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

Question 1: Diameter of the base of a cone is $\small 10.5 \hspace{1mm}cm$, and its slant height is $\small 10 \hspace{1mm}cm$. Find its curved surface area.

Answer:

Given,

Base diameter of the cone = $d=10.5\ cm$

Slant height = $l=10\ cm$

We know, Curved surface area of a cone $= \pi r l$

$\therefore$ Required curved surface area of the cone=

$\\ = \frac{22}{7}\times \frac{10.5}{2}\times10 \\ \\ = 165\ cm^2$

Question 2: Find the total surface area of a cone if its slant height is $\small 21\hspace{1mm}m$ and the diameter of its base is $\small 24\hspace{1mm}m$.

Answer:

Given,

Base diameter of the cone = $d=24\ m$

Slant height = $l=21\ cm$

We know, Total surface area of a cone = Curved surface area + Base area

$= \pi r l + \pi r^2 = \pi r (l + r)$

$\therefore$ Required total surface area of the cone=

$\\ = \frac{22}{7}\times\frac{24}{2}\times(21+12) \\ \\ =\frac{22}{7}\times\frac{24}{2}\times33 \\ = 1244.57 \ m^2$

Question 3: (i) Curved surface area of a cone is $\small 308\hspace{1mm}cm^2$, and its slant height is 14 cm. Find the radius of the base.

Answer:

Given,

The curved surface area of a cone = $\small 308\hspace{1mm}cm^2$

Slant height $= l = 14\ cm$

Let the radius of the cone be $r\ cm$

We know, the curved surface area of a cone = $\pi rl$

$\therefore$ $\\ \pi rl = 308 \\ \\ \Rightarrow \frac{22}{7}\times r\times14 = 308 \\ \Rightarrow r = \frac{308}{44} = 7$

Therefore, the radius of the cone is $7\ cm$

Question 3: (ii) Curved surface area of a cone is $\small 308\hspace{1mm}cm^2$, and its slant height is $\small 14\hspace{1mm}cm$. Find the total surface area of the cone.

Answer:

Given,

The curved surface area of a cone = $\small 308\hspace{1mm}cm^2$

Slant height $= l = 14\ cm$

The radius of the cone is $r =$ $7\ cm$

(ii) We know the total surface area of a cone = the Curved surface area + the Base area

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

$\\ = 308+\frac{22}{7}\times 7^2 $
$= 308+154 $
$= 462\ cm^2$

Therefore, the total surface area of the cone is $462\ cm^2$

Question 4: (i) A conical tent is 10 m high, and the radius of its base is 24 m. Find the slant height of the tent.

Answer:

Given,

Base radius of the conical tent = $r=24\ m$

Height of the conical tent = $h=10\ m$

$\therefore$ Slant height, $l=\sqrt{h^2+r^2}$
$\\ =\sqrt{10^2+24^2}$
$ = \sqrt{676}$
$= 26\ m$
Therefore, the slant height of the conical tent is $26\ m$

Question 4: (ii) A conical tent is 10 m high, and the radius of its base is 24 m. Find the cost of the canvas required to make the tent if the cost of $\small 1\hspace{1mm}m^2$ canvas is Rs 70.

Answer:

Given,

Base radius of the conical tent = $r=24\ m$

Height of the conical tent = $h=10\ m$

$\therefore$ Slant height = $l=\sqrt{h^2+r^2} = 26\ m$

We know the curved surface area of a cone $= \pi r l$

$\therefore$ Curved surface area of the tent

$\\ = \frac{22}{7}\times24\times26$
$=\frac{13728}{7}\ m^2$

Cost of $1\ m^2$ of canvas = $Rs.\ 70$

$\therefore$ Cost of $\frac{13728}{7}\ m^2$ of canvas

$=Rs.\ (\frac{13728}{7}\times70)$

$ = Rs.\ 137280$

Therefore, the required cost of canvas to make a tent is $Rs.\ 137280$.

Question 5: What length of tarpaulin 3 m wide will be required to make a conical tent of height 8 m and base radius 6 m? Assume that the extra length of material that will be required for stitching margins and wastage in cutting is approximately 20 cm (Use $\small \pi =3.14$ ).

