Volume & Surface Areas Questions and Answers

Answer: B

The water forms a cylinder of radius 8 cm and height 20 cm.

\[ V = \pi r^{2} h = \pi (8)^{2}(20) = 1280\pi \text{ cm}^{3} \]

\[ V = 1280 \times 3.1416 = 4021.2 \text{ cm}^{3} \]

Since \( 1000 \text{ cm}^{3} = 1 \) litre:

\[ V \approx 4.02 \text{ litres} \]

The conversion is the step most often dropped - the answer in cubic centimetres is a thousand times the answer in litres.

Answer: C

Take the original dimensions as \( 1k, 2k, 3k \).

\[ V_{\text{old}} = 1k \times 2k \times 3k = 6k^{3} \]

An increase of 100% doubles a length, and 200% triples it:

\[ 1k \to 2k, \qquad 2k \to 6k, \qquad 3k \to 9k \]

\[ V_{\text{new}} = 2k \times 6k \times 9k = 108k^{3} \]

\[ V_{\text{new}} - V_{\text{old}} = 102k^{3} = 17 \times 6k^{3} \]

So the old box falls short by 17 times its own volume. Note that a 200% increase means the length becomes three times the original, not two - that single slip turns the answer into 7.

Answer: C

Form the ratio and let the \( \pi \) cancel.

\[ \frac{S}{V} = \frac{4\pi r^{2}}{\tfrac{4}{3}\pi r^{3}} = \frac{3}{r} \]

\[ \frac{3}{r} = \frac{1}{18} \quad\Longrightarrow\quad r = 54 \text{ cm} \]

The ratio has units of inverse length, which is why the answer 1/18 cm gives a radius in centimetres. Reading the relation as \( r/3 \) instead of \( 3/r \) gives 6 cm - the second distractor.

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: D

Find the area to be covered, convert it to a weight of paper, then price it.

\[ \text{surface area of cube} = 6a^{2} = 6 \times 10^{2} = 600 \text{ m}^{2} \]

\[ \text{paper needed} = \frac{600}{20} = 30 \text{ kg} \]

\[ \text{cost} = 30 \times 60 = \text{Rs. } 1800 \]

All six faces are covered, so the factor is 6 and not 4 - using only the four side faces gives Rs. 1200.

Answer: A

Melting conserves volume, and the two shapes share the same radius \( r = 3 \) cm.

\[ V_{\text{sphere}} = \frac{4}{3}\pi r^{3} = \frac{4}{3}\pi (27) = 36\pi \text{ cm}^{3} \]

\[ V_{\text{cylinder}} = \pi r^{2} h = 9\pi h \]

\[ 9\pi h = 36\pi \quad\Longrightarrow\quad h = 4 \text{ cm} \]

Whenever a sphere is recast as a cylinder of the same radius, the height is always \( \tfrac{4}{3}r \) - here \( \tfrac{4}{3} \times 3 = 4 \).

Answer: B

The hemisphere's height is its own radius, so "same height" forces \( h = r \) for all three.

\[ V_{\text{cylinder}} = \pi r^{2} h = \pi r^{3} \]

\[ V_{\text{cone}} = \frac{1}{3}\pi r^{2} h = \frac{1}{3}\pi r^{3} \]

\[ V_{\text{hemisphere}} = \frac{2}{3}\pi r^{3} \]

\[ \text{ratio} = 1 : \frac{1}{3} : \frac{2}{3} = 3 : 1 : 2 \]

This is the classic cylinder–cone–hemisphere relation: the cone is a third of the cylinder and the hemisphere two thirds, so cone plus hemisphere exactly fills the cylinder.

Answer: C

Let the block measure \( l \times b \times h \). The three adjacent faces are

\[ P = lb, \qquad Q = bh, \qquad R = hl \]

Multiplying all three:

\[ PQR = (lb)(bh)(hl) = l^{2}b^{2}h^{2} = (lbh)^{2} \]

\[ \therefore\; S^{2} = PQR \]

Each dimension appears exactly twice in the product, which is what makes the result the square of the volume. It also gives a quick route to the volume: \( S = \sqrt{PQR} \).

Answer: C

The immersed box displaces its own volume of water, and that displaced volume spreads over the vessel's base.

\[ V_{\text{cube}} = 30^{3} = 27000 \text{ cm}^{3} \]

\[ \text{base area} = 60 \times 30 = 1800 \text{ cm}^{2} \]

\[ \text{rise} = \frac{27000}{1800} = 15 \text{ cm} \]

The cube's own 30 cm edge is not the rise: the water spreads over a base twice the cube's footprint, so it climbs only half as far.

Answer: C

Work in centimetres throughout: the cylinder is \( r = 100 \) cm, \( h = 500 \) cm.

\[ V_{\text{cyl}} = \pi (100)^{2}(500) = 5{,}000{,}000\pi \text{ cm}^{3} \]

\[ V_{\text{cone}} = \frac{1}{3}\pi (50)^{2}(50) = \frac{125000\pi}{3} \text{ cm}^{3} \]

\[ n = \frac{5{,}000{,}000\pi}{125000\pi/3} = \frac{5{,}000{,}000 \times 3}{125000} = 120 \]

Mixing metres and centimetres is the trap here - the units must be made consistent before any division.