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NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry

NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry

Edited By Komal Miglani | Updated on Mar 18, 2025 01:34 PM IST | #CBSE Class 10th

Coordinate Geometry is an integral part of Mathematics. This chapter is a link between geometry and algebra. It deals with geometrical figures in coordinate system and coordinates like the x-axis and y-axis to locate the exact position of the points in two-dimensional planes. This chapter consists of an introduction phase into coordinate geometry and gradually makes students aware of topics like distance formula and section formula. Students must practice all the NCERT Solutions and exercises to ace this chapter.

This Story also Contains
  1. NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry PDF Free Download
  2. Coordinate Geometry Class 10: Important Formulae
  3. NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry (Exercise)
  4. NCERT Solutions for Class 10 Maths - Chapter Wise
  5. NCERT Books and NCERT Syllabus
  6. Benefits of NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry
  7. NCERT solutions of class 10 subject-wise
  8. NCERT Exemplar solutions: Subject-wise
NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry
NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry

NCERT class 10 maths chapter 7 solutions can be often challenging for students who don't have their basic concepts clear. This article is for every student to understand all the Chapter 7 Maths class 10 NCERT solutions and strengthen their basic knowledge about coordinate geometry. Students must refer to the NCERT Class 10 Maths books and read all the topics to get a better understanding of the topics. This article is prepared by experienced careers360 experts to make the learning process of coordinate geometry easier for students.

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NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry PDF Free Download

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Coordinate Geometry Class 10: Important Formulae

Distance Formulae:

  • For a line defined by two points A(x1, y1) and B(x2, y2), the distance between these points can be calculated using the formula:

  • Distance AB = (x2x1)2+(y2y1)2

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Section Formula:

  • When a point P divides a line AB, with A(x1, y1) and B(x2, y2) as endpoints, in a ratio of m:n, the coordinates of point P can be found using the formula:

  • Point P = {mx2+nx1m+n,my2+ny1m+n}

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Midpoint Formula:

  • The midpoint of a line AB, defined by A(x1, y1) and B(x2, y2), can be determined using the following formula:

  • Midpoint P = {x1+x22,(y1+y22}

JEE Main Important Mathematics Formulas

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

Area of a Triangle:

  • Consider a triangle formed by points A(x1, y1), B(x2, y2), and C(x3, y3). The area of this triangle can be calculated using the formula:

Area ∆ABC = 12|x1(y2 − y3) + x2(y3 – y1) + x3(y1 – y2)|

NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry (Exercise)

Class 10 Maths ch 7 solutions Exercise: 7.1

Q1. (i) Find the distance between the following pairs of points : (2, 3), (4, 1)

Answer:

Given points: (2, 3), (4, 1)

Distance between the points will be: (x1,y1) and (x2,y2)

D=(x2x1)2+(y2y1)2

D=(42)2+(13)2=4+4=22

Q1. (ii) Find the distance between the following pairs of points : (– 5, 7), (– 1, 3)

Answer:

Given points: (– 5, 7), (– 1, 3)

Distance between the points will be: (x1,y1) and (x2,y2)

D=(x2x1)2+(y2y1)2

D=(1+5)2+(37)2=16+16=42

Q1. (iii) Find the distance between the following pairs of points :(a, b), (– a, – b)

Answer:

Given points: (a, b), (– a, – b)

Distance between the points will be: (x1,y1) and (x2,y2)

D=(x2x1)2+(y2y1)2

D=(aa)2+(bb)2=4(a2+b2)=2a2+b2

Q2. Find the distance between the points (0, 0) and (36, 15). Can you now find the distance between the two towns A and B discussed in Section 7.2?

Answer:

Given points: (0, 0) and (36, 15)

Distance between the points will be: (x1,y1) and (x2,y2)

D=(x2x1)2+(y2y1)2

D=(360)2+(150)2=1296+225=1521=39

The distance between the two towns A and B is, thus 39 km for the given town location

(0,0) and (36,15).

Q3. Determine if the points (1, 5), (2, 3) and (– 2, – 11) are collinear.

Answer:

Let the points (1, 5), (2, 3) and (– 2, – 11) be representing the vertices A, B, and C of the given triangle, respectively.

A=(1,5), B=(2,3), C=(2,11)

Therefore,

AB=(12)2+(53)2=5

BC=(2(2))2+(3(11))2=42+142=16+196=212 CA=(1(2))2+(5(11))2=32+162=9+256=265 Since these are not satisfied.

