CBSE Class 9 Maths Ganita Manjari Part 2 Ch 14 Math of Space Surface Area and Volume MCQs with Answers Set 01

Answer: (a) 150 cm²

Explanation:
1. A cube has 6 square faces, so TSA = 6a².
2. 6 × 5² = 6 × 25 = 150 cm².
3. 125 is the volume (and has the wrong unit); 25 is one face.

Teacher's Note:
1. 6 faces, each a².
2. Area units are square units.

Answer: (c) 120 cm³

Explanation:
1. Volume = l × w × h = 8 × 5 × 3 = 120.
2. Volume is measured in cubic units, so 120 cm³.
3. 158 cm² is the total surface area, 2(40 + 15 + 24).

Teacher's Note:
1. Check both the number and the unit.
2. Do not confuse volume with surface area.

Answer: (b) the length, the width and the height are all equal

Explanation:
1. A cube has l = w = h.
2. Then 2(lw + wh + hl) becomes 6l² and lwh becomes l³.
3. A square base alone (like 5 × 5 × 12) is not enough.

Teacher's Note:
1. All three dimensions equal.
2. Volume = surface area happens only for side 6.

Answer: (d) 6, 12 and 8

Explanation:
1. 6 rectangular faces.
2. 12 edges: four of each of the three lengths.
3. 8 vertices, one at each corner.

Teacher's Note:
1. Check with Euler: 6 − 12 + 8 = 2.
2. Use a real box to count.

Answer: (c) 7 cm

Explanation:
1. a³ = 343.
2. a = ∛343 = 7 cm, since 7³ = 343.
3. 49 = 7² is a face area, not a side.

Teacher's Note:
1. Learn cubes up to 10³.
2. Take the cube root, not the square root.

Answer: (a) 2πrh

Explanation:
1. Unroll the curved surface: it is a rectangle.
2. Width = base circumference 2πr; height = h.
3. Area = 2πrh. (b) is volume; (c) is total surface area.

Teacher's Note:
1. Use the unrolled label idea.
2. Curved only: no circles.

Answer: (d) S

Explanation:
1. Half a sphere has one curved surface and one flat circular face.
2. That is a hemisphere, solid S.
3. P is a cylinder, Q a cone, R the whole sphere.

Teacher's Note:
1. Hemi = half.
2. Count flat faces: a hemisphere has one.

Answer: (b) 1540 cm³

Explanation:
1. V = πr²h = \( \frac{22}{7} \) × 49 × 10.
2. = 22 × 7 × 10 = 1540 cm³.
3. 440 is the curved surface area 2πrh.

Teacher's Note:
1. Cancel 7 first.
2. 154 is the base area only.

Answer: (c) 10 cm

Explanation:
1. l² = h² + r² = 64 + 36 = 100.
2. l = 10 cm.
3. 14 adds the lengths; 48 multiplies them.

Teacher's Note:
1. r, h and l form a right triangle.
2. 6–8–10 is a Pythagorean triple.

Answer: (a) πrl

Explanation:
1. The unrolled curved surface is a sector of radius l.
2. Its arc equals the base circumference 2πr, so its area is ½ × 2πr × l = πrl.
3. (b) adds the base, giving the total surface area.

Teacher's Note:
1. Uses slant height, not height.
2. Curved only: no base.

Answer: (d) one third of the volume of the cylinder

Explanation:
1. Cone: ⅓πr²h. Cylinder: πr²h.
2. So the cone holds one third as much.
3. Experiment: three cones of salt fill the cylinder exactly.

Teacher's Note:
1. Do the salt experiment in class.
2. Same base, same height is essential.

Answer: (b) 4πr²

Explanation:
1. Surface area of a sphere = 4πr², four times a circle of the same radius.
2. (a) is the volume; an area must involve r², not r³.
3. 3πr² and 2πr² belong to the hemisphere.

Teacher's Note:
1. Area → r², volume → r³.
2. Use the string-winding activity.

Answer: (a) 3 cm

Explanation:
1. (4/3)πr³ = 36π.
2. r³ = 36 × 3/4 = 27.
3. r = 3 cm. (d) forgets the cube root.

Teacher's Note:
1. Cancel π first.
2. Cube root at the end.

Answer: (c) the flat circular face has to be counted as well

Explanation:
1. Curved surface = half of 4πr² = 2πr².
2. The flat circular face adds πr².
3. Total = 3πr².

Teacher's Note:
1. A solid hemisphere has two surfaces.
2. (d) explains 2πr² only.

Answer: (d) one half of the volume of the sphere

Explanation:
1. A hemisphere is exactly half a sphere.
2. (2/3)πr³ = ½ × (4/3)πr³.
3. But its total surface area is ¾ of the sphere's, because of the new flat face.

Teacher's Note:
1. Volume halves; surface area does not.
2. Good discussion point.

