Q1.
A cone has a radius of 7 cm and a height of 24 cm. What is its volume? (Use π = 22/7)
- 1188 cm³
- 1344 cm³
- 1078 cm³
- 1232 cm³
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Answer: D. 1232 cm³
Volume of cone = (1/3)πr²h = (1/3) × (22/7) × (7 cm)² × 24 cm = (1/3) × 22 × 7 × 24 cm³ = 22 × 7 × 8 cm³ = 1232 cm³.
Q2.
What is the ratio of the volumes of a cylinder, a cone, and a sphere, if they all have the same radius 'r' and the height of the cylinder and cone is equal to the diameter of the sphere (i.e., h = 2r)?
- 3:1:2
- 1:2:3
- 2:3:1
- 1:3:2
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Answer: A. 3:1:2
Let the radius be 'r' and height 'h = 2r'.
Volume of Cylinder (Vc) = πr²h = πr²(2r) = 2πr³
Volume of Cone (Vco) = (1/3)πr²h = (1/3)πr²(2r) = (2/3)πr³
Volume of Sphere (Vs) = (4/3)πr³
Ratio Vc : Vco : Vs = 2πr³ : (2/3)πr³ : (4/3)πr³
Multiplying by 3/(2πr³) gives 3 : 1 : 2.
Q3.
Observe the line chart showing the volume of a sphere as its radius increases. What trend does the chart illustrate regarding the relationship between the sphere's radius and its volume?
- Volume increases linearly with radius.
- Volume decreases as radius increases.
- Volume increases cubically with radius.
- Volume remains constant regardless of radius.
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Answer: C. Volume increases cubically with radius.
The formula for the volume of a sphere is V = (4/3)πr³. This shows that the volume is directly proportional to the cube of the radius (r³). The chart values (4.19, 33.5, 113, 268, 524) show a rapid, non-linear increase, consistent with a cubic relationship.
Q4.
A medicine capsule is in the shape of a cylinder with two hemispheres stuck to each of its ends. The length of the cylindrical part is 10 mm and the radius of the capsule is 3 mm. Find the total volume of the capsule. (Use π = 22/7)
- 108π mm³
- 144π mm³
- 90π mm³
- 126π mm³
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Answer: D. 126π mm³
Radius of capsule (r) = 3 mm. Length of cylindrical part (h) = 10 mm.
Volume of cylindrical part = πr²h = π(3)²(10) = 90π mm³.
Volume of two hemispheres = Volume of one sphere = (4/3)πr³ = (4/3)π(3)³ = (4/3)π(27) = 36π mm³.
Total volume of capsule = Volume of cylindrical part + Volume of two hemispheres = 90π + 36π = 126π mm³.
Q5.
A combined solid has a total volume of 1000 cm³. According to the pie chart showing the volume distribution, what is the volume contributed by the cylindrical part?
- 250 cm³
- 200 cm³
- 550 cm³
- 450 cm³
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Answer: C. 550 cm³
From the pie chart, the cylindrical part contributes 55% of the total volume.
Volume of cylindrical part = 55% of 1000 cm³ = (55/100) × 1000 cm³ = 550 cm³.
Q6.
An iron sphere of radius 12 cm is melted and recast into small spheres, each of radius 2 cm. How many small spheres can be formed?
- 36
- 72
- 108
- 216
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Answer: D. 216
Volume of large sphere = (4/3)π(12)³ cm³.
Volume of one small sphere = (4/3)π(2)³ cm³.
Number of small spheres = (Volume of large sphere) / (Volume of one small sphere) = [(4/3)π(12)³] / [(4/3)π(2)³] = (12/2)³ = 6³ = 216.
Q7.
A metallic cuboid of dimensions 4 cm x 4 cm x 2 cm is melted and recast into a single sphere. Assuming π = 3, what is the diameter of the sphere?
- 4 cm
- 2 cm
- 6 cm
- 8 cm
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Answer: A. 4 cm
Volume of cuboid = length × breadth × height = 4 cm × 4 cm × 2 cm = 32 cm³.
When recast, the volume remains the same.
Volume of sphere = (4/3)πR³.
Given V_sphere = 32 cm³ and π = 3.
(4/3) × 3 × R³ = 32
4R³ = 32
R³ = 8
R = 2 cm.
Diameter = 2R = 2 × 2 = 4 cm.
Q8.
