Rs Aggarwal 2020 2021 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 2020 2021 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 2020 2021 Solutions. All Rs Aggarwal 2020 2021 Solutions for class Class 10 Maths are prepared by experts and are 100% accurate.

= 718.67 cm3

#### Page No 786:

(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 .

#### Page No 787:

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

#### Page No 787:

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.

#### Page No 787:

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

#### Page No 787:

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.

#### Page No 787:

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.

#### Page No 787:

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

#### Page No 787:

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

#### Page No 787:

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

#### Page No 787:

So, the volume of the solid is 166.83 cm3.

#### Page No 787:

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

#### Page No 787:

So, the radius of the ice-cream cone is 3 cm.

#### Page No 787:

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

#### Page No 788:

So, the surface area of the medicine capsule is 220 mm2.

#### Page No 788:

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)

#### Page No 788:

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 =

#### Page No 788:

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

#### Page No 788:

So, the total surface area of the remaining solid is 73.92 cm2.

#### Page No 788:

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

#### Page No 788:

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.

#### Page No 788:

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

#### Page No 788:

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

#### Page No 789:

(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 ₹33.92.

#### Page No 789:

So, the surface area of the toy is 770 cm2.

#### Page No 809:

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

#### Page No 809:

So, the diameter of the sphere is 10 cm.

#### Page No 809:

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

#### Page No 809:

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

#### Page No 809:

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

#### Page No 810:

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,

#### Page No 810:

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

#### Page No 810:

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

#### Page No 810:

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

#### Page No 810:

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

#### Page No 810:

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,

#### Page No 810:

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.

#### Page No 810:

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.

#### Page No 810:

Diameter of the spherical ball= 21 cm

Volume of spherical ball
Volume of each cube

Number of cubes =

#### Page No 810:

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

#### Page No 810:

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 $=\frac{\frac{4}{3}\mathrm{\pi }×3×3×3}{\frac{4}{3}\mathrm{\pi }×\frac{3}{10}×\frac{3}{10}×\frac{3}{10}}=1000$

#### Page No 810:

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.
​Therefore,

So, the wire is 63 m long.

#### Page No 810:

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.

#### Page No 811:

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

#### Page No 811:

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

#### Page No 811:

So, the height of the rainfall is 5 cm.

#### Page No 811:

So, the height of the rainfall is 2 cm.

#### Page No 811:

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

#### Page No 811:

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.

#### Page No 811:

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

#### Page No 811:

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.

#### Page No 811:

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

#### Page No 811:

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.

#### Page No 812:

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

#### Page No 812:

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.

#### Page No 812:

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

#### Page No 812:

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 = h​
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.

#### Page No 822:

So, the capacity of the glass is 2170.67 cm3.

#### Page No 822:

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

#### Page No 823:

Hence, the cost of milk which can completely fill the container is ₹329.47.

#### Page No 823:

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 =

#### Page No 823:

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

#### Page No 823:

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

#### Page No 823:

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

#### Page No 823:

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 cm2 of metal sheet is Rs 10.
So, cost of metal used for making the bucket

#### Page No 823:

So, the height of the bucket is 15 cm.

#### Page No 823:

So, the value of r is 7 cm.

#### Page No 823:

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

Surface area of the frustum

#### Page No 824:

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

#### Page No 824:

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

#### Page No 824:

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

#### Page No 824:

So, the section is made at the height of 10 cm above the base.

#### Page No 824:

So, the length of the wire is 4480 m.

#### Page No 824:

So, the area of material used for making the fez is 710.28 cm2.

#### Page No 824:

So, the area of the tin sheet required to make the funnel is 782.57 cm2.

#### Page No 825:

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

#### Page No 826:

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

#### Page No 826:

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

#### Page No 826:

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

#### Page No 826:

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

#### Page No 826:

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

#### Page No 826:

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

#### Page No 826:

So, the height of the cylinder is 2 cm.

#### Page No 826:

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

#### Page No 826:

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

#### Page No 826:

So, the length of the wire is 84 m.

#### Page No 827:

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

#### Page No 827:

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

#### Page No 827:

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

#### Page No 827:

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

#### Page No 827:

So, the volume of the sphere is 38808 cm3.

#### Page No 827:

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

#### Page No 827:

So, the number of spherical balls obtained is 64.

