Surface Areas and Volumes — NCERT Solutions, AI Tutor & Practice

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

Joining two cubes end to end gives a cuboid with:

• length $l = 4+4 = 8\ \text{cm}$
• breadth $b = 4\ \text{cm}$
• height $h = 4\ \text{cm}$

Surface area of cuboid: $2(lb+bh+hl) = 2(8\times4 + 4\times4 + 4\times8)$ $= 2(32+16+32) = 2\times80 = 160\ \text{cm}^2$

So the surface area is $160\ \text{cm}^2$.

Radius $r = 3.5\ \text{cm}$.

Height of hemisphere $= r = 3.5\ \text{cm}$. Height of cone $h = 15.5 - 3.5 = 12\ \text{cm}$.

Slant height of cone: $l = \sqrt{r^2 + h^2} = \sqrt{3.5^2 + 12^2} = \sqrt{12.25 + 144} = \sqrt{156.25} = 12.5\ \text{cm}$

Total surface area = CSA of cone + CSA of hemisphere: $= \pi r l + 2\pi r^2 = \pi r(l + 2r)$ $= \frac{22}{7}\times 3.5 \times (12.5 + 7)$ $= \frac{22}{7}\times 3.5 \times 19.5 = 11 \times 19.5 = 214.5\ \text{cm}^2$

Total surface area $= 214.5\ \text{cm}^2$.

Side of cube $a = 7\ \text{cm}$.

Largest hemisphere has diameter equal to the side $= 7\ \text{cm}$, so radius $r = 3.5\ \text{cm}$.

Surface area of solid = Surface area of cube − circular base area of hemisphere + curved surface of hemisphere: $= 6a^2 - \pi r^2 + 2\pi r^2 = 6a^2 + \pi r^2$

Compute: $6a^2 = 6\times 49 = 294\ \text{cm}^2$ $\pi r^2 = \frac{22}{7}\times 3.5 \times 3.5 = 38.5\ \text{cm}^2$

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

Radius $r = 7\ \text{cm}$.

Height of hemisphere $= r = 7\ \text{cm}$. Height of cylinder $h = 13 - 7 = 6\ \text{cm}$.

Inner surface area = CSA of cylinder + CSA of hemisphere: $= 2\pi r h + 2\pi r^2 = 2\pi r (h + r)$ $= 2\times \frac{22}{7}\times 7 \times (6 + 7)$ $= 2\times 22 \times 13 = 572\ \text{cm}^2$

Inner surface area $= 572\ \text{cm}^2$.

Diameter $= 5\ \text{mm}$, so radius $r = 2.5\ \text{mm}$.

The two hemispheres cover the ends, occupying $r + r = 5\ \text{mm}$ of the length. Cylinder length $h = 14 - 5 = 9\ \text{mm}$.

Surface area = CSA of cylinder + 2 × CSA of hemisphere: $= 2\pi r h + 2(2\pi r^2) = 2\pi r h + 4\pi r^2 = 2\pi r(h + 2r)$ $= 2\times \frac{22}{7}\times 2.5 \times (9 + 5)$ $= 2\times \frac{22}{7}\times 2.5 \times 14 = 2\times 22 \times 2.5 \times 2 = 220\ \text{mm}^2$

Surface area $= 220\ \text{mm}^2$.

Radius $r = 4\ \text{m}$, cylinder height $h = 2.1\ \text{m}$, cone slant height $l = 2.8\ \text{m}$.

Canvas area = CSA of cylinder + CSA of cone: $= 2\pi r h + \pi r l = \pi r(2h + l)$ $= \frac{22}{7}\times 4 \times (2\times 2.1 + 2.8)$ $= \frac{22}{7}\times 4 \times (4.2 + 2.8)$ $= \frac{22}{7}\times 4 \times 7 = 88\ \text{m}^2$

Area of canvas required $= 88\ \text{m}^2$.

Radius $r = 1\ \text{cm}$, cone height $h = 1\ \text{cm}$.

Volume of solid = Volume of cone + Volume of hemisphere: $= \frac{1}{3}\pi r^2 h + \frac{2}{3}\pi r^3$ $= \frac{1}{3}\pi (1)^2(1) + \frac{2}{3}\pi (1)^3$ $= \frac{1}{3}\pi + \frac{2}{3}\pi = \pi\ \text{cm}^3$

Volume of the solid $= \pi\ \text{cm}^3$.

