If the side of a cube is doubled, how many times does its volume become?

Answer: 8 times. Thinking Process:Volume $V = a^3$.New side $2a$.New Volume $V' = (2a)^3 = 8a^3 = 8V$.

What is the ratio of volumes of two spheres if the ratio of their radii is 1:2?

Answer: 1:8. Thinking Process:Volume $\propto r^3$.Ratio = $(r_1/r_2)^3 = (1/2)^3 = 1/8$.

A cylinder and a cone have the same base radius and the same height. What is the ratio of their volumes?

Answer: 3:1. Thinking Process:Volume Cylinder = $\pi r^2 h$.Volume Cone = $\frac{1}{3} \pi r^2 h$.Ratio = $1 : 1/3 = 3:1$.

The length of the longest rod that can be placed in a room of dimensions $L, B, H$ is given by:

Answer: √(L²+B²+H²). The longest rod is the diagonal of the cuboid.Diagonal = $\sqrt{L^2 + B^2 + H^2}$.

The total surface area of a solid hemisphere of radius $r$ is:

Answer: 3πr². TSA of Hemisphere = CSA (curved part) + Base Area (flat circle).$= 2\pi r^2 + \pi r^2 = 3\pi r^2$.

If the height of a cone is doubled keeping the radius constant, its volume becomes:

Answer: 2 times. Volume $V \propto h$.If $h$ doubles, $V$ also doubles.

The number of surfaces in a cube is:

Answer: 6. A cube has 6 faces (Top, Bottom, Front, Back, Left, Right).

If the radius of a cylinder is halved and height is quadrupled, the volume will:

Answer: Remain same. Thinking Process:$V = \pi r^2 h$.New $r' = r/2, h' = 4h$.New $V' = \pi (r/2)^2 (4h) = \pi (r^2/4) (4h) = \pi r^2 h$.Volume remains unchanged.

1 Litre is equal to how many cubic centimeters?

Answer: 1000. Standard Conversion: 1 Litre = 1000 $cm^3$ (or ml).

The surface area of a sphere of radius $r$ is equal to the curved surface area of a cylinder of radius $r$ and height $h$. Then $h =$?

Answer: 2r. Thinking Process:Sphere Area = $4\pi r^2$.Cylinder CSA = $2\pi rh$.Given $4\pi r^2 = 2\pi rh \implies 2r = h$.

A room is 12 m long, 9 m broad and 8 m high. Find the length of the longest bamboo pole that can be placed in it.

Answer: 17 m. Thinking Process:Longest Pole = Diagonal.$D = \sqrt{12^2 + 9^2 + 8^2} = \sqrt{144 + 81 + 64} = \sqrt{289} = 17$.

The surface area of a sphere is 616 cm². Find its diameter.

Answer: 14 cm. Thinking Process:$4\pi r^2 = 616$.$4 \times \frac{22}{7} \times r^2 = 616$.$r^2 = \frac{616 \times 7}{88} = 49$.$r = 7$.Diameter = $2r = 14$.

The volume of a right circular cone of height 9 cm is $48\pi$ cm³. Find the diameter of its base.

Answer: 8 cm. Thinking Process:$V = \frac{1}{3}\pi r^2 h = 48\pi$.$\frac{1}{3} r^2 (9) = 48$.$3r^2 = 48 \implies r^2 = 16 \implies r=4$.Diameter = 8.

A copper sphere of diameter 18 cm is drawn into a wire of diameter 4 mm. Find the length of the wire.

Answer: 243 m. Thinking Process:Volume Sphere = Volume Wire (Cylinder).Sphere $R = 9$ cm.Wire $r = 2$ mm = 0.2 cm. Length $h$.$\frac{4}{3}\pi (9)^3 = \pi (0.2)^2 h$.$\frac{4}{3} \times 729 = 0.04 h$.$972 = 0.04 h \implies h = 97200 / 4 = 24300$ cm = 243 m.

A wall of dimensions $8m \times 6m \times 22.5cm$ is to be constructed with bricks of dimensions $25cm \times 11.25cm \times 6cm$. The number of bricks required is:

Answer: 6400. Thinking Process:Convert all to cm.Wall: $800 \times 600 \times 22.5$.Brick: $25 \times 11.25 \times 6$.Number = $V_{wall} / V_{brick}$.$= \frac{800 \times 600 \times 22.5}{25 \times 11.25 \times 6}$.$= 32 \times 100 \times 2 = 6400$.

Two cylinders have equal volume. Their heights are in the ratio 1:4. What is the ratio of their radii?

Answer: 2:1. Thinking Process:$V_1 = V_2 \implies \pi r_1^2 h_1 = \pi r_2^2 h_2$.$r_1^2 \times 1 = r_2^2 \times 4$.$r_1^2 / r_2^2 = 4/1$.$r_1/r_2 = 2/1$.