#### Page No 827:

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

#### Page No 827:

So, the number of spheres so formed is 108.

#### Page No 827:

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

#### Page No 827:

So, the length of the wire is 243 m.

#### Page No 827:

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

#### Page No 827:

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

#### Page No 827:

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

#### Page No 827:

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

#### Page No 827:

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

#### Page No 827:

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

#### Page No 828:

So, the cost of the cloth used for making the tent is ₹2750.

#### Page No 828:

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

#### Page No 828:

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

#### Page No 828:

So, the height of the cone is 14 cm.

#### Page No 828:

So, the cost of the milk which the bucket can hold is ₹702.24.

#### Page No 828:

So, the outer surface area of the building is 75.9 m2.

#### Page No 828:

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

#### Page No 828:

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.

#### Page No 828:

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

#### Page No 828:

So, the height of the bucket is 15 cm.

#### Page No 828:

Hence, the cost of the metal sheet used for making the milk container is ₹2745.60.

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

#### Page No 829:

So, the number of cones so formed is 672.

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

#### Page No 829:

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

#### Page No 829:

(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.

#### Page No 830:

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

#### Page No 830:

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

#### Page No 830:

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

#### Page No 830:

(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.

#### Page No 830:

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

#### Page No 830:

(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.

#### Page No 830:

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

#### Page No 831:

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

#### Page No 831:

(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.

#### Page No 831:

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).

#### Page No 831:

Hence, the correct answer is option (a).

#### Page No 831:

Hence, the correct answer is option (a).

#### Page No 831:

Hence, the correct answer is option (c).

#### Page No 831:

Hence, the correct answer is option (a).

#### Page No 831:

Hence, the correct answer is option (b).

#### Page No 831:

Hence, the correct answer is option (d).

#### Page No 831:

Hence, the correct answer is option (a).

#### Page No 831:

Hence, the correct answer is option (a).

#### Page No 832:

Hence, the correct answer is option (d).

#### Page No 832:

Hence, the correct answer is option (a).

#### Page No 832:

Hence, the correct answer is option (a).

#### Page No 832:

Hence, the correct answer is option (d).

#### Page No 832:

(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.

#### Page No 832:

(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.

#### Page No 832:

(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.

#### Page No 832:

(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.

#### Page No 832:

(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.

#### Page No 833:

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

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

#### Page No 833:

(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

#### Page No 833:

(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,

#### Page No 833:

(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

#### Page No 833:

(d) 17 m
Length of the longest pole that can be kept in a room =Length of the diagonal of the room

#### Page No 833:

(b) 216 cm2

Let the edge of the cube be a cm.
Then, length of the diagonal = $\sqrt{3}a$
Or,

Therefore, the total surface area of the cube = 6a2

#### Page No 833:

(b) 1176 cm2
Let the edge of the cube be a cm.
Then, volume of the cube = a3
Or,

Therefore, surface area of the cube$=6{a}^{2}$

#### Page No 833:

(c) 1728 cm3
Let the edge of the cube be a.
Total surface area of the cube = 6a2
Therefore,

Therefore, volume of the cube = a3

#### Page No 833:

(b) 6400

Volume of the wall
Volume of each brick =

Number of bricks

$=\frac{800×600×22.5}{25×11.25×6}\phantom{\rule{0ex}{0ex}}=6400$

#### Page No 833:

(b) 4 m
Area of the base of a rectangular tank

Let the depth of the water be d metres.
Then,

#### Page No 833:

Note : It should be 128 m3 instead of 12.8 m3

(b) 40 cm
Let the breadth of the wall be x cm.
Then, its height = 5x cm
and its length
= 40x cm
Hence, the volume of the wall
It is given that the volume of the wall = 128 m3.
Therefore,

#### Page No 834:

(c) $\sqrt{xyz}$
Let the length of the cuboid = l
breadth of the cuboid = b
and height of the cuboid = h
Since, the areas of the three adjacent faces are x, y and z, we have:

$lb=x\phantom{\rule{0ex}{0ex}}bh=y\phantom{\rule{0ex}{0ex}}lh=z$

Therefore,

Hence, the volume of the cuboid .

#### Page No 834:

(c) 236 cm2
Let l, b and h be the length, breadth and height of the cuboid.
Then,

Therefore,

Hence, the surface area of the cuboid is .

