Volume & Surface Areas Questions and Answers

Answer: C

A gilli is the small wooden piece struck in gilli-danda. It has a straight middle section and tapers to a point at both ends, so that a tap on either tip flips it into the air.

The middle is a cylinder and each tapered end is a cone, giving two cones and a cylinder.

A single cone would leave one end blunt, which would not let the gilli be struck from both sides.

Answer: A

A right circular cone has a circular cross-section at every level, shrinking steadily toward the apex.

A plane parallel to the base therefore cuts the cone in a circle - smaller than the base, but a circle nonetheless.

The frustum is the three-dimensional solid left below that cut, not the cross-section itself; the cross-section is a flat, two-dimensional figure.

Answer: B

The four walls form a band of height 9 m around the floor's perimeter, and the ceiling is a rectangle the same size as the floor.

\[ \text{four walls} = 2(l + b)h = 2(12 + 10) \times 9 = 2 \times 22 \times 9 = 396 \text{ m}^{2} \]

\[ \text{ceiling} = l \times b = 12 \times 10 = 120 \text{ m}^{2} \]

\[ \text{total} = 396 + 120 = 516 \text{ m}^{2} \]

The floor is excluded, so this is not the full surface area of the cuboid - that would be \( 396 + 240 = 636 \) m², which is offered as the first distractor.

Answer: B

A drinking tumbler is circular at both the rim and the base, but wider at the top than at the bottom, and neither end closes to a point.

That is a frustum of a cone - a cone with its tip sliced off by a plane parallel to the base.

A full cone would taper to a point, and a cylinder would have the same width top and bottom.

Answer: A

Melting conserves volume, so equate the sheet's volume with the bar's.

\[ V_{\text{sheet}} = \pi (25)^{2}(0.65) = \pi (625)(0.65) = 406.25\pi \text{ cm}^{3} \]

\[ V_{\text{bar}} = \pi (4)^{2} L = 16\pi L \]

\[ 16\pi L = 406.25\pi \quad\Longrightarrow\quad L = \frac{406.25}{16} = 25.39 \text{ cm} \]

The \( \pi \) cancels, so no value for it is needed. Both diameters must be halved first - using 50 and 8 directly would give the same ratio, but the thickness would then be mismatched.

Answer: C

Convert the weight to a volume using the given density, then divide out the two known dimensions.

\[ 112 \text{ kg} = 112000 \text{ g} \quad\Longrightarrow\quad V = \frac{112000}{25} = 4480 \text{ cm}^{3} \]

\[ V = l \times b \times h \quad\Longrightarrow\quad l = \frac{4480}{28 \times 5} = \frac{4480}{140} = 32 \text{ cm} \]

The kilogram-to-gram conversion is essential: the density is quoted per cubic centimetre in grams, so the weight must be in grams too.

Answer: C

Volume of a cylinder is \( \pi r^{2}h \), so form the ratio with the \( \pi \) cancelling.

\[ \frac{V_{1}}{V_{2}} = \frac{r_{1}^{2}h_{1}}{r_{2}^{2}h_{2}} = \frac{4^{2} \times 21}{7^{2} \times 8} \]

\[ = \frac{16 \times 21}{49 \times 8} = \frac{336}{392} = \frac{6}{7} \]

So the ratio is \( 6 : 7 \). The radii enter squared and the heights only to the first power - treating both alike gives \( 84 : 56 = 3 : 2 \), which is the last distractor.

Answer: C

The immersed box displaces its own volume of water, and that displaced volume spreads over the vessel's base.

\[ V_{\text{cube}} = 30^{3} = 27000 \text{ cm}^{3} \]

\[ \text{base area} = 60 \times 30 = 1800 \text{ cm}^{2} \]

\[ \text{rise} = \frac{27000}{1800} = 15 \text{ cm} \]

The cube's own 30 cm edge is not the rise: the water spreads over a base twice the cube's footprint, so it climbs only half as far.

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: C

Spin a right-angled triangle about one of its perpendicular sides. That side stays fixed as the axis, the other perpendicular side sweeps out a flat circular disc at the bottom, and the hypotenuse sweeps out a slanted surface meeting at a point on the axis.

The result is a right circular cone, with the height equal to the axis side, the base radius equal to the other perpendicular side, and the slant height equal to the hypotenuse.

Revolving a rectangle instead would give a cylinder, and revolving about the hypotenuse would give two cones joined base to base.