RS Aggarwal 2021 2022 Solutions for Class 10 Maths Chapter 17 Volumes And Surface Areas Of Solids are provided here with simple step-by-step explanations. These solutions for Volumes And Surface Areas Of Solids are extremely popular among class 10 students for Maths Volumes And Surface Areas Of Solids Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the RS Aggarwal 2021 2022 Book of class 10 Maths Chapter 17 are provided here for you for free. You will also love the ad-free experience on Meritnation’s RS Aggarwal 2021 2022 Solutions. All RS Aggarwal 2021 2022 Solutions for class 10 Maths are prepared by experts and are 100% accurate.

#### Question 1: #### Question 2: = 718.67 cm3

#### Question 4:

= 718.67 cm3

(i)

(ii)
The ratio of radius and height of a solid right-circular cone is 5:12.
Let radius, r = 5x and height, h=12x.
Volume = 314 cm3.
$⇒\frac{1}{3}\mathrm{\pi }{r}^{2}h=314\phantom{\rule{0ex}{0ex}}⇒\frac{1}{3}×3.14×{\left(5x\right)}^{2}×\left(12x\right)=314\phantom{\rule{0ex}{0ex}}⇒{x}^{3}=\frac{314×3}{3.14×5×5×12}\phantom{\rule{0ex}{0ex}}⇒{x}^{3}=1⇒x=1$
So, radius r = 5 cm and height h = 12 cm.
Using Pythagoras Theorem, slant height is given by
Total Surface Area of Cone .

#### Question 5:

(i)

(ii)
The ratio of radius and height of a solid right-circular cone is 5:12.
Let radius, r = 5x and height, h=12x.
Volume = 314 cm3.
$⇒\frac{1}{3}\mathrm{\pi }{r}^{2}h=314\phantom{\rule{0ex}{0ex}}⇒\frac{1}{3}×3.14×{\left(5x\right)}^{2}×\left(12x\right)=314\phantom{\rule{0ex}{0ex}}⇒{x}^{3}=\frac{314×3}{3.14×5×5×12}\phantom{\rule{0ex}{0ex}}⇒{x}^{3}=1⇒x=1$
So, radius r = 5 cm and height h = 12 cm.
Using Pythagoras Theorem, slant height is given by
Total Surface Area of Cone .

So, the ratio of their heights is 25:64.

#### Question 6:

So, the ratio of their heights is 25:64.

Let r, h and l be the base radius, the height and the slant height of the conical mountain, respectively.

So, the height of the mountain is 2.4 km.

#### Question 7:

Let r, h and l be the base radius, the height and the slant height of the conical mountain, respectively.

So, the height of the mountain is 2.4 km.

Let r and h be the base radius and the height of the solid cylinder, respectively.

#### Question 8:

Let r and h be the base radius and the height of the solid cylinder, respectively.

Let the radii of the given sphere and the new sphere be r and R, respectively.

So, the surface area of the new sphere is 9856 cm2.

#### Question 9:

Let the radii of the given sphere and the new sphere be r and R, respectively.

So, the surface area of the new sphere is 9856 cm2. We have,
the radii of bases of the cone and cylinder, r = 15 m,
the height of the cylinder, = 5.5 m,
the height of the tent = 8.25 m
Also, the height of the cone,

As, the width of the canvas = 1.5 m

Hence, the length of the tent used for making the tent is 825 m.

#### Question 10: We have,
the radii of bases of the cone and cylinder, r = 15 m,
the height of the cylinder, = 5.5 m,
the height of the tent = 8.25 m
Also, the height of the cone,

As, the width of the canvas = 1.5 m

Hence, the length of the tent used for making the tent is 825 m. Radius of the cylinder = 14 m
Radius of the base of the cone = 14 m
Height of the cylinder (h) = 3 m
Total height of the tent = 13.5 m
Surface area of the cylinder

Height of the cone

Surface area of the cone = $\pi r\sqrt{{r}^{2}+{h}^{2}}$

Total surface area

∴ Cost of cloth

#### Question 11: Radius of the cylinder = 14 m
Radius of the base of the cone = 14 m
Height of the cylinder (h) = 3 m
Total height of the tent = 13.5 m
Surface area of the cylinder

Height of the cone

Surface area of the cone = $\pi r\sqrt{{r}^{2}+{h}^{2}}$

Total surface area

∴ Cost of cloth Radius of the cylinder = 52.5 m
Radius of the base of the cone = 52.5 m
Slant height (l) of the cone = 53 m
Height of the cylinder (h) = 3 m
Curved surface area of the cylindrical portion
Curved surface area of the conical portion

Thus, the total area of canvas required for making the tent

#### Question 12: Radius of the cylinder = 52.5 m
Radius of the base of the cone = 52.5 m
Slant height (l) of the cone = 53 m
Height of the cylinder (h) = 3 m
Curved surface area of the cylindrical portion
Curved surface area of the conical portion

Thus, the total area of canvas required for making the tent Radius of the cylinder = 2.5 m
Height of the cylinder = 21 m

Curved surface area of the cylinder
Radius of the cone = 2.5 m
Slant height of the cone = 8 m
Curved surface area of the cone

Area of circular base
∴ Total surface area of rocket

#### Question 13: Radius of the cylinder = 2.5 m
Height of the cylinder = 21 m

Curved surface area of the cylinder
Radius of the cone = 2.5 m
Slant height of the cone = 8 m
Curved surface area of the cone

Area of circular base
∴ Total surface area of rocket So, the volume of the solid is 166.83 cm3.

#### Question 14: So, the volume of the solid is 166.83 cm3.

The radius of hemisphere as well as that of base of cone is r = 3.5 cm.
The height of toy = 15.5 cm.
The height of hemisphere, h = 3.5 cm and height of cone = 15.5$-$3.5 = 12 cm.
Using Pythagoras Theorem, slant height of cone is .
Therefore, the Total surface area of toy = Curved surface of cone + Curved surface of hemisphere

#### Question 15:

The radius of hemisphere as well as that of base of cone is r = 3.5 cm.
The height of toy = 15.5 cm.
The height of hemisphere, h = 3.5 cm and height of cone = 15.5$-$3.5 = 12 cm.
Using Pythagoras Theorem, slant height of cone is .
Therefore, the Total surface area of toy = Curved surface of cone + Curved surface of hemisphere #### Question 16:  So, the radius of the ice-cream cone is 3 cm.

#### Question 17: So, the radius of the ice-cream cone is 3 cm.

Diameter of the hemisphere = 21 cm
Therefore, radius of the hemisphere = 10.5 cm

Volume of the hemisphere

Height of the cylinder
Volume of the cylinder

Total volume

#### Question 18:

Diameter of the hemisphere = 21 cm
Therefore, radius of the hemisphere = 10.5 cm

Volume of the hemisphere

Height of the cylinder
Volume of the cylinder

Total volume #### Question 19:  So, the surface area of the medicine capsule is 220 mm2.

