NCERT Solutions for Class 10 Maths Chapter 13 Surface Areas and Volumes

# NCERT Solutions for Class 10 Maths Chapter 13 Surface Areas and Volumes

Edited By Ramraj Saini | Updated on Sep 08, 2023 06:53 PM IST | #CBSE Class 10th

## Surface Area And Volume Class 10 NCERT Solution

NCERT solution for surface area and volume class 10 are provided here. These solutions are designed to assist students in understanding and solving problems related to the surface areas and volumes of 3D shapes, such as spheres, cylinders, cones, cuboids, and combinations of any two solids keeping in mind of latest syllabus of CBSE 2023-24. In addition to this, here students will get NCERT solutions from class 6 to 12 for science and maths. NCERT Class 10 solutions Maths chapter 13 Surface Areas and Volumes consist of 5 exercises with 38 questions which will make learning easier for the students.

In NCERT solutions for class 10 maths chapter 13 Surface Areas and Volumes, a detailed explanation to each and every question is given. Refer to the NCERT Class 10 Maths books and cover all the topics to score well in the exams. Also for ease students can downlaod NCERT class 10 maths chapter 13 pdf using the link given below. Each exercises of ch 13 Maths class 10 can be downloaded and use offline. Students can get the exercise wise solutions of surface area and volume Class 10 from the following links.

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## Surface Area And Volume Class 10 NCERT Solution - Important Formulae

Total Surface Area (TSA) - Total Surface Area, denoted as TSA signifies the entire expanse covered by an object's surface. This encompasses the collective area of all its external facets. Here are some prominent geometrical figures and their corresponding TSA formulas:

• Cuboid: TSA = 2[(l x b) + (b x h) + (h x l)]

• Cube: TSA = 6a2

• Right Circular Cylinder: TSA = 2πr(h + r)

• Right Circular Cone: TSA = πr(l + r)

• Sphere: TSA = 4πr2

• Right Pyramid: TSA = LSA (Lateral Surface Area) + Area of the base

• Prism: TSA = LSA (Lateral Surface Area) × 2B (Area of the base)

• Hemisphere: TSA = 3 × π × r2

Lateral/Curved Surface Area (LSA/CSA) - The concept of Curved Surface Area (CSA) or Lateral Surface Area (LSA) comes into focus, particularly in forms such as cylinders, cones, and pyramids. CSA represents the area of the curved component or sides excluding the top and bottom facets. LSA, on the other hand, pertains to the lateral area of various shapes. The following are some key examples of these surface area measures:

• Cuboid: CSA = 2h(l + b)

• Cube: CSA = 4a2

• Right Circular Cylinder: CSA = 2πrh

• Right Circular Cone: CSA = πrl

• Right Pyramid: LSA = (1/2)× p × l

• Prism: LSA = p × h

• Hemisphere: LSA = 2 × π × r2

Volume - Volume emerges as a fundamental attribute, denoting the spatial extent occupied by an object or substance, quantified in cubic units. The following volume formulas unravel the essence of these geometric measurements:

• Cuboid: Volume = lbh

• Cube: Volume = a3

• Right Circular Cylinder: Volume = πr2h

• Right Circular Cone: Volume = (1/3)πr2h

• Sphere: Volume = (4/3)πr3

• Right Pyramid: Volume = (1/3) × Area of the base × h

• Prism: Volume = B (Area of the base) × h

Hemisphere: Volume = (2/3)× (πr3)

Free download NCERT Solutions for Class 10 Maths Chapter 13 Surface Areas and Volumes PDF for CBSE Exam.

## NCERT Solutions for Class 10 Maths Chapter 13 Surface Areas and Volumes (Intext Questions and Exercise)

Chapter 13 class 10 maths NCERT solutions Excercise: 13.1

We are given that volume of the cube $=\ 64\ cm^3$

Also, the volume of a cube is given by $=\ a^3$ ( here $a$ is the edge of the cube)

Thus : $a^3\ =\ 64$

$a\ =\ 4\ cm$

Now according to the question we have combined the two cubes that edge lengths of the formed cuboid are 4 cm, 4 cm, and 8 cm.

The surface area of a cuboid is : $=\ 2\left ( lb\ +\ bh\ +\ hl \right )$

or $=\ 2\left ( 8\times 4\ +4\times 4\ +\ 4\times 8 \right )$

or $=\ 2\left ( 80 \right )$

or $=\ 160\ cm^2$

Thus the area of the formed cuboid is 160 cm 2 .

