the number of cubic cm in its volume. Find the diameter of the sphere ?
Ans. Given, Area of sphere = Volume of Sphere
4πr2 = 4/3 πr3
where r is the radius of sphere
or, r = 3 cm [on solving]
∴ Diameter = 2r = 6 cm.
Question. If the diameter of a football is five times the diameter of a cricket ball. Then find the ratio of surface areas of football and cricket ball.
Ans. Given, diameter of football
= 5 × diameter of cricket ball If r denotes radius of a football and r’ that of a cricket ball, then we have
Question. If the height and the radius of cone is tripled, then find ratio of volume of new cone and that of original.
Ans. Let h and r be the height and radius of original cone and let h’ and r’ be the height and radius of new cone.
Hence, the ratio of new cone to the original cone is 27 : 1.
Question. Find the amount of water displaced by a solid spherical ball of diameter 4.2 cm, when it is completely immersed in water.
Ans. Amount of water displaced = Volume of solid spherical ball
∴ Volume of solid spherical ball = 4/3 πr3
r = 4.2/2 = 2.1 cm (given)
∴ Volume of solid spherical ball = 4/3 π(2.1)3
= 4/7 x 22/7 × (2.1)3 cm3
= 38808/1000 cm3
∴ Amount of water displaced = 38.808 cm3
= 38.808 ml (∴ 1 cm3 = 1 ml)
Question. Compute the curved surface area of a hemisphere whose diameter is 14 cm.
Ans. 308 cm2. 1
Detailed Solution :
Given diameter of hemisphere = 14 cm
∴ radius = 7 cm
∴ Curved surface area = 2πr2
= 2 x 2/27 × 7 × 7 = 308 cm2.
Question. Calculate the edge of the cube, if its volume is 1331 cm3.
Ans. Let the edge of cube be a, then
Volume = a3
1331 = a3
a = 3 1331
a = 11 cm
Short Answer Type Questions:
Question. The curved surface area of a right circular cylinder of height 14 cm is 88 cm2. Find the radius of the base of cylinder.
Ans. Given, CSA of a cylinder = 88 cm2
height = 14 cm
CSA of a cylinder = 2πrh
or, 88 = 2 × 22/7 × r × 14
or, r = 88 × 7/2 × 22 × 14
= 1 cm
Question. If the curved surface area of a cylinder is 4.2 cm2 and height is 5 cm, then find radius of its base and volume of the cylinder (Use π = 3.14).
Ans. Let, radius of base = r
Given, CSA, 2πrh = 94.2 cm2
or, 2 × 3.14 × r × 5 = 94.2
or, r = 94.2/10 × 3.14
= 3 cm
Now, Volume = πr2h
= 3.14 × 3 × 3 × 5
= 141.3 cm3.
Question. A solid shot-put is a metallic sphere of radius 4.9 cm. Find the volume of the shot-put.
OR
Find the mass of shot-put, if density is 7.8 gm/cm3.
Ans. Given : r = 4.9 cm
Volume V = 4/3πr3
= 4/3 × 22/7 × 4.9 × 4.9 × 4.9
= 493 cm3 (approx.)
OR
Mass of the shot-put = 7.8 × 493 = 3845.44 gm
= 3.85 kg (approx.)
Question. The radius and slant height of a cone are in the ratio 4 : 7. If its curved surface area is 792 cm2, find its radius.
Ans. Let the radius of a cone
r = 4x
and slant height
l = 7x
∴ CSA = 792 cm2
∴ πrl = 792
or, 22/7 × 4x × 7x = 792
x2 = 792 × 7/22 × 4 × 7
= 9
or, x = 3
∴ radius = 4 × 3
= 12 cm
Question. How much ice-cream can be put into a cone with base radius 3·5 cm and height 12 cm ?