#### Page No 834:

(d) 125%
Let the original edge of the cube be a units.
Then, the original surface area of the cube = 6a2 units

New edge of the cube = 150% of a
$=\frac{150a}{100}\phantom{\rule{0ex}{0ex}}=\frac{3a}{2}$
Hence, new surface area$=6×{\left(\frac{3a}{2}\right)}^{2}$
$=\frac{27{a}^{2}}{2}$
Increase in area$=\left(\frac{27{a}^{2}}{2}-6{a}^{2}\right)$
$=\frac{15{a}^{2}}{2}$
% increase in surface area$=\left(\frac{15{a}^{2}}{2}×\frac{1}{6{a}^{2}}×100\right)%$
= 125 %

#### Page No 834:

(d) 225
Volume of the cuboidal granary
Volume of each bag

Number of bags that can be stored in the cuboidal granary
$=\left(\frac{8×6×3}{0.64}\right)$
= 225

#### Page No 834:

(d) 27
Volume of the given cube
Volume of each small cube

Number of cubes formed
$=\left(\frac{6×6×6}{2×2×2}\right)$
= 27

#### Page No 834:

(c) 1000 m3
Volume of water that falls on 2 hectares of land

#### Page No 834:

(c) 1 : 9
Let the edges of the two cubes be a and b.
Then, ratio of their volumes$=\frac{{a}^{3}}{{b}^{3}}$
Therefore,

The ratio of their surface areas$=\frac{6{a}^{2}}{6{b}^{2}}$
Therefore,

Hence, the ratio of their surface areas is 1:9.

#### Page No 834:

(a) 176 cm3
Volume of the cylinder$=\mathrm{\pi }{r}^{2}h$

#### Page No 834:

(b) 2992 cm2
The total surface area of the cylinder$=2\mathrm{\pi }r+2{r}^{2}$

#### Page No 834:

(b) 396 cm3

Curved surface area of the cylinder $=2\mathrm{\pi }rh\phantom{\rule{0ex}{0ex}}$$=2×\frac{22}{7}×r×14$

Therefore,

Hence, the volume of the cylinder$=\mathrm{\pi }{r}^{2}h$

#### Page No 834:

(c) 20 cm
Curved surface area of the cylinder$=2\mathrm{\pi }rh$
$=2×\frac{22}{7}×14×h$
Therefore,

Hence, the height of the cylinder is 20 cm.

#### Page No 834:

(d) 5 : 1
Ratio of the total surface area to the lateral surface area
$=\frac{2\mathrm{\pi }r\left(h+r\right)}{2\mathrm{\pi }rh}\phantom{\rule{0ex}{0ex}}=\frac{h+r}{h}\phantom{\rule{0ex}{0ex}}=\frac{\left(20+80\right)}{20}\phantom{\rule{0ex}{0ex}}=\frac{100}{20}\phantom{\rule{0ex}{0ex}}=\frac{5}{1}\phantom{\rule{0ex}{0ex}}=5:1$
Hence, the required ratio is 5:1.

#### Page No 835:

(c) 6 m
The curved surface area of a cylindrical pillar
$=2\mathrm{\pi }rh\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$
Therefore, $2\mathrm{\pi }rh=264$
Volume of a cylinder$=\mathrm{\pi }{r}^{2}h$
Therefore, $\mathrm{\pi }{r}^{2}h=924$

Hence,

Therefore,

Hence, the height of the pillar is 6 m.

#### Page No 835:

(d) 770 cm2
Let the common multiple be x.
Let the radius of the cylinder be 2x cm and its height be 3x cm.
Then, volume of the cylinder$=\mathrm{\pi }{r}^{2}h$
$=\frac{22}{7}×{\left(2x\right)}^{2}×3x$
Therefore,

$⇒{x}^{3}=\left(\frac{7}{2}×\frac{7}{2}×\frac{7}{2}\right)\phantom{\rule{0ex}{0ex}}⇒{x}^{3}={\left(\frac{7}{2}\right)}^{3}\phantom{\rule{0ex}{0ex}}⇒x=\frac{7}{2}$
Now,
Hence, the total surface area of the cylinder$=\left(2\mathrm{\pi }rh+2\mathrm{\pi }{r}^{2}\right)\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$

#### Page No 835:

(b) 20 : 27
Let the radii of the two cylinders be 2r and 3r and their heights be 5h and 3h, respectively.

