• School
  • Engineering and Architecture
  • Management and Business Administration
  • Medicine and Allied Sciences
  • Law
  • Animation and Design
  • Media, Mass Communication and Journalism
  • Finance & Accounts
  • Computer Application and IT
  • Pharmacy
  • Hospitality and Tourism
  • Competition
  • Study Abroad
  • Arts, Commerce & Sciences
  • Learn
  • Online Courses and Certifications
NCERT Solutions for Class 10 Maths Chapter 11 Constructions

NCERT Solutions for Class 10 Maths Chapter 11 Constructions

Edited By Ramraj Saini | Updated on Sep 07, 2023 06:08 PM IST | #CBSE Class 10th
Upcoming Event
CBSE Class 10th  Exam Date : 15 Feb' 2025 - 15 Feb' 2025

Construction Class 10 NCERT Solutions

NCERT Solutions for class 10 maths construction are thoroughly explained hare. These solutions are created by expert team at Careers360 keeping in mind of latest syllabus and pattern of CBSE 2023-24. These NCERT solutions provide a step-by-step approach to solving all questions and comprehensively discussed them in easy and simple language. These NCERT solutions for class 10 are created under NCERT guidelines, are designed to aid students in preparing for their board exams.

This Story also Contains
  1. Construction Class 10 NCERT Solutions
  2. NCERT Solutions for Class 10 Maths Construction PDF Free Download
  3. NCERT Solutions for class 10 maths construction - Important Points
  4. NCERT Solutions for construction class 10 (Intext Questions and Exercise)
  5. Construction Class 10 NCERT-Topics
  6. Feature of Solutions to Class 10 Maths NCERT Chapter 11 Constructions
  7. NCERT Solutions for Class 10 Maths - Chapter Wise
  8. NCERT Solutions of Class 10 - Subject Wise
  9. NCERT Exemplar solutions - Subject Wise
  10. NCERT Books and NCERT Syllabus
NCERT Solutions for Class 10 Maths Chapter 11 Constructions
NCERT Solutions for Class 10 Maths Chapter 11 Constructions

Get free access to construction class 10 solutions on Careers360 to improve your CBSE exam preparation. Each question of the NCERT exercises is solved using diagrams with a clear step-by-step procedure for the class 10 construction solution. These NCERT solutions can help students improve their understanding and clarify any doubts they may have. Also Students can download NCERT solutions for class 10 maths chapter 11 pdf using the link given below.

Also Read,

Pearson | PTE

Register now for PTE & Unlock 20% OFF : Use promo code: 'C360SPL20'. Valid till 31st DEC'24! Trusted by 3,500+ universities globally

Apply

NCERT Solutions for Class 10 Maths Construction PDF Free Download

Download PDF

NCERT Solutions for class 10 maths construction - Important Points

Determination of a Point Dividing a Line Segment Internally:

Given a line segment AB, a point P can be determined that divides AB internally in the ratio M : N using the formula: AP/AB = M/(M+N)

Construction of a Tangent to a Circle with Known Center - A tangent can be constructed at a specific point P on a circle with a known centre O. The tangent line is perpendicular to the radius OP.

Construction of a Tangent to a Circle without Known Center - Even without knowing the exact center of the circle, you can construct a tangent at a point P on the circle using geometric methods.

Construction of Tangents from an External Point to a Circle with Known Center - Tangents can be drawn from an external point E to a circle with a known centre O. The tangent lines are equal in length and angles between the radius and tangents are right angles.

Construction of Tangents from an External Point to a Circle without Known Center - Even without knowing the circle's center, you can construct tangents from an external point E to the circle using specific techniques.

Construction of a Similar Triangle with Given Scale Factor (m/n), m < n:

To construct a triangle similar to a given triangle ABC, but scaled down with a ratio m/n, the corresponding sides of the new triangle are m times smaller than the sides of ABC.

Construction of a Similar Triangle with Given Scale Factor (m/n), m > n:

Similarly, if the scale factor m/n is greater than 1, the corresponding sides of the new triangle are m times larger than the sides of ABC.

Free download NCERT Solutions for class 10 maths construction PDF for CBSE Exam.

NCERT Solutions for construction class 10 (Intext Questions and Exercise)

NCERT Solutions for Maths Chapter 11 Construction Class 10th Excercise: 11.1

In each of the following, give the justification of the construction also:
Q1 Draw a line segment of length 7.6 cm and divide it in the ratio 5 : 8. Measure the two parts.