A circular racing track has been developed in a field. If the difference between the outer circumference and the inner circumference of the racing track is 33 m, then find the width of the track (in m). (Use pi = 22/7)
- 5 whole 1/4
- 5 whole 3/4
- 5 whole 1/5
- 4 whole 3/4
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Answer: A. 5 whole 1/4
Let the outer radius be R and inner radius be r. The width of the track is (R - r). The difference in circumferences is 2πR - 2πr = 2π(R - r) = 33. Substituting π = 22/7, we get 2 * (22/7) * (R - r) = 33. Therefore, 44/7 * (R - r) = 33, which gives (R - r) = (33 * 7) / 44 = 21/4 = 5 whole 1/4 m.
Q9.
The length of each edge of a cube is 2.6 cm. What is the total surface area (in cm squared) of the cube?
- 39.96
- 40.56
- 40.76
- 40.36
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Answer: B. 40.56
The total surface area of a cube is given by the formula 6 * (side squared). Here, the side of the cube is 2.6 cm. So, total surface area = 6 * (2.6^2) = 6 * 6.76 = 40.56 cm squared.
Q10.
A solid metallic sphere of radius 3 cm is melted and drawn into a wire of thickness 4 mm. What is the length of the wire (in m)?
- 9.25
- 7.5
- 9
- 8
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Answer: C. 9
Volume of the metallic sphere = Volume of the cylindrical wire. Radius of the sphere = 3 cm. Thickness of the wire means its diameter = 4 mm, so its radius = 2 mm = 0.2 cm. Volume of sphere = (4/3) * π * (radius cubed) = (4/3) * π * (3^3) = 36π. Volume of wire = π * (radius squared) * height = π * (0.2^2) * h = 0.04πh. Equating both volumes: 36π = 0.04πh, which gives h = 36 / 0.04 = 900 cm = 9 m.
Q11.
The area of a rectangle is 453.6 m squared. If its length is 27m, then what is the perimeter of the rectangle?
- 86.6 m
- 87.6 m
- 85.4 m
- 88.8 m
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Answer: B. 87.6 m
Area of a rectangle = length * breadth. Given, area = 453.6 m squared and length = 27 m. So, 27 * breadth = 453.6, which gives breadth = 453.6 / 27 = 16.8 m. The perimeter of a rectangle = 2 * (length + breadth) = 2 * (27 + 16.8) = 2 * 43.8 = 87.6 m.
Q12.
The ratio of the length, width and height of a cuboid is 4 : 3 : 5 and the sum of the lengths of all its edges is 144 cm. Find the total surface area of the cuboid.
- 1620 cm squared
- 1026 cm squared
- 756 cm squared
- 846 cm squared
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Answer: D. 846 cm squared
Let the length, width, and height of the cuboid be 4x, 3x, and 5x respectively. A cuboid has 12 edges (4 lengths, 4 widths, 4 heights). The sum of all edges = 4 * (length + width + height) = 144. So, 4 * (4x + 3x + 5x) = 144 => 4 * 12x = 144 => 48x = 144 => x = 3. Therefore, length = 12 cm, width = 9 cm, and height = 15 cm. Total surface area of a cuboid = 2 * (length * width + width * height + height * length) = 2 * (12 * 9 + 9 * 15 + 15 * 12) = 2 * (108 + 135 + 180) = 2 * 423 = 846 cm squared.
Q13.
The circumference of a circle is given as 308 m. What is the area of the circle? [Use pi = 22/7]
- 7646 m squared
- 7546 m squared
- 7556 m squared
- 7446 m squared
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Answer: B. 7546 m squared
Circumference of a circle = 2 * π * r = 308. Substituting π = 22/7, we get 2 * (22/7) * r = 308 => 44/7 * r = 308 => r = (308 * 7) / 44 = 49 m. Area of the circle = π * (radius squared) = (22/7) * 49 * 49 = 22 * 7 * 49 = 154 * 49 = 7546 m squared.
Q14.
The outer radius of a spherical shell is 9 cm and the thickness of the shell is 1 cm. Find the volume of the metal used for the shell (in cubic cm). (Use pi = 22/7)
- 912 whole 2/3
- 915 whole 1/3
- 909 whole 1/3
- 909 whole 2/5
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Answer: C. 909 whole 1/3
Outer radius (R) = 9 cm. Thickness = 1 cm, so inner radius (r) = 9 - 1 = 8 cm. Volume of the metal in a spherical shell = (4/3) * π * (R cubed - r cubed) = (4/3) * (22/7) * (9^3 - 8^3) = 88/21 * (729 - 512) = 88/21 * 217 = (88 * 31) / 3 = 2728 / 3 = 909 whole 1/3 cubic cm.
Q15.
The radius of the base of a conical tent is 9m and its height is 12 m, find the cost of the material needed to make it if it costs Rupees 100 per pi m squared.
