RD Sharma 2020 solution class 9 chapter 21 Surface Area and Volume of a Sphere FBQS

FBQS

Page-21.26

Question 1:

If the radius of a sphere is 2r, then its volume is _________.

Answer 1:

We know

Volume of the sphere of radius R = $\frac{4}{3}\mathrm{\pi }{R}^3$

Now,

Radius of the sphere = 2r         (Given)

∴ Volume of the sphere

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

$=\frac{4}{3}\mathrm{\pi }\times 8r^3$

$=\frac{32}{3}\mathrm{\pi }{r}^3$

Thus, the volume of the sphere is $\frac{32}{3}\mathrm{\pi }{r}^3$ cu. units.

If the radius of a sphere is 2r, then its volume is $\underline{ \frac{32}{3}\mathrm{\pi }{r}^3 \mathrm{cu}. \mathrm{units} }$.

Question 2:

The radius of a hemispherical balloon increases from 6 cm to 12 cm as air is being pumped into. The ratio of the surface areas of the balloon in two cases is __________.

Answer 2:

We know

Surface area of the hemisphere having radius r = 3$\mathrm{\pi }$r²

It is given that the radius of a hemispherical balloon increases from 6 cm to 12 cm when air is pumped into it.

Let S₁ be the surface area of the hemispherical balloon when its radius is 6 cm and S₂ be the surface area of the hemispherical balloon when its radius is 12 cm.

∴ S₁ = Surface area of the hemispherical balloon when its radius is 6 cm = 3$\mathrm{\pi }$(6 cm)²

S₂ = Surface area of the hemispherical balloon when its radius is 12 cm = 3$\mathrm{\pi }$(12 cm)²

Now,

$\frac{S_1}{S_2}=\frac{3\mathrm{\pi }{\left(6 \mathrm{cm}\right)}^2}{3\mathrm{\pi }{\left(12 \mathrm{cm}\right)}^2}$

$\Rightarrow \frac{S_1}{S_2}=\frac{6\times 6}{12\times 12}=\frac{1}{4}$

⇒ S₁ : S₂ = 1 : 4

Thus, the ratio of the surface areas of the hemispherical balloon in two cases is 1 : 4.

The radius of a hemispherical balloon increases from 6 cm to 12 cm as air is being pumped into. The ratio of the surface areas of the balloon in two cases is ___1 : 4___.

Question 3:

A cone, a hemisphere and a cylinder stand on equal bases and have the same height. The ratio of their volumes is __________.

Answer 3:

It is given that a cone, a hemisphere and a cylinder stand on equal bases and have the same height.

Let the radius of cone, hemisphere and cylinder be r units.

Radius of the cone = Radius of hemisphere = Radius of cylinder = r

Also,

Height of the cone = Height of the cylinder = Height of the hemisphere

We know that, the height of a hemisphere is same as its radius.

∴ Height of the hemisphere = r

⇒ Height of the cone = Height of the cylinder = Height of the hemisphere = r

Now,

Volume of the cone = $\frac{1}{3}\mathrm{\pi }$ × (Radius)² × Height = $\frac{1}{3}\mathrm{\pi }$ × r² × r = $\frac{1}{3}\mathrm{\pi }$r³

Volume of the hemisphere = $\frac{2}{3}\mathrm{\pi }$ × (Radius)³ = $\frac{2}{3}\mathrm{\pi }$r³

Volume of the cylinder = $\mathrm{\pi }$ × (Radius)² × Height = $\mathrm{\pi }$ × r² × r = $\mathrm{\pi }$r³

∴ Volume of the cone : Volume of the hemisphere : Volume of the cylinder

= $\frac{1}{3}\mathrm{\pi }$r³ : $\frac{2}{3}\mathrm{\pi }$r³ : $\mathrm{\pi }$r³

= $\frac{1}{3}$ : $\frac{2}{3}$ : 1

= 1 : 2 : 3

Thus, the ratio of their volumes is 1 : 2 : 3.

A cone, a hemisphere and a cylinder stand on equal bases and have the same height. The ratio of their volumes is ____1 : 2 : 3____.