Answer:

Given,

Base radius of the conical tent = $r =6\ m$

Height of the tent = $h =8\ m$

We know,

Curved surface area of a cone = $\pi rl = \pi r\sqrt{h^2 + r^2}$

$\therefore$ Area of tarpaulin required = Curved surface area of the tent

$\\ =3.14\times6\times \sqrt{8^2+ 6^2} $
$= 3.14\times6\times 10 $
$= 188.4\ m^2$

Now, let the length of the tarpaulin sheet be $x\ m$

Since $20\ cm$ is wasted, effective length = $x - 20 cm = (x - 0.2)\ m$

Breadth of tarpaulin = $3\ m$

$\\ \therefore [(x - 0.2) \times 3] = 188.4 $
$ \Rightarrow x - 0.2 = 62.8 $
$ \Rightarrow x = 63\ m$

Therefore, the length of the required tarpaulin sheet will be 63 m.

Question 6: The slant height and base diameter of a conical tomb are 25 m and 14 m, respectively. Find the cost of whitewashing its curved surface at the rate of Rs 210 per $\small 100\hspace{1mm}m^2$.

Answer:

Given, a conical tomb

The base diameter of the cone = $d =14\ m$

Slant height $= l = 25\ m$

We know the curved surface area of a cone $= \pi r l$

$\\ = \frac{22}{7}\times\frac{14}{2}\times25 $

$= 22\times25 $

$ = 550\ m^2$

Now, Cost of whitewashing per $\small 100\hspace{1mm}m^2$ = $\small Rs.\ 210$

$\therefore$ Cost of whitewashing per $\small 550\hspace{1mm}m^2$

= $\small \\ Rs. (\frac{210}{100}\times550 )$

$\small \\ = Rs.\ (21\times55 ) = Rs.\ 1155$

Therefore, the cost of whitewashing its curved surface of the tomb is $\small Rs.\ 1155$.

Question 7: A joker’s cap is in the form of a right circular cone of a base radius 7 cm and height 24 cm. Find the area of the sheet required to make 10 such caps.

Answer:

Given a right circular cone cap (which means no base)

Base radius of the cone = $r=7\ cm$

Height $= h = 24\ cm$

$\therefore l = \sqrt{h^2+r^2}$

We know the curved surface area of a right circular cone $= \pi r l$

$\therefore$ The curved surface area of a cap =

$\\ = \frac{22}{7}\times7\times\sqrt{24^2+7^2}$

$ = 22\times\sqrt{625} $

$ = 22\times25$

$ = 550\ cm^2$

$\therefore$ The curved surface area of 10 caps
= $550\times10 = 5500\ cm^2$

Therefore, the area of the sheet required for 10 caps = $5500\ cm^2$

Question 8: A bus stop is barricaded from the remaining part of the road by using 50 hollow cones made of recycled cardboard. Each cone has a base diameter of 40 cm and height 1 m. If the outer side of each of the cones is to be painted and the cost of painting is Rs 12 per $\small m^2$, what will be the cost of painting all these cones? (Use $\small \pi =3.14$ and take $\small \sqrt{1.04}=1.02$ )

Answer:

Given, hollow cone.

The base diameter of the cone = $d = 40\ cm = 0.4\ m$

Height of the cone = $h = 1\ m$

$\therefore$ Slant height = $l = \sqrt{h^2+r^2}$ $= \sqrt{1^2+0.2^2}$

We know, Curved surface area of a cone = $\pi r l = \pi r\sqrt{h^2+r^2}$

$\therefore$ The curved surface area of 1 cone
$= 3.14\times0.2\times\sqrt{1.04}$
$ = 3.14\times0.2\times1.02$
$= 0.64056\ m^2$

$\therefore$ The curved surface area of 50 cones $= (50\times0.64056)\ m^2$

$= 32.028\ m^2$

Now, the cost of painting $\small 1\ m^2$ area = $\small Rs.\ 12$

$\therefore$ Cost of the painting $32.028\ m^2$ area
$= Rs.\ (32.028\times12)$
$= Rs.\ 384.336$

Therefore, the cost of painting 50 such hollow cones is $Rs.\ 384.34$ (approx).

