Volume & Surface Areas Questions and Answers

Answer: A

The number of small balls is 8000.

Volume is conserved, and the number of balls is the ratio of the volumes, which is the cube of the ratio of radii. The large radius is 6 cm = 60 mm; the small radius is 3 mm:

\[ n = \left(\frac{60}{3}\right)^{3} = 20^{3} = 8000 \]

1000 comes from using the small diameter (6 mm) as if it were the radius, and the other options do not cube the ratio.

Answer: B

The capacity is 26.994 litres.

The volume of a frustum with end radii \(R\), \(r\) and height \(h\) is \(\frac{\pi h}{3}(R^{2} + r^{2} + Rr)\):

\[ \begin{aligned} V &= \frac{22}{7} \times \frac{63}{3} \times (225 + 64 + 120) \\ &= 66 \times 409 = 26994\ \text{cm}^{3} \end{aligned} \]

Since 1 litre = 1000 cm\(^{3}\), this is 26.994 litres. The other options are the same digits with the decimal point moved, the result of a wrong conversion.

Answer: D

The diagonal of a cube of side a is \(a\sqrt{3}\), so \(a\sqrt{3}=11\sqrt{3}\) gives a = 11 feet.

The volume is 11 cubed = 1331 cubic feet, option D. The options with the root of 3 come from multiplying the diagonal rather than the side.

The space diagonal d of a cube of side a is \(a\sqrt{3}\), which follows from Pythagoras applied twice. With d = \(11\sqrt{3}\) the side is 11 feet, and the volume is 11 x 11 x 11 = 1331 cubic feet. The options with a root of 3 would apply if the diagonal was used as a side.

Answer: C

450 books can be arranged in the box.

Work in centimetres. The box is 150 cm by 75 cm by 60 cm, and each book is 20 cm by 10 cm by 7.5 cm.

\[ \begin{aligned} \text{box} &= 150 \times 75 \times 60 = 675000\ \text{cm}^{3} \\ \text{book} &= 20 \times 10 \times 7.5 = 1500\ \text{cm}^{3} \\ \text{number} &= \frac{675000}{1500} = 450 \end{aligned} \]

The books also pack the box exactly with nothing left over: stand the 7.5 cm side along the 75 cm edge (10 books), the 10 cm side along the 150 cm edge (15 books) and the 20 cm side along the 60 cm height (3 layers), and \(10 \times 15 \times 3 = 450\). 350, 400 and 500 do not fit either the volume or any arrangement.

Answer: A

The cost is ₹ 198.

The curved surface area of a cylinder is \(2\pi r h\), which is \(\pi d h\) in terms of the diameter:

\[ \begin{aligned} \text{area} &= \frac{22}{7} \times 1 \times 4.2 = 13.2\ \text{m}^2 \\ \text{cost} &= 13.2 \times 15 = 198 \end{aligned} \]

The options 19.8 and 1.98 are the same digits with the decimal point misplaced, and 200 is a rounded figure. Only the curved surface is whitewashed, so the top and bottom circles are not added.

Answer: A

Melting conserves volume, so divide the cylinder's volume by one coin's.

\[ V_{\text{coin}} = \pi (0.75)^{2}(0.2) = \pi (0.5625)(0.2) = 0.1125\pi \text{ cm}^{3} \]

\[ V_{\text{cylinder}} = \pi (3)^{2}(8) = 72\pi \text{ cm}^{3} \]

\[ n = \frac{72\pi}{0.1125\pi} = \frac{72}{0.1125} = 640 \]

The \( \pi \) cancels, so no approximation is needed anywhere in the calculation.

Answer: C

The unsharpened body of the pencil is a straight cylinder of uniform circular cross-section. Sharpening tapers one end to a point while keeping that cross-section circular, which is exactly a cone.

So the shape is a cone mounted on a cylinder.

A hemisphere would end in a rounded dome rather than a point, and a frustum would be cut off flat instead of tapering to the graphite tip.

Answer: B

A cube of volume 27 m³ has edge \( \sqrt[3]{27} = 3 \) m, so it cuts into \( 3 \times 3 \times 3 \) unit cubes.

Only the cubes strictly inside - not touching any face - escape the paint. Removing one layer from each of the six faces leaves a cube of side

\[ 3 - 2 = 1 \]

\[ \text{unpainted cubes} = 1^{3} = 1 \]

The general result for an \( n \times n \times n \) cube is \( (n-2)^{3} \), which is why a 3-cube yields just the single central cube.

Answer: B

Rearrange the cuboid volume formula for the missing dimension.

\[ V = l \times b \times h \quad\Longrightarrow\quad h = \frac{V}{l \times b} \]

\[ h = \frac{450}{9 \times 8} = \frac{450}{72} = 6.25 \text{ cm} \]

Checking: \( 9 \times 8 \times 6.25 = 450 \) cm³, as required.

Answer: D

Diameter of largest circle = 20 cm.
∴ Area of circle = 100π cm²
∴ Area of 6 circles = 6 × 100π = 600π cm² (∵ there are six faces in a cube)
Also, Area of cube = 6 × (20)² = 2400 cm²
Area of unpainted surface = 2400 - 600π
= 2400 - 600 × 22/7
= 514.28 cm².