Then, ratio of their volumes$=\frac{\mathrm{\pi }×{\left(2\mathrm{r}\right)}^{2}×5h}{\mathrm{\pi }×{\left(3r\right)}^{2}×3h}$
$=\frac{4{r}^{2}×5}{9{r}^{2}×3}\phantom{\rule{0ex}{0ex}}=\frac{20}{27}\phantom{\rule{0ex}{0ex}}=20:27$

#### Page No 835:

(b) $\sqrt{2}:1$
Let the radii of the two cylinders be r and R and their heights be h and 2h, respectively.
Since the volumes of the cylinders are equal, therefore:

$\mathrm{\pi }×{r}^{2}×h=\mathrm{\pi }×{R}^{2}×2h$

Hence, the ratio of their radii is $\sqrt{2}:1$.

#### Page No 835:

(b) 65π cm2
Given: r = 5 cm, h = 12 cm

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

Hence, the curved surface area of the cone$=\mathrm{\pi }rl$

#### Page No 835:

(a) 28 cm
Let h be the height of the cone.
Diameter of the cone = 42 cm
Radius of the cone = 21 cm

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

Therefore,

Hence, the height of the cone is 28 cm.

#### Page No 835:

(a) $154\sqrt{5}{\mathrm{cm}}^{2}$
Area of the base of the of a right circular cone$=\mathrm{\pi }{r}^{2}$
Therefore,

Now, r = 7 cm and h = 14 cm
Then, slant height of the cone, $l=\sqrt{{r}^{2}+{h}^{2}}$

Hence, the curved surface area of the cone$=\mathrm{\pi }rl$

#### Page No 835:

(d) 72.8%
Let the original radius of the cone be r and height be h.
Then, original volume$=\frac{1}{3}\mathrm{\pi }{r}^{2}h$

Let $\frac{1}{3}\mathrm{\pi }{r}^{2}h=V$
New radius = 120% of r
$=\frac{120r}{100}\phantom{\rule{0ex}{0ex}}=\frac{6r}{5}$

New height = 120% of h
$=\frac{120h}{100}\phantom{\rule{0ex}{0ex}}=\frac{6h}{5}$

Hence, the new volume = $\frac{1}{3}\mathrm{\pi }×{\left(\frac{6\mathrm{r}}{5}\right)}^{2}×\frac{6\mathrm{h}}{5}$
$=\frac{216}{125}\left(\frac{1}{3}\mathrm{\pi }{r}^{2}h\right)\phantom{\rule{0ex}{0ex}}=\frac{216}{125}\mathrm{V}$

Increase in volume$=\left(\frac{216}{125}\mathrm{V}-\mathrm{V}\right)$

$=\frac{91\mathrm{V}}{125}$

Increase in % of the volume$=\left(\frac{91\mathrm{V}}{125}×\frac{1}{\mathrm{V}}×100\right)%$

= 72.8%

#### Page No 835:

(a) 9 : 8
Let the radii of the base of the cylinder and cone be 3r and 4r and their heights be 2h and 3h, respectively.
Then, ratio of their volumes$=\frac{\mathrm{\pi }{\left(3r\right)}^{2}×\left(2h\right)}{\frac{1}{3}\mathrm{\pi }{\left(4r\right)}^{2}×\left(3h\right)}$
$=\frac{9{r}^{2}×2×3}{16{r}^{2}×3}\phantom{\rule{0ex}{0ex}}=\frac{9}{8}\phantom{\rule{0ex}{0ex}}=9:8$

#### Page No 835:

(d) 8 cm
Radius of the cylinder = 8 cm
Height of the cylinder = 2 cm
Height of the cone = 6 cm

Volume of the cylinder  = Volume of the cone
Therefore,

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

#### Page No 835:

(b) 525 m
Area of the floor of a conical tent$=\mathrm{\pi }{r}^{2}$
Therefore,

Height of the cone = 14 m

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

Area of the canvas = Curved surface area of the conical tent

Length of the canvas

#### Page No 836:

(c) $1437\frac{1}{3}{\mathrm{cm}}^{3}$
Volume of the sphere$=\frac{4}{3}\mathrm{\pi }{r}^{3}$

