CBSE Class 10 Mathematics Surface Areas and Volumes MCQs Set 26

Answer: (c) Rs. 100

Explanation:
1. Wall area \( = 2(l+b)h \). Doubling l, b and h multiplies it by \( 2\times2 = 4 \).
2. Cost \( = 4\times25 = \) Rs. 100.

Teacher's Note:
1. Area of walls scales with the product of two lengths.
2. Cost is proportional to area.

Answer: (d) 225000 m²

Explanation:
1. Cross-section \( = 3\times1.2 = 3.6 \) m². In 30 minutes water travels 5000 m, so volume \( = 3.6\times5000 = 18000 \) m³.
2. Area irrigated \( = \frac{18000}{0.08} = 225000 \) m².

Teacher's Note:
1. Convert dm to m and km/h to m.
2. Volume = area covered × depth.

Answer: (a) 1 : 1

Explanation:
1. Here \( h = r \). Curved surface \( = 2\pi rh = 2\pi r^2 \); two ends \( = 2\pi r^2 \).
2. Ratio = 1 : 1.

Teacher's Note:
1. h : d = 1 : 2 means h = r.
2. Compare 2πrh with 2πr².

Answer: (a) 3168 cm²

Explanation:
1. Outer radius R = 12.5 cm and inner radius r = 11.5 cm.
2. Total area \( = 2\pi(R+r)h+2\pi(R^2-r^2) = 2\pi[24\times20+24] = 2\pi\times504 = 3168 \) cm² (using \( \pi = \frac{22}{7} \)).

Teacher's Note:
1. Add outer curved, inner curved and the two ring-shaped ends.
2. R² − r² = (R + r)(R − r).

Answer: (c) 550 cm²

Explanation:
1. Curved surface \( = 2\pi rh = 2\times\frac{22}{7}\times3.5\times25 = 550 \) cm².

Teacher's Note:
1. Only the curved surface is needed.
2. Use 2πrh.

Answer: (b) 531

Explanation:
1. Slant height \( l = \sqrt{20^2+2.1^2} \approx 20.11 \) cm.
2. Curved area \( = \pi rl \approx \pi\times2.1\times20.11 \approx 132.7 \) cm², grains \( \approx 4\times132.7 \approx 531 \).

Teacher's Note:
1. The grains are on the curved surface only.
2. Slant height from Pythagoras.

Answer: (a) Rs. 99000

Explanation:
1. Slant height \( l = \sqrt{7^2+24^2} = 25 \) m.
2. Canvas area \( = \pi rl = \frac{22}{7}\times7\times25 = 550 \) m², cost \( = 550\times180 = \) Rs. 99000.

Teacher's Note:
1. Curved surface area of a cone is πrl.
2. Cost = area × rate.

Answer: (b) 1 : 7

Explanation:
1. The small cone is similar with half the height, so its volume is \( \frac18 \) of the whole.
2. The ratio of the small cone to the frustum is \( 1 : 7 \).

Teacher's Note:
1. Volumes of similar solids go as the cube of corresponding lengths.
2. Frustum = whole − small cone.

Answer: (b) 1386 cm²

Explanation:
1. Total \( = 3\pi r^2 \) and curved \( = 2\pi r^2 \), so the difference is the base \( \pi r^2 = \frac{22}{7}\times441 = 1386 \) cm².

Teacher's Note:
1. The difference is the flat circular base.
2. πr² with r = 21.

Answer: (d) Rs. 61600

Explanation:
1. \( 2\pi r = 44 \Rightarrow r = 7 \) m. Curved area \( = 2\pi r^2 = 308 \) m² = 3 080 000 cm².
2. Cost \( = \frac{3\,080\,000}{100}\times2 = \) Rs. 61600.

Teacher's Note:
1. A dome is painted on the curved surface.
2. Convert m² to cm².

Answer: (b) 2541

Explanation:
1. Cube volume \( = 44^3 = 85184 \) cm³. Bullet volume \( = \frac43\times\frac{22}{7}\times2^3 = \frac{704}{21} \) cm³.
2. Number \( = 85184\times\frac{21}{704} = 2541 \).

Teacher's Note:
1. Divide the volume of the cube by the volume of one bullet.
2. Radius of a bullet is 2 cm.

