CBSE Class 10 Mathematics Surface Areas and Volumes MCQs Set 16

Mathematics Objective Questions and Answers: Chapter 12 Surface Areas and Volumes

Explore reliable objective questions for Chapter 12 Surface Areas and Volumes tailored for Class 10 learners. Utilizing these Mathematics multiple-choice formats ensures thorough preparation and strengthens problem-solving speed for upcoming school assessments.

Download Chapter 12 Surface Areas and Volumes MCQs with Answers

Access the complete set of multiple-choice questions for Chapter 12 Surface Areas and Volumes below. This focused format allows students to isolate specific topics for thorough review and uninterrupted practice alongside official CBSE textbooks.

Multiple Choice Questions

Question 1: The surface areas of two spheres are in the ratio 16 : 9. The ratio of their volumes is
(a) 64 : 27
(b) 16 : 9
(c) 4 : 3
(d) \( 16^3 : 9^3 \)
Show Answer & Explanation

Answer: (a) 64 : 27

Explanation:
1. 4πR² : 4πr² = 16 : 9, so R² : r² = 4² : 3² and R : r = 4 : 3.
2. Volumes are in the ratio R³ : r³ = 4³ : 3³ = 64 : 27.

Teacher's Note:
1. Area ratio = (radius ratio)². Volume ratio = (radius ratio)³.
2. Take the square root first.

Question 2: A solid right circular cone is cut into two parts at the middle of its height by a plane parallel to its base. The ratio of the volume of the smaller cone to the whole cone is
(a) 1 : 2
(b) 1 : 4
(c) 1 : 6
(d) 1 : 8
Show Answer & Explanation

Answer: (d) 1 : 8

Explanation:
1. The small cone is similar to the whole cone with every length halved: height h, radius r against 2h and 2r.
2. Volume ratio = (1/3)πr²h ÷ (1/3)π(2r)²(2h) = 1 ÷ 8, that is 1 : 8.

Teacher's Note:
1. Halving every length divides the volume by 2³.
2. Similar solids have volumes in the cube of the ratio of lengths.

Question 3: If the radius of the base of a right circular cylinder is halved, keeping the height the same, then the ratio of the volume of the new cylinder to that of the original cylinder is
(a) 1 : 2
(b) 2 : 1
(c) 1 : 4
(d) 4 : 1
Show Answer & Explanation

Answer: (c) 1 : 4

Explanation:
1. Original V = πr²h. New V = π(r/2)²h = πr²h/4.
2. The ratio is 1 : 4.

Teacher's Note:
1. Volume depends on the square of the radius.
2. Halving r divides V by 4.

Question 4: The radius of the largest circular cone that can be cut out from a cube of edge 4.2 cm is
(a) 4.2 cm
(b) 2.1 cm
(c) 8.4 cm
(d) 1.05 cm
Show Answer & Explanation

Answer: (b) 2.1 cm

Explanation:
1. The base of the largest cone sits on a face of the cube, so its diameter is the edge, 4.2 cm.
2. Radius = 4.2 ÷ 2 = 2.1 cm.

Teacher's Note:
1. The base circle must fit inside a square face.
2. The largest circle in a square has diameter equal to the side.

Question 5: A sphere of diameter 18 cm is dropped into a cylindrical vessel of diameter 36 cm, partly filled with water. If the sphere is completely submerged, the water level rises (in cm) by
(a) 3
(b) 4
(c) 5
(d) 6
Show Answer & Explanation

Answer: (a) 3

Explanation:
1. Volume of sphere = (4/3)π × 9³ = 972π cm³.
2. Rise h of water: π × 18² × h = 972π, so 324h = 972.
3. h = 3 cm.

Teacher's Note:
1. The volume of water displaced equals the volume of the sphere.
2. The radii are 9 cm and 18 cm.

Question 6: A solid is hemispherical at the bottom and conical (of the same radius) above it. If the curved surface areas of the two parts are equal, then the ratio of the radius to the slant height of the conical part is
(a) 2 : 1
(b) 1 : 2
(c) 1 : 4
(d) 4 : 1
Show Answer & Explanation

Answer: (b) 1 : 2

Explanation:
1. CSA of hemisphere = 2πr²; CSA of cone = πrl.
2. Equal: 2πr² = πrl, so l = 2r.
3. r : l = 1 : 2.