- Rupees 14,500
- Rupees 13,000
- Rupees 15,000
- Rupees 13,500
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Answer: D. Rupees 13,500
Radius (r) = 9 m, Height (h) = 12 m. Slant height (l) = square root of (r squared + h squared) = square root of (9 squared + 12 squared) = square root of (81 + 144) = square root of 225 = 15 m. The material needed for a conical tent corresponds to its curved surface area = π * r * l = π * 9 * 15 = 135π m squared. Cost of the material = Area * Rate = 135π * (100 / π) = 135 * 100 = Rupees 13,500.
Q16.
A right circular cone is surmounted on a hemisphere. Base radius of the cone is equal to radius of the hemisphere. The diameter of the hemisphere is 12 cm while the height of the cone is 8 cm. Find the cost of painting the compound object if it costs Rupees 25 to paint pi cm squared.
- Rupees 10,371
- Rupees 3,300
- Rupees 6,930
- Rupees 4,400
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Answer: B. Rupees 3,300
Diameter of hemisphere = 12 cm, so radius (r) = 6 cm. Height of cone (h) = 8 cm. Slant height of cone (l) = square root of (6 squared + 8 squared) = square root of 100 = 10 cm. Total surface area of the compound object = Curved surface area of cone + Curved surface area of hemisphere = π * r * l + 2 * π * (r squared) = π * 6 * 10 + 2 * π * (6 squared) = 60π + 72π = 132π cm squared. Cost of painting = 132 * 25 = Rupees 3,300.
Q17.
A semicircle has been drawn on the length of a rectangle. The area of the shaded region in the figure is: (Assume the semicircle is removed from the rectangle, length = 14 cm, breadth = 10 cm)
- 77 cm squared
- 14 cm squared
- 129 cm squared
- 63 cm squared
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Answer: D. 63 cm squared
Length of the rectangle = 14 cm, Breadth = 10 cm. Total area of the rectangle = length * breadth = 14 * 10 = 140 cm squared. A semicircle is drawn on the length, so its diameter is 14 cm, meaning its radius (r) = 7 cm. Area of the semicircle = (π * r squared) / 2 = (22/7 * 7 * 7) / 2 = 77 cm squared. Area of the shaded region = Area of rectangle - Area of semicircle = 140 - 77 = 63 cm squared.
Q18.
The area of four side walls of a cubical box is 36 cm squared. Its edge is:
- square root of 6 cm
- 3 cm
- 9 cm
- 6 cm
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Answer: B. 3 cm
The area of the four side walls of a cube (lateral surface area) is given by 4 * (edge squared). Given, 4 * (edge squared) = 36. Therefore, edge squared = 36 / 4 = 9. Taking the square root, edge = square root of 9 = 3 cm.
Q19.
If length of each side of a cube is doubled, then its volume becomes how many times the original volume?
- becomes 8 times
- is doubled
- becomes 9 times
- becomes 6 times
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Answer: A. becomes 8 times
Let the initial side of the cube be x. Original volume = x cubed. When the side is doubled, the new side becomes 2x. New volume = (2x) cubed = 8 * (x cubed). Therefore, the new volume becomes 8 times the original volume.
Q20.
The volume of a hemisphere is 19404 cm cubed. Its radius is: (Use pi = 22/7)
- 30 cm
- 19 cm
- 20 cm
- 21 cm
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Answer: D. 21 cm
Volume of a hemisphere = (2/3) * π * (radius cubed). Given, (2/3) * (22/7) * (r cubed) = 19404. This gives 44/21 * (r cubed) = 19404 => r cubed = (19404 * 21) / 44 = 441 * 21 = 21 * 21 * 21. Taking the cube root, r = 21 cm.
Q21.
The interior angle of a regular polygon is 108 degrees. The number of the sides of the polygon is:
- 5
- 360
- 15
- 108
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Answer: A. 5
Each interior angle of a regular polygon = 180 minus (each exterior angle). Given interior angle = 108 degrees, so each exterior angle = 180 - 108 = 72 degrees. The number of sides of a regular polygon = 360 divided by each exterior angle = 360 / 72 = 5 sides.
Q22.
If the base of cylinder is the same as that of a cone, and the height of the cylinder is also the same as that of the cone, then find the ratio of the volumes of the cylinder and the cone.
- 1 : 3
- 3 : 2
- 3 : 1
- 2 : 3
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Answer: C. 3 : 1
Let the common radius be r and common height be h. Volume of cylinder = π * (r squared) * h. Volume of cone = 1/3 * π * (r squared) * h. The ratio of their volumes = [π * (r squared) * h] divided by [1/3 * π * (r squared) * h] = 1 divided by (1/3) = 3 : 1.