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

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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.

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Surface Area And Volume Class 10 NCERT Solution
NCERT Solutions for Class 10 Maths Chapter 13 Surface Areas and Volumes PDF Free Download
Surface Area And Volume Class 10 NCERT Solution - Important Formulae
NCERT Solutions for Class 10 Maths Chapter 13 Surface Areas and Volumes (Intext Questions and Exercise)
NCERT Books and NCERT Syllabus
Key Features For Surface Area and Volume Class 10 Solutions
NCERT Solutions For Class 10 Maths - Chapter Wise
NCERT Exemplar Solutions - Subject Wise
NCERT Solutions of Class 10 - Subject Wise

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

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

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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

Q1 2 cubes each of volume 64 cm ³ are joined end to end. Find the surface area of the resulting cuboid.

Answer:

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 .

Q2 A vessel is in the form of a hollow hemisphere mounted by a hollow cylinder. The diameter of the hemisphere is 14 cm and the total height of the vessel is 13 cm. Find the inner surface area of the vessel.

Answer:

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 $= 308 + 264 = 572 \:cm^2.$

Q3 A toy is in the form of a cone of radius 3.5 cm mounted on a hemisphere of the same radius. The total height of the toy is 15.5 cm. Find the total surface area of the toy.

Answer:

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 = $=\ 77\ +\ 137.5\ =\ 214.5\ cm^2$

Q4 A cubical block of side 7 cm is surmounted by a hemisphere. What is the greatest diameter the hemisphere can have? Find the surface area of the solid.

Answer:

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 $= 294 + 77 - 38.5 = 332.5 \:cm^2$ .

Q5 A hemispherical depression is cut out from one face of a cubical wooden block such that the diameter l of the hemisphere is equal to the edge of the cube. Determine the surface area of the remaining solid.

Answer:

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:

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

Q6 A medicine capsule is in the shape of a cylinder with two hemispheres stuck to each of its ends (see Fig.). The length of the entire capsule is 14 mm and the diameter of the capsule is 5 mm. Find its surface area.

1636091994484

Answer:

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 :

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

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

$=\ 70 \pi$

$=\ 220\ mm^2$

Q7 A tent is in the shape of a cylinder surmounted by a conical top. If the height and diameter of the cylindrical part are 2.1 m and 4 m respectively, and the slant height of the top is 2.8 m, find the area of the canvas used for making the tent. Also, find the cost of the canvas of the tent at the rate of Rs 500 per m ² . (Note that the base of the tent will not be covered with canvas.)

Answer:

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

Q8 From a solid cylinder whose height is 2.4 cm and diameter 1.4 cm, a conical cavity of the same height and same diameter is hollowed out. Find the total surface area of the remaining solid to the nearest cm ² .

Answer:

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 .

Q9 A wooden article was made by scooping out a hemisphere from each end of a solid cylinder, as shown in Fig. 13.11. If the height of the cylinder is 10 cm, and its base is of radius 3.5 cm, find the total surface area of the article.

1636092429678

Answer:

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

Chapter 13 class 10 maths NCERT solutions Excercise: 13.2

Q1 A solid is in the shape of a cone standing on a hemisphere with both their radii being equal to 1 cm and the height of the cone is equal to its radius. Find the volume of the solid in terms of $\pi$ .

Answer:

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 :

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

Q2 Rachel, an engineering student, was asked to make a model shaped like a cylinder with two cones attached at its two ends by using a thin aluminum sheet. The diameter of the model is 3 cm and its length is 12 cm. If each cone has a height of 2 cm, find the volume of air contained in the model that Rachel made. (Assume the outer and inner dimensions of the model to be nearly the same.)

Answer:

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 $=\ 66\ cm^3$

Q3 A gulab jamun contains sugar syrup up to about 30% of its volume. Find approximately how much syrup would be found in 45 gulab jamuns, each shaped like a cylinder with two hemispherical ends with length 5 cm and diameter 2.8 cm (see Fig).

1636092497555

Answer:

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

Further, it is given that one gulab jamun contains sugar syrup upto .

So, the total volume of sugar present :

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

Q4 A pen stand made of wood is in the shape of a cuboid with four conical depressions to hold pens. The dimensions of the cuboid are 15 cm by 10 cm by 3.5 cm. The radius of each of the depressions is 0.5 cm and the depth is 1.4 cm. Find the volume of wood in the entire stand (see Fig. 13.16).

SURFACE AREAS AND VOLUMES

Answer:

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 $= 525 + 4 (0.3665) = 523.53 \:cm^3$

Q5 A vessel is in the form of an inverted cone. Its height is 8 cm and the radius of its top, which is open, is 5 cm. It is filled with water up to the brim. When lead shots, each of which is a sphere of radius 0.5 cm are dropped into the vessel, one-fourth of the water flows out. Find the number of lead shots dropped in the vessel.

Answer:

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.

