A cone is surmounted on a hemisphere as shown here. The radius of the solid formed is 6 cm and its height is 2

Exercise 2.4 · Q4

Q. A cone is surmounted on a hemisphere as shown here. The radius of the solid formed is 6 cm and its height is 21 cm. Find:

i) Curved surface area of the solid

ii) Volume of the solid. [Figure: a cone of height h sitting on top of a hemisphere of radius R, together forming one composite solid.]

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Concept understanding — Time, Work and Distance

Time, Work, and Distance — The Core Idea

Imagine you're walking to school. The faster you walk, the sooner you get there. That's the most basic link between time, speed, and distance. But in exam problems, "work" enters the picture — and it's really just a different kind of distance.

The Intuition

Distance is what you cover. Speed is how fast you cover it. Time is how long it takes.

If you walk at 4 km/h for 2 hours, you cover 8 km. That's it — the relationship is:

\text{Distance} = \text{Speed} \times \text{Time}

Now, work is like distance, but instead of metres or kilometres, you measure it in "jobs done" — a wall painted, a tank filled, a field ploughed. Rate of work is like speed: how much of the job you finish per unit time (per hour, per day). So:

\text{Work Done} = \text{Rate of Work} \times \text{Time}

The two ideas are structurally identical. Once you see that, the whole topic becomes one idea with two faces.

\text{Distance} = \text{Speed} \times \text{Time}

The Precise Statement

For distance problems: If an object moves at a constant speed $s$ , then in time $t$ it covers a distance $d = s \times t$ . If any two of these are known, the third is fixed.

For work problems: If a person or machine works at a constant rate $r$ (e.g., "1/3 of a wall per hour"), then in time $t$ the amount of work done is $W = r \times t$ . Again, any two determine the third.

Watch out

A common mistake is to treat "work" as a fixed number like 100. It isn't. Work is always 1 complete job unless stated otherwise. If a pipe fills a tank in 5 hours, its rate is $\frac{1}{5}$ tank per hour — not 5.

Why This Matters for Exams

Problems rarely give you the rate directly. They say: "A can do a job in 10 days." That means A's rate is $\frac{1}{10}$ job per day. Then they ask: "How long will A and B together take, if B alone takes 15 days?" You add rates:

\frac{1}{10} + \frac{1}{15} = \frac{3+2}{30} = \frac{5}{30} = \frac{1}{6} \text{ job per day}

So together they take 6 days. …

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A cone (radius 6 cm) sits atop a hemisphere of the same radius; total height of the composite solid is 21 cm, giving CSA ≈530.93 cm² and volume ≈1018.29 cm³.

r

The hemisphere and cone share the same radius (surmounted): $r=6$ cm.
Total height of solid $=21$ cm $=h_{\text{cone}}+r$ (hemisphere's flat face is the cone's base) $\Rightarrow h_{\text{cone}}=21-6=15$ cm.
Slant height $l=\sqrt{r^2+h^2}=\sqrt{6^2+15^2}=\sqrt{36+225}=\sqrt{261}\approx16.16$ cm.
(i) CSA of solid $=\pi rl+2\pi r^2=\pi(6)(16.16)+2\pi(36)$ .
$\pi rl=\dfrac{22}{7}\times6\times16.1555\approx304.65\text{ cm}^2$ ; $2\pi r^2=\dfrac{22}{7}\times72=\dfrac{1584}{7}\approx226.29\text{ cm}^2$ .
Total CSA $\approx304.65+226.29=530.93\text{ cm}^2$ . …

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Q.A cone is surmounted on a hemisphere as shown here. The radius of the solid formed is 6 cm and its height is 21 cm. Find:

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