Ans. r = 3·5 cm, h = 12 cm
∴ Quantity of ice-cream = 1/3πr2h
= 1/3 × 22/7 × 3·5 × 3·5 × 12 = 154 cm3
Question. How many litres of milk can a hemispherical bowl of diameter 10.5 cm hold ? (Use π = 3.14)
Ans. Given, diameter of hemispherical bowl = 10.5 cm
∴ Radius of hemispherical bowl
r = 10.5/2 = 5.25 cm
Volume of hemispherical bowl = 2/3 πr3
= 2/3 × 3.14 × (5.25)3
= 302.91
= 303 cm3
∴ Amount of milk that the hemispherical bowl can hold
= 0.303 litres
Question. A match box measures 4 cm × 2.5 cm × 1.5 cm. What will be the volume of a packet containing 12 such boxes ?
Ans. Volume of one match box = 4 × 2.5 × 1.5 cm3
= 15 cm3
Volume of 12 such boxes = 15 × 12
= 180 cm3.
Question. Find the radius of a sphere whose surface area is 616 cm2.
Ans. Surface area of sphere = 4πr2
∴ 4πr2 = 616
or, πr2 = 154
or, r2 = 154 × 7/22 (∴ π = 22/7)
∴ r = 7
Hence, the radius of sphere is 7 cm.
Question. A joker’s cap is in the form of right circular cone of base radius 7 cm and slant height 25 cm. Find the area of sheet required for 10 such caps.
OR
A joker’s cap is in the form of right circular cone of base radius 7 cm and height 24 cm. Find the area of the sheet required to make 10 such caps.
Ans. Given, l = 25 cm, r = 7 cm
Area required = C.S.A of cone
= πrl
= 22/7 × 7 × 25
= 550 cm2
Area required to make 10 such caps
= 10 × 550 = 5500 cm2
= 0.55 m2.
Question. Bhavya has a piece of canvas whose area is 552 m2. 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
amounts to approximately 2 m2. Find the volume of the tent that can be made with it. (Take π = 22/7)
Ans. Curved surface area of the tent = 552 – 2
= 550 m2
Radius (r) = 7 m
∴ π × 7 × l = 550
or, l = 25 m
∴ h = √l2 – r2
∴ h = √252 − 72
= 24 m
Volume of the tent = 1/3 × 22/7 × 7 × 7 × 24
= 1232 m3.
Long Answer Type Questions:
Question. A right triangle ABC with sides 5 cm, 12 cm and 13 cm is revolved about the side 5 cm. Find the volume of the solid so obtained. If, it is revolved about the side 12 cm, what would be the ratio of volumes of two solids obtained in two cases ?
Ans. Case I : When revolved about the side 5 cm.
Here, r = 12 cm, h = 5 cm
Volume = 1/3πr2h = 1/3π × (12)2 × 5 .
= 240π cm3
Cone is formed with radius 12 cm and height 5 cm.
Case II : When revolved about 12 cm,
r = 5 cm, h = 12 cm
∴ Volume = 1/3πr2h
= 1/3π × 5 × 5 × 12
= 100 π cm3
∴ Ratio of the two volume = 240 π : 100 π
= 12 : 5
Cone is formed with radius 5 cm and height 12 cm.
Question. The diameter of the moon is approximately onefourth the diameter of earth. What fraction of volume of earth is the volume of moon ?
Ans. Let the diameter of earth be d.
∴ The radius of the earth will be r1 = d/2
Diameter of moon will be d/4 and radius of moon
(r2) = d/8
Volume of moon = 4/3πr23
= 4/3π(d/8)3
= 1/512 ×πd3 × 4/3
Volume of earth = 4/3πr13
= 4/3π(d/2)3
= 1/8 × 4/3πd3
Volume of moon/Volume of earth = 1/512 ×πd3 × 4/3 / 1/8 × 4/3πd3
= 1/64
or, Volume of moon = 1/64 (Volume of earth)
or, Volume of earth/Volume of moon = 64/1
Question. A hemispherical dome, open at base is made from sheet of fibre. If the diameter of hemispherical dome is 80 cm and 13/170 of sheet actually used was wasted in making the dome, then find the cost of dome at the rate of Rs 35/100 cm2
Ans. Given, Diameter = 80 cm
∴ r = 40 cm
C.S.A. of the dome = 2πr2
= 2 × 22/7 × 40 × 40
C.S.A. = 70400/7 cm2
since, 13/170 of sheet was wasted,
Area of sheet wasted = 13/170 × 70400/7
= 915200/1190 cm2
∴ Total area = 70400/7 + 915200/1190
= 10826.21 cm2
Cost of sheet per square metre
= Rs 35/100
∴ Total cost of sheet
= 35/100 × 10826.21
= Rs 3789.17.
