CBSE Class 10 Mathematics Surface Areas and Volumes MCQs Set 17

Answer: (c) 99 cm³

Explanation:
1. 1. \( V = \dfrac{1}{3}\pi r^2 h = \dfrac{1}{3}\times\dfrac{22}{7}\times 3^2\times 10.5 = 99 \) cm³.

Teacher's Note:
1. Use one third of the cylinder formula with the cone's own height.
2. Each cone has height 10.5 cm, half of the tube.

Answer: (d) 452.57 cm²

Explanation:
1. 1. \( 2\pi r(h + r) = 2\times\dfrac{22}{7}\times 3\times(21+3) \).
2. 2. \( = 452.57 \) cm².

Teacher's Note:
1. Total surface area includes both circular ends.
2. Add h and r inside the bracket.

Answer: (b) 396 cm³

Explanation:
1. 1. Volume of cylinder \( = \dfrac{22}{7}\times 9\times 21 = 594 \) cm³.
2. 2. Remaining volume \( = 594 - 2\times 99 = 396 \) cm³.

Teacher's Note:
1. Subtract the volume of both cones from the cylinder.
2. The cones are identical.

Answer: (a) 56.57 cm²

Explanation:
1. 1. Each base has area \( \pi r^2 = \dfrac{22}{7}\times 9 \).
2. 2. Sum \( = 2\times\dfrac{198}{7} = 56.57 \) cm².

Teacher's Note:
1. Add two circular areas of radius 3 cm.
2. Do not include the cylinder ends.

Answer: (b) 396 cm²

Explanation:
1. 1. \( 2\pi r h = 2\times\dfrac{22}{7}\times 3\times 21 = 396 \) cm².

Teacher's Note:
1. Curved surface leaves out the two ends.
2. 2 x 22/7 x 3 x 21 = 396.

Answer: (b) 2.7 m²

Explanation:
1. 1. \( 2\pi r h = 2\times\dfrac{22}{7}\times 0.3\times 1.45 \approx 2.73 \) m².

Teacher's Note:
1. Convert 30 cm to 0.3 m before substituting.
2. Round 2.73 to 2.7.

Answer: (a) 0.6 m²

Explanation:
1. 1. \( 2\pi r^2 = 2\times\dfrac{22}{7}\times 0.09 \approx 0.57 \) m², which is about 0.6 m².

Teacher's Note:
1. Curved surface of a hemisphere is 2 pi r squared.
2. Use r = 0.3 m.

Answer: (c) 0.057 m³

Explanation:
1. 1. \( \dfrac{2}{3}\pi r^3 = \dfrac{2}{3}\times\dfrac{22}{7}\times 0.027 \approx 0.057 \) m³.

Teacher's Note:
1. Volume of a hemisphere is two thirds pi r cubed.
2. 0.3 cubed is 0.027.

Answer: (a) 5 m²

Explanation:
1. 1. Cylinder: \( 2\pi r h = 2\times\dfrac{22}{7}\times 0.28\times 2.6 \approx 4.58 \) m².
2. 2. Slant height of the cone \( l = \sqrt{0.28^2 + 0.36^2} \approx 0.456 \) m, so its curved surface is \( \pi r l \approx 0.40 \) m².
3. 3. Total \( \approx 4.97 \approx 5 \) m².

Teacher's Note:
1. Find the slant height with Pythagoras first.
2. Add the two curved areas.

Answer: (b) 0.03 m³

Explanation:
1. 1. \( \dfrac{1}{3}\pi r^2 h = \dfrac{1}{3}\times\dfrac{22}{7}\times 0.0784\times 0.36 \approx 0.03 \) m³.

Teacher's Note:
1. Use metres throughout: 28 cm = 0.28 m and 36 cm = 0.36 m.
2. The volume of a cone is one third pi r squared h.