Since the vessel consists of hemisphere and cylinder, thus its area is given by :

Area of vessel = Inner area of the cylinder(curved) + Inner area of hemisphere

The inner surface area of the hemisphere is :

$=\ 2\pi r^2$

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

or $=\ 308\ cm^2$

And the surface area of the cylinder is :

$=\ 2\pi rh$

or $=\ 2\times \frac{22}{7}\times 7\times 6$

or $=\ 264\ cm^2$

Thus the inner surface area of the vessel is $= 308 + 264 = 572 \:cm^2.$

The required surface area of the toy is given by :

Area of toy = Surface area of hemisphere + Surface area of the cone

Firstly consider the hemisphere :

The surface area of a hemisphere is $=\ 2 \pi r^2$

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

or $=\ 77\ cm^2$

Now for cone we have :

The surface area of a cone $=\ \pi rl$

Thus we need to calculate the slant of the cone.

We know that :

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

or $=\ 12^2\ +\ 3.5^2$

or $=\ \frac{625}{4}$

or $l\ =\ \frac{25}{2}\ =\ 12.5\ cm$

Thus surface area of a cone $=\ \pi rl$

or $=\ \frac{22}{7}\times 3.5\times 12.5$

or $=\ 137.5\ cm^2$

Hence the total surface area of toy = $=\ 77\ +\ 137.5\ =\ 214.5\ cm^2$

It is given that the hemisphere is mounted on the cuboid, thus the hemisphere can take on complete as its diameter (which is maximum).

Thus the greatest diameter of the hemisphere is 7 cm.

Now, the total surface area of solid = Surface area of cube + Surface area of the hemisphere - Area of the base of a hemisphere (as this is counted on one side of the cube)

The surface area of the cube is :

$=\ 6a^3$

$=\ 6\times 7^3\ =\ 294\ cm^2$

Now the area of a hemisphere is

$=\ 2\pi r^2$

$=\ 2\times \frac{22}{7}\times \left ( \frac{7}{2} \right )^2\ =\ 77\ cm^2$

And the area of the base of a hemisphere is

$=\ \pi r^2\ =\ \frac{22}{7}\times \left ( \frac{7}{2} \right )^2\ =\ 38.5\ cm^2$

Hence the surface area of solid is $= 294 + 77 - 38.5 = 332.5 \:cm^2$ .

It is given that the diameter of the hemisphere is equal to the edge length of the cube.

The total surface area of solid is given by :

The surface area of solid = Surface area of cube + Surface area of the hemisphere - Area of the base of the hemisphere

The surface area of the cube $=\ 6l^2$

And surface area of the hemisphere:

$=\ 2\pi r^2\ =\ 2\pi \left ( \frac{l}{2} \right )^2$

Area of base of the hemisphere:

$=\ \pi r^2\ =\ \pi \left ( \frac{l}{2} \right )^2$

Thus the area of solid is:

$=\ 6l^2\ +\ \pi \left ( \frac{l}{2} \right )^2\ unit^2$

It is clear from the figure that the capsule has hemisphere and cylinder structure.

The surface area of capsule = 2 (Area of the hemisphere) + Area of the cylindrical part

Area of hemisphere $=\ 2\pi r^2$

or $=\ 2\pi\times \left ( \frac{5}{2} \right )^2$

or $=\ \frac{25}{2} \pi\ mm^2$

And the area of the cylinder $=\ 2\pi rh$

or $=\ 2\pi \times \frac{5}{2}\times 9$

or $=\ 45 \pi\ mm^2$

Thus the area of the solid is :

$=\ 2\left ( \frac{25}{2} \right )\pi\ +\ 45 \pi$

$=\ 25\pi\ +\ 45 \pi$

$=\ 70 \pi$

$=\ 220\ mm^2$

The canvas will cover the cylindrical part as well as the conical part.

So, the area of canvas = Area of cylindrical part (curved) + Area of the conical part

Now, the area of the cylindrical part is $=\ 2\pi rh$

or $=\ 2\pi \times 2\times 2.1$

or $=\ 8.4 \pi\ m^2$

And the area of the cone is $=\ \pi rl$

or $=\ \pi \times 2\times 2.8$

or $=\ 5.6\pi\ m^2$

Thus, the area of the canvas $=\ 8.4\pi\ +\ 5.6\pi$

or $=\ 14\pi\ =\ 44\ m^2$

Further, it is given that the rate of canvas per m 2 is = Rs. 500.