AB+BCCA

BA+ACBC

BC+CABA

As these cases are not satisfied.
Hence the points are not collinear.

Q4. Check whether (5, – 2), (6, 4) and (7, – 2) are the vertices of an isosceles triangle.

Answer:

The distance between two points A(x1,y1) and B(x2,y2) is given by:

D=(x2x1)2+(y2y1)2

So, we have the following points: (5, – 2), (6, 4) and (7, – 2) assuming it to be the vertices of triangles A, B, and C, respectively.

AB=(56)2+(24)2=1+36=37

BC=(67)2+(4+2)2=1+36=37

CA=(57)2+(2+2)2=4+0=2

Therefore, AB = BC

Here two sides are equal in length.

Therefore, ABC is an isosceles triangle.

Q5. In a classroom, 4 friends are seated at points A, B, C and D as shown in Fig. 7.8. Champa and Chameli walk into the class and after observing for a few minutes Champa asks Chameli, “Don’t you think ABCD is a square?” Chameli disagrees. Using the distance formula, find which of them is correct.

Coordinate Geometry-Distance formula

Answer:

The coordinates of the points:

A(3,4), B(6,7), C(9,4), and D(6,1) are the positions of 4 friends.

The distance between two points A(x1,y1), and B(x2,y2) is given by:

D=(x2x1)2+(y2y1)2

Hence,

AB=(36)2+(47)2=9+9=18=32

BC=(69)2+(74)2=9+9=18=32

CD=(96)2+(41)2=9+9=18=32

AD=(36)2+(41)2=9+9=18=32

And the lengths of diagonals:

AC=(39)2+(44)2=36+0=6

BD=(66)2+(71)2=36+0=6

So, here it can be seen that all sides of quadrilateral ABCD are of the same length and diagonals are also having the same length.

Therefore, quadrilateral ABCD is a square and Champa is saying right.

Q6. (i) Name the type of quadrilateral formed, if any, by the following points, and give reasons for your answer: (– 1, – 2), (1, 0), (– 1, 2), (– 3, 0)

Answer:

Let the given points (1,2), (1,0), (1,2), and (3,0) be representing the vertices A, B, C, and D of the given quadrilateral respectively.

The distance formula:

D=(x2x1)2+(y2y1)2

AB=(11)2+(20)2=4+4=8=22

BC=(1+1)2+(02)2=4+4=8=22

CD=(1+3)2+(20)2=4+4=8=22

AD=(1+3)2+(20)2=4+4=8=22

Finding the length of the diagonals:

AC=(1+1)2+(22)2=0+16=4

BD=(1+3)2+(00)2=16+0=4

It is clear that all sides are of the same lengths and also the diagonals have the same lengths.

Hence, the given quadrilateral is a square.

Q6. (ii) Name the type of quadrilateral formed, if any, by the following points, and give reasons for your answer: (–3, 5), (3, 1), (0, 3), (–1, – 4)

Answer:

Let the given points (3,5), (3,1), (0,3), and (1,4) be representing the vertices A, B, C, and D of the given quadrilateral, respectively.

The distance formula:

D=(x2x1)2+(y2y1)2

AB=(33)2+(51)2=36+16=52=213

BC=(30)2+(13)2=9+4=13

CD=(0+1)2+(3+4)2=1+49=50=52

AD=(3+1)2+(5+4)2=4+81=85

All the sides of the given quadrilateral have different lengths.

Therefore, it is only a general quadrilateral and not a specific one like a square, rectangle, etc.

Q6. (iii) Name the type of quadrilateral formed, if any, by the following points, and give reasons for your answer: (4, 5), (7, 6), (4, 3), (1, 2)

Answer:

Let the given points (4,5), (7,6), (4,3), (1,2) represent the vertices A, B, C, and D of the given quadrilateral, respectively.

The distance formula:

D=(x2x1)2+(y2y1)2

AB=(47)2+(56)2=9+1=10

BC=(74)2+(63)2=9+9=18

CD=(41)2+(32)2=9+1=10

AD=(41)2+(52)2=9+9=18

And the diagonals:

AC=(44)2+(53)2=0+4=2

BD=(71)2+(62)2=36+16=52=213

Here we can observe that the opposite sides of this quadrilateral are of the same length.

