Volume & Surface Areas Questions and Answers

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: C

The cylinder has diameter 12 cm, so \( r = 6 \) cm, and \( h = 12 \) cm.

\[ \text{CSA of cylinder} = 2\pi r h = 2\pi (6)(12) = 144\pi \text{ cm}^{2} \]

Setting the sphere's surface area equal to it:

\[ 4\pi R^{2} = 144\pi \quad\Longrightarrow\quad R^{2} = 36 \quad\Longrightarrow\quad R = 6 \text{ cm} \]

Using the diameter 12 in place of the radius in the cylinder formula would give \( 288\pi \) and a sphere of radius \( 6\sqrt{2} \) - the reason 12 cm is on the list.

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: B

With faces \( P = lb \), \( Q = bh \), \( R = hl \), the product of all three is the square of the volume:

\[ PQR = (lbh)^{2} \quad\Longrightarrow\quad V = \sqrt{PQR} \]

\[ V = \sqrt{120 \times 72 \times 60} = \sqrt{518400} = 720 \text{ cm}^{3} \]

Forgetting the square root leaves 518,400; the dimensions themselves work out to 10, 12 and 6 cm, which do reproduce all three given face areas.

Answer: C

Melting conserves volume, and every sphere's volume is proportional to \( r^{3} \), so the \( \tfrac{4}{3}\pi \) cancels throughout.

The radii are 1.5, 2 and 2.5 cm:

\[ R^{3} = 1.5^{3} + 2^{3} + 2.5^{3} = 3.375 + 8 + 15.625 = 27 \]

\[ R = \sqrt[3]{27} = 3 \text{ cm} \]

Radii add in the cube, never directly: \( 1.5 + 2 + 2.5 = 6 \) is the first distractor and is far too large.

Answer: B

Find the wall area from the cost, then use the perimeter.

\[ \text{area of 4 walls} = \frac{24000}{20} = 1200 \text{ m}^{2} \]

The four walls form a band of height \( h \) around the floor's perimeter:

\[ \text{area} = \text{perimeter} \times h \quad\Longrightarrow\quad 1200 = 150h \]

\[ h = 8 \text{ m} \]

Writing the wall area as \( 2(l+b)h \) is the same thing, since the perimeter \( 2(l+b) \) is already given as 150 m - the individual length and breadth are never needed.

Answer: D

The hemisphere of diameter 6 cm just covers the top plate, so each plate is a disc of radius 3 cm and thickness 2 cm.

\[ V_{\text{one plate}} = \pi r^{2} h = \pi (3)^{2}(2) = 18\pi \text{ cm}^{3} \]

\[ V_{\text{14 plates}} = 14 \times 18\pi = 252\pi \text{ cm}^{3} \]

\[ V_{\text{hemisphere}} = \frac{2}{3}\pi r^{3} = \frac{2}{3}\pi (27) = 18\pi \text{ cm}^{3} \]

\[ V_{\text{total}} = 252\pi + 18\pi = 270\pi \text{ cm}^{3} \]

"Just covers" is the clue that fixes the plates' radius at 3 cm - without it the plate dimensions would be unknown.

Answer: A

Same material means the same density, so weight is proportional to volume.

\[ V_{\text{solid}} = \frac{4}{3}\pi (3)^{3} = 36\pi \text{ cm}^{3} \;\longrightarrow\; 6 \text{ kg} \]

The hollow sphere has outer radius 9 cm and inner radius 6 cm:

\[ V_{\text{hollow}} = \frac{4}{3}\pi\left(9^{3} - 6^{3}\right) = \frac{4}{3}\pi (729 - 216) = \frac{4}{3}\pi (513) = 684\pi \text{ cm}^{3} \]

\[ \text{weight} = 6 \times \frac{684\pi}{36\pi} = 6 \times 19 = 114 \text{ kg} \]

The cavity must be subtracted; treating the object as a solid 18 cm sphere would give 162 kg.

Answer: C

Find the radius from the surface area first, taking \( \pi = \tfrac{22}{7} \).

\[ 4\pi r^{2} = \frac{792}{7} \quad\Longrightarrow\quad r^{2} = \frac{792}{7} \times \frac{7}{4 \times 22} = \frac{792}{88} = 9 \]

\[ r = 3 \text{ cm} \]

\[ V = \frac{4}{3}\pi r^{3} = \frac{4}{3} \times \frac{22}{7} \times 27 = \frac{792}{7} \text{ cm}^{3} \]

The numerical coincidence is real, not a misprint: for \( r = 3 \) the surface area and volume come out to the same number, because \( \tfrac{V}{S} = \tfrac{r}{3} = 1 \).

Answer: B

The volume must stay fixed while the height grows ninefold.

\[ V = \frac{1}{3}\pi r^{2} h = \frac{1}{3}\pi r'^{2}(9h) \]

\[ r^{2} = 9r'^{2} \quad\Longrightarrow\quad r' = \frac{r}{3} \]

So the radius must be reduced to one third. Because volume depends on \( r^{2} \), a ninefold change in height is offset by a change of only \( \sqrt{9} = 3 \) in the radius - reducing it to one ninth would shrink the volume to a ninth of what it should be.