#### Page No 836:

(d) 4 : 9
Let the radii of the spheres be R and r, respectively.
Then, ratio of their volumes$=\frac{\frac{4}{3}\mathrm{\pi }{R}^{3}}{\frac{4}{3}{\mathrm{\pi r}}^{3}}$
Therefore,
$\frac{\frac{4}{3}\mathrm{\pi }{R}^{3}}{\frac{4}{3}{\mathrm{\pi r}}^{3}}=\frac{8}{27}\phantom{\rule{0ex}{0ex}}⇒\frac{{R}^{3}}{{r}^{3}}=\frac{8}{27}\phantom{\rule{0ex}{0ex}}⇒{\left(\frac{R}{r}\right)}^{3}={\left(\frac{2}{3}\right)}^{3}\phantom{\rule{0ex}{0ex}}⇒\frac{R}{r}=\frac{2}{3}$
Hence, the ratio between their surface areas$=\frac{4\mathrm{\pi }{R}^{2}}{4\mathrm{\pi }{r}^{2}}$
$=\frac{{R}^{2}}{{r}^{2}}\phantom{\rule{0ex}{0ex}}={\left(\frac{R}{r}\right)}^{2}\phantom{\rule{0ex}{0ex}}={\left(\frac{2}{3}\right)}^{2}\phantom{\rule{0ex}{0ex}}=\frac{4}{9}\phantom{\rule{0ex}{0ex}}=4:9$

#### Page No 836:

DISCLAIMER : The answer to the question does not match the options given.

External diameter = 8 cm
Internal diameter = 4 cm
Let the external and internal radii of the hollow metallic sphere be R and r, respectively.
Then,
Then, volume of the hollow sphere:
$\frac{4}{3}\mathrm{\pi }\left[{\left(R\right)}^{3}-{\left(r\right)}^{3}\right]\phantom{\rule{0ex}{0ex}}$

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

Therefore,
Volume of the hollow sphere =  Volume of the cone formed

#### Page No 836:

(a) 2.1 cm
Radius of cone = 2.1 cm
Height of cone = 8.4 cm

Volume of cone =

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

Volume of cone = Volume of sphere

Hence, the radius of the sphere is 2.1 cm.

#### Page No 836:

(a) 4158 cm2
Volume of hemisphere$=\frac{2}{3}\mathrm{\pi }{r}^{3}$
Therefore,

Hence, the total surface area of the hemisphere$=3\mathrm{\pi }{r}^{2}$

#### Page No 836:

(a) $179\frac{2}{3}{\mathrm{cm}}^{3}$
Surface area of a sphere$=4\mathrm{\pi }{r}^{2}$
Therefore,

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

#### Page No 836:

(c) (147π) cm2

Radius of the hemisphere = 7 cm
Total surface area of the hemisphere = Curved surface area of hemisphere + Area of the circle$=3\mathrm{\pi }{r}^{2}$

#### Page No 836:

(b) 80080 cm3
Let R and r be the radii of the top and base of the bucket, respectively, and let h be its height.

Then,

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

Hence, the volume of the bucket is .

#### Page No 836:

(b) 1711.3 cm2
Let R and r be the radii of the top and base of the bucket, respectively, and let h and l be its height and slant height.

Then,

$l=\sqrt{{h}^{2}+{\left(R-r\right)}^{2}}$

Surface area of the bucket$=\mathrm{\pi }\left[{r}^{2}+l\left(R+r\right)\right]$

#### Page No 836:

(d) 7920 m2
Total area of the canvas required = (Curved surface area of the cylinder) + (Curved surface area of the cone)

#### Page No 837:

(a)
Volume of the sphere$=\frac{4}{3}\mathrm{\pi }{r}^{3}$
$=\left(\frac{4}{3}\mathrm{\pi }×{\left(8\right)}^{3}\right){\text{cm}}^{3}$

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

Number of cones formed

$=\frac{4\mathrm{\pi }×8×8×8×3}{3×\mathrm{\pi }×8×8×4}\phantom{\rule{0ex}{0ex}}=8$
Hence, $\left(a\right)⇒\left(q\right)$

(b)
Volume of the earth dug out = Volume of the cylinder
$=\mathrm{\pi }{r}^{2}h\phantom{\rule{0ex}{0ex}}=\frac{22}{7}×7×7×20$
Let the height of the platform be h.
Then, volume of the platform = volume of the cuboid