Answer:

Steps of construction:-

(i) Draw a line segment AB of measurement 7.6 cm (length).

(ii) Now draw an acute angle AC with line segment AB.

(iii) Now cut 13 equal points on the line AC where the zeroth point is A.

(iv) Join the 13th point with point B. So the new line is BA 13 .

(v) Now, from point A 5 draw a line parallel to BA 13 on line AB. Name the point as D.

The point D is the required point which divides the line segment in the ratio of 5: 8.

The length of the two parts obtained is 2.9 cm and 4.7 cm for AD and DB respectively.

Constructions, 24670

Justification:- In the figure, we can see two similar triangles: \Delta ADA_5 and \Delta ABA_{13}

Thus \frac{AA_5}{A_5A_{13}}\ =\ \frac{AD}{DB}\ =\ \frac{5}{8} .

Q2 Construct a triangle of sides 4 cm, 5 cm and 6 cm and then a triangle similar to it whose sides are 2/3 of the corresponding sides of the first triangle.

Answer:

Steps of construction are:-

(i) Firstly draw a line segment AB of length 4 cm.

(ii) Now cut an arc of radius 5 cm from point A and an arc of 6 cm from point B.

(iii) Name the point of intersection of arcs to be point C.

(iv) Now join point AC and BC. Thus \Delta ABC is the required triangle.

(v) Draw a line AD which makes an acute angle with AB and is opposite of vertex C.

(vi) Cut three equal parts of line AD namely AA 1 , AA 2 , AA 3 .

(vii) Now join A 3 to B. Draw a line A 2 B' parallel to A 3 B.

(viii) And then draw a line B'C' parallel to BC.

Hence \Delta AB'C' is the required triangle.

1636090329369

Q3 Construct a triangle with sides 5 cm, 6 cm, and 7 cm and then another triangle whose sides are 7 / 5 of the corresponding sides of the first triangle.

Answer:

Steps of construction are:-

(i) Firstly draw a line segment AB of length 5 cm.

(ii) Now cut an arc of radius 6 cm from point A and an arc of 7 cm from point B.

(iii) Name the point of intersection of arcs to be point C.

(iv) Now join point AC and BC. Thus \Delta ABC is the required triangle.

(v) Draw a line AD which makes an acute angle with AB and is opposite of vertex C.

(vi) Cut seven equal parts of line AD namely AA 1 , AA 2 , AA 3 , AA 4 , AA 5 , AA 6 , AA 7 ,.

(vii) Now join A 5 to B. Draw a line A 7 B' parallel to A 5 B.

(viii) And then draw a line B'C' parallel to BC.

Hence \Delta AB'C' is the required triangle.

1636090397260

Q4 Construct an isosceles triangle whose base is 8 cm and altitude 4 cm and then another triangle whose sides are 3/ 2 times the corresponding sides of the isosceles triangle.

Answer:

Steps of construction:-

(i) Draw a line segment AB of length 8 cm.

(ii) Cut arcs taking point A and point B as the center. Draw the line to intersect on line segment AB. Mark the intersecting point as point D.

(iii) Cut arc of length 4 cm on the same line which will be the altitude of the triangle.

(iv) Name the point as C. Then \Delta ABC is the isosceles triangle.

(v) Draw a line AX which makes an acute angle with AB and is opposite of vertex C.

(vi) Cut seven equal parts of line AX namely AA 1 , AA 2 , AA 3 .

(vii) Now join A 2 to B. Draw a line A 3 B' parallel to A 2 B.

(viii) And then draw a line B'C' parallel to BC.

Hence \Delta AB'C' is the required triangle.

1636090408707

Q5 Draw a triangle ABC with side BC = 6 cm, AB = 5 cm and \angle ABC = 60°. Then construct a triangle whose sides are 3/4 of the corresponding sides of the triangle ABC.

Answer:

Steps of construction:-

(i) Draw a line segment BC with a measurement of 6 cm.

(ii) Now construct angle 60 o from point B and draw AB = 5 cm.

(iii) Join point C with point A. Thus \Delta ABC is the required triangle.

(iv) Draw a line BX which makes an acute angle with BC and is opposite of vertex A.

(v) Cut four equal parts of line BX namely BB 1 , BB 2 , BB 3 , BB 4.

(vi) Now join B 4 to C. Draw a line B 3 C' parallel to B 4 C.