Thus the required money is $=\ 500\times 44\ =\ Rs.\ 22,000$

Firstly we need to calculate the slant height of the cone :

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

or $=\ (0.7)^2\ +\ (2.4)^2$

or $l^2\ =\ 6.25$

or $l\ =\ 2.5\ cm$

Now, the total surface area of solid can be calculated as :

The surface area of solid = Surface area of cylinder + Surface area of cone + Area of base of the cylinder

The surface area of the cylinder is $=\ 2 \pi rh$

or $=\ 2 \pi \times 0.7\times 2.4$

or $=\ 10.56\ cm^2$

Now, the surface area of a cone $=\ \pi rl$

or $=\ \pi \times 0.7\times 2.5$

or $=\ 5.50\ cm^2$

And the area of the base of the cylinder is $=\ \pi r^2$

or $=\ \pi \times 0.7\times 0.7$

or $=\ 1.54\ cm^2$

Thus required area of solid = 10. 56 + 5.50 + 1.54 = 17.60 cm 2 .

Thus total surface area of remaining solid to nearest cm 2 is 18 cm 2 .

The required surface area is given by :

The surface area of article = Surface area of cylindrical part + 2 (Surface area of the hemisphere)

Now, the area of the cylinder $=\ 2\pi rh$

or $=\ 2\pi \times 3.5\times 10$

or $=\ 70\pi \ cm^2$

And the surface area of the hemisphere : $=\ 2\pi r^2$

or $=\ 2\pi \times 3.5\times 3.5$

or $=\ 24.5\pi\ cm^2$

Thus the required area $=\ 70\pi\ +\ 2(24.5\pi)\ =\ 374\ cm^2$

Chapter 13 class 10 maths NCERT solutions Excercise: 13.2

The volume of the solid is given by :

The volume of solid = Volume of cone + Volume of a hemisphere

The volume of cone :

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

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

or $=\ \frac{\pi}{3}\ cm^3$

And the volume of the hemisphere :

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

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

or $=\ \frac{2\pi}{3}\ cm^3$

Hence the volume of solid is :

$=\ \frac{\pi}{3}\ +\ \frac{2\pi}{3}\ =\ \pi\ cm^3$

The volume of air present = Volume of cylinder + 2 (Volume of a cone)

Now, the volume of a cylinder : $=\ \pi r^2h$

or $=\ \pi \left ( \frac{3}{2} \right )^2\times 8$

or $=\ 18\pi \ cm^3$

And the volume of a cone is :

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

or $=\ \frac{1}{3} \pi \times \left ( \frac{3}{2} \right )^2\times 2$

or $=\ \frac{3}{2} \pi \ cm^3$

Thus the volume of air is :

$=\ 18 \pi\ +\ 2\times \frac{3}{2} \pi \ =\ 21\pi$

or $=\ 66\ cm^3$

It is clear from the figure that gulab jamun has one cylindrical part and two hemispherical parts.

Thus, the volume of gulab jamun is = Volume of cylindrical part + 2 (Volume of the hemisphere )

Now, the volume of the cylinder is $=\ \pi r^2h$

or $=\ \pi\times 1.4^2\times 2.2$

or $=\ 13.55\ cm^3$

And the volume of a hemisphere is :

$\\=\ \frac{2}{3}\pi r^3\\\\=\ \frac{2}{3}\times \pi \times (1.4)^3\\\\=\ 5.75\ cm^3$

Thus the volume of 1 gulab jamun is $= 13.55 + 2 (5.75) = 25.05 cm^3.$

Hence the volume of 45 gulab jamun $=\ 45(25.05)\ cm^3\ =\ 1127.25\ cm^3$

Further, it is given that one gulab jamun contains sugar syrup upto $30$ $\%$ .

So, the total volume of sugar present :

$=\ \frac{30}{100}\times 1127.25\ =\ 338\ cm^3$

The volume of wood is given by = volume of the cuboid - the volume of four cones.

Firstly, the volume of cuboid : $=\ lbh$

or $=\ 15\times 10\times 3.5$

or $=\ 525\ cm^3$

And, the volume of cone :

$\\=\ \frac{1}{3} \pi r^2h\\=\ \frac{1}{3} \pi \times (0.5)^2\times 1.4\\\\=\ 0.3665\ cm^3$

Thus the volume of wood is $= 525 + 4 (0.3665) = 523.53 \:cm^3$

According to the question :

Water spilled from the container = Volume of lead balls.

Let us assume the number of lead balls to be n.

Thus the equation becomes :

$\frac{1}{4}\times Volume_{cone}\ =\ n\times \frac{4}{3} \pi r^3$

or $\frac{1}{4}\times \frac{1}{3}\pi\times5^2\times 8\ =\ n\times \frac{4}{3} \pi\times 0.5^3$

or $n\ =\ \frac{25\times 8}{16\times \left ( \frac{1}{2} \right )^3}$

or $n\ =\ 100$

Hence the number of lead shots dropped is 100.

