Surface Areas and Volumes — CBSE Class 10 Maths Textbook Solutions

Volume of each cube $= 64\text{ cm}^3$= 64 cm^3, so its edge $a = \sqrt[3]{64} = 4$a = [3]64 = 4 cm.

When the two cubes are joined end to end (face to face), the resulting solid is a cuboid of length $l = 4+4 = 8$l = 4+4 = 8 cm, breadth $b = 4$b = 4 cm, height $h = 4$h = 4 cm. The two joined faces (each $4\text{ cm}\times 4\text{ cm}$4 cm× 4 cm) are hidden inside the solid and do not contribute to the outer surface.

$\text{TSA of cuboid} = 2(lb+bh+hl) = 2(8\times 4 + 4\times 4 + 4\times 8)$TSA of cuboid = 2(lb+bh+hl) = 2(8× 4 + 4× 4 + 4× 8)
$= 2(32+16+32) = 2\times 80 = 160\text{ cm}^2$= 2(32+16+32) = 2× 80 = 160 cm^2

The surface area of the resulting cuboid is $160\text{ cm}^2$160 cm^2.

Diameter of hemisphere $= 14$= 14 cm, so radius $r = 7$r = 7 cm.

Since the total height of the vessel is 13 cm and the hemisphere itself has "height" equal to its radius ($7$7 cm), the cylindrical part has height:

$h = 13 - 7 = 6\text{ cm}$h = 13 - 7 = 6 cm

The inner surface (as seen from inside the vessel) consists of the curved surface of the cylinder plus the curved surface of the hemisphere (the vessel is open at the top, hollow, so no flat circular area is counted):

$\text{Inner surface area} = \text{CSA of cylinder} + \text{CSA of hemisphere} = 2\pi r h + 2\pi r^2 = 2\pi r (h+r)$Inner surface area = CSA of cylinder + CSA of hemisphere = 2π r h + 2π r^2 = 2π r (h+r)

$= 2 \times \frac{22}{7} \times 7 \times (6+7) = 2 \times 22 \times 13 = 572\text{ cm}^2$= 2 × 22/7 × 7 × (6+7) = 2 × 22 × 13 = 572 cm^2

The inner surface area of the vessel is $572\text{ cm}^2$572 cm^2.

Radius $r = 3.5$r = 3.5 cm. The total height of the toy is $15.5$15.5 cm, and the hemisphere contributes height equal to its radius, so the height of the cone:

$h = 15.5 - 3.5 = 12\text{ cm}$h = 15.5 - 3.5 = 12 cm

Slant height of the cone:

$l = \sqrt{r^2+h^2} = \sqrt{3.5^2 + 12^2} = \sqrt{12.25 + 144} = \sqrt{156.25} = 12.5\text{ cm}$l = √r^2+h^2 = √3.5^2 + 12^2 = √12.25 + 144 = √156.25 = 12.5 cm

Total surface area of the toy $=$= CSA of cone $+$+ CSA of hemisphere (the flat circular face where they join is hidden inside):

$\text{TSA} = \pi r l + 2\pi r^2 = \pi r (l + 2r) = \frac{22}{7} \times 3.5 \times (12.5 + 7)$TSA = π r l + 2π r^2 = π r (l + 2r) = 22/7 × 3.5 × (12.5 + 7)

$= \frac{22}{7} \times 3.5 \times 19.5 = 11 \times 19.5 = 214.5\text{ cm}^2$= 22/7 × 3.5 × 19.5 = 11 × 19.5 = 214.5 cm^2

The total surface area of the toy is $214.5\text{ cm}^2$214.5 cm^2.

The hemisphere sits on top of the cube's flat face, so its diameter cannot exceed the side of the cube. The greatest diameter it can have equals the side of the cube:

$\text{Greatest diameter} = 7\text{ cm} \Rightarrow r = 3.5\text{ cm}$Greatest diameter = 7 cm r = 3.5 cm

Surface area of the solid: The exposed surface is the total surface of the cube, minus the circular area where the hemisphere sits (which is covered, not exposed), plus the curved surface of the hemisphere.