Q6 A solid iron pole consists of a cylinder of height 220 cm and base diameter 24 cm, which is surmounted by another cylinder of height 60 cm and radius 8 cm. Find the mass of the pole, given that 1 cm 3 of iron has approximately 8g mass. (Use $\pi = 3.14$ )

Answer:

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

Q7 A solid consisting of a right circular cone of height 120 cm and radius 60 cm standing on a hemisphere of radius 60 cm is placed upright in a right circular cylinder full of water such that it touches the bottom. Find the volume of water left in the cylinder, if the radius of the cylinder is 60 cm and its height is 180 cm.

Answer:

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 :

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

Q8 A spherical glass vessel has a cylindrical neck 8 cm long, 2 cm in diameter; the diameter of the spherical part is 8.5 cm. By measuring the amount of water it holds, a child finds its volume to be 345 cm 3 . Check whether she is correct, taking the above as the inside measurements, and $\pi = 3.14$ .

Answer:

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

Q1 A metallic sphere of radius 4.2 cm is melted and recast into the shape of a cylinder of radius 6 cm. Find the height of the cylinder.

Answer:

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.

Q2 Metallic spheres of radii 6 cm, 8 cm, and 10 cm, respectively, are melted to form a single solid sphere. Find the radius of the resulting sphere.

Answer:

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.

Q3 A 20 m deep well with diameter 7 m is dug and the earth from digging is evenly spread out to form a platform 22 m by 14 m. Find the height of the platform.

Answer:

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.

Q4 A well of diameter 3 m is dug 14 m deep. The earth taken out of it has been spread evenly all around it in the shape of a circular ring of width 4 m to form an embankment. Find the height of the embankment.

Answer:

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.

Q5 A container shaped like a right circular cylinder having a diameter of 12 cm and a height 15 cm is full of ice cream. The ice cream is to be filled into cones of height 12 cm and diameter 6 cm, having a hemispherical shape on the top. Find the number of such cones which can be filled with ice cream.

Answer:

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.

Q6 How many silver coins, 1.75 cm in diameter and of thickness 2 mm, must be melted to form a cuboid of dimensions 5.5 cm × 10 cm × 3.5 cm?

Answer:

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.

Q7 A cylindrical bucket, 32 cm high and with radius of base 18 cm, is filled with sand. This bucket is emptied on the ground and a conical heap of sand is formed. If the height of the conical heap is 24 cm, find the radius and slant height of the heap.

Answer:

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 .

Q8 Water in a canal, 6 m wide and 1.5 m deep, is flowing with a speed of 10 km/h. How much area will it irrigate in 30 minutes, if 8 cm of standing water is needed?

Answer:

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 $562500 \:m^2$ .

Q9 A farmer connects a pipe of internal diameter 20 cm from a canal into a cylindrical tank in her field, which is 10 m in diameter and 2 m deep. If water flows through the pipe at the rate of 3 km/h, in how much time will the tank be filled?

Answer:

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

Q1 A drinking glass is in the shape of a frustum of a cone of height 14 cm. The diameters of its two circular ends are 4 cm and 2 cm. Find the capacity of the glass.

Answer:

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

Q2 The slant height of a frustum of a cone is 4 cm and the perimeters (circumference) of its circular ends are 18 cm and 6 cm. Find the curved surface area of the frustum.

Answer:

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

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

$=\ 48\ cm^2$

Q3 A fez , the cap used by the Turks, is shaped like the frustum of a cone (see Fig.). If its radius on the open side is 10 cm, radius at the upper base is 4 cm and its slant height is 15 cm, find the area of material used for making it .

1636092575043

Answer:

The area of material used is given by :

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

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

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

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

$=\ 710.28\ cm^2$

Q4 A container opened from the top and made up of a metal sheet, is in the form of a frustum of a cone of height 16 cm with radii of its lower and upper ends as 8 cm and 20 cm, respectively. Find the cost of the milk which can completely fill the container, at the rate of Rs 20 per litre. Also find the cost of metal sheet used to make the container, if it costs Rs 8 per 100 cm 2 . (Take $\pi = 3.14$ )

Answer:

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

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

$=\ Rs.\ 156.57$

Q5 A metallic right circular cone 20 cm high and whose vertical angle is 60° is cut into two parts at the middle of its height by a plane parallel to its base. If the frustum so obtained be drawn into a wire of diameter $\frac{1}{16}\textup{cm}$ , find the length of the wire.

Answer:

The figure for the problem is shown below :

Surface area and volume , 25864

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

Q1 A copper wire, 3 mm in diameter, is wound about a cylinder whose length is 12 cm, and diameter 10 cm, so as to cover the curved surface of the cylinder. Find the length and mass of the wire, assuming the density of copper to be 8.88 g per cm ³ .

Answer:

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.

Q2 A right triangle, whose sides are 3 cm and 4 cm (other than hypotenuse) is made to revolve about its hypotenuse. Find the volume and surface area of the double cone so formed. (Choose value of $\pi$ as found appropriate.)

