Volume & Surface Areas Questions and Answers

Answer: A

The number of small balls is 8000.

Volume is conserved, and the number of balls is the ratio of the volumes, which is the cube of the ratio of radii. The large radius is 6 cm = 60 mm; the small radius is 3 mm:

\[ n = \left(\frac{60}{3}\right)^{3} = 20^{3} = 8000 \]

1000 comes from using the small diameter (6 mm) as if it were the radius, and the other options do not cube the ratio.

Answer: B

The capacity is 26.994 litres.

The volume of a frustum with end radii \(R\), \(r\) and height \(h\) is \(\frac{\pi h}{3}(R^{2} + r^{2} + Rr)\):

\[ \begin{aligned} V &= \frac{22}{7} \times \frac{63}{3} \times (225 + 64 + 120) \\ &= 66 \times 409 = 26994\ \text{cm}^{3} \end{aligned} \]

Since 1 litre = 1000 cm\(^{3}\), this is 26.994 litres. The other options are the same digits with the decimal point moved, the result of a wrong conversion.

Answer: D

The diagonal of a cube of side a is \(a\sqrt{3}\), so \(a\sqrt{3}=11\sqrt{3}\) gives a = 11 feet.

The volume is 11 cubed = 1331 cubic feet, option D. The options with the root of 3 come from multiplying the diagonal rather than the side.

The space diagonal d of a cube of side a is \(a\sqrt{3}\), which follows from Pythagoras applied twice. With d = \(11\sqrt{3}\) the side is 11 feet, and the volume is 11 x 11 x 11 = 1331 cubic feet. The options with a root of 3 would apply if the diagonal was used as a side.

Answer: C

450 books can be arranged in the box.

Work in centimetres. The box is 150 cm by 75 cm by 60 cm, and each book is 20 cm by 10 cm by 7.5 cm.

\[ \begin{aligned} \text{box} &= 150 \times 75 \times 60 = 675000\ \text{cm}^{3} \\ \text{book} &= 20 \times 10 \times 7.5 = 1500\ \text{cm}^{3} \\ \text{number} &= \frac{675000}{1500} = 450 \end{aligned} \]

The books also pack the box exactly with nothing left over: stand the 7.5 cm side along the 75 cm edge (10 books), the 10 cm side along the 150 cm edge (15 books) and the 20 cm side along the 60 cm height (3 layers), and \(10 \times 15 \times 3 = 450\). 350, 400 and 500 do not fit either the volume or any arrangement.

Answer: A

The cost is ₹ 198.

The curved surface area of a cylinder is \(2\pi r h\), which is \(\pi d h\) in terms of the diameter:

\[ \begin{aligned} \text{area} &= \frac{22}{7} \times 1 \times 4.2 = 13.2\ \text{m}^2 \\ \text{cost} &= 13.2 \times 15 = 198 \end{aligned} \]

The options 19.8 and 1.98 are the same digits with the decimal point misplaced, and 200 is a rounded figure. Only the curved surface is whitewashed, so the top and bottom circles are not added.

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: A

Volume needs the perpendicular height, not the slant height, so find it by Pythagoras first.

\[ h = \sqrt{l^{2} - r^{2}} = \sqrt{5^{2} - 4^{2}} = \sqrt{9} = 3 \text{ cm} \]

\[ V = \frac{1}{3}\pi r^{2} h = \frac{1}{3}\pi (4)^{2}(3) = \frac{1}{3}\pi (16)(3) = 16\pi \text{ cm}^{3} \]

Using the slant height of 5 in place of \( h \) would give \( \tfrac{80}{3}\pi \) - the standard slip in every cone question that supplies \( l \) rather than \( h \).

Answer: D

A solid hemisphere has two surfaces: the curved dome and the flat circular base.

\[ \text{curved surface} = 2\pi r^{2}, \qquad \text{flat base} = \pi r^{2} \]

\[ \text{total surface area} = 2\pi r^{2} + \pi r^{2} = 3\pi r^{2} \]

The \( 2\pi r^{2} \) option is the curved surface alone - correct for a hollow bowl, but not for a solid hemisphere, which still has its flat face.

Answer: C

Surface area of a sphere is \( 4\pi r^{2} \), so it scales with the square of the radius.

\[ \frac{S_{1}}{S_{2}} = \frac{4\pi r^{2}}{4\pi (2r)^{2}} = \frac{r^{2}}{4r^{2}} = \frac{1}{4} \]

So the ratio is \( 1 : 4 \). Doubling a length quadruples an area and multiplies a volume by eight - here the volumes would be in the ratio \( 1 : 8 \).

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.