Answer: (b) 18π cm³

Explanation:
1. V = (2/3)πr³.
2. = (2/3) × π × 27 = 18π cm³.
3. 36π is the whole sphere.

Teacher's Note:
1. Cube the radius first.
2. Half of the sphere's 36π.

Answer: (a) the axis is perpendicular to the base

Explanation:
1. 'Right' means the axis stands at right angles to the base.
2. 'Circular' means the cross-section is a circle (option b).
3. A leaning cylinder is called oblique.

Teacher's Note:
1. Each word has its own job.
2. Show an oblique stack of coins.

Answer: (c) a circle

Explanation:
1. A slice parallel to the base is a cross-section.
2. It is a circle of the same radius as the base, at every height.
3. A slice along the axis gives a rectangle.

Teacher's Note:
1. Constant cross-section → V = base area × height.
2. Direction of cut matters.

Answer: (d) a cone of base radius 6 cm and height 8 cm

Explanation:
1. The side on the axis (8 cm) stays still and becomes the height.
2. The other leg (6 cm) sweeps the base, so it is the radius.
3. The hypotenuse (10 cm) becomes the slant height. (c) swaps r and h.

Teacher's Note:
1. Axis side = height.
2. Watch which side is the axis.

Answer: (b) \( \frac{1}{3} \) × base area × height

Explanation:
1. Any pyramid has volume ⅓ × base area × height, whatever the base shape.
2. A cone is a pyramid with a circular base.
3. Perimeter × height gives an area, not a volume.

Teacher's Note:
1. Same rule as the cone.
2. Check units to reject (d).

Answer: (c) Archimedes

Explanation:
1. Archimedes (about 225 BCE) found 4πr² and (4/3)πr³.
2. He compared the sphere with the cylinder that just contains it.
3. Baudhāyana is linked to the right-triangle theorem.

Teacher's Note:
1. Link to the tomb story.
2. Know one fact about each name.

Answer: (a) 616 cm²

Explanation:
1. Radius = 14 ÷ 2 = 7 cm.
2. 4πr² = 4 × \( \frac{22}{7} \) × 49 = 4 × 154 = 616 cm².
3. 2464 comes from using 14 as the radius.

Teacher's Note:
1. Halve the diameter first.
2. 154 is one great circle.

Answer: (d) 160 cm²

Explanation:
1. The cuboid is 8 cm × 4 cm × 4 cm.
2. TSA = 2(32 + 16 + 32) = 160 cm².
3. Check: two cubes give 192, minus 2 hidden faces of 16 = 160.

Teacher's Note:
1. Joined faces disappear.
2. Two methods, same answer.

Answer: (b) 64

Explanation:
1. Volume 4³ = 64 cm³; each small cube is 1 cm³.
2. So 64 small cubes (4 × 4 × 4).
3. 16 counts only one layer.

Teacher's Note:
1. Think in layers.
2. Volume ratio gives the count here.

Answer: (c) eight times as large

Explanation:
1. Volume depends on r³.
2. (2r)³ = 8r³, so the volume is 8 times.
3. The surface area would be 4 times.

Teacher's Note:
1. Scale factor cubed for volume.
2. Squared for area.

Answer: (a) the sphere

Explanation:
1. For a given volume the sphere has the least surface area.
2. Example: cube of side 6 has V = 216, S = 216; a sphere of V = 216 has r ≈ 3.72 and S ≈ 174.
3. Scaling changes both areas by the same factor, so the size does not matter.

Teacher's Note:
1. Why drops of water are round.
2. Compact shape = less surface.

Answer: (d) 12 cm

Explanation:
1. h² = l² − r² = 169 − 25 = 144.
2. h = 12 cm.
3. 8 = 13 − 5 subtracts the lengths, not their squares.

Teacher's Note:
1. 5–12–13 triple.
2. Subtract squares.

Answer: (b) 1000 cm³

Explanation:
1. 1 litre is the volume of a cube of side 10 cm.
2. 10³ = 1000 cm³.
3. Also 1 m³ = 1000 litres.

Teacher's Note:
1. Remember both conversions.
2. 1 m³ = 1 000 000 cm³.

Answer: (c) πr(2h + r)

Explanation:
1. Curved surface 2πrh plus one circle πr².
2. 2πrh + πr² = πr(2h + r).
3. (a) is for a tin closed at both ends.

Teacher's Note:
1. Count the circles: one here.
2. Factorise πr out.

Answer: (a) \( \frac{a}{2} \)

Explanation:
1. The largest sphere touches all six faces.
2. So its diameter equals the edge: 2r = a.
3. r = a/2. (d) is the sphere through the corners, which is outside the cube.

Teacher's Note:
1. Diameter = edge.
2. Inscribed vs circumscribed sphere.