NCERT Solutions for Class 9 Maths Chapter 11 Exercise 11.2 - Surface Area and Volumes

Q1 (i) Find the surface area of a sphere of radius: $\small 10.5\hspace{1mm}cm$ .

Answer:

We know,

The surface area of a sphere of radius $r$ = $4\pi r^2$

$\therefore$ Required surface area = $4\times\frac{22}{7} \times(10.5)^2$

$\\ =88\times1.5\times10.5 \\ = 1386\ cm^2$

Q1 (ii) Find the surface area of a sphere of radius: $\small 5.6\hspace{1mm}cm$

Answer:

We know,

The surface area of a sphere of radius $r$ = $4\pi r^2$

$\therefore$ Required surface area = $4\times\frac{22}{7} \times(5.6)^2$

$\\ =88\times0.8\times5.6 \\ = 394.24\ cm^2$

Q1 (iii) Find the surface area of a sphere of radius: $\small 14\hspace{1mm}cm$

Answer:

We know,

The surface area of a sphere of radius $r$ = $4\pi r^2$

$\therefore$ Required surface area = $4\times\frac{22}{7} \times(14)^2$

$\\ = 88\times2\times14 \\ = 2464 \ cm^2$

Q2 (i) Find the surface area of a sphere of diameter: 14 cm

Answer:

Given,

The diameter of the sphere = $14\ cm$

We know,

The surface area of a sphere of radius $r$ = $4\pi r^2$

$\therefore$ Required surface area = $4\times\frac{22}{7} \times\left (\frac{14}{2} \right )^2$

$= 4\times\frac{22}{7} \times\frac{14}{2}\times\frac{14}{2}$

$\\ = 22\times2\times14 \\ = 616\ cm^2$

Q2 (ii) Find the surface area of a sphere of diameter: 21 cm

Answer:

Given,

The diameter of the sphere = $21\ cm$

We know,

The surface area of a sphere of radius $r$ = $4\pi r^2$

$\therefore$ Required surface area = $4\times\frac{22}{7} \times\left (\frac{21}{2} \right )^2$

$= 4\times\frac{22}{7} \times\frac{21}{2}\times\frac{21}{2}$

$\\ = 22\times3\times21 \\ = 1386\ cm^2$

Q2 (iii) Find the surface area of a sphere of diameter: $\small 3.5\hspace{1mm}m$

Answer:

Given,

The diameter of the sphere = $3.5\ m$

We know,

The surface area of a sphere of radius $r$ = $4\pi r^2$

$\therefore$ Required surface area = $4\times\frac{22}{7} \times\left (\frac{3.5}{2} \right )^2$

$= 4\times\frac{22}{7} \times\frac{3.5}{2}\times\frac{3.5}{2}$

$\\ = 22\times0.5\times3.5 \\ = 38.5\ m^2$

Q3 Find the total surface area of a hemisphere of radius 10 cm. (Use $\small \pi =3.14$ )

Answer:

We know,

The total surface area of a hemisphere = Curved surface area of hemisphere + Area of the circular end

$= 2\pi r^2 + \pi r^2 = 3\pi r^2$

$\therefore$ The required total surface area of the hemisphere = $3\times3.14\times(10)^2$

$\\ = 942\ cm^2$

Q4 The radius of a spherical balloon increases from 7 cm to 14 cm as air is being pumped into it. Find the ratio of surface areas of the balloon in the two cases.

Answer:

Given,

$r_1 = 7\ cm$

$r_2 = 14\ cm$

We know,

The surface area of a sphere of radius $r$ = $4\pi r^2$

$\therefore$ The ratio of surface areas of the ball in the two cases = $\frac{Initial}{Final} = \frac{4\pi r_1^2}{4\pi r_2^2}$

$= \frac{r_1^2}{r_2^2}$

$\\ = \left (\frac{7}{14} \right )^2 \\ \\ = \left (\frac{1}{2} \right )^2 \\ \\ = \frac{1}{4}$

Therefore, the required ratio is $1:4$

Q5 A hemispherical bowl made of brass has inner diameter $\small 10.5\hspace{1mm}cm$ . Find the cost of tin-plating it on the inside at the rate of Rs 16 per $\small 100\hspace{1mm}cm^2$ .