Question 1: (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 = $4\times\frac{22}{7} \times(10.5)^2$

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

Question 1: (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 = $4\times\frac{22}{7} \times(5.6)^2$

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

Question 1: (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

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

$\\ = 88\times2\times14$

$= 2464 \ cm^2$

Question 2: (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

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

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

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

Question 2: (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
$=4\times\frac{22}{7} \times\left (\frac{21}{2} \right )^2$

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

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

Question 2: (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
$=4\times\frac{22}{7} \times\left (\frac{3.5}{2} \right )^2$

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

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

Question 3: 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
$=3\times3.14\times(10)^2$

$\\ = 942\ cm^2$

Question 4: 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 $1:4$

Question 5: A hemispherical bowl made of brass has an 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 $\small Rs. 27.72$

Question 6: 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 $3.5\ cm$

Question 7: The diameter of the moon is approximately one-fourth of the diameter of the earth. Find the ratio of their surface areas.

Answer:

Let the diameter of the Moon be $d_m$ and the 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 = $\frac{\text{Surface area of moon}}{\text{Surface area of Earth}}$

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

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

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

$= \frac{1}{16}$

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

Question 8: 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 the 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 $173.25\ cm^2$.

Question 9: (i) A right circular cylinder just encloses a sphere of radius $\small r$ (see Fig. $\small 13.22$ ). Find the surface area of the sphere,

Answer:

Given,

The radius of the sphere = $r$

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

Question 9: (ii) A right circular cylinder just encloses a sphere of radius $\small r$ (see Fig. $\small 13.22$ ). Find the curved surface area of the cylinder,

Answer:

Given,

The radius of the sphere = $r$

$\therefore$ The 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 $4\pi r^2$

Question 9: (iii) A right circular cylinder just encloses a sphere of radius $\small r$ (see Fig. $\small 13.22$ ). Find the ratio of the areas obtained in (i) and (ii).

Answer:

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

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

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

Question 1: (i) Find the volume of the right circular cone with radius 6 cm, height 7 cm

Answer:

Given,

Radius = $r =6\ cm$

Height = $h =7\ cm$

We know,

Volume of a right circular cone = $\frac{1}{3}\pi r^2 h$

$\therefore$ Required volume

$=\frac{1}{3}\times\frac{22}{7}\times6^2\times7$

$\\ = 22\times2\times6$

$= 264\ cm^3$

Question 1 (ii) Find the volume of the right circular cone with radius $\small 3.5$ cm, height 12 cm

Answer:

Given,

Radius = $r =3.5\ cm$

Height = $h =12\ cm$

We know,

Volume of a right circular cone = $\frac{1}{3}\pi r^2 h$

$\therefore$ Required volume

$=\frac{1}{3}\times\frac{22}{7}\times3.5^2\times12$

$\\ = 22\times0.5\times3.5\times4 $
$= 11\times14$
$ = 154\ cm^3$

Question 2: (i) Find the capacity in litres of a conical vessel with radius 7 cm, slant height 25 cm

Answer:

Given,

Radius = $r =7\ cm$

Slant height = $l = \sqrt{r^2 + h^2} = 25\ cm$

Height = $h =\sqrt{l^2-r^2}$
$ = \sqrt{25^2-7^2}$

$= \sqrt{(25-7)(25+7)}$
$ = \sqrt{(18)(32)}$

$= 24\ cm$

We know,
Volume of a right circular cone = $\frac{1}{3}\pi r^2 h$

$\therefore$ Volume of the vessel

$=\frac{1}{3}\times\frac{22}{7}\times7^2\times24$

$\\ = 22\times7\times8$
$= 154\times8 $
$= 1232\ cm^3$

$\therefore$ Required capacity of the vessel

$= \frac{1232}{1000} = 1.232\ litres$

Question 2: (ii) Find the capacity in litres of a conical vessel with height 12 cm, slant height 13 cm

Answer:

Given,

Height = $h =12\ cm$

Slant height = $l = \sqrt{r^2 + h^2} = 13\ cm$

Radius = $r =\sqrt{l^2-h^2} $
$= \sqrt{13^2-12^2}$

$= \sqrt{(13-12)(13+12)} $
$= \sqrt{(1)(25)}$

$= 5\ cm$

We know,
Volume of a right circular cone = $\frac{1}{3}\pi r^2 h$

$\therefore$ Volume of the vessel

$=\frac{1}{3}\times\frac{22}{7}\times5^2\times12$

$\\ = \frac{22}{7}\times25\times4$

$= \frac{2200}{7}\ cm^3$

$\therefore$ Required capacity of the vessel

$= \frac{2200}{7\times1000} $

$= \frac{11}{35}\ litres$

Question 3: The height of a cone is 15 cm. If its volume is 1570 $\small cm^3$, find the radius of the base. (Use $\small \pi =3.14$ )