$\text{Surface area} = \text{TSA of cube} - \pi r^2 + 2\pi r^2 = 6a^2 + \pi r^2$Surface area = TSA of cube - π r^2 + 2π r^2 = 6a^2 + π r^2

$= 6 \times 7^2 + \frac{22}{7} \times (3.5)^2 = 6 \times 49 + \frac{22}{7} \times 12.25$= 6 × 7^2 + 22/7 × (3.5)^2 = 6 × 49 + 22/7 × 12.25

$= 294 + 38.5 = 332.5\text{ cm}^2$= 294 + 38.5 = 332.5 cm^2

The greatest diameter is 7 cm and the surface area of the solid is $332.5\text{ cm}^2$332.5 cm^2.

Let the edge of the cube be $l$l, so the radius of the hemispherical depression is $\dfrac{l}{2}$l/2.

The remaining solid's surface $=$= total surface of the cube $-$- the flat circular area removed (where the depression is cut) $+$+ the curved surface of the hemispherical depression (now exposed as a concave surface):

$\text{Surface area} = 6l^2 - \pi\left(\frac{l}{2}\right)^2 + 2\pi\left(\frac{l}{2}\right)^2 = 6l^2 + \pi\left(\frac{l}{2}\right)^2$Surface area = 6l^2 - π(l/2)^2 + 2π(l/2)^2 = 6l^2 + π(l/2)^2

$= 6l^2 + \frac{\pi l^2}{4} = \frac{24l^2 + \pi l^2}{4} = \frac{l^2(24+\pi)}{4}$= 6l^2 + π l^2/4 = 24l^2 + π l^2/4 = l^2(24+π)/4

The surface area of the remaining solid is $\dfrac{1}{4}l^2(\pi + 24)$1/4l^2(π + 24) square units.

Diameter $= 5$= 5 mm, so radius $r = 2.5$r = 2.5 mm.

The total length is $14$14 mm, and the two hemispherical ends together contribute a length of $2r$2r (one radius at each end), so the length of the cylindrical part:

$h = 14 - 2(2.5) = 14 - 5 = 9\text{ mm}$h = 14 - 2(2.5) = 14 - 5 = 9 mm

Surface area $=$= CSA of cylinder $+$+ CSA of the two hemispheres (their flat faces are joined to the cylinder and hidden):

$\text{Surface area} = 2\pi r h + 2(2\pi r^2) = 2\pi r h + 4\pi r^2 = 2\pi r (h + 2r)$Surface area = 2π r h + 2(2π r^2) = 2π r h + 4π r^2 = 2π r (h + 2r)

$= 2 \times \frac{22}{7} \times 2.5 \times (9 + 5) = 2 \times \frac{22}{7} \times 2.5 \times 14$= 2 × 22/7 × 2.5 × (9 + 5) = 2 × 22/7 × 2.5 × 14

$= \frac{110}{7} \times 14 = 220\text{ mm}^2$= 110/7 × 14 = 220 mm^2

The surface area of the capsule is $220\text{ mm}^2$220 mm^2.

Diameter of cylinder $= 4$= 4 m, so radius $r = 2$r = 2 m. Height of cylindrical part $h = 2.1$h = 2.1 m. Slant height of conical part $l = 2.8$l = 2.8 m.

The canvas covers the curved surface of the cylinder and the curved surface of the cone (no base, and the top of the cylinder is covered by the cone, so it isn't counted separately):

$\text{Canvas area} = \text{CSA of cylinder} + \text{CSA of cone} = 2\pi r h + \pi r l$Canvas area = CSA of cylinder + CSA of cone = 2π r h + π r l

$\text{CSA of cylinder} = 2 \times \frac{22}{7} \times 2 \times 2.1 = \frac{2 \times 22 \times 2 \times 2.1}{7} = \frac{184.8}{7} = 26.4\text{ m}^2$CSA of cylinder = 2 × 22/7 × 2 × 2.1 = 2 × 22 × 2 × 2.1/7 = 184.8/7 = 26.4 m^2

