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

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

Having a good base in mathematics is very important from a career perspective, and class 10 is the time to build a strong foundation in it. Our expert team at Vedantu has created a detailed Solution of Chapter 13 Maths Class 10, which will clear your concepts at the root level. Once you go through the lucid explanations of Class 10 Surface Area and Volume Solutions provided by our subject matter experts, you don't need to mug up formulas. There are many tips and tricks taught here for quickly recalling important points of ch 13 Class 10 Maths. The Solution of Surface Area and Volume Class 10 is designed considering the latest CBSE curriculum; hence students can be sure of scoring high marks in maths with our solutions. Also, you can revise and solve the updated NCERT Book solutions provided by us. In the solutions, you will also find solutions for other subjects like NCERT Solutions Class 10 Science for free only at Vedantu.


Class:

NCERT Solutions for Class 10

Subject:

Class 10 Maths

Chapter Name:

Chapter 13 - Surface Areas and Volumes

Content-Type:

Text, Videos, Images and PDF Format

Academic Year:

2023-24

Medium:

English and Hindi

Available Materials:

  • Chapter Wise

  • Exercise Wise

Other Materials

  • Important Questions

  • Revision Notes

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

Exercise 13.1: This exercise involves questions that are based on finding the surface areas and volumes of cuboids and cubes. The exercise contains a total of six questions that test the student's understanding of the formulas and concepts related to finding the surface area and volume of cuboids and cubes. The solutions to each question provide step-by-step instructions on how to find the surface area and volume of the given figures.

Exercise 13.2: This exercise involves questions based on finding the surface areas and volumes of the right circular cylinders and cones. The exercise contains a total of six questions that test the student's understanding of the concepts related to the surface area and volume of cylinders and cones. The solutions to each question provide step-by-step instructions and diagrams to help students understand the concept better.

Exercise 13.3: This exercise focuses on questions based on finding the surface areas and volumes of spheres. The exercise contains a total of five questions that test the student's understanding of the formulas and concepts related to spheres and their properties. The solutions to each question provide step-by-step instructions and diagrams to help students understand the concept better.

Exercise 13.4: This exercise involves questions based on practical geometry, including constructing a frustum of a cone, finding the surface area and volume of a frustum, and finding the area of a playground in the shape of a parallelogram. The exercise contains a total of six questions that test the student's understanding of the practical applications of geometry. The solutions to each question provide step-by-step instructions and diagrams to help students understand the concept better.

Exercise 13.5: This exercise focuses on questions based on the concept of surface area and volume of combinations of solids, including cubes, cuboids, cylinders, cones, and spheres. The exercise contains a total of five questions that test the student's understanding of the formulas and concepts related to combinations of solids. The solutions to each question provide step-by-step instructions and diagrams to help students understand the concept better.


Overall, exercises in NCERT Solutions for Class 10 Maths Chapter 13 "Surface Areas and Volumes", are designed to help students develop their understanding of the concepts and formulas related to the surface areas and volumes of different geometrical shapes, and to enhance their problem-solving skills.


NCERT Solutions of Class 10 Maths Chapter 13 - Free PDF Download

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Exercise (13.1)

1. $2$ cubes of each of volume $64\text{ c}{{\text{m}}^{3}}$ are joined end to end. Find the surface area of the resulting cuboids.

Ans:

Given that,

$2$ cubes are joined end to end as given in the following diagram.

To find the surface area of the resulting cuboid.

Volume of each cube $=64\text{ c}{{\text{m}}^{3}}$

We know that,

Volume of a cube $={{a}^{3}}$

${{a}^{3}}=64$

$a=4\text{ cm}$

Thus the dimension of the resulting cuboid is of $4\text{ cm, }4\text{ cm}$ and $8\text{ cm}$when they are joined end to end. That is, $l=4\text{ cm, b=}4\text{ cm}$and $h=8\text{ cm}$

 

cuboid with a height of 8cm and a base side of 4cm

 

Then,

Surface area of cuboid \[\]

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

$=2\left( 16+32+32 \right)$

$=\sqrt{{{\left( -3 \right)}^{2}}+{{\left( 3 \right)}^{2}}}$

$=160\text{ c}{{\text{m}}^{2}}$

\[\]The surface area of the resultant cuboid is $160\text{ c}{{\text{m}}^{2}}$

 

2. A vessel is in the form of a hollow hemisphere mounted by a hollow cylinder. The diameter of the hemisphere is $14\text{ cm}$ and the total height of the vessel is $13\text{ cm}$. Find the inner surface area of the vessel. (Use $\pi =\frac{22}{7}$)

Ans:

Given that,

The diameter of the hemisphere $=14\text{ cm}$

The total height of the vessel \[\]

To find,

The inner surface area of the vessel.

Thus the radius of the hollow hemisphere $=\frac{d}{2}$

$=\frac{14}{2}$

$=7\text{ cm}$

 

A vessel with base of hollow hemisphere mounted by a hollow cylinder

 

From the diagram, it can be observed that the radius of the cylindrical part and that of the hemispherical part is the same.

Thus, height of hemispherical part $=$ Radius $=7\text{ cm}$

Height of the cylindrical part$=13-7$

$=6\text{ cm}$

Inner surface area of the vessel$=$ CSA of cylindrical part + CSA of hemispherical part

CSA of cylindrical part $=2\pi rh$

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

$=\left( 2\times 22\times 6 \right)$

$=44\times 6$

CSA of hemispherical part$=2\pi {{r}^{2}}$

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

$AB,BC,CD=\left( 2\times 22\times 7 \right)$

$=44\times 7$

Inner surface area of the vessel$=44\left( 6+7 \right)$

$=44\left( 13 \right)$

$=572\text{ c}{{\text{m}}^{2}}$

$\therefore $The inner surface area of the vessel is $572\text{ c}{{\text{m}}^{2}}$

 

3. A toy is in the form of a cone of radius $3.5\text{ cm}$ mounted on a hemisphere of the same radius. The total height of the toy is $15.5\text{ cm}$. Find the total surface area of the toy. (Use $\pi =\frac{22}{7}$)

Ans:

Given that,

Radius of the cone $=3.5\text{ cm}$

Radius of the hemisphere$=3.5\text{ cm}$

The total height of the toy$=15.5\text{ cm}$

To find,

The total surface area of the toy

 

A toy with the base of a hollow hemisphere mounted by a cone

 

From the diagram, it can be observed that the radius of both conical and hemispherical part is the same.

Height of the hemispherical part\[\]

$=\frac{7}{2}$

Height of the conical part$=15.5-3.5$

$=12\text{ cm}$

Slant height of the conical part$\left( l \right)=\sqrt{{{r}^{2}}+{{h}^{2}}}$

$=\sqrt{{{\left( \frac{7}{2} \right)}^{2}}+{{\left( 12 \right)}^{2}}}$

$=\sqrt{\left( \frac{49}{4} \right)+144}$

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

$=\frac{25}{2}D\left( -3,0 \right)$

Total surface area of the toy$=$ CSA of conical part + CSA of hemispherical part

CSA of conical part   $=\pi rl$

$=\left( \frac{22}{7}\times \frac{7}{2}\times \frac{25}{2} \right)$

=137.5

CSA of hemispherical part  $=2\pi {{r}^{2}}$

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

$=77$

Total surface area of the toy$=137.5+77$

$=214.5\text{ c}{{\text{m}}^{2}}$

$\therefore $The total surface area of the toy is $214.5\text{ c}{{\text{m}}^{2}}$.

 

4. A cubical block of side $7\text{ cm}$ is surmounted by a hemisphere. What is the greatest diameter the hemisphere can have? Find the surface area of the solid. (Use $\pi =\frac{22}{7}$)

Ans:

Given that,

Side of a cube $=7\text{ cm}$

To find,

  • The greatest diameter of the hemisphere.

  • The surface area of the solid.

 

A cubical block with surmounted by a hemisphere

 

It can be observed from the diagram that the greatest possible diameter of the hemisphere is equal to the cube’s edge.

Thus the greatest diameter of the hemispherical part$=7\text{ cm}$

So the radius of the hemispherical part $=\frac{7}{2}$

$=3.5\text{ cm}$

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

$=6{{a}^{2}}+2\pi {{r}^{2}}-\pi {{r}^{2}}$

$=6{{a}^{2}}+\pi {{r}^{2}}$

$=6{{\left( 7 \right)}^{2}}+\left( \frac{22}{7}\times {{\left( \frac{7}{2} \right)}^{2}} \right)$

$=294+38.5$

$=332.5\text{ c}{{\text{m}}^{2}}$

$\therefore $The greatest diameter of the hemisphere is $7\text{ cm}$ and the surface area of the solid is $332.5\text{ c}{{\text{m}}^{2}}$.

 

5. 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.

Ans:

Given that,

Diameter of the hemisphere = Edge of the cube

To determine,

The surface area of the remaining solid.