Answer:

Given,

Height of the cone = $h =15\ cm$

Let the radius of the base of the cone be $r\ cm$

We know,
The volume of a right circular cone

$=\frac{1}{3}\pi r^2 h$

$\therefore$ $\frac{1}{3}\times3.14\times r^2\times15 = 1570$

$\Rightarrow 3.14\times r^2\times5 = 1570$

$ \Rightarrow r^2 = \frac{1570}{15.7} $
$ \Rightarrow r^2 = 100$
$ \Rightarrow r = 10\ cm$

Question 4: If the volume of a right circular cone of height 9 cm is $\small 48\pi \hspace{1mm}cm^3$, find the diameter of its base.

Answer:

Given,

Height of the cone = $h =9\ cm$

Let the radius of the base of the cone be $r\ cm$

We know,
The volume of a right circular cone = $\frac{1}{3}\pi r^2 h$

$\therefore$ $\frac{1}{3}\times\pi\times r^2\times9 = 48\pi$

$\\ \Rightarrow 3r^2 = 48$
$ \Rightarrow r^2 = 16$
$ \Rightarrow r = 4\ cm$

Therefore, the diameter of the right circular cone is $8\ cm$

Question 5: A conical pit of top diameter $\small 3.5$ m is 12 m deep. What is its capacity in kilolitres?

Answer:

Given,

Depth of the conical pit = $h =12\ m$

The top radius of the conical pit = $r = \frac{3.5}{2}\ m$

We know,
The volume of a right circular cone = $\frac{1}{3}\pi r^2 h$

$\therefore$ The volume of the conical pit

$= \frac{1}{3}\times\frac{22}{7}\times \left (\frac{3.5}{2} \right )^2\times12$

$\\ = \frac{1}{3}\times\frac{22}{7}\times \frac{3.5\times 3.5}{4}\times12 $

$= 22\times 0.5\times 3.5 $
$ = 38.5\ m^3$

Now, $1\ m^3 = 1\ kilolitre$

$\therefore$ The capacity of the pit = $38.5\ kilolitre$

Question 6: (i) The volume of a right circular cone is $\small 9856\hspace{1mm}cm^3$. If the diameter of the base is 28 cm, find the height of the cone

Answer:

Given a right circular cone.

The radius of the base of the cone = $r = \frac{28}{2} = 14\ cm$

The volume of the cone = $\small 9856\hspace{1mm}cm^3$

(i) Let the height of the cone be $h\ m$.

We know,
The volume of a right circular cone = $\frac{1}{3}\pi r^2 h$

$\therefore$ $\frac{1}{3}\times\frac{22}{7}\times(14)^2\times h = 9856$

$\\ \Rightarrow \frac{1}{3}\times\frac{22}{7}\times14\times14\times h = 9856$

$ \Rightarrow \frac{1}{3}\times22\times2\times14\times h = 9856$

$ \Rightarrow h = \frac{9856\times3}{22\times2\times14} $

$ \Rightarrow h =48\ cm$

Therefore, the height of the cone is $48\ cm$

Question 6: (ii) The volume of a right circular cone is $\small 9856\hspace{1mm}cm^3$. If the diameter of the base is 28 cm, find the Slant height of the cone

Answer:

Given a right circular cone.

The volume of the cone = $\small 9856\hspace{1mm}cm^3$

The radius of the base of the cone = $r = \frac{28}{2} = 14\ cm$

And the height of the cone = $h = 48\ cm$

(ii) We know the slant height, $l = \sqrt{r^2+h^2}$

$ \Rightarrow l = \sqrt{14^2+48^2}$
$ \Rightarrow l = \sqrt{196+2304} = \sqrt{2500} $
$ \Rightarrow l = 50\ cm$

Therefore, the slant height of the cone is $50\ cm$.

Question 6: (iii) The volume of a right circular cone is $\small 9856\hspace{1mm}cm^3$. If the diameter of the base is 28 cm, find the curved surface area of the cone

Answer:

Given a right circular cone.