(vii) And then draw a line B'C' parallel to BC.

Hence \Delta AB'C' is the required triangle.

1636090424003

Q6 Draw a triangle ABC with side BC = 7 cm, \angle B = 45°, \angle A = 105°. Then, construct a triangle whose sides are 4/3 times the corresponding sides of D ABC.

Answer:

Steps of construction:-

(i) Draw a line segment BC.

(ii) Now draw an angle \angle B = 45 o and \angle C = 30 o and draw rays in these directions.

(iii) Name the intersection of these lines as A.

(iv) Thus \bigtriangleup ABC is the required triangle.

(v) Draw a line BX which makes an acute angle with BC and is opposite of vertex A.

(vi) Cut four equal parts of line BX namely BB 1 , BB 2 , BB 3 , BB 4.

(vii) Now join B 3 to C. Draw a line B 4 C' parallel to B 3 C.

(viii) And then draw a line B'C' parallel to BC.

Hence \Delta AB'C' is the required triangle.

1636090438268

Q7 Draw a right triangle in which the sides (other than hypotenuse) are of lengths 4 cm and 3 cm. Then construct another triangle whose sides are 5/3 times the corresponding sides of the given triangle.

Answer:

Steps of construction:-

(i) Draw a line segment AB having a length of 4 cm.

(ii) Now, construct a right angle at point A and make a line of 3 cm.

(iii) Name this point C. Thus \Delta ABC is the required triangle.

(iv) Draw a line AX which makes an acute angle with AB and is opposite of vertex C.

(v) Cut four equal parts of line AX namely AA 1 , AA 2 , AA 3 , AA 4 , AA 5 .

(vi) Now join A 3 to B. Draw a line A 5 B' parallel to A 3 B.

(vii) And then draw a line B'C' parallel to BC.

Hence \Delta AB'C' is the required triangle.

1636090455562


NCERT Solutions for Chapter 11 Class 10 construction Excercise: 11.2

Q1 In each of the following, give also the justification of the construction:
Draw a circle of radius 6 cm. From a point 10 cm away from its center, construct the pair of tangents to the circle and measure their lengths.

Answer:

Steps of construction:-

(i) Taking point O as center draw a circle of radius 6 cm.

(ii) Now, name a point P which is 10 cm away from point O. Join OP.

(iii) Draw a perpendicular bisector of OP name the intersection point of bisector and OP as O'.

(iv) Now draw a circle considering O' as center and O'P as the radius.

(v) Name the intersection point of circles as Q and R.

(vi) Join PQ and PR. These are the required tangents.

(vii) Measure lengths of PQ = 8cm and PR = 8cm

1636090832534

Q2 Construct a tangent to a circle of radius 4 cm from a point on the concentric circle of radius 6 cm and measure its length. Also, verify the measurement by actual calculation.

Answer:

Steps of constructions:-

(i) Taking point O as a center draw a circle of radius 4 cm.

(ii) Now taking O as center draw a concentric circle of radius 6 cm.

(iii) Taking any point P on the outer circle, join OP.

(iv) Draw a perpendicular bisector of OP.

(v) Name the intersection of bisector and OP as O'.

(vi) Now, draw a circle taking O' as center and O'P as the radius.

(vii) Name the intersection point of two circles as R and Q.

(viii) Join PR and PQ. These are the required tangents.

(ix) Measure the lengths of the tangents. PR = 4.47 cm and PQ = 4.47 cm.

1636090849002

Q3 Draw a circle of radius 3 cm. Take two points P and Q on one of its extended diameter each at a distance of 7 cm from its center. Draw tangents to the circle from these two points P and Q.

Answer:

Steps of construction:-

(i) Taking O as a center draw a circle of radius 3 cm.

(ii) Now draw a diameter PQ of this circle and extend it.

(iii) Mark two points R and S on the extended diameter such that OR = OS = 7 cm.

(iv) Draw the perpendicular bisector of both the lines and name their mid-points as T and U.

(v) Now, taking T and U as center draw circles of radius TR and QS.

(vi) Name the intersecting points of the circles with the first circles as V, W, X, Y.

(vii) Join the lines. These are the required tangents.

1636090862749

Q4 Draw a pair of tangents to a circle of radius 5 cm which is inclined to each other at an angle of 60°.

Answer:

Steps of construction:-

(i) Draw a circle with center O and radius 5 cm.