Teacher's Note:
1. Write both curved surface areas and equate them.
2. Cancel πr from both sides.

Question 7: A toy is in the form of a cone mounted on a hemisphere of common base radius 7 cm. If the total height of the toy is 31 cm, the height of the cone is
(a) 31 cm
(b) 38 cm
(c) 7 cm
(d) 24 cm
Show Answer & Explanation

Answer: (d) 24 cm

Explanation:
1. The hemisphere contributes a height equal to its radius, 7 cm.
2. Height of cone = 31 − 7 = 24 cm.

Teacher's Note:
1. The height of a hemisphere is its radius.
2. Total height = hemisphere height + cone height.

Question 8: If the surface area of a sphere is 616 cm², then the radius of the sphere is (Take π = 22/7.)
(a) 14 cm
(b) 7 cm
(c) 3.5 cm
(d) none of these
Show Answer & Explanation

Answer: (b) 7 cm

Explanation:
1. 4πr² = 616, so 4 × (22/7) × r² = 616.
2. r² = 616 × 7 ÷ 88 = 49, so r = 7 cm.

Teacher's Note:
1. Surface area of a sphere is 4πr².
2. 88 ÷ 7 is the value of 4π.

Question 9: The diameter of the Moon is approximately one-fourth of the diameter of the Earth. The approximate ratio of their volumes is
(a) 1 : 16
(b) 1 : 64
(c) 1 : 4
(d) 1 : 128
Show Answer & Explanation

Answer: (b) 1 : 64

Explanation:
1. If the Moon's radius is r, the Earth's radius is 4r.
2. Ratio of volumes = r³ : (4r)³ = 1 : 64.

Teacher's Note:
1. Volume of a sphere is proportional to the cube of its radius.
2. 4³ = 64.

Question 10: If the volume of a cube is 729 cm³, the length of its diagonal is
(a) \( 9\sqrt{2} \) cm
(b) \( 9\sqrt{3} \) cm
(c) 18 cm
(d) \( 18\sqrt{3} \) cm
Show Answer & Explanation

Answer: (b) \( 9\sqrt{3} \) cm

Explanation:
1. a³ = 729, so the edge a = 9 cm.
2. Diagonal of a cube = a√3 = 9√3 cm.

Teacher's Note:
1. Space diagonal = edge × √3.
2. 9³ = 729.

Question 11: The curved surface area of a right circular cone of base radius 14 cm is 440 cm². The slant height of the cone is (Take π = 22/7.)
(a) 10 cm
(b) 11 cm
(c) 12 cm
(d) 13 cm
Show Answer & Explanation

Answer: (a) 10 cm

Explanation:
1. CSA = πrl, so (22/7) × 14 × l = 440.
2. 44l = 440, so l = 10 cm.

Teacher's Note:
1. Use CSA = πrl.
2. (22/7) × 14 = 44.

Question 12: If the volumes of two cones are in the ratio 1 : 4 and their diameters are in the ratio 4 : 5, then the ratio of their heights is
(a) 1 : 5
(b) 5 : 4
(c) 5 : 16
(d) 25 : 64
Show Answer & Explanation

Answer: (d) 25 : 64

Explanation:
1. Radii are in the ratio 4 : 5; let the heights be h and H.
2. (1/3)π(4x)²h ÷ (1/3)π(5x)²H = 1/4, so (16/25)(h/H) = 1/4.
3. h/H = 25/64, that is 25 : 64.

Teacher's Note:
1. Ratio of diameters = ratio of radii.
2. Cancel π and 1/3 from both volumes.

Question 13: A cone of height 7 cm and base radius 3 cm is carved out of a rectangular block of wood 10 cm × 5 cm × 2 cm. The percentage of wood wasted is (Take π = 22/7.)
(a) 34%
(b) 46%
(c) 54%
(d) 66%
Show Answer & Explanation

Answer: (a) 34%

Explanation:
1. Volume of block = 10 × 5 × 2 = 100 cm³.
2. Volume of cone = (1/3)(22/7) × 3 × 3 × 7 = 66 cm³.
3. Wood wasted = 100 − 66 = 34 cm³, which is 34%.