The pole can be divided into one large cylinder and one small cylinder.

Thus, the volume of pole = volume of large cylinder + volume of a small cylinder

$=\ \pi r_l^2h_l\ +\ \pi r_s^2h_s$

or $=\ \pi \times 12^2\times 220\ +\ \pi \times 8^2\times 60$

or $=\ \pi \times \left ( 144\times 220\ +\ 64\times 60 \right )$

or $=\ 3.14\times 35520$

or $=\ 111532.5\ cm^3$

Now, according to question mass of the pole is :

$=\ 8\times 111532.5$

or $=\ 892262.4\ g\ =\ 892.262\ Kg$

It is clear from the question that the required volume is :

The volume of water (left) =Volume of a cylinder - Volume of solid

Now the volume of the cylinder is $=\ \pi r^2h$

or $=\ \pi\times (60)^2\times 180\ cm^3$

And the volume of solid is :

$\\=\ \frac{1}{3} \pi r^2h\ +\ \frac{2}{3}\pi r^3\\\\=\ \frac{1}{3} \times \pi \times (60)^2\times 120\ +\ \frac{2}{3}\times \pi \times (60)^3\\\\=\ \pi (60)^2\times 80\ cm^3$

Thus the volume of water left :

$\\=\ \pi (60)^2\times 180\ -\ \pi (60)^2\times 80\\\\=\ \pi (60)^2\times 100\\\\=\ 1131428.57\ cm^3\\\\=\ 1.131\ m^3$

The volume of the vessel is given by :

The volume of vessel = Volume of sphere + Volume of the cylindrical part

Now, the volume of the sphere is :

$\\=\ \frac{4}{3}\pi r^3\\\\=\ \frac{4}{3}\pi \left ( \frac{8.5}{2} \right )^3\\\\=\ 321.55\ cm^3$

And the volume of the cylinder is:-

$\\=\ \pi r^2h\\\\=\ \pi \times (1)^2\times 8\\\\=\ 25.13\ cm^3$

Thus the volume of the vessel is $= 321.55 + 25.13 = 346.68\: cm^3$

NCERT solutions for class 10 maths chapter 13 Surface Areas and Volumes Excercise: 13.3

Let us assume the height of the cylinder to be h.

Since the material is melted and recast thus its volume will remain the same.

So, Volume of sphere = Volume of obtained cylinder.

$\frac{4}{3}\pi r^3_s\ =\ \pi r^2_c h$

$\frac{4}{3}\pi \times 4.2^3\ =\ \pi \times 6^2 \times h$

$h\ =\ \frac{4}{3}\times \frac{4.2\times4.2\times 4.2}{36}$

$h\ =\ 2.74\ cm$

Hence the height of the cylinder is 2.74 cm.

According to the question, small spheres are melted and converted into a bigger sphere. Thus the sum of their volume is equal to the volume of the bigger sphere.

The volume of 3 small spheres = Volume of bigger sphere

Let us assume the radius of the bigger sphere is r.

$\frac{4}{3}\pi \left ( r^3_1\ +\ r^3_2\ +\ r^3_3 \right )\ =\ \frac{4}{3}\pi r^3$

$r^3_1\ +\ r^3_2\ +\ r^3_3 \ =\ r^3$

$r^3\ =\ 6^3\ +\ 8^3\ +\ 10^3$

$r\ =\ 12\ cm$

Hence the radius of the sphere obtained is 12 cm.

According to the question, the volume of soil dug will be equal to the volume of the platform created.

Thus we can write :

The volume of soil dug = Volume of platform

$\\\Rightarrow \pi r^2h\ =\ Area\times h\\\\\Rightarrow \pi \times \left ( \frac{7}{2} \right )^2\times 20\ =\ 22\times 14 \times h\\\\\Rightarrow h\ =\ \frac{5}{2}\ m$

Thus the height of the platform created is 2.5 m.

According to the question, the volume is conserved here :

The volume of soil dug out = Volume of the embankment made.

Let the height of the embankment is h.

$\\\pi r^2_1h'\ =\ \pi\left ( r^2_2\ -\ r^2_1 \right )h\\\\\pi \left ( \frac{3}{2} \right )^2\times 14\ =\ \pi\left ( \left ( \frac{11}{2} \right )^2\ -\ \left ( \frac{3}{2} \right )^2 \right )h\\\\h\ =\ \frac{9}{8}\ m$

Hence the height of the embankment made is 1.125 m.

