Volume & Surface Areas Questions and Answers

Answer: D

The water leaving in one minute forms a cylinder whose cross-section is the pipe's and whose length is the distance the flow covers in that minute.

\[ d = 86 \text{ mm} = 0.086 \text{ m} \quad\Longrightarrow\quad r = 0.043 \text{ m} \]

\[ \text{length per minute} = 5 \times 60 = 300 \text{ m} \]

\[ V = \pi r^{2} \times 300 = 3.1416 \times (0.043)^{2} \times 300 \]

\[ = 3.1416 \times 0.001849 \times 300 = 1.7429 \text{ m}^{3} \]

Since \( 1 \text{ m}^{3} = 1000 \) litres:

\[ V \approx 1742.87 \text{ litres} \]

The two conversions - millimetres to metres and seconds to minutes - are what set the scale of the answer.

Answer: D

The cork nose of the shuttlecock is a rounded dome - a hemisphere. The feather skirt is circular at both ends, narrow where it meets the cork and wide at the open end, closing to a point at neither - a frustum of a cone.

A sphere would make the base fully round rather than flat against the feathers, and a plain cone would taper to a point instead of flaring open.

Answer: B

When a sector of paper is rolled into a cone, the sector's radius becomes the cone's slant height - the straight distance from the apex to the rim is unchanged by the rolling.

\[ l = 10 \text{ cm} \]

The semicircle's arc becomes the base circumference, so \( \pi \times 10 = 2\pi r \) gives \( r = 5 \) cm - that 5 cm is the base radius, offered as the first distractor, not the slant height.

Answer: D

A shuttlecock has a rounded cork base and a skirt of feathers that flares outward from it.

The cork nose is a hemisphere. The feather skirt is narrow where it meets the cork and wide at the open end, with both ends circular and neither closing to a point - that is a frustum of a cone.

A full cone would come to a point at the top, and a cylinder would keep the same width all the way up; neither matches the flare.

Answer: A

The curved surface of a hemisphere is half a sphere's, so

\[ \text{CSA} = 2\pi r^{2} = 175 \]

\[ r^{2} = \frac{175}{2\pi} = \frac{175}{6.2832} = 27.85 \]

\[ r = \sqrt{27.85} = 5.28 \text{ cm} \]

Only the curved dome is asked for, so the flat base is excluded; using the total surface \( 3\pi r^{2} \) would give \( r = 4.31 \) cm instead.

Answer: C

Surface area of a sphere is \( 4\pi r^{2} \), so it scales with the square of the radius.

\[ \frac{S_{1}}{S_{2}} = \frac{4\pi r^{2}}{4\pi (2r)^{2}} = \frac{r^{2}}{4r^{2}} = \frac{1}{4} \]

So the ratio is \( 1 : 4 \). Doubling a length quadruples an area and multiplies a volume by eight - here the volumes would be in the ratio \( 1 : 8 \).

Answer: A

Halve the diameter to get the radius, then apply the volume formula with \( \pi = 3.142 \).

\[ r = \frac{8}{2} = 4 \text{ cm}, \qquad h = 16 \text{ cm} \]

\[ V = \pi r^{2} h = 3.142 \times 16 \times 16 \]

\[ V = 3.142 \times 256 = 804.352 \text{ cm}^{3} \]

Using the diameter in place of the radius would quadruple the answer to about 3217 cm³.

Answer: B

A bucket is a frustum, and its slant height, vertical height and the difference of the radii form a right triangle:

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

\[ 13^{2} = h^{2} + (9 - 4)^{2} \quad\Longrightarrow\quad 169 = h^{2} + 25 \]

\[ h^{2} = 144 \quad\Longrightarrow\quad h = 12 \text{ cm} \]

Using the sum \( 9 + 4 = 13 \) instead of the difference would make the height zero - it is the difference that gives the horizontal leg of the slant.

Answer: A

Compare volumes, since the box dimensions divide the room's cleanly.

\[ V_{\text{box}} = 4 \times 3 \times 2 = 24 \text{ cm}^{3} \]

\[ n = \frac{4080}{24} = 170 \]

A volume division only gives an upper bound in general, because packing can waste space; here it is achievable, since 24 divides 4080 exactly and the boxes can be laid in aligned layers.

Answer: A

The volume of a sphere is

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

A volume must scale with the cube of a length, so any candidate formula has to carry \( r^{3} \) - that alone eliminates the alternatives without recalling the constant.

The relation to the surface area is worth noting: \( V = \tfrac{1}{3} \times S \times r = \tfrac{1}{3} \times 4\pi r^{2} \times r \), the same one-third that appears for cones and pyramids.