Question 4:

The volume of the largest right circular cone that can be fitted in a cube whose edge is 2r equals to the volume of a hemisphere of radius __________.

Answer 4:

The largest right circular cone that can be fitted in a cube of given edge is such that the diameter of the base of the cone is equal to the edge of the cube and the height of the cone is equal to the edge of the cube.

It is given that the edge of cube is 2r.

Let R be the radius and H be the height of the largest right circular cone that can be fitted in the given cube.

∴ Diameter of the base of the cone = Edge of the cube

⇒ 2R = 2r

⇒ R = r

Height of the cone = Edge of the cube

⇒ H = 2r

∴ Volume of the largest cone that can be fitted in the given cube

$=\frac{1}{3}\mathrm{\pi }{R}^{2}H$

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

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

= Volume of the hemisphere of radius r

Thus, the volume of the largest right circular cone that can be fitted in a cube whose edge is 2r equals to the volume of a hemisphere of radius r.

The volume of the largest right circular cone that can be fitted in a cube whose edge is 2r equals to the volume of a hemisphere of radius ____r____.

Question 5:

If the ratio of the volumes of two spheres is 8 : 27, then the ratio of their surface areas is __________.

Answer 5:

Let r and R be the radii of the two spheres.

Suppose V₁ be the volume and S₁ be the surface area of the sphere of radius r & V₂ be the volume and S₂ be the surface area of the sphere of radius R.

It is given that the ratio of the volumes of two spheres is 8 : 27.

$\therefore \frac{V_1}{V_2}=\frac{8}{27}$

$\Rightarrow \frac{{\displaystyle \frac{4}{3}}\mathrm{\pi }{r}^3}{{\displaystyle \frac{4}{3}}\mathrm{\pi }{R}^3}=\frac{8}{27}$

$\Rightarrow {\left(\frac{r}{R}\right)}^3={\left(\frac{2}{3}\right)}^3$

$\Rightarrow \frac{r}{R}=\frac{2}{3}$          .....(1)

Now,

$\frac{S_1}{S_2}=\frac{4\mathrm{\pi }{r}^2}{4\mathrm{\pi }{R}^2}$

$\Rightarrow \frac{S_1}{S_2}={\left(\frac{r}{R}\right)}^2$

$\Rightarrow \frac{S_1}{S_2}={\left(\frac{2}{3}\right)}^2$            [Using (1)]

$\Rightarrow \frac{S_1}{S_2}=\frac{4}{9}$

⇒ S₁ : S₂ = 4 : 9

Thus, the ratio of the surface areas of the two spheres is 4 : 9.

If the ratio of the volumes of two spheres is 8 : 27, then the ratio of their surface areas is ___4 : 9___.

Question 6:

If the volume of a hemisphere is 18π cm³, then its total surface area is __________.

Answer 6:

Let r be the radius of the hemisphere.

Volume of the hemisphere = 18$\mathrm{\pi }$ cm³         (Given)

$\Rightarrow \frac{2}{3}\mathrm{\pi }{r}^3=18\mathrm{\pi }$

$\Rightarrow r^3=\frac{18\times 3}{2}=27$

$\Rightarrow r^3={\left(3\right)}^3$

$\Rightarrow r=3 \mathrm{cm}$

∴ Total surface area of the hemisphere

= 3$\mathrm{\pi }$r²

= 3$\mathrm{\pi }$× (3 cm)²

= 27$\mathrm{\pi }$ cm²

Thus, the total surface area of the hemisphere is 27$\mathrm{\pi }$ cm².

If the volume of a hemisphere is 18π cm³, then its total surface area is ___27$\mathrm{\pi }$ cm²___.

Question 7:

If a sphere is placed inside a right circular cylinder so as to touch the top, base and the lateral surface of the cylinder. If the radius of the sphere is r, the volume of the cylinder and the sphere are in the ratio _________.

Answer 7:

Radius of the sphere = r

∴ Volume of the sphere, V₁ = $\frac{4}{3}\mathrm{\pi }{r}^3$         .....(1)

If a sphere is placed inside a right circular cylinder so as to touch the top, base and the lateral surface of the cylinder, then the height of the cylinder is equal to the diameter of the sphere and the diameter of the base of cylinder is equal to the diameter of the sphere.

