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The Chapter-13, Surface Areas and Volumes of Class 9 NCERT Mathematics syllabus discusses the Solid Figures (Solids) such as cuboids, cylinder, etc and further establishes formulae for the Surface areas and Volumes of the same.
Class 9 Mathematics Chapter 13 Surface Areas and Volumes falls under the unit Mensuration which carries 14 marks in total with 5 questions.
The following article discusses Multiple Choice Questions related to Class 9 Mathematics Chapter 13 Surface Areas and Volumes
Find NCERT Solutions for Class-9 Mathematics here:
NCERT Solutions for Class 9 Mathematics Chapter 13
Q.1. Determine the volume and surface area of a cuboid that is 16 m long, 14 m wide, and 7 m high.
a) 1500 m³, 868 cm²
(b) 1605 m³, 848 cm²
(c) 1707 m³, 800 cm²
(d) 1568 m³, 868 cm²
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Ans: d) 1568 m³, 868 cm²
Explanation: Given: l = 16 m, b = 14 m, h = 7 m.
As we know, Volume = l*b*h
= 16*14*7
= 1568 m³
Also, Surface area = 2(lb+bh+lh)
= 2(16*14 + 14 * 7 + 16*7)
= 2* 234
= 868 cm²
Q.2. A right circular cone with a radius of 5 cm and a slant height of 13 cm has the height:
(a) 9 cm
(b) 12 cm
(c) 16 cm
(d) 14 cm
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Ans: b) 12 cm
Explanation: As we know, Slant height ( l ) = √h²+r² units.
Here, r = 5 cm and l = 13 cm, h =?
So, the formula to find height (h) is, h = √l²- r²
= √(13)²- (5)²
= √169 - 25 = √144
= 12 cm
So, the height is 12 cm.
Q.3. The length of the longest pole that can be placed in a room of dimensions (10 m × 10 m × 5 m) is
(a) 15 m
(b) 16 m
(c) 10 m
(d) 12 m
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Ans: a) 15 m
Explanation: Given, Dimension of the room is, l = 10m, b = 10 m, h = 5m
Therefore, length of the longest pole = Diagonal of a cuboid i.e., room
And Diagonal is given by the formula,
√l²+b²+h²
= √(10)²+(10)²+(5)²
= √100+100+25
= √225
= 15 m.
So, the length of the longest pole that can be placed in a room of dimensions (10 m × 10 m × 5 m) is 15m.
Q.4. The base diameter of a cone is 10.5 cm, and its slant height is 10 cm. The area of a curved surface is calculated as follows:
(a) 150 sq.cm
(b) 164.85 sq.cm
(c) 177 sq.cm
(d) 180 sq.cm
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Ans: b) 164.85 sq.cm
Explanation: Given, the base diameter of a cone (d) = 10.5 cm, then radius of the cone (r) = d/2
R = 10.5 /2
r = 5.25
Also, slant height of the cone (l) = 10 cm
Then, the curved surface area is given by formula,
πrl
= 3.14*5.25*10
= 16.485*10
= 164.85 sq.cm
So, the curved surface area of the cone is 164.85 sq. cm.
Q.5. A cone has a height of 8.4 cm and a radius of 2.1 cm at its base. It is melted and then recast into a shape. Compute the radius of the sphere's radius is
(a) 2.2 cm
(b) 2.1 cm
(c) 3.4 cm
(d) 4.6 cm
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Ans: 2.1 cm
Explanation: Given, the height of the cone = 8.4 cm, a radius of its base = 2.1 cm, the radius of the sphere =?
The formula to find the volume of the cone is 1/ 3 πr²h
So, by putting the values in the given formula we get, (1/3)π(2.1)² (8.4)
The volume of the cone = (1/3)π(2.1)² (8.4) ------------------------(1)
Now, the volume of the sphere = 4/3 πR³
4/3 πR³ = (1/3)π(2.1)² (8.4)
4R³ = (2.1)² (8.4)
R³ = ((2.1)²(8.4) ) / 4
R³ = (2.1)² (2.1)
R³ = (2.1)³
R = 2.1 cm
So, the radius of the sphere is 2.1 cm.
Q.6. The surface area of a sphere of radius 21 cm is: (Assume π = 22/7)
(a) 1386 sq.cm
(b) 1400 sq.cm
(c) 5544 sq.cm
(d) 2000 sq.cm
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Ans: c) 5544 sq.cm
Explanation: Given, radius of the sphere(r) = 21 cm.