#### Question 20: So, the surface area of the medicine capsule is 220 mm2.

The height h of cylinder = 20 cm and diameter of its base = 7 cm.
$⇒$The radius r of its base = 3.5 cm.
$⇒$Curved surface area of cylinder =

Now, curved surface area of one hemisphere

Total surface area of the article = Curved surface area of cylinder + 2(curved surface area of hemisphere)

#### Question 21:

The height h of cylinder = 20 cm and diameter of its base = 7 cm.
$⇒$The radius r of its base = 3.5 cm.
$⇒$Curved surface area of cylinder =

Now, curved surface area of one hemisphere

Total surface area of the article = Curved surface area of cylinder + 2(curved surface area of hemisphere)

The object is shown in the figure below. Radius of hemisphere = 2.1 cm
Volume of hemisphere

Height of cone = 4 cm
Volume of cone
Volume of the object

Volume of cylindrical tub

When the object is immersed in the tub, volume of water equal to the volume of the object is displaced from the tub.

Volume of water left in the tub =

#### Question 22:

The object is shown in the figure below. Radius of hemisphere = 2.1 cm
Volume of hemisphere

Height of cone = 4 cm
Volume of cone
Volume of the object

Volume of cylindrical tub

When the object is immersed in the tub, volume of water equal to the volume of the object is displaced from the tub.

Volume of water left in the tub = Volume of the solid left = Volume of cylinder - Volume of cone

The slant length of the cone, $l=\sqrt{{r}^{2}+{h}^{2}}=\sqrt{36+64}=10cm$

Total surface area of final solid = Area of base circle + Curved surface area of cylinder + Curved surface area of cone

#### Question 23: Volume of the solid left = Volume of cylinder - Volume of cone

The slant length of the cone, $l=\sqrt{{r}^{2}+{h}^{2}}=\sqrt{36+64}=10cm$

Total surface area of final solid = Area of base circle + Curved surface area of cylinder + Curved surface area of cone So, the total surface area of the remaining solid is 73.92 cm2.

#### Question 24: So, the total surface area of the remaining solid is 73.92 cm2.

So, the volume of the remaining solid is 502.04 cm3.

#### Question 25:

So, the volume of the remaining solid is 502.04 cm3.

So, the ratio of the volume of metal left in the cylinder to the volume of metal taken out in conical shape is 133 : 2.

#### Question 26:

So, the ratio of the volume of metal left in the cylinder to the volume of metal taken out in conical shape is 133 : 2. Diameter of spherical part = 21 cm
Radius of the spherical part = 10.5 cm

Volume of spherical part of the vessel
Diameter of cylinder = 4 cm
Radius of cylinder = 2 cm
Height of cylinder = 7 cm

Volume of the cylindrical part of the vessel

Total volume of the vessel

#### Question 27: Diameter of spherical part = 21 cm
Radius of the spherical part = 10.5 cm

Volume of spherical part of the vessel
Diameter of cylinder = 4 cm
Radius of cylinder = 2 cm
Height of cylinder = 7 cm

Volume of the cylindrical part of the vessel

Total volume of the vessel

Diameter of the cylindrical part =7 cm
Therefore,radius of the cylindrical part = 3.5 cm

Volume of hemisphere

Volume of the cylinder

Height of cone

Volume of the cone

Total volume

#### Question 28:

Diameter of the cylindrical part =7 cm
Therefore,radius of the cylindrical part = 3.5 cm

Volume of hemisphere

Volume of the cylinder

Height of cone

Volume of the cone

Total volume #### Question 29: (i) Disclaimer: It is written cuboid in the question but it should be cube. From the figure, it can be observed that the maximum diameter possible for such hemisphere is equal to the cube’s edge, i.e., 7 cm.

Radius (r) of hemispherical part = = 3.5 cm

Total surface area of solid = Surface area of cubical part + CSA of hemispherical part

− Area of base of hemispherical part

= 6 (Edge)2  = 6 (Edge)2 +  (ii) hence, the cost of painting the total surface area of the solid is â‚¹33.92.

#### Question 30:

(i) Disclaimer: It is written cuboid in the question but it should be cube. From the figure, it can be observed that the maximum diameter possible for such hemisphere is equal to the cube’s edge, i.e., 7 cm.

Radius (r) of hemispherical part = = 3.5 cm

Total surface area of solid = Surface area of cubical part + CSA of hemispherical part

− Area of base of hemispherical part

= 6 (Edge)2  = 6 (Edge)2 +  (ii) hence, the cost of painting the total surface area of the solid is â‚¹33.92. So, the surface area of the toy is 770 cm2.

#### Question 31: So, the surface area of the toy is 770 cm2. #### Question 32:  #### Question 1: The volume of solid metallic cuboid is
This cuboid has been recasted into solid cubes of edge 2 m whose volume is given by
Therefore, the total number of cubes so formed

#### Question 2:

The volume of solid metallic cuboid is
This cuboid has been recasted into solid cubes of edge 2 m whose volume is given by
Therefore, the total number of cubes so formed

So, the diameter of the sphere is 10 cm.

#### Question 3:

So, the diameter of the sphere is 10 cm.

So, the radius of the resulting sphere is 12 cm.

#### Question 4:

So, the radius of the resulting sphere is 12 cm.

Radius of the cone = 12 cm
Height of the cone = 24 cm

Volume

Radius of each ball = 3 cm
Volume of each ball
Total number of balls formed by melting the cone

#### Question 5:

Radius of the cone = 12 cm
Height of the cone = 24 cm

Volume

Radius of each ball = 3 cm
Volume of each ball
Total number of balls formed by melting the cone

So, the height of the cylinder is $\frac{8}{3}$ cm.

#### Question 6:

So, the height of the cylinder is $\frac{8}{3}$ cm.

Internal diameter of the hemispherical shell = 6 cm
Therefore, internal radius of the hemispherical shell = 3 cm

External diameter of the hemispherical shell = 10 cm
External radius of the hemispherical shell = 5 cm

Volume of hemispherical shell

Radius of cone = 7 cm
Let the height of the cone be h cm.
Volume of cone

The volume of the hemispherical shell must be equal to the volume of the cone. Therefore,

#### Question 7:

Internal diameter of the hemispherical shell = 6 cm
Therefore, internal radius of the hemispherical shell = 3 cm

External diameter of the hemispherical shell = 10 cm
External radius of the hemispherical shell = 5 cm

Volume of hemispherical shell

Radius of cone = 7 cm
Let the height of the cone be h cm.
Volume of cone

The volume of the hemispherical shell must be equal to the volume of the cone. Therefore,

So, the thickness of the wire is 0.2 cm or 2 mm.

#### Question 8:

So, the thickness of the wire is 0.2 cm or 2 mm.