Therefore,

Hence, $\left(b\right)⇒\left(s\right)$

(c)
Volume of the sphere
$=\frac{4}{3}\mathrm{\pi }{r}^{3}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$

$=\frac{4}{3}\mathrm{\pi }×6×6×6$

Let h be the height of the cylinder.
Then, volume of the cylinder$={\mathrm{\pi r}}^{2}\mathrm{h}\phantom{\rule{0ex}{0ex}}$
$=\mathrm{\pi }×4×4×\mathrm{h}$
Therefore,

Hence, $\left(c\right)⇒\left(p\right)$

(d)
Let the radii of the spheres be R and r respectively.
Then, ratio of their volumes$=\frac{\frac{4}{3}\mathrm{\pi }{R}^{3}}{\frac{4}{3}{\mathrm{\pi r}}^{3}}$
Therefore,
$\frac{\frac{4}{3}\mathrm{\pi }{R}^{3}}{\frac{4}{3}{\mathrm{\pi r}}^{3}}=\frac{64}{27}\phantom{\rule{0ex}{0ex}}⇒\frac{{R}^{3}}{{r}^{3}}=\frac{64}{27}\phantom{\rule{0ex}{0ex}}⇒{\left(\frac{R}{r}\right)}^{3}={\left(\frac{4}{3}\right)}^{3}\phantom{\rule{0ex}{0ex}}⇒\frac{R}{r}=\frac{4}{3}$
Hence, the ratio of their surface areas$=\frac{4\mathrm{\pi }{R}^{2}}{4\mathrm{\pi }{r}^{2}}$
$=\frac{{R}^{2}}{{r}^{2}}\phantom{\rule{0ex}{0ex}}={\left(\frac{R}{r}\right)}^{2}\phantom{\rule{0ex}{0ex}}={\left(\frac{4}{3}\right)}^{2}\phantom{\rule{0ex}{0ex}}=\frac{16}{9}\phantom{\rule{0ex}{0ex}}=16:9$
Hence, $\left(d\right)⇒\left(r\right)$

#### Page No 837:

(a)
Let R and r be the top and base of the bucket and let h be its height.
Then, R = 20 cm, r = 10 cm and h = 30 cm.
Capacity of the bucket = Volume of the frustum of the cone
$=\frac{\mathrm{\pi }h}{3}\left({R}^{2}+{r}^{2}+Rr\right)\phantom{\rule{0ex}{0ex}}$

Hence, $\left(a\right)⇒\left(q\right)$

(b)
Let R and r be the top and base of the bucket and let h be its height.
Then, R = 20 cm, r = 12 cm and h = 15 cm.
Slant height of the bucket, $l=\sqrt{{h}^{2}+{\left(R-r\right)}^{2}}$

Hence, $\left(b\right)⇒\left(s\right)$

(c)
Let R and r be the top and base of the bucket and let l be its slant height.
Then, R = 33 cm, r = 27 cm and h = 10 cm
Total surface area of the bucket$=\mathrm{\pi }\left[{R}^{2}+{r}^{2}+l\left(R+r\right)\right]$

Hence, $\left(c\right)⇒\left(p\right)$

(d)
Let the diameter of the required sphere be d.
Then, volume of the sphere$=\frac{4}{3}\mathrm{\pi }{r}^{3}$
$=\frac{4}{3}\mathrm{\pi }{\left(\frac{\mathrm{d}}{2}\right)}^{3}$
Therefore,
$\frac{4}{3}\mathrm{\pi }{\left(\frac{\mathrm{d}}{2}\right)}^{3}=\frac{4}{3}\mathrm{\pi }{\left(3\right)}^{3}+\frac{4}{3}\mathrm{\pi }{\left(4\right)}^{3}+\frac{4}{3}\mathrm{\pi }{\left(5\right)}^{3}\phantom{\rule{0ex}{0ex}}⇒\frac{4}{3}\mathrm{\pi }\frac{{\mathrm{d}}^{3}}{8}=\frac{4}{3}\mathrm{\pi }×\left[{\left(3\right)}^{3}+{\left(4\right)}^{3}+{\left(5\right)}^{3}\right]\phantom{\rule{0ex}{0ex}}⇒\frac{{\mathrm{d}}^{3}}{8}=216$
d3
= 1728
d3
= 123
d  = 12 cm

Hence, $\left(d\right)⇒\left(r\right)$

#### Page No 838:

Assertion (A):
Let R and r be the top and base of the bucket and let h be its height.
Then, R = 15 cm, r = 5 cm and h = 24 cm
Slant height, $l=\sqrt{{h}^{2}+{\left(R-r\right)}^{2}}$

Surface area of the bucket$=\mathrm{\pi }\left[{R}^{2}+{r}^{2}+l\left(R+r\right)\right]$

Thus, the area and the formula are wrong.