Let the number of cones that can be filled with ice cream be n.

Then we can write :

The volume of a cylinder containing ice cream = n ( volume of 1 ice cream cone )

$\\\pi r^2_{cy}h_{cy}\ =\ n\left ( \frac{1}{3}\pi r^2_{co}h_{co}\ +\ \frac{2}{3} \pi r^3 \right )\\\\\pi\times 6^2\times 15\ =\ n\left ( \frac{1}{3}\times \pi \times 3^2 \times 12\ +\ \frac{2}{3} \pi \times 3^3 \right )\\\\n\ =\ \frac{36\times 15}{54}\\\\n=\ 10$

Hence the number of cones that can be filled is 10.

Let us assume the number of coins that need to be melted be n.

Then we can write :

The volume of n coins = Volume of cuboid formed.

$n\left ( \pi r^2 h \right )\ =\ lbh'$

$n\left ( \pi \times \left ( \frac{1.75}{2} \right )^2\times 0.2 \right )\ =\ 5.5\times 10\times 3.5$

$n\ =\ 400$

Thus the required number of coins is 400.

According to question volume will remain constant thus we can write :

The volume of bucket = Volume of heap formed.

$\pi r^2_1h_1\ =\ \frac{1}{3}\pi r^2_2 h_2$

Let the radius of heap be r.

$\pi\times 18^2 \times 32\ =\ \frac{1}{3}\times \pi \times r^2\times 24$

$r\ =\ 18\times 2\ =\ 36\ cm$

And thus the slant height will be

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

$=\ \sqrt{36^2\ +\ 24^2}$

$=\ 12\sqrt{13}\ cm$

Hence the radius of heap made is 36 cm and its slant height is $12\sqrt{13}\ cm$ .

Speed of water is: 10 Km/hr

And the volume of water flow in 1 minute is :

$=\ 9\times \frac{10000}{60}\ =\ 1500\ m^3$

Thus the volume of water flow in 30 minutes will be : $=\ 1500\ \times 30\ =\ 45000\ m^3$

Let us assume irrigated area be A. Now we can equation the expression of volumes as the volume will remain the same.

$45000\ =\ \frac{A\times 8}{100}$

$A\ =\ 562500\ m^2$

Thus the irrigated area is $562500 \:m^2$ .

Area of the cross-section of pipe is $=\ \pi r^2$

$=\ \pi (0.1)^2\ =\ 0.01 \pi\ m^2$

Speed of water is given to be = 3 km/hr

Thus, the volume of water flowing through a pipe in 1 min. is

$=\ \frac{3000}{60}\times 0.01 \pi$

$=\ 0.5 \pi\ m^3$

Now let us assume that the tank will be completely filled after t minutes.

Then we write :

$t\times 0.5 \pi\ =\ \pi r^2h$

$t\times 0.5 \ =\ 5^2\times 2$

$t\ =\ 100$

Hence the time required for filling the tank completely in 100 minutes.

NCERT Solutions for class 10 Maths volume and surface area Excercise: 13.4

The capacity of glass is the same as the volume of glass.

Thus the volume of glass :

$=\ \frac{1}{3}\pi h\left ( r^2_1\ +\ r^2_2\ +\ r_1r_2 \right )$

$=\ \frac{1}{3}\pi h\left ( 2^2\ +\ 1^2\ +\ 2\times 1 \right )$

$=\ \frac{308}{3}\ cm^3$

= Capacity of glass

We are given the perimeter of upper and lower ends thus we can find r 1 and r 2 .

$2\pi r_1\ =\ 18$

$r_1\ =\ \frac{9}{\pi}\ cm$

And,

$2\pi r_2\ =\ 6$

$r_2\ =\ \frac{3}{\pi}\ cm$

Thus curved surface area of the frustum is given by : $=\ \pi \left ( r_1\ +\ r_2 \right )l$

$=\ \pi \left ( \frac{9}{\pi}\ +\ \frac{3}{\pi} \right )4$

$=\ 48\ cm^2$

The area of material used is given by :

Area of material = Curved surface area of a frustum of cone + Area of upper end

$=\ \pi \left ( r_1\ +\ r_2 \right )l\ +\ \pi r^2_2$

$=\ \pi \left ( 10\ +\ 4 \right )15\ +\ \pi \times 4^2$

$=\ \frac{226\times 22}{7}$

$=\ 710.28\ cm^2$

Firstly we will calculate the slant height of the cone :

$l^2\ =\ \left ( r_1^2\ -\ r^2_2 \right )\ +\ h^2$

$l^2=\ \left ( 20^2\ -\ 8^2 \right )\ +\ 16^2$

$l\ =\ 20\ cm$

Now, the volume of the frustum is :

$=\ \frac{1}{3}\pi h\left ( r^2_1\ +\ r^2_2\ +\ r_1r_2 \right )$

$=\ \frac{1}{3}\pi \times 16\left ( 20^2\ +\ 8^2\ +\ 20\times 8 \right )$

$=\ 10449.92\ cm^3$

= Capacity of the container.