Inner diameter of the bowl = 30 cm
Inner volume of the bowl = Volume of liquid

Radius of each bottle = 2.5 cm
Height = 6 cm

Volume of each bottle$={\mathrm{\pi r}}^{2}\mathrm{h}=\mathrm{\pi }×\frac{5}{2}×\frac{5}{2}×6=\frac{75\mathrm{\pi }}{2}{\mathrm{cm}}^{3}$

Total number of bottles required

#### Question 9:

Inner diameter of the bowl = 30 cm
Inner volume of the bowl = Volume of liquid

Radius of each bottle = 2.5 cm
Height = 6 cm

Volume of each bottle$={\mathrm{\pi r}}^{2}\mathrm{h}=\mathrm{\pi }×\frac{5}{2}×\frac{5}{2}×6=\frac{75\mathrm{\pi }}{2}{\mathrm{cm}}^{3}$

Total number of bottles required

Diameter of sphere = 21 cm
Radius of sphere $=\frac{21}{2}cm$

Volume of sphere

Diameter of the cone = 3.5 cm
Radius of the cone $=\frac{3.5}{2}=\frac{7}{4}cm$
Height = 3 cm

Volume of each cone

Total number of cones

#### Question 10:

Diameter of sphere = 21 cm
Radius of sphere $=\frac{21}{2}cm$

Volume of sphere

Diameter of the cone = 3.5 cm
Radius of the cone $=\frac{3.5}{2}=\frac{7}{4}cm$
Height = 3 cm

Volume of each cone

Total number of cones

Diameter of cannon ball = 28 cm
Radius of the cannon ball = 14 cm

Volume of ball

Diameter of base of cone = 35 cm

Radius of base of cone $=\frac{35}{2}cm$
Let the height of the cone be h cm.

Volume of cone

From the above results and from the given conditions,
Volume of ball = Volume of cone

#### Question 11:

Diameter of cannon ball = 28 cm
Radius of the cannon ball = 14 cm

Volume of ball

Diameter of base of cone = 35 cm

Radius of base of cone $=\frac{35}{2}cm$
Let the height of the cone be h cm.

Volume of cone

From the above results and from the given conditions,
Volume of ball = Volume of cone

Radius of sphere = 3 cm
Radius of first ball = 1.5 cm
Radius of second ball = 2 cm
Let radius of the third ball be r cm.
Volume of third ball = Volume of original sphere - Sum of volumes of other two balls

$=\frac{4}{3}\mathrm{\pi }×{3}^{3}-\left(\frac{4}{3}\mathrm{\pi }×{\frac{3}{2}}^{3}+\frac{4}{3}\mathrm{\pi }×{2}^{3}\right)\phantom{\rule{0ex}{0ex}}=\frac{4}{3}\mathrm{\pi }×3×3×3-\left(\frac{4}{3}\mathrm{\pi }×\frac{3}{2}×\frac{3}{2}×\frac{3}{2}+\frac{4}{3}\mathrm{\pi }×2×2×2\right)\phantom{\rule{0ex}{0ex}}=4\mathrm{\pi }×3×3-\left(\mathrm{\pi }×\frac{3×3}{2}+\frac{4}{3}\mathrm{\pi }×2×2×2\right)\phantom{\rule{0ex}{0ex}}=36\mathrm{\pi }-\left(\mathrm{\pi }\frac{9}{2}+\frac{32}{3}\mathrm{\pi }\right)\phantom{\rule{0ex}{0ex}}=\left(\frac{36×6-9×3-32×2}{6}\right)\mathrm{\pi }\phantom{\rule{0ex}{0ex}}=\left(\frac{216-27-64}{6}\right)\mathrm{\pi }=\frac{125\mathrm{\pi }}{6}$

Therefore,

#### Question 12:

Radius of sphere = 3 cm
Radius of first ball = 1.5 cm
Radius of second ball = 2 cm
Let radius of the third ball be r cm.
Volume of third ball = Volume of original sphere - Sum of volumes of other two balls

$=\frac{4}{3}\mathrm{\pi }×{3}^{3}-\left(\frac{4}{3}\mathrm{\pi }×{\frac{3}{2}}^{3}+\frac{4}{3}\mathrm{\pi }×{2}^{3}\right)\phantom{\rule{0ex}{0ex}}=\frac{4}{3}\mathrm{\pi }×3×3×3-\left(\frac{4}{3}\mathrm{\pi }×\frac{3}{2}×\frac{3}{2}×\frac{3}{2}+\frac{4}{3}\mathrm{\pi }×2×2×2\right)\phantom{\rule{0ex}{0ex}}=4\mathrm{\pi }×3×3-\left(\mathrm{\pi }×\frac{3×3}{2}+\frac{4}{3}\mathrm{\pi }×2×2×2\right)\phantom{\rule{0ex}{0ex}}=36\mathrm{\pi }-\left(\mathrm{\pi }\frac{9}{2}+\frac{32}{3}\mathrm{\pi }\right)\phantom{\rule{0ex}{0ex}}=\left(\frac{36×6-9×3-32×2}{6}\right)\mathrm{\pi }\phantom{\rule{0ex}{0ex}}=\left(\frac{216-27-64}{6}\right)\mathrm{\pi }=\frac{125\mathrm{\pi }}{6}$

Therefore,

External diameter of the shell = 24 cm
External radius of the shell = 12 cm
Internal diameter of the shell = 18 cm
Internal radius of the shell = 9 cm

Volume of the shell

Height of cylinder = 37 cm
Let radius of cylinder be r cm.

Volume of cylinder

Volume of the shell = Volume of cylinder

So, diameter of the base of the cylinder = 2r = 12 cm.

#### Question 13:

External diameter of the shell = 24 cm
External radius of the shell = 12 cm
Internal diameter of the shell = 18 cm
Internal radius of the shell = 9 cm

Volume of the shell

Height of cylinder = 37 cm
Let radius of cylinder be r cm.

Volume of cylinder

Volume of the shell = Volume of cylinder

So, diameter of the base of the cylinder = 2r = 12 cm.

Radius of hemisphere = 9 cm

Volume of hemisphere

Height of cone = 72 cm
Let the radius of the cone be r cm.

Volume of the cone

The volumes of the hemisphere and cone are equal.
Therefore,

The radius of the base of the cone is 4.5 cm.

#### Question 14:

Radius of hemisphere = 9 cm

Volume of hemisphere

Height of cone = 72 cm
Let the radius of the cone be r cm.

Volume of the cone

The volumes of the hemisphere and cone are equal.
Therefore,

The radius of the base of the cone is 4.5 cm.

