Class 9 Maths Ganita Manjari Chapter 14 MCQ – Math of Space: Surface Area and Volume

The volume of a cube is a³, where a is its side. Since 5³ = 125, the side is 5 cm. A cube has six equal square faces, so its total surface area is 6 × 5² = 150 cm².

A cuboid has three pairs of equal rectangular faces. Its total surface area is 2(lb + bh + hl). Substituting the dimensions gives 2(8 × 5 + 5 × 3 + 3 × 8) = 2(79) = 158 cm².

First express all dimensions in the same unit. Since 1 m = 100 cm, the tank’s volume is 100 × 50 × 40 = 2,00,000 cm³. Using 1 litre = 1,000 cm³, its capacity is 200 litres.

Joining the cubes produces a cuboid measuring 8 cm × 4 cm × 4 cm. Its total surface area is 2(32 + 16 + 32) = 160 cm². The two joined faces are inside the solid and are no longer exposed.

Small cubes with exactly two painted faces lie along the edges, excluding the corners. Each edge contains 4 − 2 = 2 such cubes. Since a cube has 12 edges, the required number is 12 × 2 = 24.

The original surface area is 6 × 4² = 96 cm². There are 4³ = 64 small cubes, each with surface area 6 cm². Their total surface area is 384 cm², giving the ratio 96 : 384 = 1 : 4.

Let the dimensions be l, b and h. The face areas are lb, bh and hl. Their product equals (lbh)² = 6 × 10 × 15 = 900. Therefore, the volume lbh is √900 = 30 cm³.

The curved surface of a cylinder opens into a rectangle whose length is the base circumference and whose breadth is the height. Thus, CSA = 2πrh = 2 × (22/7) × 7 × 10 = 440 cm².

An open container has a curved surface and one circular base. Its surface area is 2πrh + πr². Here, the curved surface area is 440 cm² and the base area is 154 cm², giving 594 cm² altogether.

The volume of a cylinder equals its circular base area multiplied by its height. Using V = πr²h, we get (22/7) × 3² × 14 = 396 cm³. The radius must be squared before multiplying by the height.

A cylinder’s volume is πr²h. Doubling the radius multiplies the base area by 4, while halving the height multiplies the volume by 1/2. The combined factor is 4 × 1/2 = 2, so the volume doubles.

The cylinder’s base diameter equals the cube’s side, so its radius is 5 cm. Its height is 10 cm, the distance between the opposite faces. Therefore, its volume is π × 5² × 10 = 250π cm³.

The cube contains 6³ = 216 cm³ of metal. Each rod requires π × 1² × 7 = 22 cm³. Since 216 ÷ 22 is between 9 and 10, only 9 complete rods can be made; some metal remains.

The radius, perpendicular height and slant height form a right triangle in a vertical section through the cone’s axis. By the Pythagoras theorem, l² = r² + h² = 25 + 144 = 169. Therefore, l = 13 cm.

The curved surface area of a cone is πrl, where l is the slant height. Substituting the values gives (22/7) × 7 × 10 = 220 cm². The circular base is excluded when calculating only the curved surface area.

A cone’s total surface area includes its curved surface and circular base. Thus, TSA = πrl + πr² = πr(l + r). Substituting the values gives (22/7) × 7 × (10 + 7) = 374 cm².

Each cap is open at its base, so only its curved surface requires paper. Its slant height is √(3² + 4²) = 5 cm. Paper for one cap is π × 3 × 5 = 15π cm²; ten caps require 150π cm².

Using V = (1/3)πr²h, we get 1,570 = (1/3) × 3.14 × r² × 15 = 15.7r². Therefore, r² = 100 and r = 10 cm. The radius is the positive square root because it represents a length.

The side of rotation becomes the cone’s height, so h = 15 cm. The other perpendicular side becomes its radius, so r = 8 cm. Its volume is (1/3) × π × 8² × 15 = 320π cm³.

At half the cone’s height, the water surface has half the full radius. Since cone volume depends on r²h, the volume fraction is (1/2)² × (1/2) = 1/8. Halving the depth does not halve the volume.

The volume of a pyramid is one-third of its base area multiplied by its perpendicular height. The square base has area 6 × 6 = 36 cm². Hence, the volume is (1/3) × 36 × 10 = 120 cm³.

A sphere’s surface area is 4πr². For radius 3 cm, this becomes 4 × π × 3² = 36π cm². A sphere has one continuous curved surface, so there are no flat circular bases to add.

A sphere’s volume is proportional to the cube of its radius. A 10% increase makes the new radius 1.1 times the original. The new volume is 1.1³ = 1.331 times the original, which represents an increase of 33.1%.

Since the volume of a sphere is (4/3)πr³, the common factor (4/3)π cancels when comparing two spheres. Their radii are therefore in the ratio of the cube roots of their volumes: ∛27 : ∛8 = 3 : 2.

A solid hemisphere has a curved surface and a flat circular face. Their areas are 2πr² and πr² respectively, giving a total of 3πr². Substituting r = 7 cm gives 3 × (22/7) × 49 = 462 cm².

A hemisphere is half a sphere, so its volume is half of (4/3)πr³, or (2/3)πr³. For radius 3 cm, the volume is (2/3) × π × 27 = 18π cm³. Surface area formulas do not give volume.

If the sphere’s radius is r, the enclosing cylinder has radius r and height 2r. Their volumes are (4/3)πr³ and 2πr³ respectively. Cancelling the common factors gives (4/3) : 2 = 2 : 3.

The displaced water has volume π × 3² × 4 = 36π cm³. Each marble has volume (4/3)π × 1³ = (4/3)π cm³. Since fully submerged marbles displace their own volume, their number is 36π ÷ [(4/3)π] = 27.

Let the uniform gap be x metres. The increase in circumference is 2π(R + x) − 2πR = 2πx. Thus, x = 1/(2π) m, approximately 15.9 cm. The sphere’s original radius cancels, so it does not affect the gap.

One scoop has volume (4/3)π × 2³ = 32π/3, approximately 33.52 cm³. Dividing 2,000 by 33.52 gives about 59.7. Therefore, 60 is the closest estimate; the question asks for an approximate count rather than an exact number of complete scoops.