 

A hemispherical depression is cut out from one face of a cubical wooden block

 

Diameter of the hemisphere = Edge of the cube \[\]

   Radius of the hemisphere $=\frac{l}{2}$

Total surface area of the solid = Surface area of cubical part + CSA of hemispherical part – Area of base of hemispherical part

$=6{{a}^{2}}+2\pi {{r}^{2}}-\pi {{r}^{2}}$

$=6{{a}^{2}}+\pi {{r}^{2}}$

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

$=6{{l}^{2}}+\frac{\pi {{l}^{2}}}{4}$

$=\frac{24{{l}^{2}}+\pi {{l}^{2}}}{4}$

$=\frac{1}{4}\left( 24+\pi  \right){{l}^{2}}\text{ uni}{{\text{t}}^{2}}$

$\therefore $The area of the remaining solid is $\frac{1}{4}\left( 24+\pi  \right){{l}^{2}}\text{ uni}{{\text{t}}^{2}}B\left( 3,1 \right)$.

 

6. A medicine capsule is in the shape of a cylinder with two hemispheres stuck to each of its ends (see the given figure). The length of the entire capsule is $14\text{ mm}$ and the diameter of the capsule is $5\text{ mm}$. Find its surface area. (Use $\pi =\frac{22}{7}$)

 

seo images

 

Ans:

Given that,

Length of the capsule $=15\text{ mm}$

Diameter of the capsule $=5\text{ mm}$

To find, 

The surface area of the capsule

From the diagram,

Radius of cylindrical part = Radius of hemispherical part

$=\frac{\text{Diameter of the capsule}}{2}$

$=\frac{5}{2}$

Length of the cylindrical part (h) = Length of the entire capsule - $2r$

$=14-2\left( \frac{5}{2} \right)$

$=9\text{ mm}$

Surface area of capsule = 2CSA of hemispherical part + CSA of cylindrical part

2CSA of hemispherical part  $=2\left( 2\pi {{r}^{2}} \right)$

$=4\pi {{\left( \frac{5}{2} \right)}^{2}}$

$=25\pi $

CSA of cylindrical part  $=2\pi rh$

$=2\pi \left( \frac{5}{2} \right)\times 9$

Surface area of the capsule$=25\pi +45\pi $

$=70\pi $

$=70\times \frac{22}{7}$

$=10\times 22$

$=220\text{ m}{{\text{m}}^{2}}$

∴The surface area of the capsule is $220\text{ m}{{\text{m}}^{2}}$.

 

7. A tent is in the shape of a cylinder surmounted by a conical top. If the height and diameter of the cylindrical part are $2.1\text{ m}$ and $4\text{ m}$ respectively, and the slant height of the top is $2.8\text{ m}$, 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.)

Ans:

Given that,

Height of the cylindrical part $=2.1\text{ m}$

Diameter of the cylindrical part $=4\text{ m}$

Radius of the cylindrical part$=2\text{ m}$

Slant height of conical part $=2.8\text{ m}$

Cost of $1\text{ }{{\text{m}}^{2}}$ canvas $=Rs.500$

 

seo images

 

Area of the canvas used = CSA of conical part + CSA os cylindrical part

$=\pi rl+2\pi rh$

$=\left( \pi \times 2\times 2.8 \right)+\left( 2\pi \times 2\times 2.1 \right)$

$=2\pi \left( 2.8+4.2 \right)$

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

$=44\text{ }{{\text{m}}^{2}}$

Cost of $1\text{ }{{\text{m}}^{2}}$ canvas $=Rs.500$

Cost of \[\] canvas$=44\times 500$

$=Rs.22,000$

$\therefore $The cost the canvas that is used to cover the tent is $Rs.22,000$.

 

8. From a solid cylinder whose height is $2.4\text{ cm}$ and diameter $1.4\text{ cm}$, 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 $c{{m}^{2}}$. (Use $\pi =\frac{22}{7}$)

Ans:

Height of the cylindrical part = Height of the conical part$=2.4\text{ cm}$

Diameter of the cylindrical part = Diameter of the conical part $=1.4\text{ cm}$

To find,

The total surface area of the remaining solid.

 

seo images
 

Diameter of the cylindrical part $=1.4\text{ cm}$

Radius of the cylindrical part \[\frac{1.4}{2}\]

$=0.7\text{ cm}$

Slant height of conical part $=\sqrt{{{r}^{2}}+{{h}^{2}}}$

$=\sqrt{0.49+5.76}$

$=3\sqrt{2}=\sqrt{6.25}$

$=2.5$

The total surface area of the solid = CSA of cylindrical part + CSA of conical part + Area of cylindrical base

$=2\pi rh+\pi rl+\pi {{r}^{2}}C\left( 4,3 \right)$

$=\left( 2\times \frac{22}{7}\times 0.7\times 2.4 \right)+\left( \frac{22}{7}\times 0.7\times 2.5 \right)+\left( \frac{22}{7}\times {{\left( 0.7 \right)}^{2}} \right)$

$=\left( 4.4\times 2.4 \right)+\left( 2.2\times 2.5 \right)+\left( 2.2\times 0.7 \right)$

$=10.56+5.5+1.54$

$=17.6\text{ c}{{\text{m}}^{2}}$

$\therefore $The total surface area of the remaining solid to the nearest $c{{m}^{2}}$ is $18\text{ c}{{\text{m}}^{2}}$.

 

9. A wooden article was made by scooping out a hemisphere from each end of a solid cylinder, as shown in the given figure. If the height of the cylinder is $10\text{ cm}$ and its base is of radius $3.5\text{ cm}$, find the total surface area of the article. (Use $\pi =\frac{22}{7}$)

Ans:

Given,

Height of the cylindrical part$=10\text{ cm}$

Radius of the cylindrical part = Radius of the hemispherical part $=3.5$

 

seo images

 

The total surface area of the article = CSA of cylindrical part + 2CSA of hemispherical part

$=2\pi rh+2\times 2\pi {{r}^{2}}$

$=\left( 2\pi \times 3.5\times 10 \right)+4\pi {{\left( 3.5 \right)}^{2}}$

$=70\pi +49\pi $

$=119\pi $

$=119\left( \frac{22}{7} \right)$

$=374\text{ c}{{\text{m}}^{2}}$

$\therefore $The total surface area of the article is $374\text{ c}{{\text{m}}^{2}}$.

 

Exercise (13.2)

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

 

seo images

Ans:

Given that,

Radius of cone and the hemisphere $=1\text{ cm}$

Height of the cone = Radius of the cone $=1\text{ cm}$

To find,

The volume of the solid in terms of $\pi $.

Volume of the given solid = Volume of the conical solid + Volume of the Hemispherical solid

$=\frac{1}{3}\pi {{r}^{2}}h+\frac{2}{3}\pi {{r}^{3}}{{\left( x-2 \right)}^{2}}+25={{\left( x+2 \right)}^{2}}+81$

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

$=\frac{1}{3}\pi +\frac{2}{3}\pi $

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

$=\pi $

$\therefore $ The volume of the solid in terms of \[\] is $\pi \text{ c}{{\text{m}}^{3}}$.

 

2. 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 $3\text{ cm}$ and its length is $x=\pm 412\text{ cm}$. If each cone has a height of $2\text{ cm}$, 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). (Use $\pi =\frac{22}{7}$)

Ans:

Given that,

Diameter of the cylindrical part $=3\text{ cm}$

Length of the cylindrical part $=12\text{ cm}$

Height of the conical part = 2cm

To find,

The volume of the air contained in the model

 

seo images

 

It can be observed from the figure that the,

Height of each conical part $\left( {{h}_{1}} \right)=2\text{ cm}$

Height of cylindrical part $\left( {{h}_{2}} \right)=12-2\times $Height of the conical part

$=12-\left( 2\times 2 \right)$

$=12-4$

$=8\text{ cm}$

Diameter of the cylindrical part $=3\text{ cm}$

Radius of the cylindrical part = Radius of the conical part

$=\frac{3}{2}$

Volume of the air in the model = Volume of cylindrical part + 2(Volume of cones)

Volume of the cylinder \[\]

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

$=\pi \times \frac{9}{4}\times 8$

$=18\pi $

Volume of $2$ cones$=2\times \frac{1}{3}\pi {{r}^{2}}h$

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

$=\frac{2}{3}\pi \times \frac{9}{4}\left( 2 \right)$

$=3\pi $

Volume of the air in cuboid $=18\pi +3\pi $

$=21\pi $

$=21\left( \frac{22}{7} \right)$

=66 cm\[^{3}\]

 

3. 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 $5\text{ cm}$ and diameter $2.8\text{ cm}$(see the given figure). (Use $\pi =\frac{22}{7}$)

 

seo images

Ans:

Given that,

The length of  the gulab jamun $=5\text{ cm}$

Diameter of the gulab jamun $=2.8\text{ cm}$

The volume of syrup in gulab jamun is $30%$ to its volume.

To find,

The volume of syrup in $45$ gulab jamun.

The diagram of gulab jamun shaped like a cylinder with two hemispherical ends is shown in the following diagram.

 

seo images

 

From the diagram,

Radius of the cylindrical part$\left( {{r}_{1}} \right)$ = Radius of hemispherical part $\left( {{r}_{2}} \right)$

$=\frac{2.8}{2}$

$=1.4\text{ cm}$

The length of the hemispherical part is the same as that of the radius of the hemispherical part.

