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NCERT Solutions for Exercise 13.1 Class 10 Maths Chapter 13 - Surface Area and Volumes

NCERT Solutions for Exercise 13.1 Class 10 Maths Chapter 13 - Surface Area and Volumes

Updated on Apr 30, 2025 05:13 PM IST | #CBSE Class 10th

Suppose a person has a cuboid and you want to paint this cuboid from the outer, and you want to know how much is required to paint this cuboid. For this, the total area of the outer layer should be calculated, and this area is called the surface area of any object. Therefore, surface area is the total outer space of any three-dimensional object. After painting this cuboid, if a person wants to fill this cuboid with water, then how much water is needed to fill this cuboid? To determine the answer to this question person should know the amount of space within this cuboid, and this amount of space is called the volume of the cuboid. Therefore, the amount of space within any three-dimensional object is called the volume.

This Story also Contains
  1. Download Free PDF of NCERT Solutions for Class 10 Maths Chapter 12 Exercise 12.1
  2. Assess NCERT Solutions for Class 10 Maths Chapter 12 Exercise 12.1
  3. Topics Covered in Chapter 12, Surface Areas and Volumes: Exercise 12.1
  4. NCERT Solutions for Class 10 Subject Wise
  5. NCERT Exemplar Solutions of Class 10 Subject Wise
NCERT Solutions for Exercise 13.1 Class 10 Maths Chapter 13 - Surface Area and Volumes
NCERT Solutions for Exercise 13.1 Class 10 Maths Chapter 13 - Surface Area and Volumes

In the NCERT Books exercise 12.1, Class 10 Maths, we have to find the surface area of the combined solids. Some of these solids are cylinders with both sides hemispheres, and cones with a hemisphere on top. Surface area can be categorised into two types of surface areas: the curved surface area and the total surface area. 10th class Maths exercise 12.1 answers are designed as per the students' demand, covering comprehensive, step-by-step solutions of every problem. Practice these questions and answers to command the concepts, boost confidence and in-depth understanding of concepts. These NCERT solutions Class 10 Maths ex 12.1 present us with the surface area of a combination of Solids. Some solids, such as the cuboid, cone, cylinder, and sphere, are recognisable to you from Class IX. You've also learned how to calculate their volumes and surface areas. In our daily lives, we come across a variety of solids that are made up of two or more of the fundamental solids.

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Download Free PDF of NCERT Solutions for Class 10 Maths Chapter 12 Exercise 12.1

Download PDF

Assess NCERT Solutions for Class 10 Maths Chapter 12 Exercise 12.1

Q1. 2 cubes, each of volume 64 cm3 are joined end to end. Find the surface area of the resulting cuboid.

Answer:

We are given that volume of the cube =64cm3

Also, the volume of a cube is given by =a3 ( here a is the edge of the cube)

Thus: a3=64

a=4cm

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

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

or =2(8×4+4×4+4×8)

or =2(80)

or =160cm2

Thus, the area of the formed cuboid is 160 cm².

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 a hemisphere and a cylinder, its area is given by :

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

The inner surface area of the hemisphere is :

=2πr2

or =2×(227)×72

or =308cm2

And the surface area of the cylinder is :

=2πrh

or =2×227×7×6

or =264cm2

Thus, the inner surface area of the vessel is =308+264=572cm2.

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 =2πr2

or =2×227×(3.5)2

or =77cm2

Now, for the cone we have:

The surface area of a cone =πrl

Thus, we need to calculate the slant of the cone.

We know that:

l2=h2+r2

or =122+3.52

or =6254

or l=252=12.5cm

Thus surface area of a cone =πrl

or =227×3.5×12.5

or =137.5cm2

Hence, the total surface area of toy = =77+137.5=214.5cm2

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

The surface area of the cube is :

=6a3

=6×73=294cm2

Now the area of a hemisphere is

=2πr2

=2×227×(72)2=77cm2

And the area of the base of a hemisphere is

=πr2=227×(72)2=38.5cm2

Hence, the surface area of the solid is =294+7738.5=332.5cm2.

