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NCERT Solutions for Class 11 Maths Chapter 11 Introduction to Three Dimensional Geometry

NCERT Solutions for Class 11 Maths Chapter 11 Introduction to Three Dimensional Geometry

Edited By Komal Miglani | Updated on Jun 21, 2025 09:01 AM IST

Have you ever wondered how large skyscrapers are designed or how the GPS tells us about our exact locations every time? To find the answer, we have to study three-dimensional geometry, where we go beyond our understanding of two-dimensional geometry to the X, Y, and Z coordinate axes. The chapter Introduction to Three-Dimensional Geometry from class 12 mathematics contains the concepts of Coordinate Axes and Coordinate Planes in Three-Dimensional Space, Coordinates of a Point in Space, Distance between Two Points and its formula, etc. Understanding these concepts will help the students grasp more advanced geometry concepts easily and enhance their problem-solving ability in real-world applications. One of the main purposes of these NCERT Class 11 solutions is to provide students with a valuable tool to ensure that they gain enough exposure to a variety of question types.

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Suggested: JEE Main: high scoring chaptersPast 10 year's papers

This Story also Contains
  1. NCERT Solution for Class 11 Maths Chapter 11 Solutions: Download PDF
  2. NCERT Solutions for Class 11 Maths Chapter 11: Exercise Questions
  3. Class 11 Maths NCERT Chapter 11: Extra Question
  4. Introduction to Three-Dimensional Geometry Class 11 Chapter 11: Topics
  5. Three-Dimensional Geometry Class 11 Solutions: Important Formulae
  6. Approach to Solve Questions of Introduction to Three-Dimensional Geometry Class 11
  7. What Extra Should Students Study Beyond NCERT for JEE?
  8. NCERT Solutions for Class 11 Mathematics - Chapter Wise
NCERT Solutions for Class 11 Maths Chapter 11 Introduction to Three Dimensional Geometry
NCERT Solutions for Class 11 Maths Chapter 11 Introduction to Three Dimensional Geometry

This article on NCERT solutions for class 11 Maths Chapter 11 Introduction to Three-Dimensional Geometry offers clear and step-by-step solutions for the exercise problems given in the NCERT book. These Introduction to Three-Dimensional Geometry class 11 solutions are made by Subject Matter Experts of Careers360, having multiple years of experience in this field, following the latest CBSE syllabus, ensuring that students can grasp the basic concepts effectively. For syllabus, notes, and PDF, refer to this link: NCERT.

NCERT Solution for Class 11 Maths Chapter 11 Solutions: Download PDF

Students who wish to access the Class 11 Maths Chapter 11 NCERT Solutions can click on the link below to download the complete solution in PDF.

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NCERT Solutions for Class 11 Maths Chapter 11: Exercise Questions

NCERT Three-Dimensional Geometry Class 11 Solutions: Exercise: 11.1
Page Number: 211
Total Questions: 4

Question 1: A point is on the x-axis. What are its y-coordinate and z-coordinates?

Answer:

Any point on x-axis has zero y coordinate and zero z coordinate.

Question 2: A point is in the XZ-plane. What can you say about its y-coordinate?

Answer:

When a point is in XZ plane, the y coordinate of this point will always be zero.

Question 3: Name the octants in which the following points lie:
(1, 2, 3), (4, –2, 3), (4, –2, –5), (4, 2, –5), (– 4, 2, –5), (– 4, 2, 5), (–3, –1, 6) (– 2, – 4, –7).

Answer:

The x-coordinate, y-coordinate, and z-coordinate of point (1, 2, 3) are all positive. Therefore, this point lies in octant I.

The x-coordinate, y-coordinate, and z-coordinate of point (4, -2, 3) are positive, negative, and positive, respectively. Therefore, this point lies in octant IV.

The x-coordinate, y-coordinate, and z-coordinate of point (4, -2, -5) are positive, negative, and negative, respectively. Therefore, this point lies in octant VIII.

