NCERT Solutions for Class 9 Maths Chapter 14 — End-of-Chapter Exercises#

$$V_{\text{container}} = 10 \times 15 \times 20$$

$$V_{\text{container}} = 3000 \text{ cm}^3$$

Assumption: a scoop is roughly a sphere of diameter 5 cm (radius 2.5 cm):

$$V_{\text{scoop}} = \dfrac{4}{3} \times 3.14 \times 2.5^3$$

$$V_{\text{scoop}} \approx 65 \text{ cm}^3$$$$\text{Scoops} \approx \dfrac{3000}{65}$$

$$\text{Scoops} \approx 46$$

∴ About 45 scoops. (Smaller scoops give more.)

We find $(a + 1)^3 - a^3$:

$$= a^3 + 3a^2 + 3a + 1 - a^3$$

$$= 3a^2 + 3a + 1$$

These are three square slabs of $a^2$ cubes (one on each of three faces), three rows of $a$ cubes along the edges between them, and 1 corner cube. For example, from $a = 2$ to $a = 3$: $12 + 6 + 1 = 19 = 27 - 8$ ✓.

Classroom: about $8 \text{ m} \times 6 \text{ m} \times 3 \text{ m} = 144 \text{ m}^3 = 1{,}44{,}000$ litres.

Water drunk: about 2.5 litres a day for 70 years:

$$2.5 \times 365 \times 70 \approx 64{,}000 \text{ litres}$$

∴ No — a lifetime of drinking fills less than half of the classroom.

Walls: perimeter $2(8 + 6) = 28$ m, height 3 m, so $84$ m², minus about 10 m² of doors and windows: about $74$ m². With walls 23 cm thick:

$$V_{\text{walls}} \approx 74 \times 0.23$$

$$V_{\text{walls}} \approx 17 \text{ m}^3$$

One brick with its mortar is about $20 \text{ cm} \times 10 \text{ cm} \times 10 \text{ cm} = 0.002$ m³:

$$\text{Bricks} \approx \dfrac{17}{0.002}$$

$$\text{Bricks} = 8500$$

∴ Roughly 8,000–9,000 bricks.

$$V_A = 50^3$$

$$V_A = 125000 \text{ mm}^3$$

$$V_B = 100 \times 10^3$$

$$V_B = 100000 \text{ mm}^3$$$$\Rightarrow 125000 > 100000$$

∴ Option A gives more chocolate (25% more). Note that 50 mm is five times 10 mm, so one big cube equals $5^3 = 125$ small cubes, not 100.

Assumptions: about 8 billion people, each standing in a space of about $0.5 \text{ m} \times 0.5 \text{ m} = 0.25$ m².

$$\text{Area} \approx 8 \times 10^9 \times 0.25$$

$$\text{Area} = 2 \times 10^9 \text{ m}^2$$

$$\text{Area} = 2000 \text{ km}^2$$

∴ About 2000 km² — a square of about 45 km on each side, a little bigger than the area of Delhi (about 1,500 km²).

$$V_{\text{glass}} = \dfrac{22}{7} \times 3.5^2 \times 12$$

$$V_{\text{glass}} = 462 \text{ cm}^3$$$$V_{\text{total}} = 1600 \times 462$$

$$V_{\text{total}} = 739200 \text{ cm}^3$$

∴ 739.2 litres of milk are needed.

The sphere’s surface is five times the cone’s curved surface, so:

$$4\pi \times 5^2 = 5 \times \pi \times 4 \times l$$

$$\Rightarrow 100\pi = 20\pi l$$

$$\Rightarrow l = 5 \text{ cm}$$$$h^2 = l^2 - r^2$$

$$h^2 = 25 - 16$$

$$h^2 = 9$$

$$\Rightarrow h = 3 \text{ cm}$$$$V = \dfrac{1}{3}\pi \times 4^2 \times 3$$

$$V = 16\pi$$

$$V \approx 50.29 \text{ cm}^3$$

Volumes of spheres are in the ratio of the cubes of their radii.

(i)

$$\dfrac{V_J}{V_E} = \left(\dfrac{69900}{6370}\right)^3$$

$$\dfrac{V_J}{V_E} \approx 10.97^3$$

$$\dfrac{V_J}{V_E} \approx 1321$$

(ii)

$$\dfrac{V_S}{V_E} = \left(\dfrac{695700}{6370}\right)^3$$

$$\dfrac{V_S}{V_E} \approx 109.2^3$$

$$\dfrac{V_S}{V_E} \approx 1{,}303{,}000$$

∴ About 1,300 Earths would fit inside Jupiter, and about 13 lakh Earths inside the Sun.