Diameter of the spherical ball= 21 cm

Volume of spherical ball
Volume of each cube

Number of cubes =

#### Question 15:

Diameter of the spherical ball= 21 cm

Volume of spherical ball
Volume of each cube

Number of cubes =

Radius of the sphere = R = 8 cm
Volume of the sphere

Radius of each new ball = r = 1 cm
Volume of each ball

Total number of new balls that can be made

#### Question 16:

Radius of the sphere = R = 8 cm
Volume of the sphere

Radius of each new ball = r = 1 cm
Volume of each ball

Total number of new balls that can be made

Radius of solid sphere = 3 cm
Volume of the sphere

Radius of each new ball = 0.3 cm
Volume of each new ball

Total number of balls $=\frac{\frac{4}{3}\mathrm{\pi }×3×3×3}{\frac{4}{3}\mathrm{\pi }×\frac{3}{10}×\frac{3}{10}×\frac{3}{10}}=1000$

#### Question 17:

Radius of solid sphere = 3 cm
Volume of the sphere

Radius of each new ball = 0.3 cm
Volume of each new ball

Total number of balls $=\frac{\frac{4}{3}\mathrm{\pi }×3×3×3}{\frac{4}{3}\mathrm{\pi }×\frac{3}{10}×\frac{3}{10}×\frac{3}{10}}=1000$

Diameter of sphere = 42 cm
Radius of sphere = 21 cm

Volume of sphere

Diameter of wire = 2.8 cm
Radius of wire = 1.4 cm
Let the length of the wire be cm.
Volume of the wire$={\mathrm{\pi r}}^{2}\mathrm{l}=\mathrm{\pi }×1.4×1.4×\mathrm{l}$

The volume of the sphere is equal to the volume of the wire.
â€‹Therefore,

So, the wire is 63 m long.

#### Question 18:

Diameter of sphere = 42 cm
Radius of sphere = 21 cm

Volume of sphere

Diameter of wire = 2.8 cm
Radius of wire = 1.4 cm
Let the length of the wire be cm.
Volume of the wire$={\mathrm{\pi r}}^{2}\mathrm{l}=\mathrm{\pi }×1.4×1.4×\mathrm{l}$

The volume of the sphere is equal to the volume of the wire.
â€‹Therefore,

So, the wire is 63 m long.

Diameter of sphere = 18 cm
Radius of the sphere = 9 cm

Volume of sphere

Length of wire = 108 m = 10800 cm
Let radius of the wire be r cm.
Volume of the wire

The volume of the sphere and the wire are the same.
Therefore,

The diameter of the wire is 0.6 cm.

#### Question 19:

Diameter of sphere = 18 cm
Radius of the sphere = 9 cm

Volume of sphere

Length of wire = 108 m = 10800 cm
Let radius of the wire be r cm.
Volume of the wire

The volume of the sphere and the wire are the same.
Therefore,

The diameter of the wire is 0.6 cm.

So, the height of the water in the cylindrical vessel is 13.5 cm.

#### Question 20:

So, the height of the water in the cylindrical vessel is 13.5 cm.

So, the time taken to empty half the tank is 16 minutes and 30 seconds.

#### Question 21:

So, the time taken to empty half the tank is 16 minutes and 30 seconds.

So, the height of the rainfall is 5 cm.

#### Question 22:

So, the height of the rainfall is 5 cm.

So, the height of the rainfall is 2 cm.

#### Question 23:

So, the height of the rainfall is 2 cm.

So, the volume of water left in the cylinder is 1.98 m3.

#### Question 24:

So, the volume of water left in the cylinder is 1.98 m3.

So, the rise in level of water in the tank in half an hour is 45 cm.

Disclaimer: The answer given in the textbook is incorrect. The same has been corrected above.

#### Question 25:

So, the rise in level of water in the tank in half an hour is 45 cm.

Disclaimer: The answer given in the textbook is incorrect. The same has been corrected above.

So, the time in which the level of water in the tank will rise by 7 cm is 1 hour.

#### Question 26:

So, the time in which the level of water in the tank will rise by 7 cm is 1 hour.

Width of the canal = 5.4 m
Depth of the canal = 1.8 m
Height of the standing water needed for irrigation = 10 cm = 0.1 m
Speed of the flowing water = 25 km/h = $\frac{25000}{60}=\frac{1250}{3}$ m/min
Volume of water flowing out of the canal in 1 min
= Area of opening of canal × $\frac{1250}{3}$
$5.4×1.8×\frac{1250}{3}$
= 4050 m3
∴ Volume of water flowing out of the canal in 40 min = 40 × 4050 m3 = 162000 m3
Now,

Thus, the area irrigated in 40 minutes is 162 hectare.

#### Question 27:

Width of the canal = 5.4 m
Depth of the canal = 1.8 m
Height of the standing water needed for irrigation = 10 cm = 0.1 m
Speed of the flowing water = 25 km/h = $\frac{25000}{60}=\frac{1250}{3}$ m/min
Volume of water flowing out of the canal in 1 min
= Area of opening of canal × $\frac{1250}{3}$
$5.4×1.8×\frac{1250}{3}$
= 4050 m3
∴ Volume of water flowing out of the canal in 40 min = 40 × 4050 m3 = 162000 m3
Now,

Thus, the area irrigated in 40 minutes is 162 hectare.

#### Question 28:

So, the rate of flow of water in the pipe is 3 km/hr.

#### Question 29:

So, the rate of flow of water in the pipe is 3 km/hr.

So, the rise in the level of water in the vessel is 5.6 cm.

Disclaimer: The diameter of the spherical marbles should be 1.4 cm instead 14 cm. The has been corrected above.

#### Question 30:

So, the rise in the level of water in the vessel is 5.6 cm.

Disclaimer: The diameter of the spherical marbles should be 1.4 cm instead 14 cm. The has been corrected above.

Diameter of each marble = 1.4 cm
Radius of each marble = 0.7 cm
Volume of each marble

The water rises as a cylindrical column.
Volume of cylindrical column filled with water

Total number of marbles

#### Question 31:

Diameter of each marble = 1.4 cm
Radius of each marble = 0.7 cm
Volume of each marble

The water rises as a cylindrical column.
Volume of cylindrical column filled with water

Total number of marbles So, the height of the embankment is $\frac{14}{3}$ m.

Value: We must lanour hard to make maximum use of the available resources.

Disclaimer: The answer provided in the textbook is incorrect. It has been corrected above.

#### Question 32: So, the height of the embankment is $\frac{14}{3}$ m.

Value: We must lanour hard to make maximum use of the available resources.

Disclaimer: The answer provided in the textbook is incorrect. It has been corrected above.

So, the rise in the level of the field is 2 m.

#### Question 33:

So, the rise in the level of the field is 2 m.

#### Question 34: #### Question 35: Diameter of cylinder (d) = 2 m
Radius of cylinder (r) = 1 m
Height of cylinder (H) = 5 m
Volume of cylindrical tank, Vc = $\mathrm{\pi }{r}^{2}H$ =

Length of the park (l) = 25 m
Breadth of park (b) = 20 m
height of standing water in the park = hâ€‹
Volume of water in the park = lbh = $25×20×h$

Now water from the tank is used to irrigate the park. So,
Volume of cylindrical tank = Volume of water in the park

Through recycling of water, better use of the natural resource occurs without wastage. It helps in reducing and preventing pollution.
It thus helps in conserving water. This keeps the greenery alive in urban areas like in parks gardens etc.