Now, the cost of 1-liter milk is Rs. 20.

Then the cost of 10.449-liter milk will be $=\ 10.45\times 20\ =\ Rs.\ 209$

The metal sheet required for the container is :

$=\ \pi (r_1\ +\ r_2)l\ +\ \pi r_2^2$

$=\ \pi (20\ +\ 8)20\ +\ \pi\times 8^2$

$=\ 624 \pi\ cm^2$

Thus cost for metal sheet is

$=\ 624 \pi\ \times \frac{8}{100}$

$=\ Rs.\ 156.57$

The figure for the problem is shown below :

Using geometry we can write :

$EG\ =\ \frac{10\sqrt{3}}{3}\ cm$ and $BD\ =\ \frac{20\sqrt{3}}{3}\ cm$

Thus the volume of the frustum is given by :

$\\=\ \frac{1}{3}\pi h\left ( r^2_1\ +\ r^2_2\ +\ r_1r_2 \right )\\\\=\ \frac{1}{3}\pi \times 10\left ( \left ( \frac{10\sqrt{3}}{3} \right )^2\ +\ \left ( \frac{20\sqrt{3}}{3} \right )^2\ +\ \left ( \frac{10\sqrt{3}}{3} \right )\times \left ( \frac{20\sqrt{3}}{3} \right ) \right )\\\\=\ \frac{22000}{9}\ cm^3$

Now, the radius of the wire is :

$=\ \frac{1}{16}\times \frac{1}{2}\ =\ \frac{1}{32}\ cm$

Thus the volume of wire is given by : $=\ \pi r^2\times l$

$=\ \pi \times \left ( \frac{1}{32} \right )^2\times l$

Now equating volume of frustum and wire, we get :

$\frac{22000}{9}\ =\ \pi\times \left ( \frac{1}{32} \right )^2\times l$

$l\ =\ 796444.44\ cm$

$l\ =\ 7964.44\ m$

Thus the length of wire drawn is 7964.44 m.

NCERT solutions for class 10 CBSE maths surface area and volume Excercise: 13.5

A number of rounds are calculated by :

$=\ \frac{Height\ of\ cylinder }{Diameter\ of\ wire}$

$=\ \frac{12 }{0.3}\ =\ 40\ rounds$

Thus the length of wire in 40 rounds will be $=\ 40\times 2\pi \times 5\ =\ 400 \pi\ cm\ =\ 12.57\ m$

And the volume of wire is: Area of cross-section $\times$ Length of wire

$=\pi \times (0.15)^2\times 1257.14$

$=\ 88.89\ cm^3$

Hence the mass of wire is. $=\ 88.89\ \times 8.88\ =\ 789.41\ gm$

The volume of the double cone will be = Volume of cone 1 + Volume of cone 2.

$=\ \frac{1}{3}\ \pi r^2h_1\ +\ \frac{1}{3}\ \pi r^2h_2$

$=\ \frac{1}{3}\ \pi \times 2.4^2\times 5$ (Note that sum of heights of both the cone is 5 cm - hypotenuse).

$=\ 30.14\ cm^3$

Now the surface area of a double cone is :

$=\ \pi rl_1\ +\ \pi rl_2$

$=\ \pi \times 2.4 \left ( 4\ +\ 3 \right )$

$=\ 52.8\ cm^2$

The total volume of the cistern is : $=\ 150\times 120\times 110\ =\ 1980000\ cm^3$

And the volume to be filled in it is $= 1980000 - 129600\ =\ 1850400\ cm^3$

Now let the number of bricks be n.

Then the volume of bricks : $= n\times 22.5\times 7.5\times 6.5 \ =\ 1096.87n\ cm^3$

Further, it is given that brick absorbs one-seventeenth of its own volume of water.

Thus water absorbed :

$=\ \frac{1}{17}\times 1096.87n\ cm^3$

Hence we write :

$1850400\ +\ \frac{1}{17}( 1096.87n)\ =\ 1096.87n$

$n\ =\ 1792.41$

Thus the total number of bricks is 1792.