Teacher's Note:
1. The block's volume is exactly 100, so percentage = cm³ wasted.
2. Find the cone volume with π = 22/7.

Question 14: The total surface area of a hemisphere of radius r is
(a) \( \pi r^2 \)
(b) \( 2\pi r^2 \)
(c) \( 3\pi r^2 \)
(d) \( \frac{2}{3}\pi r^2 \)
Show Answer & Explanation

Answer: (c) \( 3\pi r^2 \)

Explanation:
1. Curved surface = 2πr² and the flat circular base = πr².
2. Total = 2πr² + πr² = 3πr².

Teacher's Note:
1. A solid hemisphere has a curved part and a flat base.
2. Do not forget to add the base.

Question 15: A cylindrical vessel 32 cm high with base radius 18 cm is filled with sand. The sand is emptied on the ground and forms a conical heap of height 24 cm. The radius of the base of the heap is
(a) 12 cm
(b) 24 cm
(c) 36 cm
(d) 48 cm
Show Answer & Explanation

Answer: (c) 36 cm

Explanation:
1. Volume of sand = π × 18² × 32 = 10368π cm³.
2. Cone: (1/3)πR² × 24 = 8πR².
3. 8R² = 10368, so R² = 1296 and R = 36 cm.

Teacher's Note:
1. The volume of sand does not change when it is poured out.
2. Equate cylinder volume and cone volume.

Question 16: If each edge of a cube is increased by 50%, the percentage increase in its surface area is
(a) 50%
(b) 75%
(c) 100%
(d) 125%
Show Answer & Explanation

Answer: (d) 125%

Explanation:
1. Let the edge be x. Old area = 6x². New edge = 1.5x, so new area = 6 × 2.25x² = 13.5x².
2. Increase = 7.5x² ÷ 6x² × 100 = 125%.

Teacher's Note:
1. Area depends on the square of the edge.
2. 1.5² = 2.25 means 125% more.

Question 17: A cylindrical pencil sharpened at one end is a combination of
(a) a cone and a cylinder
(b) a cone and a hemisphere
(c) a hemisphere and a cylinder
(d) two cylinders
Show Answer & Explanation

Answer: (a) a cone and a cylinder

Explanation:
1. The sharpened tip is a cone and the rest is a cylinder.
2. The solid is a cone and a cylinder.

Teacher's Note:
1. Picture the pencil before and after sharpening.
2. Only the pointed end changes.

Question 18: The shape of a gilli, the small piece of wood used in the game gilli-danda, which is pointed at both ends, is a combination of
(a) two cylinders
(b) a cone and a cylinder
(c) two cones and a cylinder
(d) two cylinders and a cone
Show Answer & Explanation

Answer: (c) two cones and a cylinder

Explanation:
1. The pointed ends are cones and the middle part is a cylinder.
2. The gilli is two cones and a cylinder.

Teacher's Note:
1. Count the pointed ends: each is a cone.
2. The body between them is a cylinder.

Question 19: A right triangle with sides 3 cm, 4 cm and 5 cm is rotated about the side of 3 cm to form a cone. The volume of the cone so formed is
(a) \( 12\pi \) cm³
(b) \( 15\pi \) cm³
(c) \( 16\pi \) cm³
(d) \( 20\pi \) cm³
Show Answer & Explanation

Answer: (c) \( 16\pi \) cm³

Explanation:
1. Rotating about the side of 3 cm gives height h = 3 cm and radius r = 4 cm.
2. V = (1/3)π × 4² × 3 = 16π cm³.

Teacher's Note:
1. The axis of rotation becomes the height.
2. The other perpendicular side becomes the radius.

Question 20: A solid consists of a circular cylinder surmounted by a right circular cone. The height of the cone is h. If the total volume of the solid is 3 times the volume of the cone, then the height of the cylinder is
(a) \( \frac{2}{3}h \)
(b) \( \frac{3}{2}h \)
(c) h
(d) 2h
Show Answer & Explanation

Answer: (a) \( \frac{2}{3}h \)

Explanation:
1. Cone volume = (1/3)πr²h. Total volume = 3 × (1/3)πr²h = πr²h.
2. Cylinder volume = πr²h − (1/3)πr²h = (2/3)πr²h = πr²x.
3. So the height of the cylinder x = (2/3)h.