(ii) Now mark a point A on the circumference of the circle. And draw a line AP perpendicular to the radius OA.

(iii) Mark a point B on the circumference of the circle such that \angle AOB = 120 o . (As we know, the angle at the center is double that of the angle made by tangents).

(iv) Join B to point P.

(v) AP and BP are the required tangents.

1636090901571

Q5 Draw a line segment AB of length 8 cm. Taking A as a center, draw a circle of radius 4 cm and taking B as center, draw another circle of radius 3 cm. Construct tangents to each circle from the center of the other circle.

Answer:

Steps of construction:-

(i) Draw a line segment AB having a length of 8 cm.

(ii) Now, taking A as a center draw a circle of radius 4 cm. And taking B as a center draw a circle of radius 3 cm.

(iii) Bisect the line AB and name the mid-point as C.

(iv) Taking C as a center and AC as radius draw a circle.

(v) Name the intersection points of the circle as P, Q, R, S.

(vi) Join the lines and these are our required tangents.

1636090914889

Q6 Let ABC be a right triangle in which AB = 6 cm, BC = 8 cm and \angle B = 90 \degree . BD is perpendicular from B on AC. The circle through B, C, D is drawn. Construct the tangents from A to this circle.

Answer:

Steps of construction:-

(i) Draw a line segment BC of length 8 cm.

(ii) Construct a right angle at point B. Now draw a line of length 6 cm. Name the other point as A.

(iii) Join AC. \Delta ABC is the required triangle.

(iv) Now construct a line BD on the line segment AC such that BD is perpendicular to AC.

(v) Now draw a circle taking E as a center (E is the midpoint of line BC) and BE as the radius.

(vi) Join AE. And draw a perpendicular bisector of this line.

(vii) Name the midpoint of AE as F.

(viii) Now, draw a circle with F as center and AF as the radius.

(ix) Name the intersection point of both the circles as G.

(x) Join AG. Thus AB and AG are the required tangents.

1636090928158

Q7 Draw a circle with the help of a bangle. Take a point outside the circle. Construct the pair of tangents from this point to the circle.

Answer:

Steps of construction:-

(i) Draw a circle using a bangle.

(ii) Now draw 2 chords of this circle as QR and ST.

(iii) Take a point P outside the circle.

(iv) Draw perpendicular bisector of both the chords and let them meet at point O.

(v) Joinpoint PO.

(vi) Draw bisector of PO and name the midpoint as U.

(vii) Now, taking U as a center and UP as radius draw a circle.

(viii) Name the intersection point of both the circles as V and W.

(ix) Join PV and PW. These are the required tangents.

1636090941880

Construction Class 10 NCERT-Topics

  1. To divide a line segment in the given ratio.

  2. To construct a pair of tangents from an external point to a circle.

  3. To construct a triangle similar to a given triangle as per a given scale factor which may be greater than 1 or less than 1.

Also See

Feature of Solutions to Class 10 Maths NCERT Chapter 11 Constructions

  • Construction Class 10 is one of the scoring chapters under the geometry unit. It is a part of geometry and it holds 15 marks in the final exams. There are 2 exercises with 14 questions in this NCERT Class 10 Maths chapter 11.
  • In Class 9, we have done certain calculations and constructions using a compass and a straight edge (ruler), for example, construction of triangles, drawing the perpendicular bisector of a line segment and much more and also gave their justifications. In this NCERT Class 10 Maths chapter 11 Constructions , you will study some advanced constructions by using the concepts learnt in class 9 constructions.
  • These M aths ch 11 class 10 solutions are about the construction of a line segment and a circle, division of a line segment and constructions of tangents to a circle using an analytical approach.
  • Also, students have to write the steps of construction for each answer.
  • In exercise 11.1 of NCERT class 10th Construction Solutions, you will study how to divide a line segment and in exercise 11.2 you will study the construction of tangents to a circle.

NCERT Solutions for Class 10 Maths - Chapter Wise

Chapter No.

Chapter Name

Chapter 1

Real Numbers

Chapter 2

Polynomials

Chapter 3

Pair of Linear Equations in Two Variables

Chapter 4

Quadratic Equations

Chapter 5

Arithmetic Progressions

Chapter 6

Triangles

Chapter 7

Coordinate Geometry

Chapter 8

Introduction to Trigonometry

Chapter 9

Some Applications of Trigonometry

Chapter 10

Circles

Chapter 11

Constructions

Chapter 12

Areas Related to Circles

Chapter 13

Surface Areas and Volumes

Chapter 14

Statistics

Chapter 15

Probability

NCERT Solutions of Class 10 - Subject Wise

NCERT solutions for class 10 maths

NCERT solutions for class 10 science

How to Use NCERT Solutions for Class 10 Maths Chapter 11 Constructions?

