RS Aggarwal 2015 Solutions for Class 10 Math Chapter 19 - Volume And Surface Area Of Solids

Share with your friends

RS Aggarwal 2015 Solutions for Class 10 Math Chapter 19 Volume And Surface Area Of Solids are provided here with simple step-by-step explanations. These solutions for Volume And Surface Area Of Solids are extremely popular among class 10 students for Math Volume And Surface Area Of Solids Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the RS Aggarwal 2015 Book of class 10 Math Chapter 19 are provided here for you for free. You will also love the ad-free experience on Meritnation’s RS Aggarwal 2015 Solutions. All RS Aggarwal 2015 Solutions for class 10 Math are prepared by experts and are 100% accurate.

Page No 755:

Question 1:

A tent is in the shape of a right circular cylinder up to a height of 3 m and conical above it. The total height of the tent is 13.5 m and the radius of its base is 14 m. Find the cost of cloth required to make the tent at the rate of Rs 80 per square metre.

Answer:

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 $= 2 π r h = (2 \times \frac{22}{7} \times 14 \times 3) m^{2} = 264 m^{2}$

Height of the cone $= Total height - Height of cone = (13.5 - 3) m = 10.5 m$

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

$= π r \sqrt{r^{2} + h^{2}} = (\frac{22}{7} \times 14 \times \sqrt{14^{2} + 10 . 5^{2}}) m^{2} = (44 \times \sqrt{196 + 110.25}) m^{2} = (44 \times \sqrt{306.25}) m^{2} = (44 \times 17.5) m^{2} = 770 m^{2}$

Total surface area $= (264 + 770) m^{2} = 1034 m^{2}$

∴ Cost of cloth $= Rs 1034 \times 80 = Rs 82720$

Page No 755:

Question 2:

A circus tent is cylindrical to a height of 3 m and conical above it. If its base radius is 52.5 m and the slant height of the conical portion is 53 m, find the area of canvas needed to make the tent.

Answer:

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 $= 2 π r h = (2 \times \frac{22}{7} \times 52.5 \times 3) m^{2} = 990 m^{2}$
Curved surface area of the conical portion $= π r l = (\frac{22}{7} \times 52.5 \times 53) m^{2} = 8745 m^{2}$

Thus, the total area of canvas required for making the tent $= (8745 + 990) m^{2} = 9735 m^{2}$

Page No 755:

Question 3:

A wooden article was made by scooping out a hemisphere from each end of a cylinder, as shown in the figure. If the height of the cylinder is 20 cm and its base diameter is 7 cm, find the total surface area of the article when it is ready.

Answer:

Base diameter = 7 cm
Radius of the base = $\frac{7}{2} cm$
Height of the cylinder = 20 cm

Curved/lateral surface area of the cylinder $= 2 π r h = (2 \times \frac{22}{7} \times \frac{7}{2} \times 20) {cm}^{2} = 440 {cm}^{2}$

Curved surface area of hemisphere $= 2 {πr}^{2}$
Therefore, curved surface area of two hemispheres $= 2 \times 2 {πr}^{2} = (2 \times 2 \times \frac{22}{7} \times \frac{7}{2} \times \frac{7}{2}) {cm}^{2} = 154 {cm}^{2}$

Total surface area of the article $= (440 + 154) {cm}^{2} = 594 {cm}^{2}$

Page No 755:

Question 4:

A wooden article was made by scooping out a hemisphere from each end of a solid cylinder, as shown in the figure. If the height of the cylinder is 12 cm and its base radius is 4.2 cm, find the total surface area of the article. Also, find the volume of the wood left in the article.

Answer:

Page No 756:

Question 9:

A toy is in the form of a cone mounted on a hemisphere of diameter 7 cm. The total height of the toy is 14.5 cm. Find the volume and the total surface area of the toy.