Teacher's Note:
1. Total volume = cone volume + cylinder volume.
2. Both parts have the same base radius.

Question 21: The curved surface area of a right circular cone of height 15 cm and base diameter 16 cm is
(a) \( 60\pi \) cm²
(b) \( 68\pi \) cm²
(c) \( 120\pi \) cm²
(d) \( 136\pi \) cm²
Show Answer & Explanation

Answer: (d) \( 136\pi \) cm²

Explanation:
1. r = 8 cm and h = 15 cm, so l = √(8² + 15²) = √289 = 17 cm.
2. CSA = πrl = π × 8 × 17 = 136π cm².

Teacher's Note:
1. Find the slant height with Pythagoras theorem.
2. 8, 15, 17 is a Pythagorean triple.

Question 22: The diameter of a sphere is 6 cm. It is melted and drawn into a wire of diameter 2 mm. The length of the wire is
(a) 12 m
(b) 18 m
(c) 36 m
(d) 66 m
Show Answer & Explanation

Answer: (c) 36 m

Explanation:
1. Volume of sphere = (4/3)π × 3³ = 36π cm³.
2. Wire radius = 1 mm = 0.1 cm. π × (0.1)² × h = 36π, so h = 3600 cm.
3. Length = 36 m.

Teacher's Note:
1. Melting does not change the volume.
2. Convert 2 mm diameter to a radius of 0.1 cm before using the formula.

Question 23: A medicine capsule is in the shape of a cylinder of diameter 0.5 cm with a hemisphere stuck at each end. The length of the entire capsule is 2 cm. The capacity of the capsule is (Take π = 22/7.)
(a) 0.33 cm³
(b) 0.34 cm³
(c) 0.35 cm³
(d) 0.36 cm³
Show Answer & Explanation

Answer: (d) 0.36 cm³

Explanation:
1. r = 0.25 cm; the cylindrical part has length 2 − 2 × 0.25 = 1.5 cm.
2. Volume = πr²h + (4/3)πr³ = (22/7)[0.0625 × 1.5 + (4/3) × 0.015625] = (22/7)(0.114583).
3. Capacity ≈ 0.36 cm³.

Teacher's Note:
1. The two hemispheres together make one sphere.
2. Subtract the two radii from the total length to find the cylinder's length.

Question 24: The radii of two cylinders are in the ratio 2 : 3 and their heights are in the ratio 5 : 3. The ratio of their volumes is
(a) 27 : 20
(b) 20 : 27
(c) 4 : 9
(d) 9 : 4
Show Answer & Explanation

Answer: (b) 20 : 27

Explanation:
1. Let the radii be 2x, 3x and the heights 5y, 3y.
2. V₁ : V₂ = π(2x)²(5y) : π(3x)²(3y) = 20 : 27, that is 20 : 27.

Teacher's Note:
1. Square the radius ratio and multiply by the height ratio.
2. 4 × 5 = 20 and 9 × 3 = 27.

Question 25: Water flows at the rate of 10 m per minute through a cylindrical pipe 5 mm in diameter. How long will it take to fill a conical vessel whose base diameter is 40 cm and depth 24 cm?
(a) 48 minutes 15 seconds
(b) 51 minutes 12 seconds
(c) 52 minutes 1 second
(d) 55 minutes
Show Answer & Explanation

Answer: (b) 51 minutes 12 seconds

Explanation:
1. Pipe radius = 0.25 cm; water per minute = π × 0.25² × 1000 = 62.5π cm³.
2. Volume of vessel = (1/3)π × 20² × 24 = 3200π cm³.
3. Time = 3200 ÷ 62.5 = 51.2 minutes = 51 minutes 12 seconds.

Teacher's Note:
1. 10 m = 1000 cm of water flows each minute.
2. 0.2 of a minute is 12 seconds.