Firstly we will calculate the volume of rainfall :

Volume of rainfall :

$=\ 7280\times \frac{10}{100\times 1000}$

$=\ 0.7280\ Km ^3$

And the volume of the three rivers is :

$=\ 3\left ( 1072\times \frac{75}{1000}\times \frac{3}{1000} \right )\ Km^3$

$=\ 0.7236\ Km ^3$

It can be seen that both volumes are approximately equal to each other.

From this, we can write the values of both the radius (upper and lower) and the height of the frustum.

Thus slant height of frustum is :

$=\ \sqrt{\left ( r_1\ -\ r_2 \right )^2\ +\ h^2}$

$=\ \sqrt{\left ( 9\ -\ 4 \right )^2\ +\ 12^2}$

$=\ 13\ cm$

Now, the area of the tin shed required :

= Area of frustum + Area of the cylinder

$=\ \pi \left ( r_1\ +\ r_2 \right )l\ +\ 2\pi r_2h$

$=\ \pi \left ( 9\ +\ 4 \right )13\ +\ 2\pi \times 4\times 10$

$=\ 782.57\ cm^2$

In the case of the frustum, we can consider:- removing a smaller cone (upper part) from a larger cone.

So the CSA of frustum becomes:- CSA of bigger cone - CSA of the smaller cone

$\\=\ \pi r_1l_1\ -\ \pi r_2 (l_1\ -\ l)\\\\=\ \pi r_1\left ( \frac{lr_1}{r_1\ -\ r_2} \right )\ -\ \pi r_2\left ( \frac{r_1l}{r_1\ -\ r_2}\ -\ l \right )\\\\=\ \left ( \frac{\pi r_1^2 l}{r_1\ -\ r_2} \right )\ -\ \left ( \frac{\pi r_2^2 l}{r_1\ -\ r_2} \right )\\\\=\ \pi l\left ( \frac{r_1^2\ -\ r_2^2}{r_1\ -\ r_2} \right )\\\\=\ \pi l\left ( r_1\ +\ r_2 \right )$

And the total surface area of the frustum is = CSA of frustum + Area of upper circle and area of lower circle.

$=\ \pi l\left ( r_1\ +\ r_2 \right )\ +\ \pi r_1^2\ +\ \pi r_2^2$

Similar to how we find the surface area of the frustum.

The volume of the frustum is given by-

=Volume of the bigger cone - Volume of the smaller cone

$=\ \frac{1}{3} \pi r^2_1 h_1\ -\ \frac{1}{3} \pi r^2_2 (h_1\ -\ h)$

$=\ \frac{\pi}{3}\left ( r_1^2 h_1\ -\ r_2^2 (h_1\ -\ h) \right )$

$=\ \frac{\pi}{3}\left ( r_1^2 \left ( \frac{hr_1}{r_1\ -\ r_2} \right )\ -\ r_2^2 \left ( \frac{hr_1}{r_1\ -\ r_2}\ -\ h \right ) \right )$

$=\ \frac{\pi}{3}\left ( \frac{hr_1^3}{r_1\ -\ r_2}\ -\ \frac{hr_2^3}{r_1\ -\ r_2} \right )$

$=\ \frac{\pi}{3}h\left ( \frac{r_1^3\ -\ r_2^3}{r_1\ -\ r_2} \right )$

$=\ \frac{1}{3} \pi h\left ( r_1^2\ +\ r_2^2\ +\ r_1r_2 \right )$

## Key Features For Surface Area and Volume Class 10 Solutions

Board Exam Excellence: These ch 13 maths class 10 solutions are designed to help students excel in their class 10 board exams, providing comprehensive guidance and explanations.

Real-Life Application: Understanding the concepts of surface areas and volumes is crucial for practical scenarios. These ch 13 maths class 10 solutions enable students to calculate the dimensions of everyday objects like rectangular boxes and gas cylinders.

Versatile Usage: Students can apply the formulas and techniques learned from these ch 13 maths class 10 solutions to solve textbook questions, excel in exams, and tackle real-life problems effectively.

## NCERT Solutions For Class 10 Maths - Chapter Wise

 Chapter No. Chapter Name Chapter 1 Real Numbers Chapter 2 Polynomials Chapter 3 Pair of Linear Equations in Two Variables Chapter 4 Quadratic Equations Chapter 5 Arithmetic Progressions Chapter 6 Triangles Chapter 7 Coordinate Geometry Chapter 8 Introduction to Trigonometry Chapter 9 Some Applications of Trigonometry Chapter 10 Circles Chapter 11 Constructions Chapter 12 Areas Related to Circles Chapter 13 Surface Areas and Volumes Chapter 14 Statistics Chapter 15 Probability

## NCERT Exemplar Solutions - Subject Wise

### How to use chapter 13 class 10 maths NCERT solutions?

• First of all, memorize the formulae to calculate curved surface area, total surface area and volumes of different shapes.
• In word problems, most of the students get confused in calculating curved surface area and total surface area. So after going through some examples of the textbook, you can take an idea to solve this confusion.
• After this, you should immediately move to the next step which is practice exercises. While practicing, if you have a doubt in any question, then you can refer to NCERT solutions for class 10 Maths Chapter 13 Surface Areas and Volumes.