#### Question 1:

Diameter of cylinder (d) = 2 m
Radius of cylinder (r) = 1 m
Height of cylinder (H) = 5 m
Volume of cylindrical tank, Vc = $\mathrm{\pi }{r}^{2}H$ =

Length of the park (l) = 25 m
Breadth of park (b) = 20 m
height of standing water in the park = hâ€‹
Volume of water in the park = lbh = $25×20×h$

Now water from the tank is used to irrigate the park. So,
Volume of cylindrical tank = Volume of water in the park

Through recycling of water, better use of the natural resource occurs without wastage. It helps in reducing and preventing pollution.
It thus helps in conserving water. This keeps the greenery alive in urban areas like in parks gardens etc.

So, the capacity of the glass is 2170.67 cm3.

#### Question 2:

So, the capacity of the glass is 2170.67 cm3.

So, the total surface area of the solid frustum is 2411.52 cm2.

#### Question 3:

So, the total surface area of the solid frustum is 2411.52 cm2.

#### Question 4:

Hence, the cost of milk which can completely fill the container is â‚¹329.47.

#### Question 5:

Hence, the cost of milk which can completely fill the container is â‚¹329.47.

Let

(i)
Total metal sheet required to make the container = Curved surface area of frustum + Area of the base

The cost for 100 cm2 of sheet = Rs. 10.
The cost of 1 cm2 of sheet = .
The cost of 1961.14 cm2 of sheet =
DISCLAIMER: The answer calculated above is not matching with the answer provided in the textbook.

(ii)
The volume of frustum =

we know that 1 l=1000 cm3 or 1 cm3 = 0.001 l.
$⇒$
Volume=
Cost of milk is Rs. 35 per litre.
Hence, the cost at which this frustum can be filled =

#### Question 6:

Let

(i)
Total metal sheet required to make the container = Curved surface area of frustum + Area of the base

The cost for 100 cm2 of sheet = Rs. 10.
The cost of 1 cm2 of sheet = .
The cost of 1961.14 cm2 of sheet =
DISCLAIMER: The answer calculated above is not matching with the answer provided in the textbook.

(ii)
The volume of frustum =

we know that 1 l=1000 cm3 or 1 cm3 = 0.001 l.
$⇒$
Volume=
Cost of milk is Rs. 35 per litre.
Hence, the cost at which this frustum can be filled =

Greater radius = R = 33 cm
Smaller radius = r = 27 cm
Slant height = l = 10 cm

Using the formula for height of a frustum:

Height =  h =

Capacity of the frustum

Surface area of the frustum

#### Question 7:

Greater radius = R = 33 cm
Smaller radius = r = 27 cm
Slant height = l = 10 cm

Using the formula for height of a frustum:

Height =  h =

Capacity of the frustum

Surface area of the frustum

Greater diameter of the frustum  = 56 cm
Greater radius of the frustum = R = 28 cm
Smaller diameter of the frustum = 42 cm
Radius of the smaller end of the frustum = r = 21 cm
Height of the frustum = h = 15 cm

Capacity of the frustum

#### Question 8:

Greater diameter of the frustum  = 56 cm
Greater radius of the frustum = R = 28 cm
Smaller diameter of the frustum = 42 cm
Radius of the smaller end of the frustum = r = 21 cm
Height of the frustum = h = 15 cm

Capacity of the frustum

Greater radius of the frustum = R = 20 cm
Smaller radius of the frustum = r = 8 cm
Height of the frustum = h = 16 cm

Slant height, l, of the frustum

Surface area of the frustum

100 cm2 of metal sheet costs Rs 15.
So, total cost of the metal sheet used for fabrication

#### Question 9:

Greater radius of the frustum = R = 20 cm
Smaller radius of the frustum = r = 8 cm
Height of the frustum = h = 16 cm

Slant height, l, of the frustum

Surface area of the frustum

100 cm2 of metal sheet costs Rs 15.
So, total cost of the metal sheet used for fabrication

Greater diameter of the bucket = 30 cm
Radius of the bigger end of the bucket = R = 15 cm
Diameter of the smaller end of the bucket = 10 cm
Radius of the smaller end of the bucket = r = 5 cm
Height of the bucket = 24 cm

Slant height, l

Capacity of the frustum

A litre of milk cost Rs 20.
So, total cost of filling the bucket with milk

Surface area of the bucket

Cost of 100 cm2of metal sheet is Rs 10.
So, cost of metal used for making the bucket

#### Question 10:

Greater diameter of the bucket = 30 cm
Radius of the bigger end of the bucket = R = 15 cm
Diameter of the smaller end of the bucket = 10 cm
Radius of the smaller end of the bucket = r = 5 cm
Height of the bucket = 24 cm

Slant height, l

Capacity of the frustum

A litre of milk cost Rs 20.
So, total cost of filling the bucket with milk

Surface area of the bucket

Cost of 100 cm2of metal sheet is Rs 10.
So, cost of metal used for making the bucket

#### Question 11:

So, the height of the bucket is 15 cm.

#### Question 12:

So, the height of the bucket is 15 cm.

So, the value of r is 7 cm.

#### Question 13:

So, the value of r is 7 cm.

Greater radius = R = 33 cm
Smaller radius = r = 27 cm
Slant height = l = 10 cm

Surface area of the frustum

#### Question 14:

Greater radius = R = 33 cm
Smaller radius = r = 27 cm
Slant height = l = 10 cm

Surface area of the frustum

For the lower portion of the tent:
Diameter of the base= 20 m
Radius, R, of the base = 10 m
Diameter of the top end of the frustum = 6 m
Radius of the top end of the frustum = r = 3 m

Height of the frustum = h = 24 m

Slant height = l

For the conical part:
Radius of the cone's base = r = 3 m
Height of the cone = Total height - Height of the frustum = 28-24 = 4 m

Slant height, L, of the cone =

Total quantity of canvas = Curved surface area of the frustum + curved surface area of the conical top

#### Question 15:

For the lower portion of the tent:
Diameter of the base= 20 m
Radius, R, of the base = 10 m
Diameter of the top end of the frustum = 6 m
Radius of the top end of the frustum = r = 3 m

Height of the frustum = h = 24 m

Slant height = l

For the conical part:
Radius of the cone's base = r = 3 m
Height of the cone = Total height - Height of the frustum = 28-24 = 4 m

Slant height, L, of the cone =

Total quantity of canvas = Curved surface area of the frustum + curved surface area of the conical top

For the frustum:
Upper diameter = 14 m
Upper Radius, r = 7 m
Lower diameter = 26 m
Lower Radius, R = 13 m

Height of the frustum= h = 8 m
Slant height l =

For the conical part:
Radius of the base = r = 7 m
Slant height = L =12 m

Total surface area of the tent = Curved area of frustum + Curved area of the cone

#### Question 16:

For the frustum:
Upper diameter = 14 m
Upper Radius, r = 7 m
Lower diameter = 26 m
Lower Radius, R = 13 m

Height of the frustum= h = 8 m
Slant height l =

For the conical part:
Radius of the base = r = 7 m
Slant height = L =12 m

Total surface area of the tent = Curved area of frustum + Curved area of the cone

#### Question 17: So, the ratio of the volume of the two parts of the cone is 1 : 7.

#### Question 18: So, the ratio of the volume of the two parts of the cone is 1 : 7. So, the section is made at the height of 10 cm above the base.