The surface area of sphere = 4πr²
= 4 * 22/7 * 21* 21
= 4* 22* 3 * 21
= 5544 cm²
So, the surface area of a sphere of radius 21 cm is 5544 sq. cm.
Q.7. The radii of the two cylinders are in a 2:3 ratio, and their heights are in a 5:3 ratio. Then, find their ratio of the volume.
(a) 10: 17
(b) 20: 27
(c) 17: 27
(d) 20: 37
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Ans: b) 20: 27
Explanation: Let the radii of the cylinders be R1 and R2 and their heights are H1 and H2.
Therefore, the ratio of the radius = R1/R2 = â??
Similarly, the ratio of the height = H1/H2 = 5/3
To find, the ratio of the volumes of the cylinders = V1/V2 =?
Let us assume, V1 = Volume of the first cylinder = πR1²H1
Similarly, V2 = Volume of the second cylinder = πR2²H2
V1/V2 = πR1²H1 / πR2²H2
= R1²H1 / R2²H2
= (R1/R2)² * (H1/H2)
= 4/9 * 5/3
= 20/27
V1/V2 = 20: 27
So, the ratio of the volumes of the cylinders is 20:27
Q.8. The surface area of a cube with a 6 cm edge is:
(a) 68 sq.cm
(b) 300 sq.cm
(c) 154 sq.cm
(d) 216 sq.cm
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Ans: d) 216 sq.cm
Explanation: Given edge of the cube (a) = 6cm
The formula of the surface area of a cube = 6a²
= 6 * (6)²
= 6 * 36
= 216 sq.cm.
So, the surface area of a cube with a 6 cm edge is 216 sq. cm.
Q.9. A right circular cylinder has a curved surface area of 4.4 sq. cm. The base has a radius of 0.7 cm. What will be the height of the cylinder?
(a) 5 cm
(b) 2 cm
(c) 1 cm
(d) 3.5 cm
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Ans: c) 1 cm
Explanation: Given, curved surface area of the cylinder (a) = 4.4 sq.cm, radius of the cylinder(r) = 0.7 cm. To find, height of the cylinder(h) =?
As we know, the curved surface area of the cylinder = 2πrh
4.4 = 2 * (22/7) * 0.7 * h
4.4 = 2 * 22* 0.1 * h
4.4 / (2 * 22* 0.1) = h
4.4 / 4.4 = h
1 = h
Therefore, h = 1 cm.
The height of the cylinder is 1 cm.
Q.10. Iron weighs 50 kg per cubic meter and is 9 m long, 40 cm wide, and 20 cm deep. What will be the weight of the beam?
(a) 37 kg
(b) 36 kg
(c) 40 kg
(d) 50 kg
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Ans: (b) 36 kg
The shape of the beam is cuboid. So, the volume of a cuboid is given by, l*b*h
Given: l = 9 m,
b = 40 cm = 40 / 100 m = 0.4 m
h = 20 cm = 20 / 100 m = 0.2 m
So, putting the values in the formula we get,
l*b*h = 9*0.4*0.2 = 0.72 m
So, the volume of the beam is 0.72 m
Now, if the iron weighs 50 kg per cubic meter, the weight of the beam = 0.72 * 50 = 36 kg
Therefore, the weight of the beam = 36 kg.
Q.11. The dimensions of a rectangular sandbox are 5 m wide by 2 m long. How many cubic meters of sand is required to fill the box to a depth of 10 cm?
(a) 1
(b) 20
(c) 102
(d) 1000
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Ans: a) 1
Explanation: The box has the shape of a cuboid. So, the volume of a cuboid is given by, l*b*h.
Now, we have, l = 5m, b = 2m, h = 10 cm = 10/ 100 = 0.1 m
So, the volume of the sand that can be filled in the rectangular box = 5*2*0.1 = 10*0.1 = 1 m³.
Q.12. If the radius of a cylinder is doubled and the height is halved, the curved surface area will be
(a) halved
(b) doubled
(c) same
(d) four-time
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Ans: (c) same
Explanation: Determine the Curved Surface Area of a Cylinder using the given conditions.
Assume r is the radius of the cylinder and h is the height of the cylinder.
The cylinder's curved surface area is 2πrh.
If the radius is doubled and the height is cut in half,
R=2r, H = h/2
The cylinder's curved surface area =2πRH
= 2(2r) π (h/2)
= 2πrh
As a result, the cylinder's curved surface area will be the same.
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