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 the solid is given by :

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

The surface area of the cube =6l2

And the surface area of the hemisphere:

=2πr2=2π(l2)2

Area of the base of the hemisphere:

=πr2=π(l2)2

Thus, the area of the solid is:

=6l2+π(l2)2unit2

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 a hemisphere and cylinder structure.

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

Area of hemisphere =2πr2

or =2π×(52)2

or =252πmm2

And the area of the cylinder =2πrh

or =2π×52×9

or =45πmm2

Thus, the area of the solid is :

=2(252)π+45π

=25π+45π

=70π

=220mm2

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 =2πrh

or =2π×2×2.1

or =8.4πm2

And the area of the cone is =πrl

or =π×2×2.8

or =5.6πm2

Thus, the area of the canvas =8.4π+5.6π

or =14π=44m2

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

Thus, the required money is =500×44=Rs.22,000

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 :

l2=r2+h2

or =(0.7)2+(2.4)2

or l2=6.25

or l=2.5cm

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

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

The surface area of the cylinder is =2πrh

or =2π×0.7×2.4

or =10.56cm2

Now, the surface area of a cone =πrl

or =π×0.7×2.5

or =5.50cm2

And the area of the base of the cylinder is =πr2

or =π×0.7×0.7

or =1.54cm2

Thus required area of the solid = 10. 56 + 5.50 + 1.54 = 17.60 cm2 .

Thus total surface area of the remaining solid to the nearest cm2 is 18 cm2.

Q9. A wooden article was made by scooping out a hemisphere from each end of a solid cylinder, as shown in Fig. 12.11. 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.

2%20(1)

Answer:

The required surface area is given by :

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

Now, the area of the cylinder =2πrh

or =2π×3.5×10

or =70πcm2

And the surface area of the hemisphere: =2πr2

or =2π×3.5×3.5

or =24.5πcm2

Thus, the required area =70π+2(24.5π)=374cm2



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Topics Covered in Chapter 12, Surface Areas and Volumes: Exercise 12.1

  • Surface Area: The total outer area covered by any object is called the surface area of an object. The surface area includes: curved surface area, lateral surface area, and total surface area.
  • Curved Surface Area: The total outer curved area covered by an object is called the curved surface area, and it does not include any flat or base surface of an object.
  • Lateral Surface Area: The outer area covered by an object, without including the top and base, is called the lateral surface area of an object
  • Total Surface Area: The outer covered area of an object, including the top or curved area and the bottom of the object, is called the total surface area.
  • Volume: The space occupied within the boundaries of an object is called the volume of an object.
  • Surface Area of Combination Object: When two or more solid objects are combined, then finding the surface area of the combination of these objects. For example, combining a cone and a cylinder.
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JEE Main Important Mathematics Formulas

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NCERT Solutions for Class 10 Subject Wise

Students must check the NCERT solutions for class 10 of the Mathematics and Science Subjects.

NCERT Exemplar Solutions of Class 10 Subject Wise

Students must check the NCERT Exemplar solutions for class 10 of the Mathematics and Science Subjects.

Frequently Asked Questions (FAQs)

1. What is the curved surface area of the hemisphere ?

Curved surface area of hemisphere used is =2πr²

2. What is the curved surface area of the cylinder u?

Curved surface area of hemisphere used in  2πrl

3. What is the most commonly used value of pi used in Exercise 12.1 Class 10 Maths ?

The most commonly used value of pi used in exercise 12.1 Class 10 Maths is 227.

4. How can we use surface area in real life? Give some examples ?

We can use surface area in many places in our daily life, for example identifying the amount of paint require to paint object , determining the cost of things like clothes , covers and canvas tents 

5. What is the total number of example that have been solved before the Exercise 12.1 Class 10 Maths.

The total number of examples that have been solved before the class 10 Maths chapter 12 exercise 12.1 is four 

6. What is the total number of questions in the NCERT solutions for Class 10 Maths chapter 12 exercise?

The total number of questions in the NCERT solutions for Class 10 Maths chapter 12 exercise is nine.

