Volume & Surface Areas Questions and Answers

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.

Answer: C

The curved surface area of a cone is \( \pi r l \). Equal areas give

\[ \pi r_{1} l_{1} = \pi r_{2} l_{2} \quad\Longrightarrow\quad \frac{r_{1}}{r_{2}} = \frac{l_{2}}{l_{1}} \]

With \( l_{1} : l_{2} = 3 : 5 \):

\[ \frac{r_{1}}{r_{2}} = \frac{5}{3} \quad\Longrightarrow\quad 5 : 3 \]

The radii are in the inverse ratio of the slant heights, since their product is fixed - the cone with the shorter slant must be wider to make up the same area.

Answer: B

A sphere's surface area is

\[ S = 4\pi r^{2} \]

Note that it is exactly four times the area of a great circle (\( \pi r^{2} \)) - the flat disc you would get by slicing the sphere through its centre.

The distractors are all real formulas from elsewhere: \( 2\pi r \) is the circumference of a circle, and \( 2\pi r^{2} \) is the curved surface of a hemisphere. A surface area must be proportional to \( r^{2} \), which rules out \( 2\pi r \) on dimensional grounds alone.

Answer: A

A solid hemisphere has both the curved dome and the flat base, giving a total of \( 3\pi r^{2} \).

\[ \text{TSA} = 3 \times \frac{22}{7} \times 7^{2} = 3 \times \frac{22}{7} \times 49 \]

\[ = 3 \times 22 \times 7 = 462 \text{ cm}^{2} \]

The curved surface alone is \( 2\pi r^{2} = 308 \) cm² and the flat base \( \pi r^{2} = 154 \) cm² - that 154 is offered as a distractor.

Answer: B

The curved surface area of a cylinder is

\[ \text{CSA} = 2\pi r h \]

Halving the radius and doubling the height gives

\[ 2\pi \left(\frac{r}{2}\right)(2h) = 2\pi r h \]

The two changes cancel exactly, so the curved surface area is the same. The volume does not survive the same way: \( \pi (r/2)^{2}(2h) = \tfrac{1}{2}\pi r^{2}h \), halving, because volume depends on \( r^{2} \) while curved surface area depends on \( r \) alone.

Answer: D

Convert the diameter to metres before anything else, since the rate is per square metre.

\[ d = 40 \text{ cm} = 0.4 \text{ m} \quad\Longrightarrow\quad r = 0.2 \text{ m} \]

\[ \text{CSA} = 2\pi r h = 2 \times \frac{22}{7} \times 0.2 \times 7 = 8.8 \text{ m}^{2} \]

\[ \text{cost} = 8.8 \times 20 = \text{Rs. } 176 \]

Only the curved surface is polished, so the two circular ends are excluded; including them would add \( 2\pi r^{2} = 0.251 \) m².