Three cubes of metal with edges 3 cm, 4 cm, and 5 cm are melted to form a single cube. The lateral surface area of the new cube is:

Answer: 144 cm². Thinking Process:Total Volume = $3^3 + 4^3 + 5^3 = 27 + 64 + 125 = 216$.New Edge $a^3 = 216 \implies a = 6$.LSA = $4a^2 = 4 \times 36 = 144$.

A conical tent is to accommodate 11 persons. Each person needs 4 m² of space on the ground and 20 m³ of air to breathe. Find the height of the cone.

Answer: 15 m. Thinking Process:Base Area = $11 \times 4 = 44$. ($\pi r^2 = 44$)Volume = $11 \times 20 = 220$. ($\frac{1}{3}\pi r^2 h = 220$)Substitute base area: $\frac{1}{3} (44) h = 220$.$h = \frac{220 \times 3}{44} = 5 \times 3 = 15$.

How many spherical bullets can be made out of a solid lead cube of edge 44 cm, each bullet being 4 cm in diameter?

Answer: 2541. Thinking Process:Cube Side = 44. Bullet Radius = 2.Count = $\frac{\text{Vol Cube}}{\text{Vol Sphere}} = \frac{44^3}{\frac{4}{3}\pi (2)^3}$$= \frac{44 \times 44 \times 44 \times 3 \times 7}{4 \times 22 \times 8}$$= \frac{85184 \times 21}{704} = 121 \times 21 = 2541$.

The capacity of a cylindrical tank is 6160 m³. If the radius of its base is 14 m, the depth of the tank is:

Answer: 10 m. Thinking Process:Volume = $\pi r^2 h = 6160$.$\frac{22}{7} \times 14 \times 14 \times h = 6160$.$22 \times 2 \times 14 \times h = 6160$.$616 h = 6160 \implies h=10$.

A solid cylinder has a total surface area of 462 cm². Its curved surface area is one-third of its total surface area. Find the volume of the cylinder.

Answer: 539 cm³. Thinking Process:1. TSA ($2\pi r(h+r)$) = 462.2. CSA ($2\pi rh$) = $\frac{1}{3} \times 462 = 154$.Diff = $2\pi r^2 = 462 - 154 = 308$.$2\pi r^2 = 308 \implies r^2 = 49 \implies r=7$.From CSA: $2 \times \frac{22}{7} \times 7 \times h = 154 \implies 44h = 154 \implies h=3.5$.Volume = $\pi r^2 h = \frac{22}{7} \times 49 \times 3.5 = 22 \times 7 \times 3.5 = 539$.

A bucket is in the form of a frustum of a cone and holds 28.490 litres of water. The radii of the top and bottom are 28 cm and 21 cm respectively. Find the height of the bucket.

Answer: 15 cm. Thinking Process:Vol = 28.49 Litres = 28490 cm³.$V = \frac{1}{3}\pi h (R^2 + r^2 + Rr)$.$28490 = \frac{1}{3} \times \frac{22}{7} \times h (28^2 + 21^2 + 28\times 21)$.$28490 = \frac{22}{21} h (784 + 441 + 588) = \frac{22}{21} h (1813)$.$h = \frac{28490 \times 21}{22 \times 1813} = 15$.

The largest possible sphere is carved out of a solid wooden cube of side 21 cm. The volume of the remaining wood is:

Answer: 4410 cm³. Thinking Process:Cube Side $a=21$. Vol = $21^3 = 9261$.Sphere Diameter = 21, Radius $r = 10.5$.Vol Sphere = $\frac{4}{3} \times \frac{22}{7} \times (10.5)^3 = 4851$.Remaining = $9261 - 4851 = 4410$.

A toy is in the form of a cone surmounted on a hemisphere of same radius. The diameter of the base is 6 cm and height of the cone is 4 cm. The total surface area of the toy is:

Answer: 33π. Thinking Process:$r=3, h=4 o l=5$.TSA = CSA(Cone) + CSA(Hemisphere).$= \pi rl + 2\pi r^2$$= \pi(3)(5) + 2\pi(3)^2 = 15\pi + 18\pi = 33\pi$.

Water flows into a tank at the rate of 150 liters per minute through a pipe of cross-section area 5 cm². The speed of water flow is:

Answer: 18 km/hr. Thinking Process:Volume Rate = 150 L/min = $150000$ cm³/min.Area = 5 cm².Speed = Vol/Area = $150000/5 = 30000$ cm/min.Convert to km/hr:$30000 \times 60$ cm/hr = $1800000$ cm/hr = 18 km/hr.