Note:
Question seems to be incorrect.

#### Page No 839:

(d) Assertion (A) is false and Reason (R) is true.
Assertion (A):
Total surface area of the hemisphere$=3\mathrm{\pi }{r}^{2}$

Cost of painting at Rs 5 per cm2

= Rs 2310
Hence, Assertion (A) is false.

Reason (R): The given statement is true.

#### Page No 839:

(a) Both Assertion (A) and Reason (R) are true and Reason (R) is a correct explanation of Assertion (A).
Assertion (A):
Volume of the cuboid $=\left(l×b×h\right)$

Volume of each coin$=\mathrm{\pi }{r}^{2}h$

Number of coins

$=\frac{10×55×35×7×200×200×5}{10×10×22×175×175}\phantom{\rule{0ex}{0ex}}=400$
Hence, Assertion (A) is true.

Reason (R): The given statement is true.

#### Page No 839:

(d) Assertion (A) is false and Reason (R) is true.
Assertion (A):
Let R and r be the radii of the two spheres.
Then, ratio of their volumes$=\frac{\frac{4}{3}\mathrm{\pi }{R}^{3}}{\frac{4}{3}{\mathrm{\pi r}}^{3}}$
Therefore,
$\frac{\frac{4}{3}\mathrm{\pi }{R}^{3}}{\frac{4}{3}{\mathrm{\pi r}}^{3}}=\frac{27}{8}\phantom{\rule{0ex}{0ex}}⇒\frac{{R}^{3}}{{r}^{3}}=\frac{27}{8}\phantom{\rule{0ex}{0ex}}⇒{\left(\frac{R}{r}\right)}^{3}={\left(\frac{3}{2}\right)}^{3}\phantom{\rule{0ex}{0ex}}⇒\frac{R}{r}=\frac{3}{2}$
Hence, the ratio of their surface areas$=\frac{4\mathrm{\pi }{R}^{2}}{4{\mathrm{\pi r}}^{2}}$
$=\frac{{R}^{2}}{{r}^{2}}\phantom{\rule{0ex}{0ex}}={\left(\frac{R}{r}\right)}^{2}\phantom{\rule{0ex}{0ex}}={\left(\frac{3}{2}\right)}^{2}\phantom{\rule{0ex}{0ex}}=\frac{9}{4}\phantom{\rule{0ex}{0ex}}=9:4$
Hence, Assertion (A) is false.

Reason (R): The given statement is true.

#### Page No 839:

(c) Assertion (A) is true and Reason (R) is false.
Assertion (A):
Curved surface area of a cone$=\mathrm{\pi }r\sqrt{{r}^{2}+{h}^{2}}$

Hence, Assertion (A) is true.

Reason (R): The given statement is false.

#### Page No 849:

So, the number of solid spheres so moulded is 5.

#### Page No 849:

So, the ratio of their radii is $\sqrt{2}$ : 1.

#### Page No 849:

So, the area of the canvas required to make the tent is 7920 m2.

#### Page No 849:

Let R and r be the radii of the top and base of the bucket, respectively, and let l be its slant height.
Then, curved surface area of the bucket
= Curved surface area of the frustum of the cone

#### Page No 849:

Radius of cone = 12 cm
Height of cone = 24 cm
Volume of the metallic cone$=\frac{1}{3}\mathrm{\pi }{r}^{2}h$
$=\frac{1}{3}\mathrm{\pi }×{\left(12\right)}^{2}×24$

Volume of each spherical ball$=\frac{4}{3}\mathrm{\pi }{\mathrm{r}}^{3}$
$=\frac{4}{3}\mathrm{\pi }×{\left(3\right)}^{3}$

Number of balls formed

$=\frac{\mathrm{\pi }×12×12×24×3}{3×4×\mathrm{\pi }×3×3×3}\phantom{\rule{0ex}{0ex}}=32$