However, the diagonals are of different lengths.

Therefore, the given points are the vertices of a parallelogram.

Q7. Find the point on the x-axis which is equidistant from (2, –5) and (–2, 9).

Answer:

Let the point which is equidistant from A(2,5) and B(2,9) be X(x,0) as it lies on X-axis.

Then, we have,

Distance AX =(x2)2+(0+5)2

and Distance BX =(x+2)2+(0+9)2

According to the question, these distances are equal in length.

Hence, we have,

(x2)2+(0+5)2 =(x+2)2+(0+9)2

Squaring both sides we get,

(x2)2+25=(x+2)2+81
x24x+4+25=x2+4x+4+81
4x+4x=81+25
8x=56
x=7

Hence the point is X(7,0).

Q8. Find the values of y for which the distance between the points P(2, – 3) and Q(10, y) is 10 units.

Answer:

Given the distance between the points P(2,3) and Q(10,y) is 10 units.

The distance formula :

D=(x2x1)2+(y2y1)2

So, given PQ=10 units

PQ=(102)2+(y(3))2=10

After squaring both sides,

(102)2+(y(3))2=100

(y+3)2=10064

y+3=±6

y=63 or y=63

Therefore, the values are y=3 or y=9.

Q9. If Q(0, 1) is equidistant from P(5, –3) and R(x, 6), find the values of x. Also, find the distances QR and PR.

Answer:

Given Q(0,1) is equidistant from P(5,3) and R(x,6) .

Then, the distances PQ=RQ.

Distance PQ=(50)2+(31)2=25+16=41

Distance RQ=(x0)2+(61)2=x2+25

x2+25=41

Squaring both sides, we get,
x2+25=41

x2=16

x=±4

The points are: R(4,6) or R(4,6.)

CASE I: when R is (4,6)

The distances QR and PR.

QR=(04)2+(16)2=16+25=41

PR=(54)2+(36)2=12+(9)2=1+81=82

CASE II: when R is (4,6)

The distances QR and PR.

QR=(0(4))2+(16)2=16+25=41

PR=(5(4))2+(36)2=92+(9)2=81+81=92

Q10. Find a relation between x and y such that the point (x, y) is equidistant from the point (3, 6) and (– 3, 4).

Answer:

Let the point P(x,y) is equidistant from A(3,6) and B(3,4) .

Then, the distances AP=BP

AP=(x3)2+(y6)2 and BP=(x(3))2+(y4)2

(x3)2+(y6)2=(x(3))2+(y4)2

Squaring both sides: we obtain

(x3)2+(y6)2=(x+3)2+(y4)2

(2x)(6)+(2y10)(2)=0 [a2b2=(a+b)(ab)]

12x4y+20=0

3x+y5=0

Thus, the relation is 3x+y5=0 between x and y.

Class 10 Maths ch 7 solutions Exercise: 7.2

Q1. Find the coordinates of the point which divides the join of (–1, 7) and (4, –3) in the ratio 2 : 3.

Answer:

Let the coordinates of point P(x,y) which divides the line segment joining the points A(1,7) and B(4,3) , internally, in the ratio m1:m2 then,

Section formula: (m1x2+m2x1m1+m2,m1y2+m2y1m1+m2)

Substituting the values in the formula:

Here, m1:m2=2:3

(2(4)+3(1)2+3,2(3)+3(7)2+3)

(55,155)

(1,3)

Hence the coordinate is P(1,3).

Q2. Find the coordinates of the points of trisection of the line segment joining (4, –1) and (–2, –3).

Answer:

Let the trisection of the line segment A(4,1) and B(2,3) have the points P(x1,y1) and Q(x2,y2)

Then,

Section formula: (m1x2+m2x1m1+m2,m1y2+m2y1m1+m2)

By observation point, P divides AB internally in the ratio 1:2.

