Volume & Surface Areas Questions and Answers

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.

Answer: A

The submerged sphere displaces its own volume, which then spreads across the cylinder's cross-section.

\[ V_{\text{sphere}} = \frac{4}{3}\pi (9)^{3} = \frac{4}{3}\pi (729) = 972\pi \text{ cm}^{3} \]

\[ \text{cylinder cross-section} = \pi (18)^{2} = 324\pi \text{ cm}^{2} \]

\[ \text{rise} = \frac{972\pi}{324\pi} = 3 \text{ cm} \]

Both diameters must be halved before use; the \( \pi \) then cancels and no approximation is needed.

Answer: B

Start from the curved surface area to find the radius, taking \( \pi = \tfrac{22}{7} \).

\[ \pi r l = 308 \quad\Longrightarrow\quad \frac{22}{7} \times r \times 14 = 308 \]

\[ 44r = 308 \quad\Longrightarrow\quad r = 7 \text{ cm} \]

Now the total surface area adds the base:

\[ \text{TSA} = \pi r(l + r) = \frac{22}{7} \times 7 \times (14 + 7) = 22 \times 21 = 462 \text{ cm}^{2} \]

So the true statement is that the total surface area is 462 cm².

Checking the others: the radius is 7 cm, not 14 - 14 cm is the slant height. The height is \( \sqrt{l^{2} - r^{2}} = \sqrt{196 - 49} = \sqrt{147} \approx 12.1 \) cm, which is the square root of 147, not its square.

Answer: B

A cone has a single apex and a base bounded by a simple closed curve - in the familiar right circular cone, that curve is a circle.

The distinction from a pyramid is exactly the base: a pyramid's base is a polygon, made of straight edges, while a cone's is a smooth closed curve.

A triangle is a plane figure, not a solid, and a dice is a cube with no apex at all.

Answer: A

The curved surface needs the slant height, so find it by Pythagoras first.

\[ l = \sqrt{r^{2} + h^{2}} = \sqrt{5^{2} + 12^{2}} = \sqrt{25 + 144} = \sqrt{169} = 13 \text{ cm} \]

\[ \text{CSA} = \pi r l = 3.1416 \times 5 \times 13 \]

\[ = 204.2 \approx 204 \text{ cm}^{2} \]

5–12–13 is a Pythagorean triple, which is why the slant height comes out whole. Including the base would add \( \pi r^{2} = 78.5 \) cm² and give the total surface area of about 283 cm².

Answer: B

For a frustum the slant height spans the difference of the radii horizontally and the height vertically.

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

\[ l = \sqrt{4^{2} + (5-2)^{2}} = \sqrt{16 + 9} = \sqrt{25} = 5 \text{ cm} \]

It is the difference of the radii, not their sum, that forms the horizontal leg - using \( 5+2 \) would give \( \sqrt{65} \), the third option.