Answer:

A figure representing the toy is shown below:

Height of cone = 14.5 cm - 3.5 cm = 11 cm
The slant height of the conical part can be expressed using Pythagoras' theorem as shown:

$l = \sqrt{{(\frac{7}{2})}^{2} + {(11)}^{2}} = 11.54 c m$

Volume of the toy = Volume of the conical part + Volume of the hemisphere
Therefore,
$Volume of the toy = \frac{2}{3} {πr}^{3} + \frac{1}{3} {πr}^{2} h = \frac{1}{3} {πr}^{2} (2 r + h) = \frac{1}{3} π {(\frac{7}{2})}^{2} (2 \times \frac{7}{2} + 11) = 231 {cm}^{3}$

Total area of the toy = Curved surface area of the cone + Curved surface area of the hemisphere.
Therefore,

$Total surface area of the toy = 2 {πr}^{2} + πrl = πr (2 r + l) = π \times (\frac{7}{2}) \times (7 + 11.54) = 203.94 {cm}^{2}$

Page No 756:

Question 10:

A tent is in the form of a right circular cylinder surmounted by a cone. The diameter of the cylinder is 24 m. The height of the cylindrical portions is 11 m while the vertex of the cone is 16 m above the ground. Find the area of the canvas required for the tent.

Answer:

The tent consists of a cylinder, topped by a cone.
Diameter of cylinder = 24 m
Radius of cylinder = 12 m
Total height of the tent = 16 m
Height of cylindrical portion = 11 m
Height of cone $= 16 - 11 = 5 m$

Curved surface area of cylinder $= 2 πrh = 2 \times \frac{22}{7} \times 12 \times 11 = 829.714 m^{2}$
Curved surface area of cone $= πrl = πr \sqrt{r^{2} + h^{2}} = \frac{22}{7} \times 12 \sqrt{12 \times 12 + 5 \times 5} = 37.714 \times \sqrt{169} = 490.282 m^{2}$
Area of canvas required for tent = Total surface area $= 829.714 + 490.282 = 1319.996 m^{2} \approx 1320 m^{2}$

Page No 756:

Question 11:

A vessel is in the form of a hemispherical bowl surmounted by a hollow cylinder. The diameter of the hemisphere is 21 cm and the total height of the vessel is 14.5 cm. Find its capacity.

Answer:

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

Volume of the hemisphere $= \frac{2}{3} {πr}^{3} = \frac{2}{3} \times \frac{22}{7} \times 10.5 \times 10.5 \times 10.5 = 2425.5 {cm}^{3}$

Height of the cylinder $= Total height of the vessel-Radius of the hemisphere = 14.5 - 10.5 = 4 cm$
Volume of the cylinder $= {πr}^{2} h = \frac{22}{7} \times 10.5 \times 10.5 \times 4 = 1386 {cm}^{3}$

Total volume $= V olume of hemispherical part + V olume of cylinder = 1386 + 2425.5 = 3811.5 {cm}^{3}$

Page No 756:

Question 12:

The adjoining figure represents a solid consisting of a cylinder surmounted by a cone at one end and a hemisphere at the other. Find the volume of the solid.

Answer:

Page No 757:

Question 18:

A juice seller serves his customers units a glass as shown in the figure. The inner diameter of the cylindrical glass is 5 cm but the bottom of the glass has a hemispherical portion raised, which reduces the capacity of the glass. If the height of the glass is 10 cm, find the apparent capacity of the glass and its actual capacity.

A metallic right circular cone is 20 m high and has a vertical angle of 60°. It is cut into two parts at the middle of its height by a plane parallel to the base. If the frustum so obtained is drawn into a wire of diameter $\frac{1}{16}$ cm, find the length of the wire.

Answer:

Consider the frustum created by cutting the original cone.
Consider the vertical axis.
Vertical angle of the cone = 60^o
Angle of the central axis with slant height = 30^o

Height of the cone = 20 cm
Height of the smaller cone = Height of the frustum = 10 cm

In the smaller cone,
$\frac{Radius}{Height} = \tan 30 ° \frac{Radius}{10} = \frac{1}{\sqrt{3}} Radius = \frac{10}{\sqrt{3}} cm$

Height of the original cone = 20 m

$\frac{Radius}{Height} = \tan 30 ° \frac{Radius}{20} = \frac{1}{\sqrt{3}} R = \frac{20}{\sqrt{3}} cm$

Volume of the frustum
$= \frac{1}{3} π h (R^{2} + r^{2} + R r) = \frac{1}{3} π \times 10 ({(\frac{20}{\sqrt{3}})}^{2} + {(\frac{10}{\sqrt{3}})}^{2} + \frac{20 \times 10}{\sqrt{3} \times \sqrt{3}}) = \frac{10 π}{3} (\frac{400}{3} + \frac{100}{3} + \frac{200}{3}) = \frac{10 π}{3} \times \frac{700}{3} {cm}^{3} = \frac{7000 π}{9} {cm}^{3}$