The smallest cylinder enclosing a sphere of radius $r$ has radius $r$ and height $2r$:

$$V_{\text{cylinder}} = \pi r^2 \times 2r$$

$$V_{\text{cylinder}} = 2\pi r^3$$$$\dfrac{V_{\text{sphere}}}{V_{\text{cylinder}}} = \dfrac{\frac{4}{3}\pi r^3}{2\pi r^3}$$

$$\dfrac{V_{\text{sphere}}}{V_{\text{cylinder}}} = \dfrac{2}{3}$$

Hence shown. (Archimedes was so proud of this result that he asked for a sphere inside a cylinder to be carved on his tomb.)

Let the radius be $R$ and the height of the second string $h$. The second string is 1 m longer, so:

$$2\pi(R + h) - 2\pi R = 1$$

$$\Rightarrow 2\pi h = 1$$

$$\Rightarrow h = \dfrac{1}{2\pi}$$

$$\Rightarrow h \approx 0.159 \text{ m}$$

The radius $R$ cancels out! So for the Earth, the Moon, Jupiter and a volleyball, the string is the same height above the surface: about 16 cm.

Surprising? Most people expect that 1 extra metre spread round the huge Earth would lift the string by almost nothing, but enough to notice on a volleyball. In fact the extra height depends only on the extra length: every extra 1 m of circumference always adds $\dfrac{1}{2\pi}$ m to the radius.

The square’s corners lie on the circle of radius $r$, so its diagonals are diameters, of length $2r$. The area of a square is half the product of its diagonals:

$$\text{Base area} = \dfrac{1}{2} \times 2r \times 2r$$

$$\text{Base area} = 2r^2$$

The pyramid’s height is $h$:

$$V_{\text{pyramid}} = \dfrac{1}{3} \times 2r^2 \times h$$

$$V_{\text{pyramid}} = \dfrac{2}{3}r^2 h$$$$\dfrac{V_{\text{pyramid}}}{V_{\text{cylinder}}} = \dfrac{\frac{2}{3}r^2 h}{\pi r^2 h}$$

$$\dfrac{V_{\text{pyramid}}}{V_{\text{cylinder}}} = \dfrac{2}{3\pi}$$

∴ The ratio is $2 : 3\pi$, about $0.21$.

$$(l + 1)wh - lwh = wh$$

∴ (c) $wh$ cubic units — one extra slab of area $wh$ and thickness 1.

We find the change $\pi(r + 1)^2 h - \pi r^2 h$:

$$= \pi h\bigl((r + 1)^2 - r^2\bigr)$$

$$= \pi h(2r + 1)$$

$$= 2\pi rh + \pi h$$

∴ (e) $2\pi rh + \pi h$ cubic units.

We find the decrease $\dfrac{4}{3}\pi r^3 - \dfrac{4}{3}\pi (r - 1)^3$:

$$= \dfrac{4}{3}\pi\bigl(r^3 - (r^3 - 3r^2 + 3r - 1)\bigr)$$

$$= \dfrac{4}{3}\pi(3r^2 - 3r + 1)$$

∴ The volume decreases by $\dfrac{4}{3}\pi(3r^2 - 3r + 1)$ cubic units (for $r \geq 1$).

$$a = \sqrt[3]{V}$$

$$a = V^{1/3}$$

$$S = 6a^2$$

$$S = 6V^{2/3}$$

$$6a^2 = S$$

$$\Rightarrow a = \sqrt{\dfrac{S}{6}}$$

$$V = a^3$$

$$V = \left(\dfrac{S}{6}\right)^{3/2}$$

The water must rise by $20 - 16 = 4$ cm. The marbles must push up this much water:

$$V_{\text{rise}} = \pi \times 4^2 \times 4$$

$$V_{\text{rise}} = 64\pi \text{ cm}^3$$

$$V_{\text{marble}} = \dfrac{4}{3}\pi \times 1^3$$

$$V_{\text{marble}} = \dfrac{4}{3}\pi \text{ cm}^3$$$$\text{Marbles} = \dfrac{64\pi}{\frac{4}{3}\pi}$$

$$\text{Marbles} = 48$$

∴ The crow should drop 48 marbles.

Measure the ink tube (refill). A typical refill has an inner diameter of about 2 mm and an ink column about 9 cm long. Treating it as a cylinder with $r = 0.1$ cm and $h = 9$ cm:

$$V \approx 3.14 \times 0.1^2 \times 9$$

$$V \approx 0.28 \text{ cm}^3$$

∴ About 0.3 cm³ (0.3 mL) of ink. Your own measurements may differ a little.

$$V_{\text{dough}} = \dfrac{4}{3} \times 3.14 \times 6^3$$

$$V_{\text{dough}} \approx 904 \text{ cm}^3$$

Assumption: a chapati is a thin cylinder of radius 8 cm and thickness 0.2 cm:

$$V_{\text{chapati}} \approx 3.14 \times 8^2 \times 0.2$$

$$V_{\text{chapati}} \approx 40 \text{ cm}^3$$$$\text{Chapatis} \approx \dfrac{904}{40}$$

$$\text{Chapatis} \approx 22$$

∴ About 20–22 chapatis.