#### Question 19: So, the section is made at the height of 10 cm above the base. So, the length of the wire is 4480 m.

#### Question 20: So, the length of the wire is 4480 m. So, the area of material used for making the fez is 710.28 cm2.

#### Question 21: So, the area of material used for making the fez is 710.28 cm2. So, the area of the tin sheet required to make the funnel is 782.57 cm2.

#### Question 22: So, the area of the tin sheet required to make the funnel is 782.57 cm2.

Let ABC be a right circular cone of height 3h and base radius r. This cone is cut by two planes such that AQ = QP = PO = h. Since $∆\mathrm{ABO}~∆\mathrm{AEP}$   (AA Similarity)

Also, $∆\mathrm{ABO}~∆\mathrm{AGQ}$     (AA Similarity)

Now,
Volume of cone AGF,

Voulme of the frustum GFDE,

Voulme of the frustum EDCB,

∴ Required ratio = ${V}_{1}:{V}_{2}:{V}_{3}=\frac{1}{27}\mathrm{\pi }{r}^{2}h:\frac{7}{27}\mathrm{\pi }{r}^{2}h:\frac{19}{27}\mathrm{\pi }{r}^{2}h=1:7:19$

#### Question 1:

Let ABC be a right circular cone of height 3h and base radius r. This cone is cut by two planes such that AQ = QP = PO = h. Since $∆\mathrm{ABO}~∆\mathrm{AEP}$   (AA Similarity)

Also, $∆\mathrm{ABO}~∆\mathrm{AGQ}$     (AA Similarity)

Now,
Volume of cone AGF,

Voulme of the frustum GFDE,

Voulme of the frustum EDCB,

∴ Required ratio = ${V}_{1}:{V}_{2}:{V}_{3}=\frac{1}{27}\mathrm{\pi }{r}^{2}h:\frac{7}{27}\mathrm{\pi }{r}^{2}h:\frac{19}{27}\mathrm{\pi }{r}^{2}h=1:7:19$

So, the amount of water that runs into the sea per minute is 3150 m3.

#### Question 2:

So, the amount of water that runs into the sea per minute is 3150 m3.

So, the surface area of the cube is 486 cm2.

#### Question 3:

So, the surface area of the cube is 486 cm2.

So, the number of cubes that can be put in the cubical box is 1000.

#### Question 4:

So, the number of cubes that can be put in the cubical box is 1000.

So, the edge of the new cube so formed is 12 cm.

#### Question 5:

So, the edge of the new cube so formed is 12 cm.

So, the volume of the resulting cuboid is 625 cm3.

#### Question 6:

So, the volume of the resulting cuboid is 625 cm3.

So, the ratio of the surface areas of the given cubes is 4 : 9.

#### Question 7:

So, the ratio of the surface areas of the given cubes is 4 : 9.

So, the height of the cylinder is 2 cm.

#### Question 8:

So, the height of the cylinder is 2 cm.

So, the radius of the base of the cylinder is 14 cm.

#### Question 9:

So, the radius of the base of the cylinder is 14 cm.

So, the ratio of the volumes of the given cylinders is 20 : 27.

#### Question 10:

So, the ratio of the volumes of the given cylinders is 20 : 27.

So, the length of the wire is 84 m.

#### Question 11:

So, the length of the wire is 84 m.

So, the slant height of the given cone is 91 cm.

#### Question 12:

So, the slant height of the given cone is 91 cm.

So, the radius of the base of the cone is 8 cm.

#### Question 13:

So, the radius of the base of the cone is 8 cm.

So, the number of cones that will be needed to store the water is 3.

#### Question 14:

So, the number of cones that will be needed to store the water is 3.

So, the curved surface area of the sphere is 1386 cm2.

#### Question 15:

So, the curved surface area of the sphere is 1386 cm2.

So, the volume of the sphere is 38808 cm3.

#### Question 16:

So, the volume of the sphere is 38808 cm3.

So, the ratio of the volumes of the given spheres is 8 : 125.

#### Question 17:

So, the ratio of the volumes of the given spheres is 8 : 125.

So, the number of spherical balls obtained is 64.

#### Question 18:

So, the number of spherical balls obtained is 64.

So, the number of lead shots that can be made from the cuboid is 84000.

#### Question 19:

So, the number of lead shots that can be made from the cuboid is 84000.

So, the number of spheres so formed is 108.

#### Question 20:

So, the number of spheres so formed is 108.

So, the radius of the base of the cone is 2.4 cm.

#### Question 21:

So, the radius of the base of the cone is 2.4 cm.

So, the length of the wire is 243 m.

#### Question 22:

So, the length of the wire is 243 m.

So, the slant height of the frustum is 10 cm.

#### Question 23:

So, the slant height of the frustum is 10 cm.

So, the ratio of the volume of the cube to that of the sphere is 6 : $\mathrm{\pi }$.

#### Question 24:

So, the ratio of the volume of the cube to that of the sphere is 6 : $\mathrm{\pi }$.

So, the ratio of the volumes of the cylinder, the cone and the sphere is 3 : 1 : 2.

#### Question 25:

So, the ratio of the volumes of the cylinder, the cone and the sphere is 3 : 1 : 2.

So, the surface area of the resulting cuboid is 250 cm2.

#### Question 26:

So, the surface area of the resulting cuboid is 250 cm2.

So, the edge of the new cube so formed is 6 cm.

#### Question 27:

So, the edge of the new cube so formed is 6 cm.

So, the width of the wire is $\frac{8}{15}$ cm.

#### Question 28:

So, the width of the wire is $\frac{8}{15}$ cm.

So, the cost of the cloth used for making the tent is â‚¹2750.

#### Question 29:

So, the cost of the cloth used for making the tent is â‚¹2750. So, the volume of wood in the toy is $\frac{616}{3}$ cm3 or 205.33 cm3.

#### Question 30: So, the volume of wood in the toy is $\frac{616}{3}$ cm3 or 205.33 cm3.

Hence, the edges of the given three metallic cubes are 6 cm, 8 cm and 10 cm.

#### Question 31:

Hence, the edges of the given three metallic cubes are 6 cm, 8 cm and 10 cm.

So, the height of the cone is 14 cm.

#### Question 32:

So, the height of the cone is 14 cm.

So, the cost of the milk which the bucket can hold is â‚¹702.24.

#### Question 33:

So, the cost of the milk which the bucket can hold is â‚¹702.24. So, the outer surface area of the building is 75.9 m2.