Length of each hemispherical part $=1.4\text{ cm}$

Height of the cylindrical part$=5-2\times $Length of hemispherical part

$=5-2\left( 1.4 \right)$

$=5-2.8$

$=2.2\text{ cm}$

Volume of one gulab jamun = Volume of cylindrical part$+2$(Volume of hemispherical part

Volume of cylindrical part$=\pi {{r}^{2}}h$

$=\pi {{\left( 1.4 \right)}^{2}}\left( 2.2 \right)$

$=\frac{22}{7}{{\left( 1.4 \right)}^{2}}\times 2.2$

$=13.552$

$2$(Volume of hemispherical part) $=2\times \frac{2}{3}\pi {{r}^{3}}$

$=\frac{4}{3}\pi {{r}^{3}}$

$=\frac{4}{3}\pi {{\left( 1.4 \right)}^{3}}$

$=\frac{4}{3}\times \frac{22}{7}\times {{\left( 1.4 \right)}^{3}}$

$=11.498$

Volume of one gulab jamun$=13.552+11.498$

$=25.05\text{ c}{{\text{m}}^{3}}$

Volume of $45$ gulab jamuns$=45\times 25.05$

$=1,127.25\text{ c}{{\text{m}}^{3}}$

olume of sugar syrup $=30%$ of volume

$=\frac{30}{100}\times 1127.25$

$=338.17\text{ c}{{\text{m}}^{3}}$

$\therefore $The volume of sugar syrup found in $45$ gulab jamuns is approximately $338\text{ c}{{\text{m}}^{3}}$.

 

4. A pen stand made of wood is in the shape of a cuboid with four conical depressions to hold pens. The dimensions of the cuboids are $15\text{ cm}$ by $10\text{ cm}$ by $3.5\text{ cm}$. The radius of each of the depressions is $0.5\text{ cm}$ and the depth is $1.4\text{ cm}$. Find the volume of wood in the entire stand (see the following figure. (Use $\pi =\frac{22}{7}$)

 

(Image will be Uploaded Soon)

 

Ans:

Length of the cuboid$=15\text{ cm}$

Breadth of the cuboid$=10\text{ cm}$

Height of the cuboid$=3.5\text{ cm}$

Radius of conical depression $=0.5\text{ cm}$

Height of conical depression $=1.4\text{ cm}$

To find,

The volume of wood in entire stand

 

seo images

 

Volume of the wood = Volume of cuboid - $4\times $Volume of cones

Volume of cuboid $=lbh$

$=15\times 10\times 3.5$

$=525\text{ c}{{\text{m}}^{3}}$

$4\times $Volume of cones$=4\times \frac{1}{3}\pi {{r}^{2}}h$

$=\frac{4}{3}\times \frac{22}{7}\times {{\left( \frac{1}{2} \right)}^{2}}\times 1.4$

$=1.47\text{ c}{{\text{m}}^{3}}$

Volume of the wood$=525-1.47$

$=523.53\text{ c}{{\text{m}}^{3}}$

$\therefore $The volume of the wood in the entire stand is $523.53\text{ c}{{\text{m}}^{3}}$.

 

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

Ans:

Given that,

Height of the conical vessel$\left( h \right)=8\text{ cm}$

Radius of conical vessel$\left( {{r}_{1}} \right)=5\text{ cm}$

Radius of lead shots $\left( {{r}_{2}} \right)=0.5\text{ cm}$

One-fourth of water flows out from the vessel.

To find,

The number of lead shots dropped in the vessel.

 

seo images

 

Let the number of lead shots that has been dropped in the vessel be $n$.

Volume of water flows out = Volume of lead shots dropped in the vessel

$\frac{1}{4}\times $Volume of the cone $=n\times \frac{4}{3}\pi {{r}_{2}}^{3}$

$\frac{1}{4}\times \frac{1}{3}\pi {{r}_{1}}^{2}h=n\times \frac{4}{3}\pi {{r}_{2}}^{3}$

${{r}_{1}}^{2}h=16n{{r}_{2}}^{3}$

Substituting the values we known, we obtain,

${{5}^{2}}\left( 8 \right)=n\left( 16\times {{\left( 0.5 \right)}^{3}} \right)$

$n=\frac{200}{16\times {{\left( 0.5 \right)}^{3}}}$

$n=100$

$\therefore $The number of lead shots dropped in the vessel is $100$.

 

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

Ans:

Let there be two cylinders, one is of larger and the other is smaller

Given that,

Height of the larger cylinder ${{h}_{1}}=220\text{ cm}$

Diameter of the larger cylinder ${{d}_{1}}=24\text{ cm}$

Height of the smaller cylinder${{h}_{2}}=60\text{ cm}$

Radius of the smaller cylinder ${{r}_{2}}=8\text{ cm}$

To find,

The mass of iron pole.

 

seo images

 

Radius of larger cylinder $\left( {{r}_{1}} \right)=\frac{24}{2}$

$=12\text{ cm}$

Total volume of pole = Volume of the larger cylinder + Volume of the smaller cylinder

Volume of the larger cylinder$=\pi {{r}_{1}}^{2}{{h}_{1}}$

$=\pi {{\left( 12 \right)}^{2}}\times 220$

$=31,680\pi $

Volume of the smaller cylinder$=\pi {{r}_{2}}^{2}{{h}_{2}}$

$=3840\pi $

Volume of the iron pole $=31,680\pi +3,840\pi $

$=35,520\pi $

$=35,520\left( 3.14 \right)$

$=111,532.8\text{ c}{{\text{m}}^{3}}$

Mass of $1\text{ c}{{\text{m}}^{3}}$iron$=8\text{ g}$

Mass of $111,532.8\text{ c}{{\text{m}}^{3}}$ iron $=111,532.8\times 8$

$=892262.4\text{ g}$

We know that $1000\text{ g}=1\text{ kg}$

$892262.4\text{ g}=892262.4\times 1000\text{ kg}$

$=892.262\text{ kg}$

$\therefore $The mass of the iron is $892.262\text{ kg}$.

 

7. A solid consisting of a right circular cone of height $120\text{ cm}$ and radius $60\text{ cm}$ standing on a hemisphere of radius $60\text{ cm}$ is placed upright in a right circular cylinder full of water such that it touches the bottom. Find the volume of the water left in the cylinder, if the radius of the cylinder is $60\text{ cm}$ and its height is $180\text{ cm}$. (Use $\pi =\frac{22}{7}$).

Ans:

Given that,

A solid with a right circular cone and a hemisphere.

Height of the conical part of the cylinder $=120\text{ cm}$

Radius of the conical part of the cylinder$=60\text{ cm}$

Radius of hemispherical part of the cylinder $=60\text{ cm}$

Radius of the outer cylinder $=60\text{ cm}$

Height of the outer cylinder $=180\text{ cm}$

To find, 

The volume of the water left in the cylinder.

 

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Volume of the water left = Volume of cylinder – Volume of the solid

Volume of the cylinder$=\pi {{r}^{2}}h$

$=\pi {{\left( 60 \right)}^{2}}\left( 180 \right)$

$=\pi \times 3600\times 180$

$=648,000\pi \text{ c}{{\text{m}}^{3}}$

Volume of the solid = Volume of cone + Volume of the hemisphere

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

$=\frac{1}{3}\pi {{\left( 60 \right)}^{2}}120+\frac{2}{3}\pi {{\left( 60 \right)}^{3}}$

$=\frac{1}{3}\pi \left( 432,000 \right)+\frac{2}{3}\pi \left( 216,000 \right)$

$=288,000\pi $

Volume of the water left$=648,000\pi -288,000\pi $

$=360,000\pi $

$=360,000\times \frac{22}{7}$

$=11311428.57\text{ c}{{\text{m}}^{3}}$

$=1.131\text{ }{{\text{m}}^{3}}$

$\therefore $The volume of the water left in the cylinder is $1.131\text{ }{{\text{m}}^{3}}$.

 

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

Ans:

Given that,

Height of the cylindrical part $=8\text{ cm}$

Diameter of cylindrical neck $=2\text{ cm}$

Diameter of spherical glass vessel $=8.5\text{ cm}$

Volume of the water that the vessel holds $=345\text{ c}{{\text{m}}^{3}}$

To find it the above given volume is correct.

 

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Volume of the vessel = Volume of sphere + Volume of cylinder

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

$=\frac{4}{3}\times 3.14\times {{\left( \frac{8.5}{2} \right)}^{3}}$

$=321.392\text{ c}{{\text{m}}^{3}}$

Volume of cylinder $=\pi {{r}^{2}}h$

$=3.14\times {{\left( 1 \right)}^{2}}\times 8$

$=25.12\text{ c}{{\text{m}}^{3}}$

Volume of the vessel $=321.392+25.12$

$=346.51\text{ c}{{\text{m}}^{3}}$

$\therefore $The volume of the vessel is $346.51\text{ c}{{\text{m}}^{3}}$ and hence the child is wrong.

 

Exercise (13.3)

1. A metallic sphere of radius $4.2\text{ cm}$ is melted and recast into the shape of a

cylinder of radius $6\text{ cm}$. Find the height of the cylinder.

Ans:

Radius of the sphere $=4.2\text{ cm}$

Radius of the cylinder $=6\text{ cm}$

To find,

The height of the cylinder.

 

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Let the height of the cylinder be $h$.

Since the spherical object has been melted and recast into a cylinder, The volumes of both the solids will be equal.

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

$\frac{4}{3}{{\left( 4.2 \right)}^{3}}=\left( {{6}^{2}} \right)h$

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

$h=2.74\text{ cm}$

$\therefore $The height of the hollow cylinder is $2.74\text{ cm}$.

