Volume & Surface Areas Questions and Answers

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.

Answer: A

A cone's lateral surface is what the cardboard has to supply, so find the radius first.

\[ r = \sqrt{l^{2} - h^{2}} = \sqrt{25^{2} - 24^{2}} = \sqrt{625 - 576} = \sqrt{49} = 7 \text{ cm} \]

\[ \text{CSA} = \pi r l = \frac{22}{7} \times 7 \times 25 = 550 \text{ cm}^{2} \]

\[ \text{sheet area} = 8 \times 550 = 4400 \text{ cm}^{2} \]

A cone rolled from a sheet has no base, so only the curved surface counts - adding \( \pi r^{2} \) would overstate each cone by 154 cm².

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: D

Total surface area of one block = 2(lb + bh + lh)
= [2(22.5 ×10 + 10 × 7.5 + 22.5 × 7.5)] cm
= 2(225 + 75 + 168.75) cm²
= (2 × 468.75) cm²
= 937.5 cm²
Let n blocks can be painted out by the paint of the container.
Area of n bricks = (n ×937.5) cm² = 937.5n cm²
Area that can be painted by the paint of the container = 93.75 m2= 937500 cm²
 937500 = 937.5n
n = 1000
Therefore, 100 blocks can be painted out by the paint of the container

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.