$\text{CSA of cone} = \frac{22}{7} \times 2 \times 2.8 = \frac{123.2}{7} = 17.6\text{ m}^2$CSA of cone = 22/7 × 2 × 2.8 = 123.2/7 = 17.6 m^2

$\text{Total canvas area} = 26.4 + 17.6 = 44\text{ m}^2$Total canvas area = 26.4 + 17.6 = 44 m^2

Cost of the canvas at Rs 500 per $\text{m}^2$m^2:

$\text{Cost} = 44 \times 500 = \text{Rs } 22000$Cost = 44 × 500 = Rs 22000

The area of canvas needed is $44\text{ m}^2$44 m^2 and it costs Rs 22000.

Diameter $= 1.4$= 1.4 cm, so radius $r = 0.7$r = 0.7 cm; height $h = 2.4$h = 2.4 cm (the cavity has the same height and diameter as the cylinder, so its apex just touches the opposite flat face).

Slant height of the conical cavity:

$l = \sqrt{r^2+h^2} = \sqrt{0.7^2+2.4^2} = \sqrt{0.49+5.76} = \sqrt{6.25} = 2.5\text{ cm}$l = √r^2+h^2 = √0.7^2+2.4^2 = √0.49+5.76 = √6.25 = 2.5 cm

The remaining solid's total surface consists of: the curved surface of the cylinder (outside), the flat circular base at the bottom (untouched, since the cone's apex meets it only at a single point), and the curved (inner) surface of the conical cavity carved out from the top:

$\text{TSA} = 2\pi r h + \pi r^2 + \pi r l$TSA = 2π r h + π r^2 + π r l

$2\pi r h = 2 \times \frac{22}{7} \times 0.7 \times 2.4 = \frac{73.92}{7} = 10.56\text{ cm}^2$2π r h = 2 × 22/7 × 0.7 × 2.4 = 73.92/7 = 10.56 cm^2

$\pi r^2 = \frac{22}{7} \times 0.49 = 1.54\text{ cm}^2$π r^2 = 22/7 × 0.49 = 1.54 cm^2

$\pi r l = \frac{22}{7} \times 0.7 \times 2.5 = \frac{38.5}{7} = 5.5\text{ cm}^2$π r l = 22/7 × 0.7 × 2.5 = 38.5/7 = 5.5 cm^2

$\text{TSA} = 10.56 + 1.54 + 5.5 = 17.6\text{ cm}^2 \approx 18\text{ cm}^2$TSA = 10.56 + 1.54 + 5.5 = 17.6 cm^2 18 cm^2

The total surface area of the remaining solid, to the nearest $\text{cm}^2$cm^2, is $18\text{ cm}^2$18 cm^2.

Radius $r = 3.5$r = 3.5 cm, height $h = 10$h = 10 cm.

The total surface consists of: the curved surface of the cylinder (unchanged, still exposed all around), plus the curved surfaces of the two hemispherical cavities scooped out at each end (each flat circular end of the cylinder is replaced by a concave hemispherical surface of the same radius):

$\text{TSA} = \text{CSA of cylinder} + 2 \times \text{CSA of hemisphere} = 2\pi r h + 2(2\pi r^2) = 2\pi r h + 4\pi r^2$TSA = CSA of cylinder + 2 × CSA of hemisphere = 2π r h + 2(2π r^2) = 2π r h + 4π r^2

$= 2\pi r (h + 2r) = 2 \times \frac{22}{7} \times 3.5 \times (10+7)$= 2π r (h + 2r) = 2 × 22/7 × 3.5 × (10+7)

$= 2 \times \frac{22}{7} \times 3.5 \times 17 = 22 \times 17 = 374\text{ cm}^2$= 2 × 22/7 × 3.5 × 17 = 22 × 17 = 374 cm^2

The total surface area of the article is $374\text{ cm}^2$374 cm^2.

Radius $r = 1$r = 1 cm. The cone's height equals its radius, so $h = 1$h = 1 cm.