7. What kind of questions are asked in 10 Maths chapter 12 exercise 12.1?

There are many basic question in which we have to find the total surface of combination of solid or real life objects such as tents , capsules toys etc but some are direct that only ask about the solid  

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Questions related to CBSE Class 10th

Have a question related to CBSE Class 10th ?

Hello

Since you are a domicile of Karnataka and have studied under the Karnataka State Board for 11th and 12th , you are eligible for Karnataka State Quota for admission to various colleges in the state.

1. KCET (Karnataka Common Entrance Test): You must appear for the KCET exam, which is required for admission to undergraduate professional courses like engineering, medical, and other streams. Your exam score and rank will determine your eligibility for counseling.

2. Minority Income under 5 Lakh : If you are from a minority community and your family's income is below 5 lakh, you may be eligible for fee concessions or other benefits depending on the specific institution. Some colleges offer reservations or other advantages for students in this category.

3. Counseling and Seat Allocation:

After the KCET exam, you will need to participate in online counseling.

You need to select your preferred colleges and courses.

Seat allocation will be based on your rank , the availability of seats in your chosen colleges and your preferences.

4. Required Documents :

Domicile Certificate (proof that you are a resident of Karnataka).

Income Certificate (for minority category benefits).

Marksheets (11th and 12th from the Karnataka State Board).

KCET Admit Card and Scorecard.

This process will allow you to secure a seat based on your KCET performance and your category .

check link for more details

https://medicine.careers360.com/neet-college-predictor

Hope this helps you .

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

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

hello Zaid,

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

best of luck!

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

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

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

Option 1)

0.34\; J

Option 2)

0.16\; J

Option 3)

1.00\; J

Option 4)

0.67\; J

A person trying to lose weight by burning fat lifts a mass of 10 kg upto a height of 1 m 1000 times.  Assume that the potential energy lost each time he lowers the mass is dissipated.  How much fat will he use up considering the work done only when the weight is lifted up ?  Fat supplies 3.8×107 J of energy per kg which is converted to mechanical energy with a 20% efficiency rate.  Take g = 9.8 ms−2 :

Option 1)

2.45×10−3 kg

Option 2)

 6.45×10−3 kg

Option 3)

 9.89×10−3 kg

Option 4)

12.89×10−3 kg

 

An athlete in the olympic games covers a distance of 100 m in 10 s. His kinetic energy can be estimated to be in the range

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

A particle is projected at 600   to the horizontal with a kinetic energy K. The kinetic energy at the highest point

Option 1)

K/2\,

Option 2)

\; K\;

Option 3)

zero\;

Option 4)

K/4

In the reaction,

2Al_{(s)}+6HCL_{(aq)}\rightarrow 2Al^{3+}\, _{(aq)}+6Cl^{-}\, _{(aq)}+3H_{2(g)}

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 .

How many moles of magnesium phosphate, Mg_{3}(PO_{4})_{2} will contain 0.25 mole of oxygen atoms?

Option 1)

0.02

Option 2)

3.125 × 10-2

Option 3)

1.25 × 10-2

Option 4)

2.5 × 10-2

If we consider that 1/6, in place of 1/12, mass of carbon atom is taken to be the relative atomic mass unit, the mass of one mole of a substance will

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.

With increase of temperature, which of these changes?

Option 1)

Molality

Option 2)

Weight fraction of solute

Option 3)

Fraction of solute present in water

Option 4)

Mole fraction.

Number of atoms in 558.5 gram Fe (at. wt.of Fe = 55.85 g mol-1) is

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

A pulley of radius 2 m is rotated about its axis by a force F = (20t - 5t2) newton (where t is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is 10 kg m2 , the number of rotations made by the pulley before its direction of motion if reversed, is

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

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