Surface Areas and Volumes

All

Guides

Practice

Learn

Surface Area and Volume Class 10 NCERT Solutions Chapter 13 are made strictly in accordance with the CBSE Curriculum and the exam pattern. NCERT Solutions for Class 10 Maths Chapter 13 - Surface Areas and Volumes are curated by our team of subject experts at Toppr in a detailed manner. The NCERT textbook questions are answered in a way to provide you with a better understanding of the concepts in a systematic and step-by-step manner. It also contains the appropriate diagrams to explain the concepts to the students. As Class 10 exams are Board exams, these solutions will not only help the students in preparing for the board exams but also for the Olympiads. NCERT Solutions provided by Toppr are the best study material to excel in the exams. Also, the MCQs and long and short questions are all answered according to the weightage and the exam pattern. With the help of NCERT Solutions for Class 10 Maths Chapter 13 Surface Area and Volume Class 10 Solutions, you can also test your subject knowledge and analyze your shortcomings and work on them before the exams. These are the best resources designed after proper study and research and study to help the students in scoring good marks.

Table of Content

Exercise: 13.1

Question 1

Two cubes each of volume 64 $cm_{3}$ are joined end to end. Find the surface area of the resulting cuboid.

Solution

Verified by Toppr

The volume of cube $64cm_{3}$

Side of cube $364 =4cm$

Length of resulting cuboid $4+4=8$

Surface area

$=2(lb+hl+bh)$

$=2(4(4)+4(8)+8(4))=2(16+32+32)=2(80)=160cm_{2}$

Question 2

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.

Solution

Verified by Toppr

Radius=$7$cm

Height of the cylindrical portion$13−7=6$cm.

Area of a Curved surface of cylindrical portion is,

=$2πrh$

=$2×722 ×7×6$

=$264cm_{2}$

Area of a curved of hemispherical portion is

=$2πr_{2}$

=$2×722 ×7×7$

=$308cm_{2}$

$∴$ total surface area is $=308+264=572cm_{2}$.

Height of the cylindrical portion$13−7=6$cm.

Area of a Curved surface of cylindrical portion is,

=$2πrh$

=$2×722 ×7×6$

=$264cm_{2}$

Area of a curved of hemispherical portion is

=$2πr_{2}$

=$2×722 ×7×7$

=$308cm_{2}$

$∴$ total surface area is $=308+264=572cm_{2}$.

Question 3

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

Solution

Verified by Toppr

Total surface area of toy $=$ Curved surface area of cone $+$ surface area of hemisphere

Curved surface area of cone $=$ $πrl$

Where $r=3.5cm$, Height $=15.5−3.5=12cm$

And hence, $l=12.5cm$ (by using formula $l_{2}=h_{2}+b_{2}$)

Therefore C.S.A. of cone $=π×3.5×12.5$

$=137.5cm_{2}$

Surface area of hemisphere $=2πr_{2}$

$=2×π×(3.5)_{2}$

$=77cm_{2}$

Hence T.S.A of toy $=77+137.5=214.5cm_{2}$

Question 5

A hemispherical depression is cut out from one face of the 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.

Solution

Verified by Toppr

Consider the diagram shown below.

$=6l_{2}−π(2l )_{2}+2π(2l )_{2}$

It is given that a hemisphere of radius $2l $ is cut out from the top face of the cuboidal wooden block.

Therefore, surface area of the remaining solid

$=$ surface area of the cuboidal box whose each edge is of length $l−$ Area of the top of the hemispherical part $+$ curved surface area of the hemispherical part

$=6l_{2}−πr_{2}+2πr_{2}$

$=6l_{2}−π(2l )_{2}+2π(2l )_{2}$

$=6l_{2}−4πl_{2} +2πl_{2} $

$=4l_{2} (24+π)sq.units$

Question 6

A medicine capsule is in the shape of a cylinder with two hemispheres stuck to each of its ends .The length of the entire capsule is $14mm$ and the diameter of the capsule is $5mm$. Find its surface area. (Use $π=722 $).