 

2. Metallic spheres of radii $6\text{ cm, }8\text{ cm}$ and $10\text{cm}$respectively, are melted to

form a single solid sphere. Find the radius of the resulting sphere.

Ans:

Let ${{r}_{1}},{{r}_{2,}}{{r}_{3}}$ denote the radii of the first, second and third sphere respectively.

Given that,

Radius of the first sphere ${{r}_{1}}=6\text{ }cm$

Radius of the second sphere ${{r}_{2}}=8\text{ cm}$

Radius of the third sphere ${{r}_{3}}=10\text{ cm}$

To find,

The radius of the resultant sphere.

 

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Let the radius of the resultant sphere be $r$.

Since these spheres have been recast, the volume of the resultant solid will be the same as that of the volume of these three spheres.

Sum of the volume of three spheres = Volume of the resultant sphere

$\frac{4}{3}\pi \left( {{r}_{1}}^{3}+{{r}_{2}}^{3}+{{r}_{3}}^{3} \right)=\frac{4}{3}\pi {{r}^{3}}$

$\left( {{6}^{3}}+{{8}^{3}}+{{10}^{3}} \right)={{r}^{3}}$

${{r}^{3}}=216+512+1000$

${{r}^{3}}=1728$

$r=12\text{ cm}$

$\therefore $The radius of the resultant solid sphere is $12\text{ cm}$

 

3. A $20\text{ m}$deep well with diameter $7\text{ m}$ is dug and the earth from digging is

evenly spread out to form a platform $22\text{ m}$by $14\text{ m}$. Find the height of the

platform. (Use $\pi =\frac{22}{7}$)

Ans:

Given that,

Diameter of the well $=7\text{ m}$

Height of the well $=20\text{ m}$

It forms the platform of $22\text{ m}$ by $14\text{ m}$.

To find,

The height of the platform.

 

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Radius of the well $=\frac{7}{2}\text{ m}$

Let $H$ denotes the height of the platform

Volume of the soil dug from the well will be equal to the volume of the soil on the platform.

Volume of the soil from well = Volume of the soil to make platform

$\pi {{r}^{2}}h=lbH$

$\pi {{\left( \frac{7}{2} \right)}^{2}}\left( 20 \right)=22\times 14\times H$

$H=\frac{22\times 49\times 20}{7\times 4\times 22\times 14}$

$H=\frac{5}{2}\text{ m}$

$=2.5\text{ m}$

$\therefore $The height of the platform is $2.5\text{ m}$

 

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

Ans:

Diameter of the well $=3\text{ m}$

Height of the well $=14\text{ m}$

Width of embankment $=4\text{ m}$

To find the height of the embankment.

 

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It can be observed from the figure that the embankment will in the shape of cylindrical shape with outer radius ${{r}_{2}}=4+\frac{3}{2}=\frac{11}{2}$ and inner radius ${{r}_{1}}=\frac{3}{2}\text{ m}$.

Let the height of embankment be ${{h}_{2}}$.

Volume of the soil from well = Volume of the soil in embankment 

$\pi {{r}^{2}}{{h}_{1}}=\pi \left( {{r}_{2}}^{2}-{{r}_{1}}^{2} \right){{h}_{2}}$

${{\left( \frac{3}{2} \right)}^{2}}\left( 14 \right)=\left[ {{\left( \frac{11}{2} \right)}^{2}}-{{\left( \frac{3}{2} \right)}^{2}} \right]{{h}_{2}}$

$\frac{14\times 9}{4}=\frac{112}{4}{{h}_{2}}$

${{h}_{2}}=\frac{9}{8}$

${{h}_{2}}=1.125\text{ m}$

$\therefore $The height of  the embankment is $1.125\text{ m}$.

 

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

Ans:

Height of the cylinder ${{h}_{1}}=15\text{ cm}$

Diameter of the cylinder ${{d}_{1}}=12\text{ cm}$

Height of cone with hemispherical top ${{h}_{2}}=12\text{ cm}$

Diameter of cone with hemispherical top ${{d}_{2}}=6\text{ cm}$

To find,

The number of cones to be filled with ice cream

 

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Hence Volume of the ice cream in the cylinder is to be filled in cones.

So,

Volume of ice cream in cylinder = $n\times $Volume of cones

Volume of ice cream in cylinder$=\pi {{r}_{1}}^{2}{{h}_{1}}$

$=\pi \left( {{6}^{2}} \right)\left( 15 \right)$

Volume of ice cream cones = Volume of 1 ice cream cone + Volume of hemispherical shape on the top

$=n\times \frac{1}{3}\pi {{r}_{2}}^{2}{{h}_{2}}+\frac{2}{3}\pi {{r}_{2}}^{3}$

$=n\times \pi \left( \frac{1}{3}\times {{\left( 3 \right)}^{2}}\times 12+\left( \frac{2}{3}\times {{\left( 3 \right)}^{3}} \right) \right)$

By equating both, we get,

$n=\frac{{{6}^{2}}\times 15}{\frac{1}{3}\times 9\times 12+\frac{2}{3}\times {{\left( 3 \right)}^{2}}}$

$n=\frac{36\left( 15 \right)\left( 3 \right)}{108\left( 54 \right)}$

$n=10$

$\therefore $The number of ice cream cones that can be filled with the ice cream in the container is $10$.

 

6. How many silver coins, $1.75\text{ cm}$in diameter and of thickness $2\text{ mm}$, must be melted to form a cuboid of dimensions $5.5\text{ cm}\times 10\text{ cm}\times 3.5\text{ cm}$?

Ans:

Diameter of the silver coin $=1.75\text{ cm}$

Thickness of the silver coin$=2\text{ mm}$

Dimension of a cuboid$=5.5\times 10\times 3.5$

To find,

The number of silver coins melted to form the cuboid of above dimension

 

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Radius of silver coin $=\frac{1.75}{2}$

$=0.875\text{ cm}$

The volume of the cuboid formed is equal to the number of coins melted to form it.

Volume of the cuboid = Volume of the $n$ coins melted

$lbh=n\times \left( \pi {{r}^{2}}{{h}_{1}} \right)$

$5.5\times 10\times 3.5=n\times \pi {{\left( 0.875 \right)}^{2}}\times 0.2$

$n=\frac{55\times 10\times 3.5}{{{\left( 0.875 \right)}^{2}}\times \left( 0.2 \right)\times \frac{22}{7}}$

$n=400$

$\therefore $The number of coins required to from a cuboid is $400$.

 

7. A cylindrical bucket, $32\text{ cm}$ high and with radius of base $18\text{ cm}$, 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 $24\text{ cm}$. Find the radius and slant height of the heap.

Ans:

Given that,

Height of the cylindrical bucket ${{h}_{1}}=32\text{ cm}$

Radius of cylindrical bucket with circular end ${{r}_{1}}=18\text{ cm}$

Height of conical heap formed ${{h}_{2}}=24\text{ cm}$

To find,

The radius and slant height of the heap

 

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The volume of the sand in the cylindrical bucket is equal; to the volume of the sand that forms the conical heap.

So,

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

${{\left( 18 \right)}^{2}}\times 32=\frac{1}{3}\times {{r}_{2}}^{2}\times 24$

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

${{r}_{2}}^{2}={{18}^{2}}\times {{2}^{2}}$

${{r}_{2}}=36\text{ cm}$

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

$=\sqrt{{{\left( 36 \right)}^{2}}+{{\left( 24 \right)}^{2}}}$

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

$\therefore $The radius of the conical heap is $36\text{ cm}$ and that of slant height is $12\sqrt{13}\text{ cm}$.

 

8. Water in canal, $6\text{ m}$ wide and $1.5\text{ m}$ deep, is flowing with a speed of $10\text{ km/hr}$. How much area will it irrigate in $30$ minutes, if $8\text{ cm}$ Is standing water needed?

Ans:

Cross section of the canal $=6\text{ m}\times \text{1}\text{.5 m}$

Flowing with the speed of $10\text{ km/h}$

To find,

The area it can irrigate in $30$ minutes if the standing water needed is $8\text{ cm}$

 

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Area of cross section of canal $=6\times 1.5$

$=9\text{ }{{\text{m}}^{2}}$

Speed of the water $=10\text{ km/h}$

$=\frac{10000}{60}\text{ m/min}$

Volume of the water that flows in one minute from canal $=9\times \frac{10000}{60}$

$=1500\text{ }{{\text{m}}^{3}}$

Volume of the water that flows in $30$ minute from canal $=30\times 1500$

$=45000\text{ }{{\text{m}}^{3}}$

 

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Let the irrigated area be $A$. Volume of water irrigating the required area will be equal to the volume of water that flowed in $30$ minutes from the canal.