#### Question 34: So, the outer surface area of the building is 75.9 m2.

So, the diameter of the solid sphere is 42 cm.

#### Question 35:

So, the diameter of the solid sphere is 42 cm. So, the total surface area of the toy is 214.5 cm2.

Disclaimer: The answer given in the textbook is incorrect. The same has been corrected above.

#### Question 36: So, the total surface area of the toy is 214.5 cm2.

Disclaimer: The answer given in the textbook is incorrect. The same has been corrected above.

Disclaimer: The answer of the total surface area given in the textbook is incorrect. The same has been corrected above.

#### Question 37:

Disclaimer: The answer of the total surface area given in the textbook is incorrect. The same has been corrected above.

So, the height of the bucket is 15 cm.

#### Question 38:

So, the height of the bucket is 15 cm.

Hence, the cost of the metal sheet used for making the milk container is â‚¹2745.60.

Disclaimer: The answer given in the textbook is incorrect. The same has been corrected above.

#### Question 39:

Hence, the cost of the metal sheet used for making the milk container is â‚¹2745.60.

Disclaimer: The answer given in the textbook is incorrect. The same has been corrected above.

So, the number of cones so formed is 672.

Disclaimer: The answer given in the textbook is incorrect. The same has been corrected above.

#### Question 40:

So, the number of cones so formed is 672.

Disclaimer: The answer given in the textbook is incorrect. The same has been corrected above.

#### Question 1:

(a) a cylinder and a cone
A cylindrical pencil sharpened at one end is a combination of a cylinder and a cone.

#### Question 2:

(a) a cylinder and a cone
A cylindrical pencil sharpened at one end is a combination of a cylinder and a cone.

(b) frustum of a cone and a hemisphere
A shuttlecock used for playing badminton is a combination of frustum of a cone and a hemisphere.

#### Question 3:

(b) frustum of a cone and a hemisphere
A shuttlecock used for playing badminton is a combination of frustum of a cone and a hemisphere.

(c) a cylinder and frustum of a cone
A funnel is a combination of a cylinder and frustum of a cone.

#### Question 4:

(c) a cylinder and frustum of a cone
A funnel is a combination of a cylinder and frustum of a cone.

(a) a sphere and a cylinder
A surahi is a combination of a sphere and a cylinder.

#### Question 5:

(a) a sphere and a cylinder
A surahi is a combination of a sphere and a cylinder.

(b) frustum of a cone
The shape of a glass (tumbler) is usually in the form of frustum of a cone.

#### Question 6:

(b) frustum of a cone
The shape of a glass (tumbler) is usually in the form of frustum of a cone.

(c) two cones and a cylinder
The shape of the gilli used in a gilli-danda game is a combination of two cones and a cylinder.

#### Question 7:

(c) two cones and a cylinder
The shape of the gilli used in a gilli-danda game is a combination of two cones and a cylinder.

(a) a hemisphere and a cone
A plumbline (sahul) is a combination of a hemisphere and a cone.

#### Question 8:

(a) a hemisphere and a cone
A plumbline (sahul) is a combination of a hemisphere and a cone.

(d) frustum of a cone
A cone is cut by a plane parallel to its base and the upper part is removed. The part that is left is called frustum of a cone.

#### Question 9:

(d) frustum of a cone
A cone is cut by a plane parallel to its base and the upper part is removed. The part that is left is called frustum of a cone.

(c) remain unaltered
During conversion of a solid from one shape to another, the volume of the new shape will remain unaltered.

#### Question 10:

(c) remain unaltered
During conversion of a solid from one shape to another, the volume of the new shape will remain unaltered.

(c) circle
In a right circular cone, the cross-section made by a plane parallel to the base is a circle.

#### Question 11:

(c) circle
In a right circular cone, the cross-section made by a plane parallel to the base is a circle.

(b) 21 cm
Volume of the cuboid $=\left(l×b×h\right)$ =
Let the radius of the sphere be r cm.
Volume of the sphere$=\frac{4}{3}\mathrm{\pi }{r}^{3}$

The volume of the sphere and the cuboid are the same.
Therefore,

Hence, the radius of the sphere is 21 cm.

#### Question 12:

(b) 21 cm
Volume of the cuboid $=\left(l×b×h\right)$ =
Let the radius of the sphere be r cm.
Volume of the sphere$=\frac{4}{3}\mathrm{\pi }{r}^{3}$

The volume of the sphere and the cuboid are the same.
Therefore,

Hence, the radius of the sphere is 21 cm.

Since, the diameter of the base of the largest cone that can be cut out from the cube = Edge of the cube = 4.2 cm

So, the radius of the base of the largest cone = $\frac{4.2}{2}$ = 2.1 cm

Hence, the correct answer is option (a).

#### Question 13:

Since, the diameter of the base of the largest cone that can be cut out from the cube = Edge of the cube = 4.2 cm

So, the radius of the base of the largest cone = $\frac{4.2}{2}$ = 2.1 cm

Hence, the correct answer is option (a).

Hence, the correct answer is option (a).

#### Question 14:

Hence, the correct answer is option (a).

Hence, the correct answer is option (a).

#### Question 15:

Hence, the correct answer is option (a).

Hence, the correct answer is option (c).

#### Question 16:

Hence, the correct answer is option (c).

Hence, the correct answer is option (a).

#### Question 17:

Hence, the correct answer is option (a).

Hence, the correct answer is option (b).

#### Question 18:

Hence, the correct answer is option (b).

Hence, the correct answer is option (d).

#### Question 19:

Hence, the correct answer is option (d).

Hence, the correct answer is option (a).

#### Question 20:

Hence, the correct answer is option (a).

Hence, the correct answer is option (a).

#### Question 21:

Hence, the correct answer is option (a). Hence, the correct answer is option (d).

#### Question 22: Hence, the correct answer is option (d).

Hence, the correct answer is option (a).

#### Question 23:

Hence, the correct answer is option (a).

Hence, the correct answer is option (a).

#### Question 24:

Hence, the correct answer is option (a).

Hence, the correct answer is option (d).

#### Question 25:

Hence, the correct answer is option (d).

(c) 363
The edge of the cubical ice-cream brick = a = 22 cm
Volume of the cubical ice-cream brick$={\left(a\right)}^{3}$

Radius of an ice-cream cone = 2 cm
Height of an ice-cream cone = 7 cm
Volume of each ice-cream cone$=\frac{1}{3}\mathrm{\pi }{r}^{2}h$

Number of ice-cream cones

$=\frac{22×22×22×3×7}{22×2×2×7}\phantom{\rule{0ex}{0ex}}=363$
Hence, the number of ice-cream cones is 363.

