Volume & Surface Areas Questions and Answers

Answer: C

The maximum volume is \(\frac{1}{3}\pi r^{3}\).

The largest cone that fits in a solid hemisphere of radius r has its base on the flat face of radius r and its height equal to r.

\[ V=\frac{1}{3}\pi r^{2}\cdot r=\frac{1}{3}\pi r^{3} \]

\(\frac{2}{3}\pi r^{3}\) is the volume of the hemisphere itself, and \(\frac{4}{3}\pi r^{3}\) is a full sphere.

The cone with the greatest volume has its base on the flat face of the hemisphere and its apex at the top of the curved surface, which makes the height equal to the radius. A hemisphere has volume 2/3 pi r cubed, so the cone is exactly half of it: 1/3 pi r cubed.

Answer: A

The radius is 6.31 cm.

The volume of the new sphere equals the sum of the three volumes, so the cubes of the radii add up.

\[ R^{3}=2^{3}+3^{3}+6^{3}=8+27+216=251 \]

\[ R=\sqrt[3]{251}\approx6.31\text{ cm} \]

6.35, 6.55 and 6.15 cm would have cubes of about 256, 281 and 233.

You can check 6.31: its cube is 6.31 × 6.31 × 6.31, about 251.2, which is almost exactly 251. For 6.35 the cube is about 256, for 6.55 about 281 and for 6.15 about 233, which are all too far from 251.

Answer: C

The volume is \(4077.\overline{3}\ \text{cm}^{3}\).

The metal is the difference between the outer and inner spheres. Using 22/7 for pi:

\[ V=\frac{4}{3}\cdot\frac{22}{7}\,(10^{3}-3^{3}) \]

\[ V=\frac{88}{21}\times973=4077.\overline{3} \]

The other options differ in the decimal part and do not match.

The shell is hollow, so the metal fills the space between the radius 3 cm and radius 10 cm spheres. The difference of the cubes is 1000 − 27 = 973 and the volume is 4/3 × 22/7 × 973 = 4077.33 cm cubed, which is 4077 with the 3 repeating.

Answer: A

The total cost is Rs. 3,78,000.

The bottom is 24 × 12 = 288 m². The sides are the four walls: 2(24 + 12) × 3 = 216 m². The tiled area is 504 m². The printed rate is Rs. 750 per square metre (the question says m³, which is a slip).

\[ 504\times750=378000 \]

5,94,000 and 5,88,000 would include the top surface, and 3,96,000 does not come from these areas.

The pool is open at the top, so only the bottom and four walls are tiled. The long walls are 24 × 3 × 2 = 144 and the short walls are 12 × 3 × 2 = 72, so the walls are 216 m squared. With the bottom, 288, the total area is 504 m squared.

Answer: B

The perimeter of the base is 26.4 cm. The volume of a cylinder is \(\pi r^{2}h\), so \[ \pi r^{2} = \frac{554.4}{10} = 55.44 \] \[ \begin{aligned} r^{2} &= 55.44\times\frac{7}{22} \\ &= 17.64 \Rightarrow r \\ &= 4.2\text{ cm} \end{aligned} \] The perimeter (circumference) of the base is \[ \begin{aligned} 2\pi r &= 2\times\frac{22}{7}\times 4.2 \\ &= 26.4\text{ cm} \end{aligned} \] The base radius of 4.2 cm gives 26.4 cm, so 17.6 cm, 22 cm and 44 cm are not the circumference of that circle. Check: \(\frac{22}{7}\times 4.2^{2}\times 10 = 554.4\), which is the given volume.

Answer: D

The volume of the cone is \(1680\pi\) cm³. The base radius is 12 cm. The curved surface area is \(\pi r l\), so \[ \begin{aligned} \pi\times 12\times l &= 444\pi \Rightarrow l \\ &= 37\text{ cm} \end{aligned} \] The height follows from the right triangle: \[ \begin{aligned} h &= \sqrt{37^{2} - 12^{2}} \\ &= \sqrt{1369 - 144} \\ &= \sqrt{1225} \\ &= 35 \end{aligned} \] \[ \begin{aligned} V &= \frac{1}{3}\pi r^{2}h \\ &= \frac{1}{3}\pi\times 144\times 35 \\ &= 1680\pi \end{aligned} \] The options 840, 1120 and 1400 times \(\pi\) correspond to heights of 17.5, 23.3 and 29.2, which do not fit a slant height of 37.

Answer: D

The lateral surface area is \(0.25\,x^{2}\).

The base is a square of perimeter \(x\), so each side is \(x/4\). The slant height is half the perimeter, \(x/2\). The lateral surface consists of four congruent triangles, each with base \(x/4\) and height equal to the slant height:

\[ \begin{aligned} A &= 4 \times \frac{1}{2} \times \frac{x}{4} \times \frac{x}{2} \\ &= \frac{x^{2}}{4} = 0.25\,x^{2} \end{aligned} \]

\(0.50\,x^{2}\) forgets the half in the triangle area, and \(x^{2}\) uses the full perimeter as the base of each face. \(0.75\,x^{2}\) matches no consistent working.

Answer: D

The order is (iv), (iii), (ii), (i).

• (iv) sphere: \(\frac{4}{3}\pi(7)^{3} \approx 1437\ \text{cm}^{3}\)
• (iii) cylinder: \(\pi(7)^{2}(7) \approx 1078\ \text{cm}^{3}\)
• (ii) cube: \(8^{3} = 512\ \text{cm}^{3}\)
• (i) cuboid: \(10 \times 8 \times 6 = 480\ \text{cm}^{3}\)

From largest to smallest this is sphere, cylinder, cube, cuboid. The other orders put the cuboid or cube ahead of the curved solids, whose radius 7 cm gives them much larger volumes.

Answer: D

The minimum volume is \(8r^{3}\).

The smallest cube that can hold a sphere of radius \(r\) is the one the sphere just touches on all six faces, so the cube’s edge equals the sphere’s diameter, \(2r\). Its volume is

\[ (2r)^{3} = 8r^{3} \]

The options \(r^{3}\) and \(r^{3}/8\) come from taking the edge as \(r\) or \(r/2\), which are too small to hold the sphere: the sphere’s own volume, \(\tfrac{4}{3}\pi r^{3} \approx 4.19r^{3}\), is already larger than either. \(2r^{3}\) is also smaller than the sphere itself, so it cannot be a box that encloses it.

Answer: A

The ratio is \(1/\pi\).

Joining the two 54 cm edges makes a tube 54 cm long whose circumference is the 4 cm edge, so its radius is \(r = 4/2\pi = 2/\pi\) cm:

\[ \begin{aligned} V_{tube} &= \pi r^{2} h \\ &= \pi \cdot \frac{4}{\pi^{2}} \cdot 54 = \frac{216}{\pi} \end{aligned} \]

The sheet's area is \(54 \times 4 = 216\ \text{cm}^{2}\). A cube with that surface area has \(6a^{2} = 216\), so \(a = 6\) cm and \(V_{cube} = 216\ \text{cm}^{3}\). The ratio is \(\frac{216/\pi}{216} = \frac{1}{\pi}\).

The trap is to roll the sheet the other way (circumference 54 cm, length 4 cm), which gives a far larger tube; the stem joins the longer edges, so the circumference is the short side. The other ratios do not follow from either reading.