Let the length of the wire be x cm.
Diameter $= \frac{1}{16} c m$
Radius $= \frac{1}{32} c m$

Volume of the wire $= π \times {(\frac{1}{32})}^{2} \times x {cm}^{3}$

Volume of the wire = Volume of the frustum
Therefore,
$\frac{7000 π}{9} = π \times {(\frac{1}{32})}^{2} \times x \Rightarrow x = \frac{7000 \times 32 \times 32}{9} \Rightarrow x = 796444 cm = 7964.44 m$

So, the length of the wire drawn is 7964.44 m.

Page No 780:

Question 9:

A milk container is in the form of frustum of a cone of height 18 cm with radii of its upper and lower ends being 8 cm and 32 cm, respectively. Find the amount of milk that can be used to completely fill the container and the cost of doing so at the rate of Rs 20 per litre.

Answer:

Greater radius of the frustum = R = 32 cm
Smaller radius of the frustum = r = 8 cm
Height of the frustum = h = 18 cm

Capacity of the frustum
$= \frac{1}{3} π h (R^{2} + r^{2} + R r) = \frac{1}{3} \times π \times 18 (32^{2} + 8^{2} + 32 \times 8) = \frac{22}{7} \times 6 \times 1344 = 25344 {cm}^{3} = 25.344 litres$

Cost of 1 l is Rs 20.
So, total cost of filling the container with milk $= 25.344 \times 20 = Rs 506.88$

Page No 782:

Question 1:

A cylindrical pencil sharpened at one end is a combination of
(a) a cylinder and a cone
(b) a cylinder and frustum of a cone
(c) a cylinder and a hemisphere
(d) two cylinders

Answer:

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

Page No 782:

Question 2:

A shuttlecock used for playing badminton is a combination of

(a) cylinder and a hemisphere
(b) frustum of a cone and a hemisphere
(c) a cone and a hemisphere
(d) a cylinder and a sphere

Answer:

(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 782:

Question 3:

A funnel is a combination of

(a) a cylinder and a cone
(b) a cylinder and a hemisphere
(c) a cylinder and frustum of a cone
(d) a cone and hemisphere

Answer:

Page No 783:

Question 9:

During conversion of a solid from one shape to another, the volume of the new shape will
(a) decrease
(b) increase
(c) remain unaltered
(d) be doubled

Answer:

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

Page No 783:

Question 10:

In a right circular cone, the cross-section made by a plane parallel to the base is a
(a) sphere
(b) hemisphere
(c) circle
(d) a semicircle

Answer:

Page No 788:

Question 61:

Match the following columns:

Column I	Column II
(a) A solid metallic sphere of radius 8 cm is melted and the material is used to make solid right cones with height 4 cm and base radius of 8 cm. How many cones are formed?	(p) 18
(b) A 20-m-deep well with diameter 14 m is dug up and the earth from digging is evenly spread out to form a platform 44 m by 14 m. The height of the platform is ...........m.	(q) 8
(c) A sphere of radius 6 cm is melted and recast in the shape of a cylinder of radius 4 cm. Then, the height of the cylinder is ......... cm.	(r) 16 : 9
(d) The volumes of two spheres are in the ratio 64 : 27. The ratio of their surface areas is ....... .	(s) 5

Answer:

(a)
Volume of the sphere $= \frac{4}{3} π r^{3}$
$= (\frac{4}{3} π \times {(8)}^{3}) {cm}^{3}$

Volume of each cone $= \frac{1}{3} π r^{2} h$
$= \frac{1}{3} π \times {(8)}^{2} \times 4 {cm}^{3}$

Number of cones formed $= \frac{Volume of the sphere}{Volume of each cone}$

$= \frac{4 π \times 8 \times 8 \times 8 \times 3}{3 \times π \times 8 \times 8 \times 4} = 8$
Hence, $(a) \Rightarrow (q)$

(b)
Volume of the earth dug out = Volume of the cylinder
$= π r^{2} h = \frac{22}{7} \times 7 \times 7 \times 20$
Let the height of the platform be h.
Then, volume of the platform = volume of the cuboid
$= (44 \times 14 \times h) m^{3}$
Therefore,