$\text{Volume of solid} = \text{Volume of cone} + \text{Volume of hemisphere} = \frac{1}{3}\pi r^2 h + \frac{2}{3}\pi r^3$Volume of solid = Volume of cone + Volume of hemisphere = 1/3π r^2 h + 2/3π r^3

$= \frac{1}{3}\pi (1)^2(1) + \frac{2}{3}\pi (1)^3 = \frac{\pi}{3} + \frac{2\pi}{3} = \frac{3\pi}{3} = \pi$= 1/3π (1)^2(1) + 2/3π (1)^3 = π/3 + 2π/3 = 3π/3 = π

The volume of the solid is $\pi\text{ cm}^3$π cm^3.

Diameter $= 3$= 3 cm, so radius $r = 1.5$r = 1.5 cm. Each cone has height $2$2 cm, so the two cones together take up $4$4 cm of the total length, leaving the cylindrical part with height:

$h_{\text{cyl}} = 12 - 2(2) = 8\text{ cm}$h_cyl = 12 - 2(2) = 8 cm

$\text{Volume of air} = \text{Volume of cylinder} + 2 \times \text{Volume of one cone}$Volume of air = Volume of cylinder + 2 × Volume of one cone

$= \pi r^2 h_{\text{cyl}} + 2 \times \frac{1}{3}\pi r^2 h_{\text{cone}} = \pi r^2\left[h_{\text{cyl}} + \frac{2}{3}h_{\text{cone}}\right]$= π r^2 h_cyl + 2 × 1/3π r^2 h_cone = π r^2[h_cyl + 2/3h_cone]

$= \frac{22}{7} \times (1.5)^2 \times \left[8 + \frac{2}{3}(2)\right] = \frac{22}{7} \times 2.25 \times \left[8 + \frac{4}{3}\right]$= 22/7 × (1.5)^2 × [8 + 2/3(2)] = 22/7 × 2.25 × [8 + 4/3]

$= \frac{22}{7} \times 2.25 \times \frac{28}{3} = \frac{22 \times 2.25 \times 28}{21}$= 22/7 × 2.25 × 28/3 = 22 × 2.25 × 28/21

$= \frac{1386}{21} = 66\text{ cm}^3$= 1386/21 = 66 cm^3

The volume of air contained in the model is $66\text{ cm}^3$66 cm^3.

Diameter $= 2.8$= 2.8 cm, so radius $r = 1.4$r = 1.4 cm. Total length $= 5$= 5 cm, and the two hemispherical ends together take up $2r = 2.8$2r = 2.8 cm, so the cylindrical part has length:

$h = 5 - 2(1.4) = 5 - 2.8 = 2.2\text{ cm}$h = 5 - 2(1.4) = 5 - 2.8 = 2.2 cm

Volume of one gulab jamun:

$V = \pi r^2 h + \frac{4}{3}\pi r^3 = \pi r^2\left(h + \frac{4}{3}r\right)$V = π r^2 h + 4/3π r^3 = π r^2(h + 4/3r)

$= \frac{22}{7} \times (1.4)^2 \times \left(2.2 + \frac{4}{3}\times 1.4\right) = \frac{22}{7} \times 1.96 \times (2.2 + 1.8\overline{6})$= 22/7 × (1.4)^2 × (2.2 + 4/3× 1.4) = 22/7 × 1.96 × (2.2 + 1.86)

$= \frac{22}{7} \times 1.96 \times 4.0\overline{6} \approx 6.16 \times 4.0667 \approx 25.05\text{ cm}^3$= 22/7 × 1.96 × 4.06 6.16 × 4.0667 25.05 cm^3

Volume of 45 gulab jamuns:

$45 \times 25.05 \approx 1127.28\text{ cm}^3$45 × 25.05 1127.28 cm^3

Syrup is about 30% of this volume:

$0.30 \times 1127.28 \approx 338.18\text{ cm}^3$0.30 × 1127.28 338.18 cm^3

The amount of syrup in 45 gulab jamuns is approximately $338\text{ cm}^3$338 cm^3.

