Volume & Surface Areas Questions and Answers

Answer: A

A surahi is a water pot with a rounded, bulbous body and a long straight neck.

The body is a sphere and the neck is a cylinder.

A hemisphere would leave the body flat on one side, and a cone would taper the neck rather than keeping it a uniform tube.

Answer: D

Diameter of largest circle = 20 cm.
∴ Area of circle = 100π cm²
∴ Area of 6 circles = 6 × 100π = 600π cm² (∵ there are six faces in a cube)
Also, Area of cube = 6 × (20)² = 2400 cm²
Area of unpainted surface = 2400 - 600π
= 2400 - 600 × 22/7
= 514.28 cm².

Answer: A

Joining two hemispheres along their bases produces a complete sphere; the two flat circular faces are sealed inside and no longer form part of the surface.

\[ 2 \times 2\pi r^{2} = 4\pi r^{2} \]

which is precisely a sphere's surface area. Adding the two flat faces would give \( 6\pi r^{2} \) - correct for two separate hemispheres, but not for the joined solid.

Answer: B

The largest cone that fits inside a cube has its base inscribed in one face and its apex at the centre of the opposite face.

The base circle is limited by the face's width, so its diameter equals the cube's edge:

\[ d = 4.2 \text{ cm} \quad\Longrightarrow\quad r = \frac{4.2}{2} = 2.1 \text{ cm} \]

The cone's height is the full 4.2 cm edge, but the question asks only for the radius. The 4.2 option confuses the diameter with the radius.

Answer: A

Divide the granary's volume by the space one bag needs.

\[ V = 12 \times 6 \times 5 = 360 \text{ m}^{3} \]

\[ n = \frac{360}{0.48} = 750 \]

The division by a decimal is where slips creep in: \( \tfrac{360}{0.48} = \tfrac{36000}{48} = 750 \), which is cleaner to compute after multiplying both numbers by 100.

Answer: D

The men displace their own volume of water, and that volume spreads over the pool's whole surface.

\[ V_{\text{displaced}} = 150 \times 8 = 1200 \text{ m}^{3} \]

\[ \text{pool area} = 90 \times 40 = 3600 \text{ m}^{2} \]

\[ \text{rise} = \frac{1200}{3600} = \frac{1}{3} \text{ m} = 33.33 \text{ cm} \]

The answer is asked for in centimetres, so the metre figure must be multiplied by 100 - the whole point of the awkward-looking 33.33.

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: 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: 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: B

For the sphere to be enclosed snugly, it must touch the curved wall all the way round, so its radius equals the cylinder's radius \( r \).

\[ \text{diameter of sphere} = 2r \]

The condition \( h > 2r \) simply guarantees the cylinder is tall enough to contain that sphere; the height itself never limits the fit, so it does not appear in the answer.