Students must complete the NCERT Class 10 Maths syllabus as soon as possible.

## NCERT Solutions of Class 10 - Subject Wise

1. What is the weightage of the NCERT solutions for class 10 Maths chapter 13 Surface Areas and Volumes for CBSE board exam ?

Two chapters - area related to circles and surface areas and volumes have combined has 10 marks weightage in the CBSE class 10 maths final board exam. The syllabus of the chapter is similar to the NCERT syllabus. The students can follow NCERT book and NCERT exemplar problem to get a good score in the board exam.

2. How can I score full marks in a class test for surface area volume class 10?

To achieve full marks in a class test for Class 10 Maths Chapter 13, utilizing the surface area and volume class 10 solutions provided on Careers360 website can be very helpful. These solutions are an essential tool for quick and easy revision during class tests and exams. Additionally, these solutions for class 10 maths surface area and volume serve as the best study material for Class 10 students.

3. Where can I find the complete solutions of NCERT class 10 maths ?

Here you will get the detailed NCERT solutions for class 10 maths  by clicking on the link. On opening the link all the chapters are listed. Students can select the chapter for which they required the solution. All the exercise of the chapter are solved with all the necessary steps. Students can download NCERT class 10 Maths chapter 13 pdf  using the link given above in this article and study both online and offline mode.

4. How are surface area and volume class 10 ncert solutions helpful in CBSE exams?

NCERT Solutions for Class 10 Maths Chapter 13 provide answers with detailed explanations that align with the latest CBSE syllabus. Practicing these questions will ensure that students are well-prepared for any type of question that may appear in the CBSE exams. Interested students can study both online and offline surface area and volume class 10 pdf at Careers360 website.

5. What are the main concepts covered in surface area and volume formulas class 10 ncert solutions?

The main concepts covered in volume surface area class 10 include the surface area of a combination of solids, the volume of a combination of solids, the conversion of solid shapes, and the frustum of a cone. To get indepth understanding students can practice surface areas and volumes class 10 solutions

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### Questions related to CBSE Class 10th

Have a question related to CBSE Class 10th ?

Hello

The Sadhu Ashram in Aligarh is located in Chhalesar . The ashram is open every day of the week, except for Thursdays . On Mondays, Wednesdays, and Saturdays, it's open from 8:00 a.m. to 7:30 p.m., while on Tuesdays and Fridays, it's open from 7:30 a.m. to 7:30 p.m. and 7:30 a.m. to 6:00 a.m., respectively . Sundays have varying hours from 7:00 a.m. to 8:30 p.m. . You can find it at Chhalesar, Aligarh - 202127 .

Yes, scoring above 80% in ICSE Class 10 exams typically meets the requirements to get into the Commerce stream in Class 11th under the CBSE board . Admission criteria can vary between schools, so it is advisable to check the specific requirements of the intended CBSE school. Generally, a good academic record with a score above 80% in ICSE 10th result is considered strong for such transitions.

hello Zaid,

Yes, you can apply for 12th grade as a private candidate .You will need to follow the registration process and fulfill the eligibility criteria set by CBSE for private candidates.If you haven't given the 11th grade exam ,you would be able to appear for the 12th exam directly without having passed 11th grade. you will need to give certain tests in the school you are getting addmission to prove your eligibilty.

best of luck!

According to cbse norms candidates who have completed class 10th, class 11th, have a gap year or have failed class 12th can appear for admission in 12th class.for admission in cbse board you need to clear your 11th class first and you must have studied from CBSE board or any other recognized and equivalent board/school.

You are not eligible for cbse board but you can still do 12th from nios which allow candidates to take admission in 12th class as a private student without completing 11th.

A block of mass 0.50 kg is moving with a speed of 2.00 ms-1 on a smooth surface. It strikes another mass of 1.00 kg and then they move together as a single body. The energy loss during the collision is

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 Option 1) 0.02 Option 2) 3.125 × 10-2 Option 3) 1.25 × 10-2 Option 4) 2.5 × 10-2

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