Volume of the cuboid:

$V_{\text{cuboid}} = 15 \times 10 \times 3.5 = 525\text{ cm}^3$V_cuboid = 15 × 10 × 3.5 = 525 cm^3

Volume of one conical depression (radius $= 0.5$= 0.5 cm, depth $= 1.4$= 1.4 cm):

$V_{\text{cone}} = \frac{1}{3}\pi r^2 h = \frac{1}{3} \times \frac{22}{7} \times (0.5)^2 \times 1.4 = \frac{1}{3} \times \frac{22}{7} \times 0.25 \times 1.4$V_cone = 1/3π r^2 h = 1/3 × 22/7 × (0.5)^2 × 1.4 = 1/3 × 22/7 × 0.25 × 1.4

$= \frac{1}{3} \times \frac{7.7}{7} = \frac{1}{3} \times 1.1 = 0.3\overline{6}\text{ cm}^3$= 1/3 × 7.7/7 = 1/3 × 1.1 = 0.36 cm^3

Volume of 4 depressions:

$4 \times 0.3\overline{6} = 1.4\overline{6}\text{ cm}^3$4 × 0.36 = 1.46 cm^3

Volume of wood in the stand:

$V_{\text{wood}} = V_{\text{cuboid}} - 4V_{\text{cone}} = 525 - 1.46\overline{6} = 523.53\text{ cm}^3$V_wood = V_cuboid - 4V_cone = 525 - 1.466 = 523.53 cm^3

The volume of wood in the entire pen stand is $523.53\text{ cm}^3$523.53 cm^3.

Volume of water in the cone (filled to the brim):

$V_{\text{cone}} = \frac{1}{3}\pi r^2 h = \frac{1}{3}\pi (5)^2(8) = \frac{200\pi}{3}\text{ cm}^3$V_cone = 1/3π r^2 h = 1/3π (5)^2(8) = 200π/3 cm^3

The water that flows out equals one-fourth of this volume, and this displaced water equals the total volume of the lead shots dropped in:

$V_{\text{displaced}} = \frac{1}{4} \times \frac{200\pi}{3} = \frac{50\pi}{3}\text{ cm}^3$V_displaced = 1/4 × 200π/3 = 50π/3 cm^3

Volume of one lead shot (sphere of radius $0.5$0.5 cm):

$V_{\text{shot}} = \frac{4}{3}\pi (0.5)^3 = \frac{4}{3}\pi \times 0.125 = \frac{\pi}{6}\text{ cm}^3$V_shot = 4/3π (0.5)^3 = 4/3π × 0.125 = π/6 cm^3

Number of lead shots $n$n:

$n \times \frac{\pi}{6} = \frac{50\pi}{3}$n × π/6 = 50π/3
$n = \frac{50\pi}{3} \times \frac{6}{\pi} = \frac{300}{3} = 100$n = 50π/3 × 6/π = 300/3 = 100

The number of lead shots dropped into the vessel is 100.

Larger cylinder: base diameter $= 24$= 24 cm, so $r_1 = 12$r_1 = 12 cm, $h_1 = 220$h_1 = 220 cm.

Smaller cylinder: $r_2 = 8$r_2 = 8 cm, $h_2 = 60$h_2 = 60 cm.

$\text{Volume} = \pi r_1^2 h_1 + \pi r_2^2 h_2$Volume = π r_1^2 h_1 + π r_2^2 h_2

$\pi r_1^2 h_1 = 3.14 \times 144 \times 220 = 3.14 \times 31680 = 99475.2\text{ cm}^3$π r_1^2 h_1 = 3.14 × 144 × 220 = 3.14 × 31680 = 99475.2 cm^3

$\pi r_2^2 h_2 = 3.14 \times 64 \times 60 = 3.14 \times 3840 = 12057.6\text{ cm}^3$π r_2^2 h_2 = 3.14 × 64 × 60 = 3.14 × 3840 = 12057.6 cm^3

$\text{Total volume} = 99475.2 + 12057.6 = 111532.8\text{ cm}^3$Total volume = 99475.2 + 12057.6 = 111532.8 cm^3

Mass (at 8 g per $\text{cm}^3$cm^3):

$\text{Mass} = 111532.8 \times 8 = 892262.4\text{ g} = 892.2624\text{ kg}$Mass = 111532.8 × 8 = 892262.4 g = 892.2624 kg

The mass of the pole is approximately 892.26 kg.