Hence, m:n=1:2

Substituting the values in the equation we get;

P(1(2)+2(4)1+2,1(3)+2(1)1+2)

P(2+83,323)

P(2,53)

And by observation point Q, divides AB internally in the ratio 2:1

Hence, m:n=2:1

Substituting the values in the equation above, we get

Q(2(2)+1(4)2+1,2(3)+1(1)2+1)

Q(4+43,613)

Q(0,73)

Hence, the points of trisections are P(2,53) and Q(0,73)

Q3. To conduct Sports Day activities, in your rectangular-shaped school ground ABCD, lines have been drawn with chalk powder at a distance of 1m each. 100 flower pots have been placed at a distance of 1m from each other along AD, as shown in Fig. 7.12. Niharika runs 14th the distance AD on the 2nd line and posts a green flag. Preet runs 15th the distance AD on the eighth line and posts a red flag. What is the distance between both flags? If Rashmi has to post a blue flag exactly halfway between the line segment joining the two flags, where should she post her flag

Coordinate Geometry-Section formula

Answer:

Niharika posted the green flag at the distance P, i.e.,

14×100 m=25 m from the starting point of 2nd line.

Therefore, the coordinates of this point P are (2,25).

Similarly, Preet posted red flag at 15 of the distance Q i.e.,

15×100 m=20 m from the starting point of 8th line.

Therefore, the coordinates of this point Q are (8,20).

The distance PQ is given by,

PQ=(82)2+(2520)2=36+25=61m

and the point at which Rashmi should post her Blue Flag is the mid-point of the line joining these points. Let this point be R(x,y).

Then, by Section Formula,

P(x,y)=(m1x2+m2x1m1+m2,m1y2+m2y1m1+m2)

x=2+82, y=25+202

x=5, y=22.5

Therefore, Rashmi should post her Blue Flag at 22.5 m on the 5th line.

Q4. Find the ratio in which the line segment joining the points (– 3, 10) and (6, – 8) is divided by (– 1, 6).

Answer:

Let the ratio be: k:1

Then, By section formula:

P(x,y)=(kx2+x1k+1,ky2+y1k+1)

Given point P(x,y)=(1,6)

1=6k3k+1

k1=6k3

k=27

Hence, the point P divides the line AB in the ratio 2:7.

Q5. Find the ratio in which the line segment joining A(1, – 5) and B(– 4, 5) is divided by the x-axis. Also, find the coordinates of the point of division.

Answer:

Let the point on the x-axis be P(x,0) and it divides it in the ratio k:1.

Then, we have

Section formula:

P(x,y)=(kx2+x1k+1,ky2+y1k+1)

ky2+y1k+1=0

k=y1y2

Hence, the value of k will be: k=55=1

Therefore, the x-axis divides the line in the ratio 1:1 and the point will be,

Putting the value of k=1 in the section formula.

P(x,0)=(x2+x12,0)

P(x,0)=(142,0)=(32,0)

Q6. If (1, 2), (4, y), (x, 6) and (3, 5) are the vertices of a parallelogram taken in order, find x and y.

Answer:

Let the given points A(1,2), B(4,y), C(x,6), D(3,5)

Since the diagonals of a parallelogram bisect each other. Intersection point O of diagonal AC and BD also divides these diagonals.

Therefore, O is the mid-point of AC and BD.

The coordinates of the point O when it is the mid-point of AC.

(1+x2,2+62)(x+12,4)

The coordinates of the point O when it is the mid-point of BD.

(4+32,5+y2)(72,5+y2)

Since both coordinates are of the same point O.

Therefore,

x+12=72 and 4=5+y2

Or,

x=6 and y=3

Q7. Find the coordinates of point A, where AB is the diameter of a circle whose centre is (2, – 3) and B is (1, 4).

Answer:

As the centre point C(2,3) will be the mid-point of the diameter AB.

Then, the coordinates of point A will be A(x,y).

Given point B(1,4) .

Therefore,

(2,3)=(x+12,y+42)

x+12=2 and y+42=3

x=3 and y=10 .

Therefore, the coordinates of A are (3,10).

Q8. If A and B are (- 2, - 2) and (2, - 4), respectively, find the coordinates of P such that AP = 3 AB 7 and P lies on the line segment AB.

Answer:

From the figure:

Coordinate Geometry exercise

As AP=37AB

PB=47AB hence the ratio is 3:4,

Now, from the section formula, we can find the coordinates of Point P.

Section Formula:

P(x,y)=(m1x2+m2x1m1+m2,m1y2+m2y1m1+m2)

P(x,y)=(3(2)+4(2)3+4,3(4)+4(2)3+4)

P(x,y)=(687,1287)

P(x,y)=(27,207)

Q9. Find the coordinates of the points which divide the line segment joining A (– 2, 2) and B(2, 8) into four equal parts.

Answer:

From the figure:

Coordinate Geometry exercise

Points C, D, and E divide the line segment AB into four equal parts.