Solution

Verified by Toppr

Given:

Length of Capsule (l) = $14$ mm

Therefore, Radius of each Hemisphere = Radius of Cylinder = r = $25 $ = 2.5 mm

Length of Capsule (l) = $14$ mm

Diameter of Capsule = Diameter of Cylinder $=5$ mm

Radius = $2Diameter $

Therefore, Radius of each Hemisphere = Radius of Cylinder = r = $25 $ = 2.5 mm

Length of Cylinder = AB = Total length of Capsule - Radius of left Hemisphere - Radius of Right Hemisphere

$=14−2.5−2.5=9$ mm

Surface Area of Capsule = Curved Surface Area of Cylinder + Surface Area of Left Hemisphere + Surface Area of Right Hemisphere

$=2πrl+2πr_{2}+2πr_{2}$

$=2πrl+4πr_{2}$

$=2×722 ×2.5×9+4×722 ×2.5_{2}$

$=722 [45+25]=722 ×70=220$ mm$_{2}$

Question 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 $2.1$ 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.) (Use $π=722 $).

Solution

Verified by Toppr

Given:

Height (h) of the cylindrical part = $2.1$ m

Diameter of the cylindrical part = $4$ m

Radius of the cylindrical part = $2$ m

Slant height (l) of conical part = $2.8$ m

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

Height (h) of the cylindrical part = $2.1$ m

Diameter of the cylindrical part = $4$ m

Radius of the cylindrical part = $2$ m

Slant height (l) of conical part = $2.8$ m

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

$=πrl+2πrh$

$=π×2×2.8+2π×2×2.1$

$=2π[2.8+4.2]$

$=2π[2.8+4.2]$

$=2×722 ×7$

$=44m_{2}$

Cost of $1$ m $_{2}$ canvas $=500$ rupees

Cost of $44$ m $_{2}$ canvas $=44×500=22000$ rupees.

Cost of $44$ m $_{2}$ canvas $=44×500=22000$ rupees.

Therefore, it will cost $22000$ rupees for making such a tent.

Question 8

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 $cm_{2}$. (Use $π=722 $).

Solution

Verified by Toppr

Given:

Height (h) of the conical part = Height (h) of the cylindrical part $=2.4$ cm

Diameter of the cylindrical part $=1.4$ cm

Radius = $2Diameter $

Radius(r) of the cylindrical part $=0.7$ cm

Slant height (l) of conical part = $r_{2}+h_{2} $

Height (h) of the conical part = Height (h) of the cylindrical part $=2.4$ cm

Diameter of the cylindrical part $=1.4$ cm

Radius = $2Diameter $

Radius(r) of the cylindrical part $=0.7$ cm

Slant height (l) of conical part = $r_{2}+h_{2} $

$=0.7_{2}+2.4_{2} =0.49+5.76 =6.25 =2.5$

Total surface area of the remaining solid = CSA of cylindrical part + CSA of conical part + Area of cylindrical base

$=2πrh+πrl+πr_{2}$

$=2×722 ×0.7×2.4+722 ×0.7×2.5+722 ×0.7×0.7$

$=4.4×2.4+2.2×2.+2.2×0.7$

$=10.56+5.50+1.54=17.60$ cm$_{2}$

The total surface area of the remaining solid to the nearest $cm_{2}$ is $18$ $cm_{2}$

Exercise: 13.2

Question 1

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

Solution

Verified by Toppr

We have

Radius of the cone $=r_{1}=1cm$

Height of the cone $=h_{1}=1cm$

Radius of the hemisphere $=r_{2}=1cm$

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

$=31 πr_{1}h_{1}+32 πr_{2}$

Radius of the cone $=r_{1}=1cm$

Height of the cone $=h_{1}=1cm$

Radius of the hemisphere $=r_{2}=1cm$

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

$=31 πr_{1}h_{1}+32 πr_{2}$

$=(31 π×1_{2}×1+32 π×1_{3})cm_{3}$

$=πcm_{3}$

Question 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 aluminium sheet. The diameter of the model is $3cm$ and its length is $12cm$. If each cone has a height of $2cm$, 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.)