Volume of the cylinder (radius $R = 60$R = 60 cm, height $H = 180$H = 180 cm):

$V_{\text{cyl}} = \pi R^2 H = \frac{22}{7} \times 60^2 \times 180 = \frac{22}{7} \times 3600 \times 180 = \frac{22}{7} \times 648000 = \frac{14256000}{7}\text{ cm}^3$V_cyl = π R^2 H = 22/7 × 60^2 × 180 = 22/7 × 3600 × 180 = 22/7 × 648000 = 14256000/7 cm^3

Volume of the solid (cone + hemisphere, both of radius $r = 60$r = 60 cm, cone height $=120$=120 cm):

$V_{\text{solid}} = \frac{1}{3}\pi r^2 (120) + \frac{2}{3}\pi r^3 = \pi r^2\left[\frac{120}{3} + \frac{2}{3}(60)\right] = \pi r^2 (40+40) = \pi r^2 (80)$V_solid = 1/3π r^2 (120) + 2/3π r^3 = π r^2[120/3 + 2/3(60)] = π r^2 (40+40) = π r^2 (80)

$= \frac{22}{7} \times 3600 \times 80 = \frac{22}{7} \times 288000 = \frac{6336000}{7}\text{ cm}^3$= 22/7 × 3600 × 80 = 22/7 × 288000 = 6336000/7 cm^3

Volume of water left:

$V_{\text{water}} = V_{\text{cyl}} - V_{\text{solid}} = \frac{14256000 - 6336000}{7} = \frac{7920000}{7} = 1131428.57\text{ cm}^3$V_water = V_cyl - V_solid = 14256000 - 6336000/7 = 7920000/7 = 1131428.57 cm^3

Converting to $\text{m}^3$m^3 (since $1\text{ m}^3 = 10^6\text{ cm}^3$1 m^3 = 10^6 cm^3):

$V_{\text{water}} \approx 1.131\text{ m}^3$V_water 1.131 m^3

The volume of water left in the cylinder is approximately $1.131\text{ m}^3$1.131 m^3.

Cylindrical neck: diameter $= 2$= 2 cm, so radius $r_1 = 1$r_1 = 1 cm, length $h = 8$h = 8 cm.

$V_{\text{neck}} = \pi r_1^2 h = 3.14 \times 1^2 \times 8 = 25.12\text{ cm}^3$V_neck = π r_1^2 h = 3.14 × 1^2 × 8 = 25.12 cm^3

Spherical part: diameter $= 8.5$= 8.5 cm, so radius $r_2 = 4.25$r_2 = 4.25 cm.

$V_{\text{sphere}} = \frac{4}{3}\pi r_2^3 = \frac{4}{3} \times 3.14 \times (4.25)^3$V_sphere = 4/3π r_2^3 = 4/3 × 3.14 × (4.25)^3

$(4.25)^3 = 4.25 \times 4.25 \times 4.25 = 18.0625 \times 4.25 = 76.765625$(4.25)^3 = 4.25 × 4.25 × 4.25 = 18.0625 × 4.25 = 76.765625

$V_{\text{sphere}} = \frac{4}{3} \times 3.14 \times 76.765625 \approx 4.1867 \times 76.765625 \approx 321.39\text{ cm}^3$V_sphere = 4/3 × 3.14 × 76.765625 4.1867 × 76.765625 321.39 cm^3

Total volume:

$V_{\text{total}} = V_{\text{neck}} + V_{\text{sphere}} = 25.12 + 321.39 \approx 346.51\text{ cm}^3$V_total = V_neck + V_sphere = 25.12 + 321.39 346.51 cm^3

The actual volume is about $346.51\text{ cm}^3$346.51 cm^3, not $345\text{ cm}^3$345 cm^3 as the child measured.

The child's answer is not correct. The correct volume is approximately $346.51\text{ cm}^3$346.51 cm^3.