Answer:

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 :

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

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

$=\ 52.8\ cm^2$

Q3 A cistern, internally measuring 150 cm × 120 cm × 110 cm, has 129600 cm ³ of water in it. Porous bricks are placed in the water until the cistern is full to the brim. Each brick absorbs one-seventeenth of its own volume of water. How many bricks can be put in without overflowing the water, each brick being 22.5 cm × 7.5 cm × 6.5 cm?

Answer:

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.

Q4 In one fortnight of a given month, there was a rainfall of 10 cm in a river valley. If the area of the valley is 7280 km ² , show that the total rainfall was approximately equivalent to the addition to the normal water of three rivers each 1072 km long, 75 m wide and 3 m deep.

Answer:

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.

Q5 An oil funnel made of tin sheet consists of a 10 cm long cylindrical portion attached to a frustum of a cone. If the total height is 22 cm, the diameter of the cylindrical portion is 8 cm and the diameter of the top of the funnel is 18 cm, find the area of the tin sheet required to make the funnel (see Fig.).

1648793390695

Answer:

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

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

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

$=\ 782.57\ cm^2$

Q6 Derive the formula for the curved surface area and total surface area of the frustum of a cone, given to you in Section 13.5, using the symbols as explained.

Answer:

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.

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

Q7 Derive the formula for the volume of the frustum of a cone, given to you in Section 13.5, using the symbols as explained.

Answer:

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

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

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

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

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

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

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

NCERT Books and NCERT Syllabus

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

Also Refer,

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

Frequently Asked Questions (FAQs)

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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sadhu ashram ramgart road aligarh

If you're looking for directions or steps to reach Sadhu Ashram on Ramgart Road in Aligarh, here’s how you can get there:

Steps to Reach Sadhu Ashram, Ramgart Road, Aligarh:

Starting Point:
- Determine your starting point in Aligarh or the nearby area.
Use Google Maps:
- Open Google Maps on your phone or computer.
- Enter "Sadhu Ashram, Ramgart Road, Aligarh" as your destination.
- Follow the navigation instructions provided by Google Maps.
By Local Transport:
- Auto-rickshaw/Taxi: Hire an auto-rickshaw or taxi and inform the driver about your destination. Most local drivers should be familiar with Sadhu Ashram.
- Bus: Check if there are any local buses that operate on Ramgart Road. Ask the bus conductor if the bus stops near Sadhu Ashram.
Landmarks to Look For:
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Ask for Directions:
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If you need detailed directions from a specific location or more information about Sadhu Ashram, feel free to ask

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sai raja 01 September,2024

show the previous year exams ?

Hello Aspirant, Hope your doing great, your question was incomplete and regarding what exam your asking.

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SWETCHCHA MISKA 10 July,2024

as an icse student in class 10 is above 80 percent good enough to get into commerce in 11th cbse

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.

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Manasvini Gupta 23 July,2024

i had cleared 10th from cbse in 2022 2 years ago now i wish to apply for 12th as private candidate in 2025. can i do so?

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!

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Ashvadha 03 July,2024

i had cleared 10th in 2022 from cbse can i apply for 12th as private candidate this year?

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.

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Central Board of Secondary Education Class 10th Examination

Applications Open

Option 1) $0.34\; J$	Option 2) $0.16\; J$
Option 3) $1.00\; J$	Option 4) $0.67\; J$

Option 1) $2,000 \; J - 5,000\; J$	Option 2) $200 \, \, J - 500 \, \, J$
Option 3) $2\times 10^{5}J-3\times 10^{5}J$	Option 4) $20,000 \, \, J - 50,000 \, \, J$

Option 1) $K/2\,$	Option 2) $\; K\;$
Option 3) $zero\;$	Option 4) $K/4$

Option 1) $11.2\, L\, H_{2(g)}$ at STP is produced for every mole $HCL_{(aq)}$ consumed	Option 2) $6L\, HCl_{(aq)}$ is consumed for ever $3L\, H_{2(g)}$ produced
Option 3) $33.6 L\, H_{2(g)}$ is produced regardless of temperature and pressure for every mole Al that reacts	Option 4) $67.2\, L\, H_{2(g)}$ at STP is produced for every mole Al that reacts .

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

Option 1) decrease twice	Option 2) increase two fold
Option 3) remain unchanged	Option 4) be a function of the molecular mass of the substance.

Option 1) Molality	Option 2) Weight fraction of solute
Option 3) Fraction of solute present in water	Option 4) Mole fraction.

Option 1) twice that in 60 g carbon	Option 2) 6.023 × 10²²
Option 3) half that in 8 g He	Option 4) 558.5 × 6.023 × 10²³

Option 1) less than 3	Option 2) more than 3 but less than 6
Option 3) more than 6 but less than 9	Option 4) more than 9

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

Surface Area And Volume Class 10 NCERT Solution

Pearson | PTE

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

Surface Area And Volume Class 10 NCERT Solution - Important Formulae

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

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NCERT Books and NCERT Syllabus

Key Features For Surface Area and Volume Class 10 Solutions

NCERT Solutions For Class 10 Maths - Chapter Wise

NCERT Exemplar Solutions - Subject Wise

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