Each half is a hollow hemisphere. Measure the outer radius of the flesh and its thickness. For example, with outer radius 6 cm and flesh 1 cm thick (inner radius 5 cm), the flesh in one half is:

$$V \approx \dfrac{2}{3} \times 3.14 \times (6^3 - 5^3)$$

$$V \approx \dfrac{2}{3} \times 3.14 \times 91$$

$$V \approx 190 \text{ cm}^3$$

So the whole coconut has about 380 cm³ of flesh.

The book has about 160 pages, that is, about 80 sheets, each about $20 \text{ cm} \times 27 \text{ cm} = 0.054$ m². Laid out sheet by sheet:

$$\text{Area} \approx 80 \times 0.054$$

$$\text{Area} \approx 4.3 \text{ m}^2$$

(About 8.6 m² if both sides of every page are laid face up.) A classroom floor is about $8 \times 6 = 48$ m², so no — the pages would cover only a small part of the floor.

Book: about $27 \text{ cm} \times 20 \text{ cm} \times 1 \text{ cm} = 540$ cm³.

Storeroom: $15 \times 20 \times 30 = 9000$ ft³. Since 1 ft ≈ 30.48 cm, 1 ft³ ≈ 28,317 cm³:

$$V_{\text{room}} \approx 9000 \times 28317$$

$$V_{\text{room}} \approx 2.55 \times 10^8 \text{ cm}^3$$$$\text{Books} \approx \dfrac{2.55 \times 10^8}{540}$$

$$\text{Books} \approx 4{,}72{,}000$$

∴ About 4.7 lakh books at most, if packed perfectly with no gaps.

Assumptions: about 8 billion people, each with a volume of about 0.07 m³ (a 70 kg person, about as dense as water), packed with no gaps:

$$V \approx 8 \times 10^9 \times 0.07$$

$$V = 5.6 \times 10^8 \text{ m}^3$$

$$\text{Side} = \sqrt[3]{5.6 \times 10^8}$$

$$\text{Side} \approx 824 \text{ m}$$

∴ A cube only about 800 m on each side — less than a kilometre! (With room to stand, it would be larger, but still surprisingly small.)

Let the thickness be $t$ km. The water fills the space between spheres of radii $R = 6371$ and $R + t$:

$$\dfrac{4}{3}\pi\bigl((R + t)^3 - R^3\bigr) = 1.38 \times 10^9$$

$$\Rightarrow t = \sqrt[3]{R^3 + \dfrac{3 \times 1.38 \times 10^9}{4\pi}} - R$$

Since the layer is very thin, a simpler approximation is surface area × thickness = volume:

$$t \approx \dfrac{1.38 \times 10^9}{4\pi R^2}$$

$$t \approx \dfrac{1.38 \times 10^9}{4 \times 3.1416 \times 6371^2}$$

$$t \approx \dfrac{1.38 \times 10^9}{5.1 \times 10^8}$$

$$t \approx 2.7 \text{ km}$$

(The exact expression gives 2.70 km too.) ∴ The water layer would be about 2.7 km thick.

Making a solid flatter increases its surface area while its volume can shrink. Here is one example.

Start with a cube of side 2 cm: $V = 8$ cm³, $S = 24$ cm². We want a cuboid with $V = 4$ cm³ and $S = 48$ cm². Try a square base $x \times x$ and height $z$:

$$x^2 z = 4 \quad \cdots (1)$$

$$2x^2 + 4xz = 48 \quad \cdots (2)$$

From equation (1), $z = \dfrac{4}{x^2}$. Substituting in equation (2):

$$2x^2 + \dfrac{16}{x} = 48$$

$$\Rightarrow x^3 - 24x + 8 = 0$$

Solving (by trial or with a calculator): $x \approx 4.72$ cm, and then $z = \dfrac{4}{x^2} \approx 0.18$ cm.

∴ A cube of $2 \times 2 \times 2$ cm becomes a thin cuboid of about 4.72 cm × 4.72 cm × 0.18 cm: its volume is halved (8 → 4 cm³) while its surface area is doubled (24 → 48 cm²). Any cuboid can be treated the same way.

1. Split the house into simple shapes — each room as a cuboid, a sloping roof as a triangular prism, a water tank as a cylinder.
2. Measure (or pace out) the length, width and height of each part. Round to convenient numbers.
3. Find each volume: for example, a room 4 m × 3.5 m × 3 m has volume $42$ m³.
4. Add the volumes of all the parts.
5. Present a sketch with the measurements, the formula used for each part, and the total. Note which approximations you made (wall thickness ignored, roof treated as flat, etc.).

A typical 2-room house of about 40 m² floor area and 3 m height has a volume of about 120 m³.