Question 26: To find the total surface area of an article made by digging out a hemisphere from each end of a solid cylinder, the formula to use is
(a) C.S.A. of the cylinder − 2 (C.S.A. of the hemisphere)
(b) C.S.A. of the cylinder + C.S.A. of the hemisphere
(c) C.S.A. of the cylinder + 2 (C.S.A. of the hemisphere)
(d) C.S.A. of the cylinder − C.S.A. of the hemisphere
Show Answer & Explanation

Answer: (c) C.S.A. of the cylinder + 2 (C.S.A. of the hemisphere)

Explanation:
1. Digging a hemisphere removes a flat circular end of the cylinder but exposes a curved hollow of area 2πr².
2. So the total surface = curved surface of the cylinder + 2 curved surfaces of hemispheres, which is option (c).

Teacher's Note:
1. The two flat circular ends disappear, replaced by hollow curved surfaces.
2. Added areas, not subtracted.

Question 27:
Assertion (A): A hemisphere of radius 7 cm is to be painted on the outside. The total cost of painting at Rs. 5 per cm² is Rs. 2300.
Reason (R): The total surface area of a hemisphere is 3πr².
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.
Show Answer & Explanation

Answer: (d) A is false, but R is true.

Explanation:
1. Total surface area of the hemisphere = 3πr² = 3 × (22/7) × 7 × 7 = 462 cm², so the Reason is true.
2. Cost = 462 × 5 = Rs. 2310, not Rs. 2300, so the Assertion is false.
3. Answer: (d).

Teacher's Note:
1. Compute the painted area and the cost before judging the Assertion.
2. The outside of a solid hemisphere includes the flat base.

Question 28: A mathematics teacher took her Class 10 students on an educational trip to the Red Fort. She pointed out that the monument has combinations of solid shapes: two cylindrical pillars and two hemispherical domes at the corners. The formula for the volume of a cylindrical pillar is
(a) \( \pi r^2 h \)
(b) \( \pi r l \)
(c) \( \pi r (l + r) \)
(d) \( 2\pi r \)
Show Answer & Explanation

Answer: (a) \( \pi r^2 h \)

Explanation:
1. A cylinder of base radius r and height h has volume (base area) × height = πr²h.

Teacher's Note:
1. Volume = area of base × height.
2. πrl and πr(l + r) are cone areas; 2πr is a circumference.

Question 29: A mathematics teacher took her Class 10 students on an educational trip to the Red Fort. She pointed out that the monument has combinations of solid shapes: two cylindrical pillars and two hemispherical domes at the corners. The lateral (curved) surface area of the two pillars, if the height of each pillar is 7 m and the radius of its base is 1.4 m, is (Take π = 22/7.)
(a) 112.3 m²
(b) 123.2 m²
(c) 90 m²
(d) 345.2 m²
Show Answer & Explanation

Answer: (b) 123.2 m²

Explanation:
1. Curved surface area of one pillar = 2πrh = 2 × (22/7) × 1.4 × 7 = 61.6 m².
2. For two pillars: 2 × 61.6 = 123.2 m².

Teacher's Note:
1. Lateral area means the curved surface, without the top and bottom.
2. Multiply by 2 for two pillars.

Question 30: A mathematics teacher took her Class 10 students on an educational trip to the Red Fort. She pointed out that the monument has combinations of solid shapes: two cylindrical pillars and two hemispherical domes at the corners. The volume of a hemisphere of base radius 3.5 m is (Take π = 22/7.)
(a) 85.9 m³
(b) 80 m³
(c) 98 m³
(d) 89.83 m³
Show Answer & Explanation

Answer: (d) 89.83 m³

Explanation:
1. Volume = (2/3)πr³ = (2/3) × (22/7) × 3.5³ = (2/3)(22/7)(42.875).
2. = 89.83 m³.

Teacher's Note:
1. Volume of a hemisphere is half that of a sphere.
2. 3.5³ = 42.875.

Question 31: A mathematics teacher took her Class 10 students on an educational trip to the Red Fort. She pointed out that the monument has combinations of solid shapes: two cylindrical pillars and two hemispherical domes at the corners. The ratio of the sum of the volumes of two hemispheres of radius 1 cm each to the volume of a sphere of radius 2 cm is
(a) 1 : 1
(b) 1 : 8
(c) 8 : 1
(d) 1 : 16
Show Answer & Explanation

Answer: (b) 1 : 8

Explanation:
1. Two hemispheres of radius 1 cm = one sphere of radius 1 cm, volume (4/3)π × 1³.
2. Sphere of radius 2 cm: (4/3)π × 2³ = 8 × (4/3)π.
3. Ratio = 1 : 8.