A cone is cut by a plane parallel to the base at 1/2 of its height. The ratio of volume of the frustum to the volume of the original cone is:

Answer: 7:8. Thinking Process:Top small cone is similar to original large cone.Height ratio = 1:2. Volume ratio = $1^3 : 2^3 = 1:8$.Frustum Vol = Total - Small = $8-1=7$.Ratio Frustum:Total = 7:8.

A hollow cylindrical tube 20 cm long is made of iron and its external and internal diameters are 8 cm and 6 cm respectively. The volume of iron used in making the tube is:

Answer: 440 cm³. Thinking Process:$R=4, r=3, h=20$.$V = \pi h (R^2 - r^2) = \frac{22}{7} \times 20 \times (16 - 9)$.$= \frac{22}{7} \times 20 \times 7 = 440$.

A sphere and a cylinder have equal radii and equal volumes. The ratio of the diameter of the sphere to the height of the cylinder is:

Answer: 3:2. Thinking Process:$\frac{4}{3}\pi r^3 = \pi r^2 h$.$\frac{4}{3} r = h$.We need ratio Diameter ($2r$) : $h$.$2r : \frac{4}{3}r = 2 : \frac{4}{3} = 6:4 = 3:2$.

The base of a right prism is an equilateral triangle of side 6 cm. If the volume of the prism is $108\sqrt{3}$ cm³, its height is:

Answer: 12 cm. Thinking Process:Base Area = $\frac{\sqrt{3}}{4} (6)^2 = 9\sqrt{3}$.Volume = Base Area $\times$ Height.$108\sqrt{3} = 9\sqrt{3} \times h$.$h = 108/9 = 12$.

The areas of three adjacent faces of a cuboid are 12 cm², 20 cm², and 15 cm². The volume of the cuboid is:

Answer: 60 cm³. Thinking Process:Areas are $LB=12, BH=20, HL=15$.Multiply all: $(LB)(BH)(HL) = (LBH)^2$.$V^2 = 12 \times 20 \times 15 = 3600$.$V = \sqrt{3600} = 60$.

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Mensuration 3D (Volume & Surface Area)

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Expert Answer & Key Takeaways

Mastering Mensuration 3D (Volume & Surface Area) is essential for high-fidelity technical architecture and senior engineering roles in 2026.

1. Cube & Cuboid (घन और घनाभ)

Cuboid:

Volume: $L \times B \times H$
Total Surface Area (TSA): $2(LB + BH + HL)$
Lateral Surface Area (LSA): $2(L+B) \times H$
Diagonal: $\sqrt{L^2 + B^2 + H^2}$

Cube:

Volume: $a^3$
TSA: $6a^2$
LSA: $4a^2$
Diagonal: $a\sqrt{3}$

Example:

Q: Find the volume of a cube with side 5 cm.

Solution:

V = 5^3 = 125 \text{ cm}^3

2. Cylinder (बेलन)

Solid Cylinder:

Volume: $\pi r^2 h$
Curved Surface Area (CSA): $2\pi rh$
Total Surface Area (TSA): $2\pi r(h+r)$

Hollow Cylinder:

Volume: $\pi h(R^2 - r^2)$
TSA: $2\pi(R+r)(h+R-r)$

3. Cone (शंकु)

Cone:

Slant Height ( $l$ ): $\sqrt{h^2 + r^2}$
Volume: $\frac{1}{3} \pi r^2 h$
Curved Surface Area (CSA): $\pi rl$
Total Surface Area (TSA): $\pi r(l+r)$

4. Sphere & Hemisphere (गोला और अर्धगोला)

Sphere:

Volume: $\frac{4}{3} \pi r^3$
Surface Area: $4\pi r^2$

Hemisphere:

Volume: $\frac{2}{3} \pi r^3$
Curved Surface Area (CSA): $2\pi r^2$
Total Surface Area (TSA): $3\pi r^2$

Example:

Q: Find the volume of a sphere with radius 3 cm.

Solution:

V = \frac{4}{3} \times \frac{22}{7} \times 3^3 = \frac{4}{3} \times \frac{22}{7} \times 27 = 36\pi \approx 113.14

5. Frustum of a Cone (छिन्नक)

Volume:

\frac{1}{3} \pi h (R^2 + r^2 + Rr)

CSA:

\pi l (R+r)

where

l = \sqrt{h^2 + (R-r)^2}

TSA:

\pi l (R+r) + \pi R^2 + \pi r^2

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Mensuration 3D (Volume & Surface Area)

Expert Answer & Key Takeaways

1. Cube & Cuboid (घन और घनाभ)

2. Cylinder (बेलन)

3. Cone (शंकु)

4. Sphere & Hemisphere (गोला और अर्धगोला)

5. Frustum of a Cone (छिन्नक)

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