Volume & Surface Areas Questions and Answers

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.

Answer: D

One litre is 1000 cubic centimetres.

\[ 900{,}000 \text{ cm}^{3} = \frac{900{,}000}{1000} = 900 \text{ litres} \]

Dividing by 100 or by 10,000 gives 9000 and 90, the two distractors either side. It helps to remember that a cubic metre - 1,000,000 cm³ - is exactly 1000 litres, so 900 litres is a little under a cubic metre.

Answer: A

Let the second cone have radius \( r \) and height \( h \). The first then has radius \( 2r \) and height \( \tfrac{h}{2} \).

\[ V_{1} = \frac{1}{3}\pi (2r)^{2}\left(\frac{h}{2}\right) = \frac{1}{3}\pi \cdot 4r^{2} \cdot \frac{h}{2} = \frac{2}{3}\pi r^{2}h \]

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

\[ \frac{V_{1}}{V_{2}} = \frac{2/3}{1/3} = 2 \quad\Longrightarrow\quad 2 : 1 \]

Doubling the radius quadruples the volume while halving the height only halves it, so the net effect is a doubling - radius counts twice over, height only once.

Answer: A

Solve the surface-area formula for \( r \), taking \( \pi = \tfrac{22}{7} \).

\[ 4\pi r^{2} = 616 \]

\[ r^{2} = \frac{616 \times 7}{4 \times 22} = \frac{4312}{88} = 49 \]

\[ r = 7 \text{ cm} \]

Using \( 2\pi r^{2} \) - the hemisphere formula - would give \( r = 7\sqrt{2} \approx 9.9 \) cm instead.

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

Recasting a solid rearranges the same material into a different shape. No matter is added or removed, so the volume is conserved:

\[ V_{\text{new}} = V_{\text{old}} \]

This is the principle behind every "melted and recast" problem - it is the equation you set up first. The surface area, by contrast, does change, and usually increases when a compact shape is drawn out into a long thin one.