Volume of the water flowing in $30$ minutes from the canal = Volume of the water irrigating the required area

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

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

$A=562500\text{ }{{\text{m}}^{2}}$

$\therefore $The area irrigated in $30$ minutes is $562500\text{ }{{\text{m}}^{2}}$

 

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

Ans:

Given That,

Diameter of pipe$\left( {{d}_{1}} \right)=20\text{ cm}$

Diameter of cylindrical tank $\left( {{d}_{2}} \right)=10\text{ m}$

Height of the cylindrical tank ${{h}_{2}}=2\text{ m}$

Speed of water $=30\text{ km/h}$

To find,

The time at which the tank will be filled

 

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Consider an area of cross section of the pipe as shown above:

Radius of pipe in $m\left( {{r}_{1}} \right)=\frac{20}{2\times 100}$

$=0.1\text{ m}$

Radius of cylindrical tank ${{r}_{2}}=\frac{10}{2}$

$=5\text{ m}$

Area of cross section of the pipe $=\pi {{r}_{1}}^{2}$

$=\pi \times {{\left( 0.1 \right)}^{2}}$

$=0.01\pi \text{ }{{\text{m}}^{2}}$

Speed of water $=30\text{ km/h}$

$=\frac{3000}{60}\text{ m/min}$

$=50\text{ m/min}$

 

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Volume of water flows in $1\text{ min}$ from the pipe $=50\times 0.01\pi $

$=0.5\pi \text{ }{{\text{m}}^{3}}$

Volume of water flows in $t$minute from the pipe $=t\times 0.5\pi $

Let us assume that the tank has been filled completely. Volume of water filled in $t$ minutes in the tank is equal to the volume of water flowed out from the pipe in $t$ minutes.

So,

$t\times 0.5\pi =\pi \times {{r}_{2}}^{2}\times {{h}_{2}}$

$t=\frac{{{5}^{2}}\times 2}{0.5}$

$t=100$ minutes

$\therefore $The tank will be filled in $100$ minutes.

 

Exercise (13.4)

1.A drinking glass is in the shape of a frustum of a cone of height $14\text{ cm}$. The diameters of its two circular ends are $4\text{ cm}$ and $2\text{ cm}$. Find the capacity of the glass. (Use $\pi =\frac{22}{7}$)

 

Ans:

Given that,

Height of the conical glass $h=14\text{ cm}$

Diameter of upper base of circular part ${{d}_{1}}=4\text{ cm}$

Diameter of lower base of the circular part ${{d}_{2}}=2\text{ cm}$

To find,

The capacity of the glass

 

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Radius of upper base of glass ${{r}_{1}}=\frac{4}{2}$

$=2\text{ cm}$

Radius of lower base of glass ${{r}_{2}}=\frac{2}{2}$

$=1\text{ cm}$

Volume of the conical glass $=\frac{1}{3}\pi h\left( {{r}_{1}}^{2}+{{r}_{2}}^{2}+{{r}_{1}}{{r}_{2}} \right)$

$=\frac{1}{3}\pi \times 14\left[ {{\left( 2 \right)}^{2}}+{{\left( 1 \right)}^{2}}+\left( 2\times 1 \right) \right]$

$=\frac{1}{3}\times \frac{22}{7}\times 14\times 7$

$=\frac{308}{3}$

$=102\frac{2}{3}\text{ c}{{\text{m}}^{3}}$

$\therefore $The capacity of the drinking glass is $102\frac{2}{3}\text{ c}{{\text{m}}^{3}}$.

 

2. The slant height of a frustum of a cone is $4\text{ cm}$ and the perimeters (circumference) of its circular ends are $18\text{ cm}$ and $6\text{ cm}$. Find the curved surface area of the frustum.

 

Ans:

Slant height of the frustum $l=4\text{ cm}$

Perimeter of upper circular part $2\pi {{r}_{1}}=18$

Perimeter of lower circular part $2\pi {{r}_{2}}=6$

To find, 

The curved surface area of the frustum

 

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$2\pi {{r}_{1}}=18$

${{r}_{1}}=\frac{9}{\pi }$

$2\pi {{r}_{2}}=6$

${{r}_{2}}=\frac{6}{\pi }$

Curved surface area of frustum $=\pi \left( {{r}_{1}}+{{r}_{2}} \right)l$

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

$=\pi \times \frac{48}{\pi }$

$=48\text{ c}{{\text{m}}^{2}}$

The curved surface area off the frustum is $48\text{ c}{{\text{m}}^{2}}$.

 

3. A fez, the cap used by the Turks, is shaped like the frustum of a con (see the figure given below). If its radius on the open side is $10\text{ cm}$, radius at the upper base is $4\text{ cm}$ and its slant height is $15\text{ cm}$, find the area of the material used for making it. (Use $\pi =\frac{22}{7}$)

 

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Ans:

Radius of lower circular end of the frustum${{r}_{1}}=10\text{ cm}$

Radius of upper circular end of the frustum ${{r}_{2}}=4\text{ cm}$

Slant height of the frustum $l=15\text{ cm}$

To find,

The area of the material used for making it

 

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Area of the material used = CSA of the frustum + Area of the upper circle

CSA of the frustum $=\pi \left( {{r}_{1}}+{{r}_{2}} \right)l$

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

$=\pi \left( 14 \right)15$

$=210\pi $

Area of the upper circle $=\pi {{r}_{2}}^{2}$

$=\pi {{\left( 4 \right)}^{2}}$

$=16\pi $

Area of the material used $=210\pi +16\pi $

$=226\pi $

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

$=710\frac{2}{7}\text{ c}{{\text{m}}^{2}}$

$\therefore $The area of the material used for making the cap is $710\frac{2}{7}\text{ c}{{\text{m}}^{2}}$.

 

4. 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 $16\text{ cm}$ with radii of its lower and upper ends as $8\text{ cm}$ and $20\text{ cm}$respectively. Find the cost of the milk which can completely fill the container, at the rate of $Rs.20$ per liter. Also find the cost of metal sheet ,if it costs $Rs.8$ per $100\text{ c}{{\text{m}}^{2}}$. (Take $\pi =3.14$)

Ans:

Height of the frustum $h=16\text{ cm}$

Radius of the upper circular end ${{r}_{1}}=20\text{ cm}$

Radius of the lower circular end ${{r}_{2}}=8\text{ cm}$

Cost of milk per liter $=Rs.20$

Cost of metal sheet per $\text{100 c}{{\text{m}}^{3}}=Rs.8$

To find,

The cost of metal sheet that fill the container completely

 

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

$=\sqrt{{{\left( {{r}_{1}}-{{r}_{2}} \right)}^{2}}+{{h}^{2}}}$

$=\sqrt{{{\left( 20-8 \right)}^{2}}+{{\left( 16 \right)}^{2}}}$

$=\sqrt{{{\left( 12 \right)}^{2}}+{{\left( 16 \right)}^{2}}}$

$=\sqrt{144+256}$

$=\sqrt{400}$

$=20\text{ cm}$

The capacity of the container is equal to the volume of the frustum.

Volume of frustum $=\frac{1}{3}\pi h\left[ {{r}_{1}}^{2}+{{r}_{2}}^{2}+{{r}_{1}}{{r}_{2}} \right]$

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

$=\frac{1}{3}\pi \times 16\left( 400+64+160 \right)$

$=\frac{1}{3}\times 3.14\times 16\times 624$

$=10,449.92\text{ c}{{\text{m}}^{3}}$

We know that $1000\text{ c}{{\text{m}}^{3}}=1\text{ l}$

So, Volume of the frustum $=10.45$liters

Cost of one liter milk $=Rs.20$

Cost of $10.45$ liter milk $=10.45\times 20$

$=Rs.209$

Area of the metal sheet used to make the container $=\pi \left( {{r}_{1}}+{{r}_{2}} \right)l+\pi {{r}_{2}}^{2}$

$=\pi \left( 20+8 \right)20+\pi {{\left( 8 \right)}^{2}}$

$=560\pi +64\pi $

$=624\times 3.14$

Cost of $100\text{ c}{{\text{m}}^{2}}$ sheet $=Rs.8$

Cost of $624\pi \text{ c}{{\text{m}}^{2}}$ sheet $=624\times 3.14\times \frac{8}{100}$

$=Rs.156.75$

$\therefore $The cost of the milk that can completely fill the container is $Rs.209$

The cost of the metal sheet that is used to make the container is $Rs.156.75$.

 

5. A metallic right circular cone $20\text{ cm}$ high and whose vertical angle is $60{}^\circ $ is cut into two parts at the middle of its height by a plane parallel to its base. If the frustum so obtained is drawn into a middle of a diameter $\frac{1}{16\text{ }}\text{ cm}$, find the length of the wire. (Use $\pi =\frac{22}{7}$)

Ans:

Given that,

Height of the cone$=20\text{ cm}$

Diameter of the wire $=\frac{1}{16}\text{ cm}$

To find, 

The length of the wire

 

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Consider $\vartriangle AEG$,

$\tan 30{}^\circ =\frac{EG}{AG}$

$EG=\frac{10}{\sqrt{3}}$

$EG=\frac{10\sqrt{3}}{3}\text{ cm}$

Consider $\vartriangle ABD$,

$\tan 30{}^\circ =\frac{BD}{AD}$

$BD=\frac{20}{\sqrt{3}}$

$BD=\frac{20\sqrt{3}}{3}\text{ cm}$

Thus the radius of the upper end of the frustum ${{r}_{1}}=\frac{10\sqrt{3}}{3}\text{ cm}$

Radius of the lower end of the container ${{r}_{2}}=\frac{20\sqrt{3}}{3}\text{ cm}$

Height of the container $h=10\text{ cm}$

Volume of the frustum $=\frac{1}{3}\pi h\left( {{r}_{1}}^{2}+{{r}_{2}}^{2}+{{r}_{1}}{{r}_{2}} \right)$

$=\frac{1}{3}\pi \times 10\left[ {{\left( \frac{10\sqrt{3}}{3} \right)}^{2}}+{{\left( \frac{20\sqrt{3}}{3} \right)}^{2}}+\left( \frac{10\sqrt{3}\times 20\sqrt{3}}{3\times 3} \right) \right]$

$=\frac{10}{3}\pi \left[ \frac{100}{3}+\frac{400}{3}+\frac{200}{3} \right]$

$=\frac{10}{3}\times \frac{22}{7}\times \frac{700}{3}$

$=\frac{22,000}{9}\text{ c}{{\text{m}}^{3}}$

Volume of the frustum is $\frac{22,000}{9}\text{ c}{{\text{m}}^{3}}$

Radius of the wire $r=\frac{1}{16}\times \frac{1}{2}$

$=\frac{1}{32}\text{ cm}$

Let us consider the length of the wire to be $l$.