  • Before coming to this NCERT Construction Class 10 Maths chapter 3, please ensure that you have done the construction chapter of Class 9.

  • Learn some basic constructions like angles, triangles, and circles using compass and scale.

  • After this, go through some examples available in Cass 10 Maths Construction.

  • After covering the above processes, you can come to the practice exercises.

  • While practising the question, If you feel some trouble in any problem then take the help of NCERT solutions for Class 10 Maths Chapter 11 Constructions.

  • NCERT Solutions for Class 10 Maths Chapter 3 pdf download will be available soon. Till then you can download the webpage to practice the questions offline.

NCERT Exemplar solutions - Subject Wise

NCERT Books and NCERT Syllabus

Keep working hard & happy learning!

Frequently Asked Questions (FAQs)

1. What are the important topics of class 10 maths chapter 11 constructions ?

NCERT Class 10 Maths chapter 11 constructions dividing a line segment in the given ratio, construct a pair of tangents from an external point to a circle, construct a triangle similar to a given triangle as per a given scale factor are the important topics in this chapter.

2. How does the NCERT solutions are helpful in CBSE board exam?

In order to perform well in the CBSE board exam, one must know the NCERT syllabus very well. NCERT solutions can help you, if stuck in NCERT problems while solving them. Also, NCERT solutions are provided by the experts who knows how to answer in the board exam in order to get good marks. For more problems refer to NCERT exemplar questions for Construction Class 10th chapter. For ease download NCERT solutions for class 10 maths chapter 11 pdf and study both online and offline mode.

3. What are the benefits of practicing NCERT solution for construction chapter class 10?

Using the NCERT Solutions for chapter construction class 10 can give students a preview of the types of questions that may appear on their board exams, helping them to prepare effectively. These construction solutions class 10 are valuable resources that provide students with all the necessary information in a clear and concise format. The solutions cover all of the topics outlined in the NCERT syllabus set by the CBSE board.

4. Can you list the topics covered in NCERT Solutions for ch 11 construction class 10th?

The NCERT Solutions for Maths chapter 11 construction class 10th covers topics such as an introduction to constructions, the division of a line segment and construction of tangents to a circle, and a summary of all concepts presented in the chapter. By reviewing these solutions, students can clarify any doubts they have and practice additional questions.

5. Is the only way to access NCERT Solutions for Class 10 Maths Chapter 11 through the internet?

To make it easier for students to learn, the solutions have been provided in PDF format, which students can download for free, and refer to them even without the internet. These NCERT Solutions for the construction of triangles class 10 can also be accessed online via the Careers360 website.

Also Read

Articles

Upcoming School Exams

Bihar Open Board 10th Examination

Application Date:11 November,2024 - 10 January,2025

Bihar Open Board 12th Examination

Application Date:11 November,2024 - 10 January,2025

National Means Cum-Merit Scholarship

Admit Card Date:27 November,2024 - 22 December,2024

Madhya Pradesh State Open School 12th Examination

Admit Card Date:13 December,2024 - 06 January,2025

Gujarat Board Higher Secondary School Certificate

Late Fee Application Date:13 December,2024 - 22 December,2024

View All School Exams

Certifications By Top Providers

Edx
 1115 courses
Coursera
 816 courses
Udemy
 394 courses
Futurelearn
 295 courses
Harvard University, Cambridge
 98 courses
University of Michigan, Ann Arbor
 83 courses

Explore Top Universities Across Globe

University of Essex, Colchester
  Wivenhoe Park Colchester CO4 3SQ
University College London, London
  Gower Street, London, WC1E 6BT
The University of Edinburgh, Edinburgh
  Old College, South Bridge, Edinburgh, Post Code EH8 9YL
University of Bristol, Bristol
  Beacon House, Queens Road, Bristol, BS8 1QU
University of Delaware, Newark
  Newark, DE 19716
University of Nottingham, Nottingham
  University Park, Nottingham NG7 2RD

Related E-books & Sample Papers

Questions related to CBSE Class 10th

Have a question related to CBSE Class 10th ?
sadhu ashram ramgart road aligarh

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

Steps to Reach Sadhu Ashram, Ramgart Road, Aligarh:

  1. Starting Point:

    • Determine your starting point in Aligarh or the nearby area.