Teacher's Note:
1. Two hemispheres make one sphere.
2. Volume depends on the cube of the radius.

Question 32: In December 2020 it rained heavily in Hyderabad. Anil observed that each raindrop was in the shape of a hemisphere surmounted by a cone of the same radius 1 mm. The volume of one such drop is 3.14 mm³. Anil collected rain water in a pot of capacity 1099 cm³. (Use π = 3.14 and √2 = 1.4.) The total height of the drop is
radius r = 1 mmh (a) 1 mm
(b) 2 mm
(c) 3 mm
(d) 4 mm
Show Answer & Explanation

Answer: (b) 2 mm

Explanation:
1. Let the cone have height h mm: (2/3)π(1)³ + (1/3)π(1)²h = 3.14.
2. (2 + h) × (3.14/3) = 3.14, so 2 + h = 3 and h = 1 mm.
3. Total height = radius of hemisphere + height of cone = 1 + 1 = 2 mm.

Teacher's Note:
1. The hemisphere adds a height equal to its radius.
2. Use the given volume to find the height of the cone first.

Question 33: In December 2020 it rained heavily in Hyderabad. Anil observed that each raindrop was in the shape of a hemisphere surmounted by a cone of the same radius 1 mm. The volume of one such drop is 3.14 mm³. Anil collected rain water in a pot of capacity 1099 cm³. (Use π = 3.14 and √2 = 1.4.) The curved surface area of the drop is
radius r = 1 mmh (a) 8.74 mm²
(b) 9.12 mm²
(c) 10.68 mm²
(d) 12.54 mm²
Show Answer & Explanation

Answer: (c) 10.68 mm²

Explanation:
1. Slant height of the cone = √(1² + 1²) = √2 = 1.4 mm.
2. CSA = 2πr² + πrl = πr(2r + l) = 3.14 × 1 × (2 + 1.4) = 3.14 × 3.4.
3. = 10.68 mm² (approximately).

Teacher's Note:
1. The curved surface of the drop is the hemisphere's curved surface plus the cone's.
2. The circle where they meet is not part of the outer surface.

Question 34: In December 2020 it rained heavily in Hyderabad. Anil observed that each raindrop was in the shape of a hemisphere surmounted by a cone of the same radius 1 mm. The volume of one such drop is 3.14 mm³. Anil collected rain water in a pot of capacity 1099 cm³. (Use π = 3.14 and √2 = 1.4.) As the drop fell into the pot it changed into a sphere. The radius of this sphere (in mm) was
radius r = 1 mmh (a) \( \left(\frac{3}{4}\right)^{1/3} \)
(b) \( \left(\frac{4}{3}\right)^{1/3} \)
(c) \( 3^{1/3} \)
(d) \( 4^{1/3} \)
Show Answer & Explanation

Answer: (a) \( \left(\frac{3}{4}\right)^{1/3} \)

Explanation:
1. Volume of sphere = volume of drop: (4/3)πR³ = 3.14.
2. With π = 3.14, (4/3)R³ = 1, so R³ = 3/4.
3. R = (3/4)^(1/3) mm.

Teacher's Note:
1. The volume stays the same when the shape changes.
2. Cancel 3.14 from both sides.

Question 35: In December 2020 it rained heavily in Hyderabad. Anil observed that each raindrop was in the shape of a hemisphere surmounted by a cone of the same radius 1 mm. The volume of one such drop is 3.14 mm³. Anil collected rain water in a pot of capacity 1099 cm³. (Use π = 3.14 and √2 = 1.4.) The number of such drops needed to fill the pot completely is
radius r = 1 mmh (a) 260000
(b) 280000
(c) 320000
(d) 350000
Show Answer & Explanation

Answer: (d) 350000

Explanation:
1. Volume of pot = 1099 cm³ = 1099 × 1000 mm³ = 1099000 mm³.
2. Number of drops = 1099000 ÷ 3.14 = 350000.