Let R be the radius and H be the height of the cylinder.

Height of the cylinder = Diameter of the sphere

⇒ H = 2r      .....(2)

Diameter of the base of cylinder = Diameter of the sphere

⇒ 2R = 2r

⇒ R = r        .....(3)

∴ Volume of the cylinder, V₂ = $\mathrm{\pi }$R²H = $\mathrm{\pi }$ × r² × 2r = 2$\mathrm{\pi }$r³           .....(4)             [Using (2) and (3)]

Now,

$\frac{V_2}{V_1}=\frac{2\mathrm{\pi }{r}^3}{{\displaystyle \frac{4}{3}}\mathrm{\pi }{r}^3}=\frac{3}{2}$

⇒ V₂ : V₁ = 3 : 2

Thus, the ratio of volume of cylinder to volume of sphere is 3 : 2.

If a sphere is placed inside a right circular cylinder so as to touch the top, base and the lateral surface of the cylinder. If the radius of the sphere is r, the volume of the cylinder and the sphere are in the ratio ___3 : 2___.

Question 8:

If the volume and surface area of a sphere are numerically the same, then its radius is __________.

Answer 8:

Let r be the radius of sphere.

Volume of the sphere = $\frac{4}{3}\mathrm{\pi }$r³ cu. units

Surface area of the sphere = 4$\mathrm{\pi }$r² sq. units

It is given that the volume and surface area of sphere are numerically same.

$\therefore \frac{4}{3}\mathrm{\pi }{r}^3=4\mathrm{\pi }{r}^2$

⇒ r = 3 units

Thus, the radius of sphere is 3 units.

If the volume and surface area of a sphere are numerically the same, then its radius is ___3 units___.

Question 9:

A right circular cylinder and a sphere have the same volume and same radius. The ratio of the areas of their curved surfaces is ________.

Answer 9:

Let the radius of right circular cylinder and sphere be r.

Suppose h be the height of the cylinder.

∴ Volume of the sphere = $\frac{4}{3}\mathrm{\pi }{r}^3$

Volume of the cylinder = $\mathrm{\pi }{r}^{2}h$

Now,

Volume of the cylinder = Volume of the sphere            (Given)

$\Rightarrow \mathrm{\pi }{r}^{2}h=\frac{4}{3}\mathrm{\pi }{r}^3$

$\Rightarrow h=\frac{4}{3}r$         .....(1)

Now,

Curved surface area of the cylinder, S₁ = 2$\mathrm{\pi }$rh

Curved surface area of the sphere, S₂ = 4$\mathrm{\pi }$r²

$\therefore \frac{S_1}{S_2}=\frac{2\mathrm{\pi }rh}{4\mathrm{\pi }{r}^2}$

$\Rightarrow \frac{S_1}{S_2}=\frac{2\mathrm{\pi }r\times {\displaystyle \frac{4}{3}}r}{4\mathrm{\pi }{r}^2}$        [From (1)]

$\Rightarrow \frac{S_1}{S_2}=\frac{2}{3}$

⇒ S₁ : S₂ = 2 : 3

Thus, the ratio of the curved surface of cylinder to the curved surface area of sphere is 2 : 3.

A right circular cylinder and a sphere have the same volume and same radius. The ratio of the areas of their curved surfaces is ___2 : 3___.

Question 10:

If the radius of a sphere is doubled, then the percentage increase in the surface area is _________.

Answer 10:

Let the radius of the original sphere be r.

∴ Surface area of the original sphere, S₁ = 4$\mathrm{\pi }$r²

If the radius of the sphere is doubled, then the radius of the new sphere is 2r.

∴ Surface area of the new sphere, S₂ = 4$\mathrm{\pi }$(2r)² = 16$\mathrm{\pi }$r²

Now,

Percentage increase in the surface area of sphere

$=\frac{\mathrm{Increase} \mathrm{in} \mathrm{surface} \mathrm{area} \mathrm{of} \mathrm{sphere}}{\mathrm{Surface} \mathrm{area} \mathrm{of} \mathrm{original} \mathrm{sphere}}\times 100\%$

$=\frac{S_2-S_1}{S_1}\times 100\%$

$=\frac{16\mathrm{\pi }{r}^2-4\mathrm{\pi }{r}^2}{4\mathrm{\pi }{r}^2}\times 100\%$

$=\frac{12\mathrm{\pi }{r}^2}{4\mathrm{\pi }{r}^2}\times 100\%$

$=300\%$

Thus, the percentage increase in the surface area of the sphere is 300%.