Volume of the wire $=$ Area of the cross section $\times $ Length

$=\pi {{r}^{2}}\times l$

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

Volume of the frustum = Volume of the wire 

$\frac{22000}{9}=\frac{22}{7}\times \frac{1}{1024}\times l$

$l=796444.44\text{ cm}$

$l=7964.44\text{ m}$

$\therefore $The length of the wire is $7964.44\text{ m}$.

 

Exercise (13.5)

1.A copper wire $3\text{ mm}$ in diameter, is wound about a cylinder whose length is $12\text{ cm}$, diameter $10\text{ cm}$, so as to cover the curved surface area of the cylinder. Find the length and mass of the wire, assuming the density of copper to be $8.88\text{ g}$ per $c{{m}^{3}}$.

Ans:

Diameter of the copper wire $=3\text{ mm}$

Length of the cylinder $=12\text{ cm}$

Diameter of the cylinder $=10\text{ cm}$

Density of the copper per $c{{m}^{3}}=8.88g$

 

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We have to note that a round of wire can cover the cylinder of height $3\text{ mm}$.

Length of the wire in a round = Circumference of the base cylinder

$=2\pi r$

$=2\pi \left( 5 \right)$

$=10\pi $

Number of rounds $=\frac{\text{height of cylinder }}{\text{Diameter of the wire}}$

$=\frac{12}{0.3}$

$=40$ rounds

Length of the wire in $40$ rounds $=40\times 10\pi $

$=400\times \frac{22}{7}$

$=1257.14\text{ cm}$

$=12.57\text{ m}$

Radius of the wire $=\frac{0.3}{2}$

$=0.15$

Volume of the wire = Area of the cross section $\times $ Length of the wire

$=\pi {{\left( 0.15 \right)}^{2}}\times 1257.14$

$=88.898\text{ c}{{\text{m}}^{3}}$

Mass of the wire = Volume $\times $ density

$=88.898\times 8.88$

$=789.41\text{ g}$

$\therefore $The length of the wire is $12.57\text{ m}$

The mass of the wire is $789.41\text{ g}$

 

2. A right triangle whose sides are $3\text{ cm}$ and $4\text{ cm}$(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).

Ans:

Given that,

The sides of a triangle are $3,4$.

To find,

The volume and surface area of the double cone.

 

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The double cone formed by rotating the right triangle about its hypotenuse  is shown in the above figure.

By Pythagorean theorem,

$AC=\sqrt{{{3}^{2}}+{{4}^{2}}}$

$=\sqrt{25}$

$=5\text{ cm}$

Area of $\vartriangle ABC=\frac{1}{2}\times AB\times AC$

$\frac{1}{2}\times AC\times OB=\frac{1}{2}\left( 4 \right)\left( 3 \right)$

$\frac{5}{2}OB=6$

$OB=\frac{12}{5}$

$OB=2.4\text{ cm}$

Volume of double cone = Volume of cone 1+ Volume of cone 2

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

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

$=\frac{1}{3}\times 3.14\times {{\left( 2.4 \right)}^{2}}\left( OA+AC \right)$

$=\frac{1}{3}\times 3.14\times {{\left( 2.4 \right)}^{2}}5$

$=30.14\text{ c}{{\text{m}}^{3}}$

Surface area of the double cone $=\pi r{{l}_{1}}+\pi r{{l}_{2}}$

$=\pi r\left( {{l}_{1}}+{{l}_{2}} \right)$

$=\pi r\left( 4+3 \right)$

$=3.14\times 2.4\times 7$

$=52.75\text{ c}{{\text{m}}^{2}}$

$\therefore $The volume of the double cone is $30.14\text{ c}{{\text{m}}^{3}}$.

The surface area of the double cone is $52.75\text{ c}{{\text{m}}^{2}}$

 

3. A cistern, internally measuring 150 cm x 120 cm x 110 cm, has $129600\text{ c}{{\text{m}}^{3}}$ 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 $22.5\text{ cm}\times 7.5\text{ cm}\times 6.5\text{ cm}$?

Ans:

Given that,

Dimension of cistern $=150\times 120\times 110$

Volume of water in cistern $=129600\text{ c}{{\text{m}}^{3}}$

Dimension of brick $=22.5\times 7.5\times 6.5$

To find, 

The number of bricks to be put in the cistern without the overflow of water

Volume of cistern$=150\times 120\times 110$

$=1980000\text{ c}{{\text{m}}^{3}}$

Volume of bricks to be filled in cistern $=1980000-129600$

$=1850400\text{ c}{{\text{m}}^{3}}$

Let $n$ be the number of bricks that is to be placed in the cistern.

Volume of $n$bricks $=n\times 22.5\times 7.5\times 6.5$

$=1096.875n$

As each brick absorbs one-seventeenth of its volume, therefore, volume absorbed by these bricks are,

$=\left( \frac{n}{17} \right)1096.875$

$1850400+\left( \frac{n}{17} \right)1096.875=1096.875n$

$\frac{16n}{17}1096.875=1850400$

$n=1792.41$

$\therefore 1792$ bricks can be placed in a cistern without the overflow of water.

 

4. In One fortnight of the given month, there was a rainfall of 10 cm in a river valley. If the area of the valley is 97280 km², show that the total rainfall was approximately equivalent to the addition to the normal water of the three rivers each x$1072\text{ km}$ long,$75\text{ m}$ wide and $3\text{ m}$ deep. 

Ans:

Given that,

Area of the valley$=97280\text{ k}{{\text{m}}^{2}}$

Amount of rise in water level in valley $h=10\text{ cm}$

Dimension of rivers$=1072\times 75\times 3$

To show that the total rainfall is approximately equivalent to the addition to normal water of three rivers.

 Amount of rise in the water $h=10\text{ cm}$

$=\frac{10}{100000}\text{ km}$

$=\frac{1}{10000}\text{ km}$

Amount of rainfall in $14$ days$=A\times h$

$=97820\times \frac{1}{10000}$

$=9.828\text{ k}{{\text{m}}^{3}}$

Amount of rainfall in a day$=\frac{9.828}{14}$

$=0.702\text{ k}{{\text{m}}^{3}}$

Volume of water in three rivers$=3lbh$

$=3\left( 1072\text{ km}\times 75\text{ m}\times 3\text{ m} \right)$

$=3\left( 1072\text{ km}\times \frac{75}{1000}\text{ km}\times \frac{3}{1000}\text{ km} \right)$

$=3\left( 0.2412 \right)$

$=0.7236\text{ k}{{\text{m}}^{3}}$

This shows that the rainfall is approximately equal to the amount of water in three rivers.

 

5. 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 x$22\text{ cm}$, diameter of the cylindrical portion is $8\text{ cm}$ and the diameter of the top of the funnel is $18\text{ cm}$, find the area of the tin sheet required to make the funnel(see the given figure).

 

seo images

 

Ans:

Height of a cylindrical part ${{h}_{1}}=10\text{ cm}$

Total height of the funnel $h=22\text{ cm}$

Diameter of the circular portion in conical part${{d}_{2}}=18\text{ cm}$

Diameter of the circular portion in cylindrical part ${{d}_{1}}=8\text{ cm}$

To find,

The area of the tin sheet required to make the funnel


seo images


Total height of the frustum ${{h}_{2}}=22-10$

$=12\text{ cm}$

Slant height of the frustum $\left( l \right)=\sqrt{{{\left( {{r}_{2}}-{{r}_{1}} \right)}^{2}}+{{h}^{2}}}$

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

$=\sqrt{169}$

$=13\text{ cm}$

Area of the tin sheet required = CSA of frustum part +CSA of cylindrical part

$=\pi \left( {{r}_{2}}+{{r}_{1}} \right)l+2\pi {{r}_{1}}{{h}_{1}}$

$=\left( \frac{22}{7}\times 169 \right)+\left( \frac{22}{7}\times 2\times 4\times 10 \right)$

$=\frac{22}{7}\left( 169+80 \right)$

$=782\frac{4}{7}\text{ c}{{\text{m}}^{2}}$

Area of the tin sheet required to make the funnel is $782\frac{4}{7}\text{ c}{{\text{m}}^{2}}$.

 

6. Derive the formula for the curved surface area and total surface area of the frustum of the cone.

Ans:

To derive the formula for CSA and TSA of the cone

 

seo images

 

Let $ABC$ be a cone. A frustum $DECB$is cut by a plane parallel to its base. Let ${{r}_{1}},{{r}_{2}}$ be the radii of the ends of the frustum of the cone and $h$ be the height of the frustum of the cone.