Answer:

Given,

The inner radius of the hemispherical bowl = $r = \frac{10.5}{2}\ cm$

We know,

The curved surface area of a hemisphere = $2\pi r^2$

$\therefore$ The surface area of the hemispherical bowl = $2\times\frac{22}{7}\times\left (\frac{10.5}{2} \right )^2$

$=11\times1.5\times10.5$

$= 173.25 \ cm^2$

Now,

Cost of tin-plating $\small 100\hspace{1mm}cm^2$ = Rs 16

$\therefore$ Cost of tin-plating $\small 33\hspace{1mm}cm^2$ = $\small \\ Rs. \left (\frac{16}{100}\times173.25 \right )$

$\small = Rs. 27.72$

Therefore, the cost of tin-plating it on the inside is $\small Rs. 27.72$

Q6 Find the radius of a sphere whose surface area is $\small 154\hspace{1mm}cm^2$ .

Answer:

Given,

The surface area of the sphere = $\small 154\hspace{1mm}cm^2$

We know,

The surface area of a sphere of radius $r$ = $4\pi r^2$

$\therefore 4\pi r^2 = 154$

$\\ \Rightarrow 4\times\frac{22}{7}\times r^2 = 154$

$\\ \Rightarrow r^2 = \frac{154\times7}{4\times22} = \frac{7\times7}{4}$

$\\ \Rightarrow r = \frac{7}{2}$

$\\ \Rightarrow r = 3.5\ cm$

Therefore, the radius of the sphere is $3.5\ cm$

Q7 The diameter of the moon is approximately one fourth of the diameter of the earth. Find the ratio of their surface areas.

Answer:

Let diameter of Moon be $d_m$ and diameter of Earth be $d_e$

We know,

The surface area of a sphere of radius $r$ = $4\pi r^2$

$\therefore$ The ratio of their surface areas = $\frac{Surface\ area\ of\ moon}{Surface\ area\ of\ Earth}$

$= \frac{4\pi \left (\frac{d_m}{2} \right )^2}{4\pi \left (\frac{d_e}{2} \right )^2}$

$= \frac{d_m^2}{d_e^2}$

$=\left ( \frac{\frac{1}{4}d_e}{d_e} \right )^2$

$= \frac{1}{16}$

Therefore, the ratio of the surface areas of the moon and earth is $= 1:16$

Q8 A hemispherical bowl is made of steel, $\small 0.25\hspace{1mm}cm$ thick. The inner radius of the bowl is 5 cm. Find the outer curved surface area of the bowl.

Answer:

Given,

The inner radius of the bowl = $r_1 = 5\ cm$

The thickness of the bowl = $\small 0.25\hspace{1mm}cm$

$\therefore$ Outer radius of the bowl = (Inner radius + thickness) =

$r_2 = 5+0.25 = 5.25\ cm$

We know, Curved surface area of a hemisphere of radius $r$ = $2\pi r^2$

$\therefore$ The outer curved surface area of the bowl = $2\pi r_2^2$

$= 2\times\frac{22}{7}\times (5.25)^2$

$= 2\times\frac{22}{7}\times5.25\times5.25 = 173.25\ cm^2$

Therefore, the outer curved surface area of the bowl is $173.25\ cm^2$

Q9 (i) A right circular cylinder just encloses a sphere of radius $\small r$ (see Fig. $\small 11.10$ ). Find surface area of the sphere,

Answer:

Given,

The radius of the sphere = $r$

$\therefore$ Surface area of the sphere = $4\pi r^2$

Q9 (ii) A right circular cylinder just encloses a sphere of radius $\small r$ (see Fig. $\small 11.10$ ). Find curved surface area of the cylinder,

Answer:

Given,

The radius of the sphere = r

\thereforeThe surface area of the sphere = $4\pi r^2$

According to the question, the cylinder encloses the sphere.

Hence, the diameter of the sphere is the diameter of the cylinder.

Also, the height of the cylinder is equal to the diameter of the sphere.

We know, the curved surface area of a cylinder = $2\pi rh$

= $2\pi r(2r) = 4\pi r^2$

Therefore, the curved surface area of the cylinder is $4\pi r^2$

Q9 (iii) A right circular cylinder just encloses a sphere of radius $\small r$ (see Fig. $\small 11.10$ ). Find ratio of the areas obtained in (i) and (ii).

Answer:

The surface area of the sphere = $4\pi r^2$

And, Surface area of the cylinder = $4\pi r^2$

So, the ratio of the areas = $\frac{4\pi r^2}{4\pi r^2} = 1$