#### Question 26:

(c) 363
The edge of the cubical ice-cream brick = a = 22 cm
Volume of the cubical ice-cream brick$={\left(a\right)}^{3}$

Radius of an ice-cream cone = 2 cm
Height of an ice-cream cone = 7 cm
Volume of each ice-cream cone$=\frac{1}{3}\mathrm{\pi }{r}^{2}h$

Number of ice-cream cones

$=\frac{22×22×22×3×7}{22×2×2×7}\phantom{\rule{0ex}{0ex}}=363$
Hence, the number of ice-cream cones is 363.

(c) 11200
Volume of wall =
$\frac{1}{8}\mathrm{th}$ of the wall is covered with mortar.
So,
Volume of the wall filled with bricks
Volume of each brick

Number of bricks used to construct the wall

$=\frac{7×270×300×350×32}{8×9×225×35}$
= 11200

Hence, the number of bricks used to construct the wall is 11200.

#### Question 27:

(c) 11200
Volume of wall =
$\frac{1}{8}\mathrm{th}$ of the wall is covered with mortar.
So,
Volume of the wall filled with bricks
Volume of each brick

Number of bricks used to construct the wall

$=\frac{7×270×300×350×32}{8×9×225×35}$
= 11200

Hence, the number of bricks used to construct the wall is 11200.

(a) 2 cm
Let the diameter of each sphere be d cm.
Let r and be the radii of the sphere and the cylinder, respectively,
and h be the height of the cylinder.

Therefore,

Hence, the diameter of each sphere is 2 cm.

#### Question 28:

(a) 2 cm
Let the diameter of each sphere be d cm.
Let r and be the radii of the sphere and the cylinder, respectively,
and h be the height of the cylinder.

Therefore,

Hence, the diameter of each sphere is 2 cm.

(b) 32.7 litres
Let R and r be the radii of the top and base of the bucket, respectively, and let h be its height.

Then,

Capacity of the bucket = Volume of the frustum of the cone

Hence, the capacity of the bucket is 32.7 litres.

#### Question 29:

(b) 32.7 litres
Let R and r be the radii of the top and base of the bucket, respectively, and let h be its height.

Then,

Capacity of the bucket = Volume of the frustum of the cone

Hence, the capacity of the bucket is 32.7 litres.

(d) 4950 cm2
Let the radii of the top and bottom of the bucket be R and r and let its slant height be l.
Then,
Curved surface area of the bucket$=\mathrm{\pi }l\left(R+r\right)$

Hence, the curved surface area of the bucket is 4950 cm2.

#### Question 30:

(d) 4950 cm2
Let the radii of the top and bottom of the bucket be R and r and let its slant height be l.
Then,
Curved surface area of the bucket$=\mathrm{\pi }l\left(R+r\right)$

Hence, the curved surface area of the bucket is 4950 cm2.

(b) 16 : 9
Let the radii of the spheres be R and r.
Then, ratio of their volumes
$=\frac{\frac{4}{3}\mathrm{\pi }{R}^{3}}{\frac{4}{3}{\mathrm{\pi r}}^{3}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$

Therefore,

$⇒{\left(\frac{R}{r}\right)}^{3}={\left(\frac{4}{3}\right)}^{3}\phantom{\rule{0ex}{0ex}}⇒\frac{R}{r}=\frac{4}{3}$
Hence, the ratio of their surface areas$=\frac{4\mathrm{\pi }{R}^{2}}{4{\mathrm{\pi r}}^{2}}$

$=\frac{{R}^{2}}{{r}^{2}}\phantom{\rule{0ex}{0ex}}={\left(\frac{R}{r}\right)}^{2}\phantom{\rule{0ex}{0ex}}={\left(\frac{4}{3}\right)}^{2}\phantom{\rule{0ex}{0ex}}=\frac{16}{9}\phantom{\rule{0ex}{0ex}}=16:9$

#### Question 31:

(b) 16 : 9
Let the radii of the spheres be R and r.
Then, ratio of their volumes
$=\frac{\frac{4}{3}\mathrm{\pi }{R}^{3}}{\frac{4}{3}{\mathrm{\pi r}}^{3}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$

Therefore,

$⇒{\left(\frac{R}{r}\right)}^{3}={\left(\frac{4}{3}\right)}^{3}\phantom{\rule{0ex}{0ex}}⇒\frac{R}{r}=\frac{4}{3}$
Hence, the ratio of their surface areas$=\frac{4\mathrm{\pi }{R}^{2}}{4{\mathrm{\pi r}}^{2}}$

$=\frac{{R}^{2}}{{r}^{2}}\phantom{\rule{0ex}{0ex}}={\left(\frac{R}{r}\right)}^{2}\phantom{\rule{0ex}{0ex}}={\left(\frac{4}{3}\right)}^{2}\phantom{\rule{0ex}{0ex}}=\frac{16}{9}\phantom{\rule{0ex}{0ex}}=16:9$

(a) 142296
Since $\frac{1}{8}\mathrm{th}$ of the cube remains unfulfilled,
volume of the cube =

Space filled in the cube

Volume of each marble$=\frac{4}{3}\mathrm{\pi }{r}^{3}$

Therefore, number of marbles required$=\left(\frac{7×1331×24×7}{11}\right)$
= 142296

#### Question 32:

(a) 142296
Since $\frac{1}{8}\mathrm{th}$ of the cube remains unfulfilled,
volume of the cube =

Space filled in the cube

Volume of each marble$=\frac{4}{3}\mathrm{\pi }{r}^{3}$

Therefore, number of marbles required$=\left(\frac{7×1331×24×7}{11}\right)$
= 142296

(b) 14 cm
Let the internal and external radii of the spherical shell be r and R, respectively.
Volume of the spherical shell$=\frac{4}{3}\mathrm{\pi }\left({R}^{3}-{r}^{3}\right)$

Volume of the cone$=\frac{1}{3}\mathrm{\pi }{r}^{2}h$

Therefore,

#### Question 33:

(b) 14 cm
Let the internal and external radii of the spherical shell be r and R, respectively.
Volume of the spherical shell$=\frac{4}{3}\mathrm{\pi }\left({R}^{3}-{r}^{3}\right)$

Volume of the cone$=\frac{1}{3}\mathrm{\pi }{r}^{2}h$

Therefore,

(d) 0.36 cm3

= 0.25 cm

Let the length of the cylindrical part of the capsule be x cm.

Then,

$0.25+x+0.25=2\phantom{\rule{0ex}{0ex}}⇒0.5+x=2\phantom{\rule{0ex}{0ex}}⇒x=1.5$

Hence, the capacity of the capsule
$=\left(2×\frac{2}{3}\mathrm{\pi }r\right)$3+πr2h

3
3= 0.36 cm3

#### Question 34:

(d) 0.36 cm3

= 0.25 cm

Let the length of the cylindrical part of the capsule be x cm.

Then,

$0.25+x+0.25=2\phantom{\rule{0ex}{0ex}}⇒0.5+x=2\phantom{\rule{0ex}{0ex}}⇒x=1.5$

Hence, the capacity of the capsule
$=\left(2×\frac{2}{3}\mathrm{\pi }r\right)$3+πr2h

3