  2. Use Google Maps:

    • Open Google Maps on your phone or computer.
    • Enter "Sadhu Ashram, Ramgart Road, Aligarh" as your destination.
    • Follow the navigation instructions provided by Google Maps.

  3. By Local Transport:

    • Auto-rickshaw/Taxi: Hire an auto-rickshaw or taxi and inform the driver about your destination. Most local drivers should be familiar with Sadhu Ashram.
    • Bus: Check if there are any local buses that operate on Ramgart Road. Ask the bus conductor if the bus stops near Sadhu Ashram.

  4. Landmarks to Look For:

    • As you approach Ramgart Road, look for local landmarks that might help you confirm you’re on the right path, such as known shops, temples, or schools nearby.

  5. Ask for Directions:

    • If you're unsure, ask locals for directions to Sadhu Ashram on Ramgart Road. It's a known location in the area.

  6. Final Destination:

    • Once you reach Ramgart Road, Sadhu Ashram should be easy to spot. Look for any signage or ask nearby people to guide you to the exact location.

If you need detailed directions from a specific location or more information about Sadhu Ashram, feel free to ask

Read Complete Answer
sai raja 01 September,2024
show the previous year exams ?

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

Read Complete Answer
SWETCHCHA MISKA 10 July,2024
as an icse student in class 10 is above 80 percent good enough to get into commerce in 11th cbse

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

Read Complete Answer
Manasvini Gupta 23 July,2024
i had cleared 10th from cbse in 2022 2 years ago now i wish to apply for 12th as private candidate in 2025. can i do so?

hello Zaid,

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

best of luck!

Read Complete Answer
Ashvadha 03 July,2024
i had cleared 10th in 2022 from cbse can i apply for 12th as private candidate this year?

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

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

Read Complete Answer
sadafreen356 03 July,2024
View All

Popular CBSE Class 10th Questions

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
Read More

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

 
Read More

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
Read More

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
Read More

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 .
Read More

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
Read More

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.
Read More

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.
Read More

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
Read More

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
Read More

Colleges After 12th

Applications for Admissions are open.

JEE Main Important Physics formulas
Apply

As per latest 2024 syllabus. Physics formulas, equations, & laws of class 11 & 12th chapters

JEE Main Important Chemistry formulas
Apply

As per latest 2024 syllabus. Chemistry formulas, equations, & laws of class 11 & 12th chapters

TOEFL ® Registrations 2024
Apply

Accepted by more than 11,000 universities in over 150 countries worldwide

Pearson | PTE
Apply

Register now for PTE & Unlock 20% OFF : Use promo code: 'C360SPL20'. Valid till 31st DEC'24! Trusted by 3,500+ universities globally

JEE Main high scoring chapters and topics
Apply

As per latest 2024 syllabus. Study 40% syllabus and score upto 100% marks in JEE

JEE Main Important Mathematics Formulas
Apply

As per latest 2024 syllabus. Maths formulas, equations, & theorems of class 11 & 12th chapters

View All Application Forms
News and Notifications
Board Exam Date 2025 Live: Bihar Class 10, 12 model papers out, time table soon; updates on CBSE, UP
December 06, 2024 - 10:42 PM IST
CBSE Class 10 board exams from February 15 to March 18; download date sheet 2025 PDF
November 20, 2024 - 10:02 PM IST
CBSE Date Sheet 2025 Live: CBSE Class 10, 12 board exam timetable soon at cbse.gov.in; practical dates
November 14, 2024 - 03:21 PM IST
CBSE Date Sheet 2025 Live: CBSE to announce Class 10, 12 board exam dates soon at cbse.gov.in; sample paper
November 13, 2024 - 07:53 AM IST
CBSE Board Exam 2025: Class 10, 12 practical exam dates out at cbse.gov.in; passing marks
October 24, 2024 - 09:54 PM IST
CBSE urges schools to complete 100% registration of Class 9, 11 students by October 16; late fee for delays
October 11, 2024 - 08:56 AM IST
CBSE sample papers 2024-25 for Class 10, 12 out at cbseacademic.nic.in; marking scheme
September 05, 2024 - 07:38 PM IST

Stay up-to date with CBSE Class 10th News

Back to top