Solution

Verified by Toppr

$Step 1: Finding the radius and height of cone and cylinder$

$Radius of the conical and cylindrical part(r)=1.5cm$

$Height of the cylindrical part(H)=8cm$

$Height of the conical part(h)=2cm$

$Step 2: Finding the volume of air in the model$

$Volume of air in the model=Volume of cylinder +2×Volume of cone$

$Volume of cylinder=πr_{2}H$

$=722 ×(1.5)_{2}×8$

$Volume of cone=31 πr_{2}h$

$=31 ×722 ×(1.5)_{2}×2$

$Volume of air=722 ×(1.5)_{2}×8+2×31 ×722 ×(1.5)_{2}×2≈66cm_{3}$

$Hence , the volume of air in the model is 66cm_{3}$

Question 3

A gulab jamun, contains sugar syrup up to about 30% of its volume. Find the 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 .

Solution

Verified by Toppr

$Step 1: Find out the volume.$

$Givenh=5cm$

$Length of cylinderh=5−2.8=2.2$

$Volume of gulab jamun = volume of cylinder + volume of 2 hemisphere$

$=3πr_{2}h+34 πr_{3}$

$=722 ×1.4×1.4×3(3×2.2+4×1.4) $

$=36.16(12.2) $

$=375.152 $

$=25.05cm$

$There are 45 gulab jamuns.$

$Step 2: Find the volume of 45 gulab jamuns.$

$Volume = 45×25.05$

$=1127.2830percentage of syrup$

$=1127.28×10030 $

$=338.184cm$

$Hence, there are338.184cm of syrup be found in45gulab jamuns.$

View more

Practice more questions

- Class 10 NCERT Solutions are provided in a step-by-step manner with appropriate diagrams.
- The solutions provide a better understanding of the subject and concepts.
- These are curated by the experts after thorough research.
- These solutions provide excellent study material for revision.
- They are the best means to evaluate your preparations and overcome your shortcomings.
- The Class 10 NCERT Solutions will help the students in board exams as well as Olympiads.
- These are absolutely free to download.

Related Chapters

- Chapter 1 : Real NumbersChapter 2 : PolynomialsChapter 3 : Pair of Linear Equations in Two VariablesChapter 4 : Quadratic EquationsChapter 5 : Arithmetic ProgressionChapter 6 : TrianglesChapter 7 : Coordinate GeometryChapter 8 : Introduction to TrigonometryChapter 9 : Some Applications of TrigonometryChapter 10 : CirclesChapter 11 : ConstructionsChapter 12 : Area Related to CirclesChapter 14 : StatisticsChapter 15 : Probability

Question 1. How is the surface area of a solid which is a combination of solid shapes calculated?

Answer. In order to find out the surface area of a solid which is a combination of solid shapes, firstly, the surface area of individual solid shapes needs to be calculated separately.

Question 2. How is the volume of a combination of solids calculated?

Answer. The volume of a combination of solids is calculated by adding the volumes of individual solids. For example, a 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 the cuboid and the volume of the pyramid.

Question 3. Does the volume of a solid change when we convert it into another solid by the method of melting or remoulding?

Answer. No, the volume of a solid does not change when we convert it into another solid by the method of melting or remoulding, despite the change in its shape.

Question 4. What do you understand by the Frustum of a cone?

Answer. ‘Frustum’ is a Latin word that means ‘piece cut off’. The plural of the frustum is ‘frusta’. When we slice or cut a given cone through it with a plane parallel to its base and remove the cone that is formed on one side of that plane, the part that is now leftover on the other side of the plane is called a frustum of the cone. The frustum of the cone has two circular ends with different radii.

Related Chapters

- Chapter 1 : Real NumbersChapter 2 : PolynomialsChapter 3 : Pair of Linear Equations in Two VariablesChapter 4 : Quadratic EquationsChapter 5 : Arithmetic ProgressionChapter 6 : TrianglesChapter 7 : Coordinate GeometryChapter 8 : Introduction to TrigonometryChapter 9 : Some Applications of TrigonometryChapter 10 : CirclesChapter 11 : ConstructionsChapter 12 : Area Related to CirclesChapter 14 : StatisticsChapter 15 : Probability