Consider $\vartriangle ABG$ and $\vartriangle ADF$,

$DF\parallel BG$

$\therefore \vartriangle ABG\sim \vartriangle ADF$

$\frac{DF}{BG}=\frac{AF}{AG}=\frac{AD}{AB}$

$\frac{{{r}_{2}}}{{{r}_{1}}}=\frac{{{h}_{1}}-h}{{{h}_{1}}}=\frac{{{l}_{1}}-l}{{{l}_{1}}}$

$\frac{{{r}_{2}}}{{{r}_{1}}}=1-\frac{h}{{{h}_{1}}}=1-\frac{l}{{{l}_{1}}}$

$1-\frac{l}{{{l}_{1}}}=\frac{{{r}_{2}}}{{{r}_{1}}}$

$\frac{l}{{{l}_{1}}}=1-\frac{{{r}_{2}}}{{{r}_{1}}}$

$=\frac{{{r}_{1}}-{{r}_{2}}}{{{r}_{1}}}$

By taking reciprocal and solving for ${{l}_{1}}$, we get,

${{l}_{1}}=l\frac{{{r}_{1}}}{{{r}_{1}}-{{r}_{2}}}$

CSA of frustum DECB = CSA of con $ABC-$CSA of cone $ADE$

$=\pi {{r}_{1}}{{l}_{1}}-\pi {{r}_{2}}\left( {{l}_{1}}-l \right)$

$=\pi {{r}_{1}}\left( \frac{l{{r}_{1}}}{{{r}_{1}}-{{r}_{2}}} \right)-\pi {{r}_{2}}\left( \frac{l{{r}_{1}}}{{{r}_{1}}-{{r}_{2}}}-l \right)$

On solving this, we get,

$=\frac{\pi {{r}_{1}}^{2}l}{{{r}_{1}}-{{r}_{2}}}-\frac{\pi {{r}_{2}}^{2}l}{{{r}_{1}}-{{r}_{2}}}$

Taking $\pi $ as common, we get,

$=\pi l\left[ \frac{{{r}_{1}}^{2}-{{r}_{2}}^{2}}{{{r}_{1}}-{{r}_{2}}} \right]$

CSA of frustum $=\pi \left( {{r}_{1}}+{{r}_{2}} \right)l$

Total surface area of frustum = CSA of frustum + Area of upper circular ends + area of lower circular end

$=\pi l\left( {{r}_{1}}+{{r}_{2}} \right)+\pi {{r}_{1}}^{2}+\pi {{r}_{2}}^{2}$

$=\pi \left( l\left( {{r}_{1}}+{{r}_{2}} \right)+{{r}_{1}}^{2}+{{r}_{2}}^{2} \right)$

Hence the formula for the curved surface area and the total surface area of the frustum has been derived.

 

7. Derive the formula for the volume of the frustum of the cone.

Ans:

To derive the formula for the volume of the frustum of a cone.

 

seo images

 

Let x$ABC$ be a cone. A frustum $DECB$is cut by a plane parallel to its base. Let ${{r}_{1}},{{r}_{2}}$ be the radii of the ends of the frustum of the cone and $h$ be the height of the frustum of the cone.

Consider $\vartriangle ABG$ and $\vartriangle ADF$,

$DF\parallel BG$

$\therefore \vartriangle ABG\sim \vartriangle ADF$

$\frac{DF}{BG}=\frac{AF}{AG}=\frac{AD}{AB}$

$\frac{{{r}_{2}}}{{{r}_{1}}}=\frac{{{h}_{1}}-h}{{{h}_{1}}}=\frac{{{l}_{1}}-l}{{{l}_{1}}}$

$\frac{{{r}_{2}}}{{{r}_{1}}}=1-\frac{h}{{{h}_{1}}}=1-\frac{l}{{{l}_{1}}}$

$1-\frac{h}{{{h}_{1}}}=\frac{{{r}_{2}}}{{{r}_{1}}}$

$\frac{h}{{{h}_{1}}}=1-\frac{{{r}_{2}}}{{{r}_{1}}}$

$=\frac{{{r}_{1}}-{{r}_{2}}}{{{r}_{1}}}$

${{h}_{1}}=h\frac{{{r}_{1}}}{{{r}_{1}}-{{r}_{2}}}$

Volume of the frustum of the cone = Volume of cone ABC -  Volume of the cone ADE

$=\frac{1}{3}\pi {{r}_{1}}^{2}{{h}_{1}}-\frac{1}{3}\pi {{r}_{2}}^{2}\left( {{h}_{1}}-h \right)$

$=\frac{\pi }{3}\left[ {{r}_{1}}^{2}{{h}_{1}}-{{r}_{2}}^{2}\left( {{h}_{1}}-h \right) \right]$

Substitute ${{h}_{1}}=h\frac{{{r}_{1}}}{{{r}_{1}}-{{r}_{2}}}$

$=\frac{\pi }{3}\left[ {{r}_{1}}^{2}\left( \frac{h{{r}_{1}}}{{{r}_{1}}-{{r}_{2}}} \right)-{{r}_{2}}^{2}\left( \frac{h{{r}_{1}}}{{{r}_{1}}-{{r}_{1}}}-h \right) \right]$

$=\frac{\pi }{3}\left[ \frac{h{{r}_{1}}^{3}}{{{r}_{1}}-{{r}_{2}}}-\frac{h{{r}_{2}}^{3}}{{{r}_{1}}-{{r}_{2}}} \right]$

$=\frac{\pi }{3}h\left[ \frac{\left( {{r}_{1}}-{{r}_{2}} \right)}{{{r}_{1}}-{{r}_{2}}}\left( {{r}_{1}}^{2}+{{r}_{2}}^{2}+{{r}_{1}}{{r}_{2}} \right) \right]$

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

Thus the formula for the volume of the cone has been derived.

 

NCERT Solutions of Class 10 Maths Chapter 13 - Free PDF Download

You can opt for Chapter 13 - Surface Areas and Volumes NCERT Solutions for Class 10 Maths PDF for upcoming Exams  and also You can Find the Solutions of All the Maths Chapters below.


NCERT Solutions for Class 10 Maths


 When you need a quick revision of Class 10 Surface Area and Volume you can refer to the NCERT Solutions Class 10 Maths Chapter 13 PDF which is now available at the official website of Vedantu. You can download on your device or print out these solutions for quick and easy reference during exam times.


List of Exercises in class 10 Maths Chapter 13:

Chapter 13 Surface Areas and Volumes All Exercises in PDF Format

Exercise 13.1

9 Question and Solutions

Exercise 13.2

8 Questions and Solutions

Exercise 13.3

9 Questions and Solutions

Exercise 13.4

5 Questions and Solutions

Exercise 13.5

7 Questions and Solutions

 

Overview of the Exercises Covered in NCERT Solutions for Class 10 Maths Chapter 13 Surface Areas & Volumes

There are five exercises covered in Chapter 13 of Class 10 Maths. The exercises comprise sums of various levels of difficulty to assess students’ understanding of the concepts of surface areas and volumes of solids. Several formulas are used to solve the sums given in the NCERT Class 10 Maths Chapter 13. For example, there are formulae to calculate the volumes and surface areas of cones, cuboids, cylinders, spheres, and hemispheres. 


Students must learn and practice these formulae thoroughly so that they can apply the formulae to solve the sums of Ex.-13.1, 13.2, 13.3, 13.4, and 13.5. Exercise 13.1 consists of 9 sums, Exercise 13.2 consists of 8 sums, Exercise 13.3 consists of 9 sums, Exercise 13.4 consists of 5 sums, and Exercise 13.5 consists of 7 sums. 


While solving these sums students will have to compute the total and lateral surface areas and volumes of various combinations of shapes and solids. These types of sums are frequently asked in the exam. For example, students may encounter a sum in the CBSE Class 10 Term 2 Maths exam from Chapter 13, which requires computing the lateral surface area of a funnel placed upon a cylindrical vessel and its capacity. 


In fact, the sums covered in the exercises of Class 10 Maths Chapter 13 will teach such applications for multiple real-world scenarios. For example, they can use the formula of the volume of a cylinder to find the capacity of a cylindrical water tank. With a thorough understanding of all sums given in the NCERT Solutions for Class 10 Maths Chapter 13 Surface Areas and Volumes, students will be able to solve all types of sums asked in the CBSE Term 2 Class 10 Maths examination from this chapter.


Chapter 13 - Surface Area and Volumes

13.1 Introduction

In the introduction part of Class 10 Chapter 13, you would recall what you studied in class IX about solids like a cube, cylinder, cuboid, etc. You had learned the formulas for finding the surface area and volumes of these solids earlier.

 

We come across multiple objects in our daily lives that are a combination of many of these shapes. For example, a truck carrying oil which is in the shape of a cylinder that has 2 hemispheres at its end. In the Surface Area Volume Class 10 you would learn how to calculate the surface area and volumes of such solids which are a combination of two or more solid shapes.

