Exercise-11.2, Class 9th, Maths, Chapter 11, NCERT

Useful formula (used throughout): Surface area of a sphere $=4\pi r^2$
Curved surface area of a hemisphere $=2\pi r^2$
Total surface area of a hemisphere $=3\pi r^2$
Use $\pi ={\displaystyle \frac{22}{7}}$ unless stated otherwise.

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Q1. Find surface area of a sphere of radius:

(i) $r=10.5$

$$\text{SA}=4\pi r^2=4\cdot \frac{22}{7}\cdot (10.5{)}^2\mathrm{.}$$

Calculate: ${10.5}^2=110.25$
$4\pi ={\displaystyle \frac{88}{7}}$. So

$$\text{SA}=\frac{88}{7}\times 110.25=88\times \frac{110.25}{7}=88\times 15.75=1386{\text{cm}}^2\mathrm{.}$$

(ii) $r=5.6$

$$\text{SA}=4\cdot \frac{22}{7}\cdot (5.6{)}^2\mathrm{.}$$

${5.6}^2=31.36$

$\frac{88}{7}\times 31.36=88\times \frac{31.36}{7}=88\times 4.48=394.24{\text{cm}}^2\mathrm{}$

(iii) $r=14$

$$\text{SA}=4\cdot \frac{22}{7}\cdot {14}^2\mathrm{}$$

${14}^2=196$. $\frac{88}{7}\times 196=88\times 28=2464{\text{cm}}^2\mathrm{}$

Answers Q1: (i) $1386{\text{cm}}^2$, (ii) $394.24{\text{cm}}^2$, (iii) $2464{\text{cm}}^2\mathrm{}$

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Q2. Surface area of a sphere of diameter:

(i) $d=14$ cm $\Rightarrow r=7$ cm

$$\text{SA}=4\cdot \frac{22}{7}\cdot 7^2=4\cdot \frac{22}{7}\cdot 49=\frac{88}{7}\times 49=88\times 7=616{\text{cm}}^2\mathrm{}$$

(ii) $d=21$ cm $\Rightarrow r=10.5$ → same as Q1(i): $1386{\text{cm}}^2\mathrm{}$

(iii) $d=3.5$ $\Rightarrow r=1.75$

$$\text{SA}=4\cdot \frac{22}{7}\cdot (1.75{)}^2\mathrm{.}$$

${1.75}^2=3.0625$
$\frac{\mathrm{}}{}$

$\frac{88}{7}\times 3.0625=88\times \frac{3.0625}{7}=88\times 0.4375=38.5{\text{m}}^2\mathrm{}$

Answers Q2: (i) $616{\text{cm}}^2$, (ii) $1386{\text{cm}}^2$, (iii) $38.5{\text{m}}^2\mathrm{}$

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Q3. Total surface area of a hemisphere of radius $10$ cm. (Use $\pi =3.14$)

Total surface area (hemisphere) $=3\pi r^2$

$$=3\times 3.14\times {10}^2=3\times 3.14\times 100=3\times 314=942{\text{cm}}^2\mathrm{}$$

Answer Q3: $942{\text{cm}}^2\mathrm{}$

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Q4. Radius increases from $7$ cm to $14$ cm. Ratio of surface areas?

Surface area $\propto r^2$. So ratio

$$\frac{4\pi (14{)}^2}{4\pi (7{)}^2}=\frac{{14}^2}{7^2}=\frac{196}{49}=4:1.$$

Answer Q4: $4:1$.

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Q5. Hemispherical bowl, inner diameter $=10.5$ cm ⇒ inner radius $r=5.25$ cm. Tin-plating inside at `₹16 per 100 cm².

Inside area to be tin-plated = curved surface area of hemisphere $=2\pi r^2$ with $\pi ={\displaystyle \frac{22}{7}}$

$$2\cdot \frac{22}{7}\cdot (5.25{)}^2\mathrm{}$$

${5.25}^2=27.5625$$\frac{44}{7}\times 27.5625=44\times \frac{27.5625}{7}=44\times 3.9375=173.25{\text{cm}}^2\mathrm{}$

Cost $={\displaystyle \frac{173.25}{100}}\times 16=1.7325\times 16=27.72\text{ rupees}\mathrm{}$

Answer Q5: Area $=173.25{\text{cm}}^2$. Cost $=\text{₹}27.72$

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Q6. Find radius of a sphere whose surface area is $154{\text{cm}}^2$

Use $4\pi r^2=154$ and $\pi ={\displaystyle \frac{22}{7}}$

$$r^2=\frac{154}{4\pi }=\frac{154}{4\cdot (22\mathrm{/}7)}=\frac{154}{88\mathrm{/}7}=\frac{154\times 7}{88}=\frac{1078}{88}=\mathrm{12.25.}$$

So $r=\sqrt{12.25}=3.5\text{ cm}\mathrm{}$

Answer Q6: $r=3.5\text{ cm}\mathrm{<span\; style="font-size:\; revert;\; font-family:\; var(--wp--preset--font-family--manrope);\; letter-spacing:\; -0.1px;">.</span>}$

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Q7. Moon diameter is one-fourth that of Earth. Ratio of their surface areas?

If $D_{\text{moon}}=\frac{1}{4}{D}_{\text{earth}}$,  radii ratio $=\frac{1}{4}$. Surface area ratio $={\left(\frac{1}{4}\right)}^2=\frac{1}{16}$

Answer Q7: $1:16$ (moon : earth) or moon’s surface area is ${\displaystyle \frac{1}{16}}$ of Earth’s.

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Q8. Hemispherical bowl made of steel, thickness $0.25$ cm, inner radius $=5$ cm. Find outer curved surface area.

Outer radius $R=5+0.25=5.25$ cm. Outer curved surface area of hemisphere $=2\pi R^2$

$$2\cdot \frac{22}{7}\cdot (5.25{)}^2=173.25{\text{cm}}^2\mathrm{.}$$

Answer Q8: Outer curved surface area $=173.25{\text{cm}}^2\mathrm{}$

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Q9. A right circular cylinder just encloses a sphere of radius $r$. Find:

(i) Surface area of the sphere $=4\pi r^2\mathrm{}$

(ii) Curved surface area of the cylinder: cylinder radius $=r$, height $=2r$, so curved area $=2\pi rh=2\pi r\times 2r=4\pi r^2\mathrm{}$

(iii) Ratio of (i) to (ii) $={\displaystyle \frac{4\pi r^2}{4\pi r^2}}=1:1$

Answer Q9: (i) $4\pi r^2$ (ii) $4\pi r^2$ (iii) Ratio $1:1$