$\frac{22}{7} \times 7 \times 7 \times 20 = 44 \times 14 \times h \Rightarrow 3080 = 616 \times h \Rightarrow h = \frac{3080}{616} \Rightarrow h = 5 m$

Hence, $(b) \Rightarrow (s)$

(c)
Volume of the sphere
$= \frac{4}{3} π r^{3}$

$= \frac{4}{3} π \times 6 \times 6 \times 6$

Let h be the height of the cylinder.
Then, volume of the cylinder $= {πr}^{2} h$
$= π \times 4 \times 4 \times h$
Therefore,

$\frac{4}{3} π \times 6 \times 6 \times 6 = π \times 4 \times 4 \times h \Rightarrow \frac{4}{3} \times 6 \times 6 \times 6 = 4 \times 4 \times h \Rightarrow 228 = 16 \times h \Rightarrow h = \frac{228}{16} \Rightarrow h = 18 cm$

Hence, $(c) \Rightarrow (p)$

(d)
Let the radii of the spheres be R and r respectively.
Then, ratio of their volumes $= \frac{\frac{4}{3} π R^{3}}{\frac{4}{3} {πr}^{3}}$
Therefore,
$\frac{\frac{4}{3} π R^{3}}{\frac{4}{3} {πr}^{3}} = \frac{64}{27} \Rightarrow \frac{R^{3}}{r^{3}} = \frac{64}{27} \Rightarrow {(\frac{R}{r})}^{3} = {(\frac{4}{3})}^{3} \Rightarrow \frac{R}{r} = \frac{4}{3}$
Hence, the ratio of their surface areas $= \frac{4 π R^{2}}{4 π r^{2}}$
$= \frac{R^{2}}{r^{2}} = {(\frac{R}{r})}^{2} = {(\frac{4}{3})}^{2} = \frac{16}{9} = 16 : 9$
Hence, $(d) \Rightarrow (r)$

Page No 789:

Question 62:

Match the following columns:

Column I	Column II
(a) The radii of the circular ends of a bucket, in the form of the frustum of a cone of height 30 cm, are 20 cm and 10 cm respectively. The capacity of the bucket is ........cm³.	(p) 2418π
(b) The radii of the circular ends of a conical bucket of height 15 cm are 20 and 12 cm respectively. The slant height of the bucket is ........ cm.	(q) 22000
(c) The radii of the circular ends of a solid frustum of a cone are 33 cm and 27 cm and its slant height is 10 cm. The total surface area of the bucket is .........cm².	(r) 12
(d) Three solid metallic spheres of radii 3 cm, 4 cm and 5 cm are melted to form a single solid sphere. The diameter of the resulting sphere is ........ cm.	(s) 17

Answer:

Radius of cylinder = $\frac{12}{2} = 6 cm$
Height of cylinder = 15 cm
Volume of cylinder $= π r^{2} h$
   $= (\frac{22}{7} \times 6 \times 6 \times 15) {cm}^{3} [d = 12 cm, r \Rightarrow \frac{12}{2} cm = 6 cm] = \frac{11880}{7} {cm}^{3}$

Radius of the hemisphere = 3 cm

Height of the cone = Total height of the toy - Radius of the hemisphere

                              $= (12 - 3) cm = 9 cm$

Volume of each toy = (Volume of the hemisphere) + (Volume of the cone)
$= \frac{2}{3} π r^{3} + \frac{1}{3} π r^{2} h = \frac{1}{3} π r^{2} (2 r + h) = \frac{1}{3} \times \frac{22}{7} \times {(3)}^{2} \times [(2 \times 3) + 9] {cm}^{3} = (\frac{1}{3} \times \frac{22}{7} \times 9 \times 15) {cm}^{3} = \frac{990}{7} {cm}^{3}$

Number of toys $= \frac{Volume of the cylinder}{Volume of each toy}$

   $= (\frac{11880 \times 7}{7 \times 990}) = 12$

Page No 801:

Question 20:

The radii of two cylinders are in the ratio 2 : 3 and their heights are in the ratio 5 : 3. Find the ratio of their volumes.

Answer:

Let the radii of the two cylinders be 2r and 3r and their heights be 5h and 3h respectively.

Then, ratio of their volumes $= \frac{π \times 4 r^{2} \times 5 h}{π \times 9 r^{2} \times 3 h}$

$= \frac{20}{27} = 20 : 27$

View NCERT Solutions for all chapters of Class 10