If the radius of a sphere is doubled, then the percentage increase in the surface area is ___300%___.

Question 11:

If a solid hemisphere of radius 8 cm is melted and recast into n spheres of radius 2 cm each, then n = _________.

Answer 11:

Radius of the solid hemisphere, R = 8 cm

Radius of each sphere, r = 2 cm

It is given that, solid hemisphere of radius 8 cm is melted and recast into n spheres of radius 2 cm each.

∴ n × Volume of each sphere = Volume of the solid hemisphere

$\Rightarrow n\times \frac{4}{3}\mathrm{\pi }{r}^3=\frac{2}{3}\mathrm{\pi }{R}^3$

$\Rightarrow n=\frac{1}{2}\left(\frac{R^3}{r^3}\right)$

$\Rightarrow n=\frac{1}{2}{\left(\frac{R}{r}\right)}^3$

$\Rightarrow n=\frac{1}{2}\times {\left(\frac{8 \mathrm{cm}}{2 \mathrm{cm}}\right)}^3$                   (R = 8 cm and r = 2 cm)

$\Rightarrow n=\frac{1}{2}\times 64$

$\Rightarrow n=32$

Thus, the value of n is 32.

If a solid hemisphere of radius 8 cm is melted and recast into n spheres of radius 2 cm each, then n = ___32___.

Question 12:

If a solid sphere of radius 4 cm is melted and recast into n solid hemispheres of radius 2 cm each, then n = _________.

Answer 12:

Radius of the solid sphere, R = 4 cm

Radius of each solid hemisphere, r = 2 cm

It is given that, the solid sphere of radius 4 cm is melted and recast into n solid hemispheres of radius 2 cm each.

∴ n × Volume of each solid hemisphere = Volume of the sphere

$\Rightarrow n\times \frac{2}{3}\mathrm{\pi }{r}^3=\frac{4}{3}\mathrm{\pi }{R}^3$

$\Rightarrow n=2\times \frac{R^3}{r^3}$

$\Rightarrow n=2\times {\left(\frac{R}{r}\right)}^3$

$\Rightarrow n=2\times {\left(\frac{4 \mathrm{cm}}{2 \mathrm{cm}}\right)}^3$           (R = 4 cm and r = 2 cm)

$\Rightarrow n=2\times 8$

$\Rightarrow n=16$

Thus, the value of n is 16.

If a solid sphere of radius 4 cm is melted and recast into n solid hemispheres of radius 2 cm each, then n = ___16___.

Question 13:

Fifteen identical spheres are made by melting a solid cylinder of radius 10 cm and height 5.4 cm, the diameter of each sphere is _________.

Answer 13:

Let the radius of each sphere be r.

Radius of the cylinder, R = 10 cm

Height of the cylinder, H = 5.4 cm

It is given that, 15 identical spheres are made by melting a solid cylinder of radius 10 cm and height 5.4 cm.

∴ 15 × Volume of each sphere = Volume of the cylinder

$\Rightarrow 15\times \frac{4}{3}\mathrm{\pi }{r}^3=\mathrm{\pi }{R}^{2}H$

$\Rightarrow r^3=\frac{{\left(10 \mathrm{cm}\right)}^2\times 5.4 \mathrm{cm}}{20}$

$\Rightarrow r^3=27 {\mathrm{cm}}^3$

$\Rightarrow r^3={\left(3 \mathrm{cm}\right)}^3$

$\Rightarrow r=3 \mathrm{cm}$

∴ Diameter of each sphere = 2r = 2 × 3 cm = 6 cm

Thus, the diameter of each sphere is 6 cm.

Fifteen identical spheres are made by melting a solid cylinder of radius 10 cm and height 5.4 cm, the diameter of each sphere is ___6 cm___.