Now, from the section formula, we can find the coordinates of Points C, D, and E.

Section Formula:

P(x,y)=(m1x2+m2x1m1+m2,m1y2+m2y1m1+m2)

Here point D divides the line segment AB into two equal parts hence

D(x2,y2)=(2+22,2+82)

D(x2,y2)=(0,5)

Now, point C divides the line segment AD into two equal parts hence

C(x1,y1)=(2+02,2+52)

C(x2,y2)=(1,72)

Also, point E divides the line segment DB into two equal parts hence

E(x1,y1)=(2+02,8+52)

E(x2,y2)=(1,132)

Q10. Find the area of a rhombus if its vertices are (3, 0), (4, 5), (– 1, 4) and (– 2, – 1) taken in order.

Answer:

From the figure:

Rhombus-Coordinate Geometry

Let the vertices of the rhombus be:

A(3,0), B(4,5), C(1,4), D(2,1)

Area of the rhombus ABCD is given by:

=12× Product of length of diagonals

Hence we have to find the lengths of the diagonals AC and BD of the rhombus.

The distance formula:

D=(x2x1)2+(y2y1)2

Length of the diagonal AC:

AC=(3(1))2+(04)2=16+16=42

Length of the diagonal BD:

BD=(4(2))2+(5(1))2=36+36=62

Thus, the area will be,

=12×(AC)×(BD)

=12×(42)×(62)=24 square units

To ease the learning for students, the below links of the exercises can be also used.

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

Areas Related to Circles

Chapter 12

Surface Areas and VolMaths

Chapter 13

Statistics

Chapter 14

Probability

NCERT Books and NCERT Syllabus

Here is the latest NCERT syllabus which is very useful for students before strategizing their study plan.
Also, links to some reference books which are important for further studies.

Benefits of NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry

Coordinate geometry is not only a useful chapter in class 10 but also for higher studies and competitive exams. Strengthening basic concepts is a necessity for students so that later they do not face any difficulties solving coordinate geometry questions in higher studies or competitive examinations.
These are some further benefits for students listed below.

  • Students can study strategically at their own pace after accessing Class 10 Maths NCERT Solutions chapter 7. This will boost their confidence to attempt other questions from this chapter.
  • Class 10 Maths Chapter 7 NCERT solutions are solved by subject-matter experts and are very reliable at the same time. The solutions provide shortcuts as well as detailed explanations with necessary formulae that will help students to understand the answers better.
  • These solutions will help students manage their time efficiently in this chapter and understand which questions are easier to approach and which are time-consuming. This will be helpful during the exam.
  • NCERT solutions for class 10 maths chapter 7 Coordinate Geometry is designed to give the students step-by-step solutions for a particular question.
  • In Coordinate Geometry, students will study the algebraic forms of geometric figures.

NCERT solutions of class 10 subject-wise

NCERT Exemplar solutions: Subject-wise

Frequently Asked Questions (FAQs)

1. What are the important topics covered in Class 10 Maths Chapter 7?

In class 10 Maths Chapter 7 following topics of coordinate geometry are covered.

  • Distance formula
  • Section formula
  • Midpoint formula
  • Area of a triangle using coordinate geometry
  • Collinearity of three points
  • Various questions and their solutions
2. How to find the distance between two points in Coordinate Geometry?

Let the two points in a Cartesian plane be (x1,y1) and (x2,y2).

So, their distance = (x2x1)2+(y2y1)2

3. What is the formula for the section formula in Coordinate Geometry?

If a point P(x,y) divides the line segment joining A(x1,y1) and B(x2,y2) in the ratio m:n, then the coordinates of P are:

x=mx2+nx1m+n,y=my2+ny1m+n

4. How to prove that three points are collinear using Coordinate Geometry?

The formula of the area of a triangle can be used to prove that three points A(x1,y1),B(x2,y2) and C(x3,y3) are collinear.

If the area of the triangle formed by these three points is zero(0), then the points are collinear.

Area of a triangle =12|x1(y2y3)+x2(y3y1)+x3(y1y2)|

5. What is the midpoint formula in Class 10 Maths Chapter 7?

Let the two points in a Cartesian plane be (x1,y1) and (x2,y2).
Then their midpoint M(x,y)=M(x1+x22,y1+y22)

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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.

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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 :

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Income Certificate (for minority category benefits).

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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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