Cone: $r = 5\ \text{cm}$, $h = 8\ \text{cm}$.

Volume of water in cone: $V = \frac{1}{3}\pi r^2 h = \frac{1}{3}\pi (25)(8) = \frac{200}{3}\pi\ \text{cm}^3$

Water that flows out $= \dfrac{1}{4}V = \dfrac{1}{4}\times \dfrac{200}{3}\pi = \dfrac{50}{3}\pi\ \text{cm}^3$.

Volume of one lead shot (sphere), $r_s = 0.5\ \text{cm}$: $= \frac{4}{3}\pi r_s^3 = \frac{4}{3}\pi (0.5)^3 = \frac{4}{3}\pi \times 0.125 = \frac{0.5}{3}\pi\ \text{cm}^3$

Number of shots: $n = \frac{\text{water out}}{\text{one shot}} = \frac{\tfrac{50}{3}\pi}{\tfrac{0.5}{3}\pi} = \frac{50}{0.5} = 100$

So $100$ lead shots were dropped.

When recast, volume is conserved.

Volume of sphere $=$ Volume of cylinder: $\frac{4}{3}\pi R^3 = \pi r^2 h$

Here $R = 6\ \text{cm}$, $r = 4\ \text{cm}$. $\frac{4}{3}\pi (6)^3 = \pi (4)^2 h$ $\frac{4}{3}\times 216 = 16 h$ $288 = 16 h$ $h = 18\ \text{cm}$

Height of the cylinder $= 18\ \text{cm}$.

Cylinder: radius $R = 6\ \text{cm}$, height $H = 15\ \text{cm}$. $V_{cyl} = \pi R^2 H = \pi (36)(15) = 540\pi\ \text{cm}^3$

Each ice-cream serving: cone radius $r = 3\ \text{cm}$, cone height $h = 12\ \text{cm}$, hemisphere radius $3\ \text{cm}$.

Volume of cone $= \dfrac{1}{3}\pi r^2 h = \dfrac{1}{3}\pi (9)(12) = 36\pi$

Volume of hemisphere $= \dfrac{2}{3}\pi r^3 = \dfrac{2}{3}\pi (27) = 18\pi$

Total per serving $= 36\pi + 18\pi = 54\pi\ \text{cm}^3$.

Number of cones: $n = \frac{540\pi}{54\pi} = 10$

So $10$ cones can be filled.

Lower cylinder: radius $r_1 = 12\ \text{cm}$, height $h_1 = 220\ \text{cm}$. $V_1 = \pi r_1^2 h_1 = 3.14 \times 144 \times 220 = 99475.2\ \text{cm}^3$

Upper cylinder: radius $r_2 = 8\ \text{cm}$, height $h_2 = 60\ \text{cm}$. $V_2 = \pi r_2^2 h_2 = 3.14 \times 64 \times 60 = 12057.6\ \text{cm}^3$

Total volume: $V = V_1 + V_2 = 99475.2 + 12057.6 = 111532.8\ \text{cm}^3$

Total volume of the pole $= 111532.8\ \text{cm}^3$.

Well: radius $r = 1.5\ \text{m}$, depth $= 14\ \text{m}$.

Volume of earth dug out: $V = \pi r^2 h = \frac{22}{7}\times (1.5)^2 \times 14 = \frac{22}{7}\times 2.25 \times 14 = 99\ \text{m}^3$

Embankment is a hollow ring: inner radius $= 1.5\ \text{m}$, width $= 4\ \text{m}$, so outer radius $R = 1.5 + 4 = 5.5\ \text{m}$.

Base area of ring: $= \pi(R^2 - r^2) = \frac{22}{7}(5.5^2 - 1.5^2) = \frac{22}{7}(30.25 - 2.25) = \frac{22}{7}\times 28 = 88\ \text{m}^2$

Let height of embankment $= H$. Since volume is conserved: $88 \times H = 99$ $H = \frac{99}{88} = 1.125\ \text{m}$

Height of the embankment $= 1.125\ \text{m}$.