Teacher's Note:
1. Convert cm³ to mm³ by multiplying by 1000.
2. Divide the pot volume by the volume of one drop.

Question 36: A right triangle with sides 3 cm and 4 cm (enclosing the right angle) is revolved about its hypotenuse. The volume of the double cone so formed is (use π = 3.14)
(a) 30.144 cm³
(b) 31.4 cm³
(c) 37.68 cm³
(d) 25.12 cm³
Show Answer & Explanation

Answer: (a) 30.144 cm³

Explanation:
1. Hypotenuse = 5 cm. The common base radius is the altitude to the hypotenuse: r = (3 × 4) ÷ 5 = 2.4 cm.
2. The two cones have heights that add up to 5 cm, so volume = (1/3)πr² × 5 = (1/3)π(5.76)(5) = 9.6π.
3. = 9.6 × 3.14 = 30.144 cm³.

Teacher's Note:
1. The two cones share the same circular base; their heights add up to the hypotenuse.
2. The radius is the perpendicular from the right angle to the hypotenuse.

Question 37: The volume of a right circular cone of height 24 cm is 1232 cm³. Its curved surface area is (Take π = 22/7.)
(a) 550 cm²
(b) 154 cm²
(c) 616 cm²
(d) 704 cm²
Show Answer & Explanation

Answer: (a) 550 cm²

Explanation:
1. V = (1/3)πr²h = 1232, so (1/3)(22/7)r²(24) = 1232 and r² = 49, r = 7 cm.
2. Slant height l = √(7² + 24²) = 25 cm.
3. CSA = πrl = (22/7) × 7 × 25 = 550 cm².

Teacher's Note:
1. Find the radius from the volume first.
2. 7, 24, 25 is a Pythagorean triple.

Question 38: The radius and height of a right circular cone are 12 cm and 9 cm respectively. The curved surface area of the cone is (Take π = 22/7.)
(a) 565.71 cm²
(b) 452.57 cm²
(c) 339.43 cm²
(d) 678.86 cm²
Show Answer & Explanation

Answer: (a) 565.71 cm²

Explanation:
1. Slant height l = √(12² + 9²) = 15 cm.
2. CSA = πrl = (22/7) × 12 × 15 = 3960 ÷ 7 = 565.71 cm².

Teacher's Note:
1. 12, 9, 15 is a multiple of the 3, 4, 5 triple.
2. Use the slant height, not the height, in πrl.

Question 39: The volume of a sphere and of a cylinder are equal, and the diameter of the base of the cylinder is equal to the diameter of the sphere. The ratio of the radius of the base to the height of the cylinder is
(a) 3 : 4
(b) 4 : 3
(c) 2 : 3
(d) 1 : 2
Show Answer & Explanation

Answer: (a) 3 : 4

Explanation:
1. Let the common radius be r and the cylinder's height h. (4/3)πr³ = πr²h, so h = 4r/3.
2. r : h = r : (4r/3) = 3 : 4.

Teacher's Note:
1. Equal diameters mean equal radii.
2. Cancel πr² from both sides.

Question 40: The volume of a right circular cylinder is 352 cm³ and its height is 7 cm. The radius of the base is (Take π = 22/7.)
(a) 2 cm
(b) 4 cm
(c) 3.5 cm
(d) 7 cm
Show Answer & Explanation

Answer: (b) 4 cm

Explanation:
1. πr²h = 352, so (22/7) × r² × 7 = 352.
2. 22r² = 352, so r² = 16 and r = 4 cm.

Teacher's Note:
1. Cancel 7 with the denominator of 22/7.
2. Take the positive square root.

Free study material for Mathematics

Multiple Choice Questions (MCQs) for Class 10 Mathematics Chapter 12 Surface Areas and Volumes

Download Multiple Choice Questions: Chapter 12 Surface Areas and Volumes (Class 10 Mathematics)

Review structured objective questions for Class 10 Mathematics Chapter 12 Surface Areas and Volumes. Built according to official CBSE guidelines, these MCQ sets support daily revision and core concept reinforcement.

Concept Clarification for Chapter 12 Surface Areas and Volumes

Each question includes structured solution keys mapped directly to standard CBSE textbooks, helping students evaluate their reasoning and correct mistakes early in their revision.

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FAQs

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