The radius of the base of the cone = $r = \frac{28}{2} = 14\ cm$

And Slant height of the cone = $l = 50\ cm$

(iii) We know,

The curved surface area of a cone = $\pi r l$

$\therefore$ Required curved surface area

$=\frac{22}{7}\times14\times50$

$\\ = 22\times2\times50$
$= 2200\ cm^2$

Question 7: A right triangle ABC with sides 5 cm, 12 cm and 13 cm is revolved about the side 12 cm. Find the volume of the solid so obtained.

Answer:

When a right-angled triangle is revolved about the perpendicular side, a cone is formed whose,

Height of the cone = Length of the axis= $h = 12\ cm$

Base radius of the cone = $r = 5\ cm$

And, Slant height of the cone = $l = 13\ cm$

We know,

The volume of a cone = $\frac{1}{3}\pi r^2 h$

The required volume of the cone formed

$=\frac{1}{3}\times\pi\times5^2\times12$

$\\ = \pi\times25\times4 $
$ = 100\pi\ cm^3$

Therefore, the volume of the solid cone obtained is $100\pi\ cm^3$.

Question 8: If the triangle ABC in Question 7 above is revolved about the side 5 cm, then find the volume of the solid so obtained. Find also the ratio of the volumes of the two solids obtained in Questions 7 and 8.

Answer:

When a right-angled triangle is revolved about the perpendicular side, a cone is formed whose,

Height of the cone = Length of the axis= $h = 5\ cm$

Base radius of the cone = $r = 12\ cm$

And, Slant height of the cone = $l = 13\ cm$

We know,

The volume of a cone = $\frac{1}{3}\pi r^2 h$

The required volume of the cone formed

$=\frac{1}{3}\times\frac{22}{7}\times12^2\times5$

$\\ = \pi\times4\times60$
$ = 240\pi\ cm^3$

Now, the Ratio of the volumes of the two solids

$\\ = \frac{100\pi}{240\pi}$

$\\ = \frac{5}{12}$

Therefore, the required ratio is $5:12$.

Question 9: A heap of wheat is in the form of a cone whose diameter is $\small 10.5$ m and height is 3 m. Find its volume. The heap is to be covered by canvas to protect it from rain. Find the area of the canvas required.

Answer:

Given,

Height of the conical heap = $h = 3\ m$

Base radius of the cone = $r = \frac{10.5}{2}\ m$

We know,

The volume of a cone = $\frac{1}{3}\pi r^2 h$

The required volume of the cone formed

$=\frac{1}{3}\times\frac{22}{7}\times\left (\frac{10.5}{2} \right )^2\times3$

$\\ = 22\times\frac{1.5\times10.5}{4}$

$ = 86.625\ m^3$

Now,

The slant height of the cone, $l = \sqrt{r^2+h^2}$

$\Rightarrow l = \sqrt{3^2+5.25^2} = \sqrt{9+27.5625} \approx 6.05$

We know the curved surface area of a cone = $\pi r l$

The required area of the canvas to cover the heap

$=\frac{22}{7}\times\frac{10.5}{2}\times6.05$

$= 99.825\ m^2$

Question 1: (i) Find the volume of a sphere whose radius is 7 cm

Answer:

Given,

The radius of the sphere = $r = 7\ cm$

We know, Volume of a sphere = $\frac{4}{3}\pi r^3$

The required volume of the sphere

$=\frac{4}{3}\times\frac{22}{7}\times (7)^3$

$\\ = \frac{4}{3}\times22\times 7\times 7$

$\\ = \frac{4312}{3}$

$\\ = 1437\frac{1}{3}\ cm^3$

Question 1 (ii) Find the volume of a sphere whose radius is $\small 0.63\hspace{1mm}m$

Answer:

Given,

The radius of the sphere = $r = 0.63\ m$

We know, Volume of a sphere = $\frac{4}{3}\pi r^3$

The required volume of the sphere

$=\frac{4}{3}\times\frac{22}{7}\times (0.63)^3$

$\\ = 4\times22\times 0.03\times 0.63\times 0.63$

$\\ = 1.048\ m^3$

$\\ = 1.05\ m^3$ (approx)

Question 2: (i) Find the amount of water displaced by a solid spherical ball of a diameter of 28 cm

Answer:

The solid spherical ball will displace water equal to its volume.