Answer: (a) \( 4:4:\sqrt5 \)

Explanation:
1. Height = 2r. Sphere \( 4\pi r^2 \); cylinder \( 2\pi r\cdot2r = 4\pi r^2 \); cone \( \pi r\sqrt{r^2+4r^2} = \sqrt5\pi r^2 \).
2. Ratio \( = 4:4:\sqrt5 \).

Teacher's Note:
1. Height equal to the diameter means h = 2r.
2. Slant height of the cone is r√5.

Answer: (a) 3 cm

Explanation:
1. Sphere area \( = 4\pi\times25 = 100\pi \), so the cone's curved area is \( 20\pi = \pi\times4\times l \Rightarrow l = 5 \).
2. Height \( = \sqrt{25-16} = 3 \) cm.

Teacher's Note:
1. Find slant height from the curved area.
2. Then use Pythagoras.

Answer: (b) 7 cm

Explanation:
1. Length of wire \( = 2(15+7) = 44 \) cm \( = 2\pi r \Rightarrow r = 7 \) cm.

Teacher's Note:
1. The wire length equals the perimeter of the rectangle.
2. Circumference = 2πr.

Answer: (c) 15 cm

Explanation:
1. One serving \( = \frac13\pi(3)^2(12)+\frac23\pi(3)^3 = 36\pi+18\pi = 54\pi \) cm³, so 10 servings \( = 540\pi \).
2. \( \pi\times36\times h = 540\pi \Rightarrow h = 15 \) cm.

Teacher's Note:
1. Each serving = cone + hemisphere.
2. Equate total volumes.

Answer: (b) \( 2.25\pi \)

Explanation:
1. Let the cone's height be h and radius \( \sqrt2h \). The cube of side s has its top vertices on the cone: the top square's half-diagonal \( \frac{s}{\sqrt2} = \sqrt2h\left(1-\frac sh\right) \).
2. This gives \( s = \frac{2h}{3} \). Ratio \( = \frac{\frac13\pi\cdot2h^3}{8h^3/27} = \frac{9\pi}{4} = 2.25\pi \).

Teacher's Note:
1. The cube sits on the base with its top face touching the slant surface.
2. Use similar triangles for the radius at height s.

Answer: (c) \( \frac21 \)

Explanation:
1. \( \frac13\pi r^2h = \frac12\times\frac43\pi r^3 \Rightarrow h = 2r \).
2. So the ratio h : r = 2 : 1.

Teacher's Note:
1. Equate the volumes.
2. Same radius cancels.

Answer: (c) \( \sqrt{\frac6\pi} \)

Explanation:
1. \( 4\pi r^2 = 6a^2 \Rightarrow \frac ar = \sqrt{\frac{2\pi}{3}} \).
2. \( \frac{V_s}{V_c} = \frac{\frac43\pi r^3}{a^3} = \frac{4\pi}{3}\left(\frac{3}{2\pi}\right)^{3/2} = \sqrt{\frac6\pi} \).

Teacher's Note:
1. Express a in terms of r.
2. Then compare the volumes.

Answer: (a) \( \frac38 \)

Explanation:
1. Slant height is 10 cm. The largest sphere in the cone (touching the base and the slant surface) has radius \( \frac{rh}{r+l} = \frac{48}{16} = 3 \) cm.
2. Sphere volume \( = 36\pi \) and cone volume \( = \frac13\pi\cdot36\cdot8 = 96\pi \). Fraction \( = \frac{36}{96} = \frac38 \).

Teacher's Note:
1. The inradius of the axial section is the sphere's radius.
2. Fraction = sphere volume ÷ cone volume.

Answer: (a) \( 400\pi(5+2\sqrt6) \) sq mm

Explanation:
1. \( 4\pi r^2 = 4\pi(\sqrt2+\sqrt3)^2 = 4\pi(5+2\sqrt6) \) cm².
2. Converting to mm² (1 cm² = 100 mm²): \( 400\pi(5+2\sqrt6) \) sq mm.

Teacher's Note:
1. Expand (√2 + √3)² first.
2. Convert cm² to mm² by × 100.

Answer: (c) \( 3\sqrt[3]{3/2} \) cm

Explanation:
1. Cylinder volume \( = \pi\times9\times6 = 54\pi \). Sphere: \( \frac43\pi R^3 = 54\pi \Rightarrow R^3 = \frac{81}{2} \).
2. \( R = 3\sqrt[3]{\frac32} \) cm.

Teacher's Note:
1. Volume is conserved.
2. 27 × 3/2 = 81/2.