 

13.2 Surface Area of a Combination of Solids

The outer part of any 3-D figure is the surface area of that figure. To find out the surface area of a solid which is a combination of solid shapes, we would need to find out the surface area of individual solid shapes separately to find the surface area of the entire 3-D solid shape. Let us clarify this with an example:

 

(image will be uploaded soon)

 

The solid in the figure above is a combination of a cone, cylinder, and hemisphere. So TSA or total surface area of the solid = Curved surface area of (Cone + Cylinder + Hemisphere) = \[\pi r \times \sqrt{(l^{2} + r^{2})} (cone) + 2\pi rl (cylinder) + 2\pi r^{2} (hemisphere)\].

 

Here l = 5 cms and r = radius of the hemisphere = 6/2 = 3 cms.

So, Total surface area of the solid = \[\pi (66 + 3 + \sqrt{34})\]

 

13.3 Volume of a Combination of Solids

The volume of solids by joining two or more basic solids is the sum of the volumes of individual solids. Let us understand this with an example:

 

(image will be uploaded soon)

 

The solid in the above figure is made up of two solids, i.e. a cuboid and a pyramid. So the total volume of the solid is obtained by adding up the volume of these two constituent solids.

Volume of cuboid = l * b * h = 4 * 4 * 5 = 80.

Volume of pyramid = ⅓ * area of the base * height = ⅓ * (4 * 4) * 6 = 32.

Volume of the solid = 80 + 32 = 112 inches3.

 

13.4 Conversion of a Solid From One Form to Another

All Important Formulas to Remember before Solving NCERT Class 10 Maths Chapter 13 - Surface Areas and Volumes

Below given is the list of Surface areas of 3D objects. 

S.No.

Shape

Formula

1

Total surface area of sphere = Curved surface area of sphere

4πr2

2

Total surface area of cone

πr (r + l)

3

Curved surface area of cone

πrl

4

Total surface area of cube

6 (side)2

5

Total surface area of cuboid

2 (lb + bh + hl)

6

Total surface area of cylinder

2πr(r + h)

7

Curved surface area of cylinder

2πrh


Below given is the list of volumes of 3D objects. 

S.No.

Shape

Formula

1

Volume of sphere 

43πr3

2

Volume of hemisphere

23πr3

3

Volume of cone

πr2h3

4

Volume of cube

(side)3

5

Volume of cuboid

lbh

6

Volume of cylinder

πr2h


Where r is the radius of the object, l (for cone) is the slant height and l = length, b = breadth, h= height for the cuboid.

These are the basic formulas that you must remember. In the questions asked in these chapters, you might need to use more than one formula at a time, so you can not skip them. 


When we convert a solid from one form to another by the method of melting or remoulding, then the volume of the solid stays the same, despite the change in shape. We will see an example to understand why:

 

(image will be uploaded soon)

 

In the figure above, if there is water in the cuboid shape (which has dimensions of 20 * 22) which is transferred into a cylinder that has a height of 3.5 m and 2 m then what is the height of the water level in the cuboid shape if the water (once transferred into the cylinder) fills the cylinder to the brim?

Solution - From the theorem on volumes, we know that volume of the water in the cylinder and the volume of the water in the cuboid would be the same. If the water fills the cylinder to the brim then:

Volume of water in cylinder = π * r² * h = 22/7 * 1 * 3.5 = 11.

Volume of water in cuboid = l * b * h = 20 * 22 * h.

SInce 20 * 22 * h  = 3.5 * π, we get h = 11/(20 * 22) = 2.5 cm.

 

Key Features of NCERT Solutions for Class 9 Maths Chapter 13

Many students often find it difficult to remember important Mathematical formulas and even in solving textbook exercises. Therefore, referring to these solved NCERT Solutions will help them prepare better for the exams.  Some of the important features of using NCERT Solutions for Class 9 Maths Chapter 13 are given below:

  1. The formulas and numericals included in this NCERT Solutions are very important and will help the students in various competitive exams like JEE and Olympiads.

  2. The NCERT Solutions for Class 9 Maths provides the students with several formulas that are useful to solve the exercises. They can list down these formulas and revise them on a regular basis.  Doing this will help students solve the exercises more efficiently.

  3. These solutions are solved in a very simple and step-by-step manner to help the students to understand and solve the problems easily. 

  4. It includes solved numericals of all the exercises that are included in the NCERT textbook. These solutions are solved by the expert teachers keeping in mind the basic guidelines of the CBSE Board.


NCERT Solutions of Class 10 Maths Chapter 13 - Free PDF Download

You can opt for Chapter 13 - Surface Areas and Volumes NCERT Solutions for Class 10 Maths PDF for upcoming Exams  and also You can Find the Solutions of All the Maths Chapters below.


NCERT Solutions for Class 10 Maths


Other Related Links


Chapter 13 on Surface Areas and Volumes is a very interesting and scoring chapter from the examination point of view. Students can apply the knowledge of computing the surface areas and volumes of various solids like frustums, cones, cylinders, spheres, cubes, etc., for various real-world objects, as well. Students should download and practice the NCERT Solutions for Class 10 Maths Chapter 13 Surface Areas and Volumes and aim to score a full mark in their Maths Term 2 exam.


Conclusion

The NCERT Solutions for Class 10 Maths Chapter 13 - Surface Areas and Volumes, provided by Vedantu, is a valuable tool for Class 10 students. It helps introduce Maths concepts in an accessible manner. The provided solutions and explanations simplify complex ideas, making it easier for Class 10 Students to understand the material. By using Vedantu's resources, Students can develop a deeper understanding of NCERT concepts. These solutions are a helpful aid for grade 10 students, empowering them to excel in their studies and develop a genuine appreciation for Constructions.

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

1. List out all the formulas of Class 9 Maths Chapter 13 Surface Area and Volume

All the important formulas that are included in Class 9 Maths Chapter 13 Surface Area and Volume are listed in the table below:

NCERT Class 9 Maths Chapter 13 Surface Area and Volume Formulas

Objects

Total Surface Area

Curved Surface Area

Volume

Cube

6a2

4a2

a2

Cuboid

2(lb+bh+hl)

2(l+b)h

lbh

Right-circular Cone

πr(l+r)

πrl

⅓πr2h

Right-circular Cylinder

2πr(h+r)

2πrh

π2h

Sphere

4πr2

4πr2

4/3πr3

Hollow Hemisphere

2πr2

2πr2

2/3πr3

Solid Hemisphere

3πr2

2πr2

2/3πr3

2. What are the important topics that come under NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes?

The important topics that are included under NCERT Class 9 Maths Chapter 13 Surface Areas and Volumes are mentioned below:

  • Curved surface area, total surface area and volume of cubes and cuboids

  • Surface area and volume of a right circular cone and right-circular cylinder.

  • The curved surface area of a sphere and a hemisphere

  • Surface area and volume of a sphere, hollow hemisphere and solid hemisphere

  • Curved surface area and volume of a right circular cylinder and cone.

All the above-mentioned topics are very important for the CBSE Class 9 final examination. Students must practise these topics and formulas regularly for better performance. 

3. Where can I download NCERT Solutions for Class 9 Maths Chapter 13 pdf for free?

You can download free PDFs of NCERT Solutions for Class 9 Maths Chapter 13 by registering yourself on the Vedantu app. We provide free PDF downloads of solved NCERT Solutions which are created by expert teachers in a very simple and step-by-step manner.

4. How can Vedantu help me in scoring good grades in Class 10 Maths?

Class 10 holds an important place in the life of every student. Securing good grades on 10th boards is what every student wishes for. Therefore, it is important to have the best study partner for exam preparations. Given that, Vedantu comes with exclusive study materials developed by a team of experts. Also, you will get NCERT Solutions for Class 10 Maths, mock tests, and previous year papers for Class 10 Maths.

5. Is Class 10 Maths Chapter 13 important for Boards?

Class 10 Maths Chapter 13 is essential for Boards. This chapter may present you with long word-problem-type questions. Since it is fairly interesting and fun to solve, all you need to do is recall the key formulas of surface areas and volumes and their applications. You can use Vedantu's study materials to help you understand all of the ideas. You will be given a series of questions to aid you with quality practices.

6. How is Chapter 13 ‘Surface area and volume’ helpful in real life?

Chapter 13 ‘Surface area and volume’ is not just a chapter that teaches you a few formulas but if you understand the concepts thoroughly, you can apply them in real-world situations as well. In our daily life, we apply them unknowingly. For example, filling oil in a can, covering the floor with tiles, making a box out of cardboard, or wrapping a box with gift paper, all come under surface area and volume.  

7. Are PDFs of NCERT Solutions for Class 10 Maths Chapter 13 available online?

Yes, PDFs of NCERT Solutions for Class 10 Maths Chapter 13 are available on the online platform of Vedantu (vedantu.com). It is freely accessible for download from the website as well as the Vedantu app. The NCERT solutions are offered in an easy-to-understand and detailed format. Subject matter experts have ensured that the finest answers are curated for students. You can browse over these solutions and get clarity on any sort of query or doubt.

8. How should I prepare myself for Class 10 CBSE Boards?

The Class 10 boards are essential in determining your specializing subject for later. Hence, we recommend that you take the following steps to ensure good scores in the Class 10 Boards:

  •  Follow the CBSE syllabus.

  •  Choose books wisely.

  •  Focus on important chapters and topics.

  •  Once the subject is prepared, revise and practice on a regular basis.

  •  Solve the sample papers and also previous years' exam papers.