Given,

The radius of the sphere = $r = \frac{28}{2}\ cm = 14\ cm$

We know, Volume of a sphere = $\frac{4}{3}\pi r^3$

$\therefore$ The required volume of the sphere

$=\frac{4}{3}\times\frac{22}{7}\times (14)^3$

$\\ = \frac{4}{3}\times22\times 2\times 14\times 14$

$\\ = \frac{34469}{3} $

$ = 11489\frac{2}{3}\ cm^3$

Therefore, the amount of water displaced will be $11489\frac{2}{3}\ cm^3$.

Question 2: (ii) Find the amount of water displaced by a solid spherical ball of diameter $\small 0.21\hspace{1mm}m$

Answer:

The solid spherical ball will displace water equal to its volume.

Given,

The radius of the sphere = $r = \frac{0.21}{2}\ m$

We know, Volume of a sphere = $\frac{4}{3}\pi r^3$

$\therefore$ The required volume of the sphere

$=\frac{4}{3}\times\frac{22}{7}\times \left(\frac{0.21}{2} \right )^3$

$\\ = 4\times22\times \frac{0.01\times 0.21\times 0.21}{8}$

$\\ = 11\times 0.01\times 0.21\times 0.21$

$\\ =0.004851\ m^3$

Therefore, the amount of water displaced will be $0.004851\ m^3$.

Question 3: The diameter of a metallic ball is $\small 4.2\hspace{1mm}cm$. What is the mass of the ball, if the density of the metal is $\small 8.9\hspace{1mm}g$ per $\small cm^3$?

Answer:

Given,

The radius of the metallic sphere = $r = \frac{4.2}{2}\ cm = 2.1\ cm$

We know, Volume of a sphere = $\frac{4}{3}\pi r^3$

$\therefore$ The required volume of the sphere

$=\frac{4}{3}\times\frac{22}{7}\times 2.1^3$

$ = 4\times22\times 0.1\times 2.1\times 2.1$

$ =38.808\ cm^3$

Now, the density of the metal is $\small 8.9\hspace{1mm}g$ per $\small cm^3$, which means,

Mass of $\small 1\ cm^3$ of the metallic sphere = $\small 8.9\hspace{1mm}g$

Mass of $38.808\ cm^3$ of the metallic sphere = $\small (8.9\times38.808)\ g$

$\small \approx 345.39\ g$

Question 4: The diameter of the moon is approximately one-fourth of the diameter of the earth. What fraction of the volume of the Earth is the volume of the moon?

Answer:

Given,

Let $d_e$ be the diameter of Earth.

$\therefore$ The diameter of the Moon = $d_m = \frac{1}{4}d_e$

We know the volume of a sphere

$=\frac{4}{3}\pi r^3$

$ =\frac{4}{3}\pi \left (\frac{d}{2} \right )^3$

$ = \frac{1}{6}\pi d^3$

$\therefore$ The ratio of the volumes

$=\frac{\text{Volume of the Earth}}{\text{Volume of the Moon}}$

$= \frac{\frac{1}{6}\pi d_e^3}{\frac{1}{6}\pi d_m^3} $

$= \frac{ d_e^3}{(\frac{d_e}{4})^3} $

$= 64: 1$

Therefore, the required ratio of the volume of the moon to the volume of the earth is $1: 64$.

Question 5: How many litres of milk can a hemispherical bowl of diameter $\small 10.5\hspace{1mm}cm$ hold?

Answer:

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

We know, Volume of a hemisphere = $\frac{2}{3}\pi r^3$

The volume of the given hemispherical bowl = $\frac{2}{3}\times\frac{22}{7}\times \left (\frac{10.5}{2} \right )^3$

$= \frac{2}{3\times8}\times22\times1.5\times10.5\times10.5$

$= 303.1875\ cm^3$

The capacity of the hemispherical bowl

$= \frac{303.1875}{1000} \approx 0.303\ litres$ (approx).

Question 6: A hemispherical tank is made up of an iron sheet 1 cm thick. If the inner radius is 1 m, then find the volume of the iron used to make the tank.