Question. The water for a industry is stored in a hemispherical tank of internal diameter 14 m. The tank contains 40 kilolitres of water. Water is pumped into the tank to fill it to full capacity. Calculate the volume of water pumped into the tank.
Ans. Volume of hemispherical tank
= 2/3πr3
= 2/3 × 22/7 × 7 × 7 × 7 m3
= 2156/3 m3
= 718.67 m3
= 718.67 kl.
Volume of water already present = 40 Kilolitres
∴ Volume to be pumped = 718.67 – 40
= 678.67 m3.
Question. The capacity of a cuboidal tank is 50,000 litres of water. Find the breadth of the tank, if its length and depth are respectively 2.5 m and 10 m.
Ans. Volume of cuboidal tank = 50000 litres
= 5 × 104 litres
= 5 × 104 × 103 cm3
[1 litre = 1000 cm3]
Given,
Length of cuboidal tank = 2.5 m
= 250 cm
Depth of cuboidal tank = 10 m = 1000 cm
∴ Volume of tank = length × breadth × depth
or, 5 × 104 × 103 = 250 × breadth × 1000
or, breadth = 5 × 104 × 103/250 × 1000
= 200 cm = 2 m.
Question. A cube and cuboid have the same volume. The dimensions of the cuboid are in the ratio of 1 : 2 : 4. If the difference between the cost of polishing the cuboid and the cube at the rate of Rs 5 per m2 is Rs 80, find the edge of the cube.
Ans. Let the dimensions of cuboid be x, 2x and 4x then volume of cube = volume of cuboid
or, a3 = x × 2x × 4x [a is the side of a cube]
or, a3 = 8x3
or, a = 2x
Difference in S.A. of cuboid and cube
= S.A. of cuboid – S.A. of cube
= 2(x·2x + 2x·4x + 4x·x) – 6(2x)2
= 2(2x2 + 8x2 + 4x2) – 6 × 4x2
= 28x2 – 24x2 = 4x2
∴ Difference in cost of polishing
= 5 × 4x2 = 20x2 = 80
or, x2 = 4
or, x = 2 m
∴ Edge of cube = 2x = 4 m.
Question. A village has a population of 4000 people. 60 litres of water is required per person per day. The village tanker of water is cuboidal in shape with dimensions 48 m × 27 m × 5 m which is completely filled with water. For how many days the water of this is sufficient ?
Ans. The given tank is cuboidal in shape having its length
(l) as 48 m, breadth (b) as 27 m, and height (h) as 5 m
∴ Volume of tank = l × b × h
= 48 × 27 × 5
= 6480 m3
= 6480000 litres
Total water consumption in village per day
= 4000 × 60
= 240000 litres
∴ Number of days for which water is sufficient
= 6480000/240000
= 27 days.
Question. Curved surface of cylindrical reservoir 12 m deep is plastered from inside with concrete mixture at rate of Rs 15 per m2 . If the total payment made is of Rs 5652, then find the capacity of this reservoir in litres. (User π = 3.14)
Ans. Height of cylindrical reservoir = 12 m
Total cost of plastering it from inside = Rs 5652
Cost of 1 m2 = Rs 15
∴ Surface area plastered = Total Cost/Cost / m2
2πrh = 5652/15
∴ 2 × 3.14 × r × 12 = 5652/15
∴ r = 5652/3.14 × 15 × 2 × 12
≈ 5 m
∴ Volume = πr2h
= 3.14 × 5 × 5 × 12
= 942 m3
= 942000 litres