#### Page No 849:

Volume of hemispherical bowl$=\frac{2}{3}\mathrm{\pi }{\mathrm{r}}^{3}$

Radius of each bottle = $\frac{5}{2}\mathrm{cm}$

Height of each bottle = 6 cm

Volume of each bottle$=\mathrm{\pi }{r}^{2}h$

Number of bottles required

$=\frac{2×\mathrm{\pi }×15×15×15×2×2}{3×\mathrm{\pi }×5×5×6}\phantom{\rule{0ex}{0ex}}=60$

#### Page No 849:

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

Height of cone =  3 cm

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

Number of cones

$=\frac{4×\mathrm{\pi }×21×21×3×20×20}{3×2×2×2×\mathrm{\pi }×35×3×3×3}\phantom{\rule{0ex}{0ex}}=504$

#### Page No 849:

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

Let the length of the wire be h cm. Then,
Volume of the wire$=\mathrm{\pi }{r}^{2}h$

Therefore,

Hence, the length of the wire is 63 m.

#### Page No 849:

Let R and r be the radii of the top and base, respectively, of the drinking glass and let its height be h.

Then,

Capacity of the glass = Capacity of the frustum of the cone

#### Page No 849:

Volume of the cube = a3

Therefore,

Each side of the cube = 4 cm

Then,
Length of the cuboid
Breadth of the cuboid = 4 cm
Height of the cuboid = 4 cm

Total surface area of the cuboid $=2\left(lb+bh+lh\right)$

#### Page No 849:

Let the radius of the cylinder be 2x cm and its height be 3x cm.

Then, volume of the cylinder$=\mathrm{\pi }{r}^{2}h$
$=\frac{22}{7}×{\left(2x\right)}^{2}×3x$
Therefore,

$⇒{x}^{3}=\left(\frac{7}{2}×\frac{7}{2}×\frac{7}{2}\right)\phantom{\rule{0ex}{0ex}}⇒{x}^{3}={\left(\frac{7}{2}\right)}^{3}\phantom{\rule{0ex}{0ex}}⇒x=\frac{7}{2}$
Now,
Hence, the total surface area of the cylinder:

$\left(2\mathrm{\pi }rh+2\mathrm{\pi }{r}^{2}\right)\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$
​

#### Page No 849:

Radius of the hemisphere = Radius of the cone = 7 cm

Height of the cone

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

Total surface area of the toy = (Curved surface area of the hemisphere) + (Curved surface area of the cone)

#### Page No 850:

Radius of hemisphere = 9 cm
Volume of hemisphere$=\frac{2}{3}\mathrm{\pi }{\mathrm{r}}^{3}$
$=\left(\frac{2}{3}\mathrm{\pi }×9×9×9\right){\text{cm}}^{3}$
Height of each bottle = 4 cm
Volume of each bottle$=\mathrm{\pi }{r}^{2}h$

Number of bottles
$=\frac{2\mathrm{\pi }×9×9×9×2×2}{3×\mathrm{\pi }×3×3×4}\phantom{\rule{0ex}{0ex}}=54$

#### Page No 850:

Surface area of the sphere$=4\mathrm{\pi }{r}^{2}$
Surface area of the cube$=6{a}^{2}$
Therefore,

Ratio of their volumes$=\frac{4}{3}\mathrm{\pi }{r}^{3}}{{a}^{3}}=\frac{4\mathrm{\pi }{r}^{3}}{3{a}^{3}}$

Thus, the ratio of their volumes is .

#### Page No 850:

Let R and r be the radii of the top and base of the frustum of the cone, respectively, and its slant height be l.

Then,

Curved surface area of the frustum$=\mathrm{\pi }l\left(R+r\right)$

#### Page No 850:

Radius of the hemispherical end = 7 cm
Height of the hemispherical end = 7 cm
Height of the cylindrical part
Surface area of the solid = 2(curved surface area of the hemisphere) + (curved surface area of the cylinder)
$=\left[2\left(2\mathrm{\pi }{r}^{2}\right)+2\mathrm{\pi }rh\right]$

#### Page No 850:

So, the total surface area of the remaining solid is 1381.6 cm2.

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

#### Page No 850:

So, the length of the pipe is 112 m.