Answer:

Given,

Inner radius of the hemispherical tank = $r_1 = 1\ m$

Thickness of the tank = $1\ cm = 0.01\ m$

$\therefore$ Outer radius = Internal radius + thickness = $r_2 = (1+0.01)\ m = 1.01\ m$

We know, Volume of a hemisphere = $\frac{2}{3}\pi r^3$

$\therefore$ Volume of the iron used = Outer volume - Inner volume

$= \frac{2}{3}\pi r_2^3 - \frac{2}{3}\pi r_1^3$

$= \frac{2}{3}\times\frac{22}{7}\times (1.01^3 - 1^3)$

$= \frac{44}{21}\times0.030301$

$= 0.06348\ m^3$ (approx)

Question 7: Find the volume 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 = $4\pi r^2$

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

$\\ \Rightarrow 4\times\frac{22}{7}\times r^2 = 14\times11$

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

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

$ \Rightarrow r = 3.5\ cm$

$\therefore$ The volume of the sphere

$=\frac{4}{3}\pi r^3$

$= \frac{4}{3}\times\frac{22}{7}\times (3.5)^3$

$= 179\frac{2}{3}\ cm^3$

Question 8: (i) A dome of a building is in the form of a hemisphere. From inside, it was white-washed at the cost of $\small Rs\hspace{1mm} 4989.60$. If the cost of white-washing is Rs 20 per square metre, find the inside surface area of the dome

Answer:

Given,

$\small Rs\hspace{1mm}20$ is the cost of white-washing $1\ m^2$ of the inside area

$\small Rs\hspace{1mm}4989.60$ is the cost of white-washing of:

$\frac{1}{20}\times4989.60\ m^2 = 249.48\ m^2$ of inside area

(i) Therefore, the surface area of the inside of the dome is $249.48\ m^2$

Question 8: (ii) A dome of a building is in the form of a hemisphere. From inside, it was white-washed at the cost of $\small Rs\hspace{1mm}4989.60$. If the cost of whitewashing is Rs 20 per square meter, find the volume of the air inside the dome.

Answer:

Let the radius of the hemisphere be $r\ m$

Inside the surface area of the dome = $249.48\ m^2$

We know the surface area of a hemisphere = $2\pi r^2$

$\\ \therefore 2\pi r^2 = 249.48 $

$ \Rightarrow r^2 = \frac{249.48\times7}{2\times22} $

$ \Rightarrow r = 6.3\ m$

$\therefore$ The volume of the hemisphere

$=\frac{2}{3}\pi r^3$

$= \frac{2}{3}\times\frac{22}{7}\times (6.3)^3$

$= 523.908\ m^3$

Question 9: (i) Twenty-seven solid iron spheres, each of radius r and surface area S, are melted to form a sphere with surface area $\small S'$. Find the radius $\small r'$ of the new sphere

Answer:

Given,

The radius of a small sphere = $r$

The radius of the bigger sphere = $r'$

$\therefore$ The volume of each small sphere= $\frac{4}{3}\pi r^3$

And, Volume of the big sphere of radius $r'$ = $\frac{4}{3}\pi r'^3$

According to the question,

$27\times\frac{4}{3}\pi r^3=\frac{4}{3}\pi r'^3$

$\\ \Rightarrow r'^3 = 27\times r^3 $
$ \Rightarrow r' = 3\times r$

$\therefore r' = 3r$

Question 9: (ii) Twenty-seven solid iron spheres, each of radius r and surface area S, are melted to form a sphere with surface area $\small S'$. Find the ratio of S and $\small S'$.

Answer:

Given,

The radius of a small sphere = $r$

The surface area of a small sphere = $S$

The radius of the bigger sphere = $r'$

The surface area of the bigger sphere = $S'$

And, $r' = 3r$

We know the surface area of a sphere = $4\pi r^2$

$\therefore$ The ratio of their surface areas

$=\frac{4\pi r'^2}{4\pi r^2}$

$\\ = \frac{ (3r)^2}{ r^2} \\ = 9$

Therefore, the required ratio is $1:9$.

Question 10: A capsule of medicine is in the shape of a sphere of diameter $\small 3.5\hspace{1mm}mm$. How much medicine (in $\small mm^3$ ) is needed to fill this capsule?

Answer:

Given,

The radius of the spherical capsule = $r =\frac{3.5}{2}$

$\therefore$ The volume of the capsule = $\frac{4}{3}\pi r^3$

$= \frac{4}{3}\times\frac{22}{7}\times(\frac{3.5}{2})^3$

$= \frac{4}{3}\times22\times\frac{0.5\times3.5\times3.5}{8}$

$= 22.458\ mm^3 \approx 22.46\ mm^3$ (approx)

Therefore, approximately $22.46\ mm^3$ of medicine is needed to fill the capsule.