The x-coordinate, y-coordinate, and z-coordinate of point (4, 2, -5) are positive, positive, and negative, respectively. Therefore, this point lies in octant V.

The x-coordinate, y-coordinate, and z-coordinate of point (-4, 2, -5) are negative, positive, and negative, respectively. Therefore, this point lies in octant VI.

The x-coordinate, y-coordinate, and z-coordinate of point (-4, 2, 5) are negative, positive, and positive, respectively. Therefore, this point lies in octant II.

The x-coordinate, y-coordinate, and z-coordinate of point (-3, -1, 6) are negative, negative, and positive,e respectively. Therefore, this point lies in octant III.

The x-coordinate, y-coordinate, and z-coordinate of point (2, -4, -7) are positive, negative, and negative, respectively. Therefore, this point lies in octant VIII.

Question 4(i): Fill in the blanks: The x-axis and y-axis taken together determine a plane known as_______.

Answer:

The x-axis and y-axis taken together determine a plane known as XY Plane.

Question 4(ii): Fill in the blanks: The coordinates of points in the XY-plane are of the form _______.

Answer:

The coordinates of points in the XY-plane are of the form (x, y, 0 ).

Question 4(iii): Fill in the blanks: Coordinate planes divide the space into ______ octants.

Answer:

Coordinate planes divide the space into eight octants.

NCERT Three-Dimensional Geometry Class 11 Solutions: Exercise: 11.2
Page Number: 213
Total Questions: 5

Question 1(i): Find the distance between the following pairs of points: (2, 3, 5) and (4, 3, 1)

Answer:

The distance between two points (x1,y1,z1) and (x2,y2,z2) is given by

d=(x2x1)2+(y2y1)2+(z2z1)2

So, the distance between (2, 3, 5) and (4, 3, 1) is given by

d=(42)2+(33)2+(15)2

=4+0+16

=20

=25

Question 1(ii): Find the distance between the following pairs of points: (–3, 7, 2) and (2, 4, –1)

Answer:

The distance between two points (x1,y1,z1) and (x2,y2,z2) is given by

d=(x2x1)2+(y2y1)2+(z2z1)2

So, distance between (–3, 7, 2) and (2, 4, –1) is given by

d=(2(3))2+(47)2+(12)2

d=25+9+9

d=43

Question 1(iii): Find the distance between the following pairs of points:(–1, 3, – 4) and (1, –3, 4)

Answer:

The distance between two points (x1,y1,z1) and (x2,y2,z2) is given by

d=(x2x1)2+(y2y1)2+(z2z1)2

So, distance between (–1, 3, – 4) and (1, –3, 4) is given by

d=(1(1))2+(33)2+(4(4))2

=4+36+64

=104

=226

Question 1(iv): Find the distance between the following pairs of points: (2, –1, 3) and (–2, 1, 3).

Answer:

The distance between two points (x1,y1,z1) and (x2,y2,z2) is given by

d=(x2x1)2+(y2y1)2+(z2z1)2

So, the distance between (2, –1, 3) and (–2, 1, 3).) is given by

d=(2(2))2+(1(1))2+(33)2

=16+4+0

=20

=25

Question 2: Show that the points (–2, 3, 5), (1, 2, 3) and (7, 0, –1) are collinear.

Answer:

Given,three points A=(–2, 3, 5), B=(1, 2, 3) and C=(7, 0, –1)

The distance between two points (x1,y1,z1) and (x2,y2,z2) is given by

d=(x2x1)2+(y2y1)2+(z2z1)2

The distance AB :

AB=(1(2))2+(23)2+(35)2

=(3)2+(1)2+(2)2

=9+1+4

=14

The distance BC:

BC=(71)2+(02)2+(13)2

=36+4+16

=56

=214

The distance CA

CA=(7(2))2+(03)2+(15)2

=81+9+36

=126

=314

As we can see here,

AB+BC=14+214=314=AC

Hence, we can say that points A, B, and C are colinear.

Question 3(i): Verify the following: (0, 7, –10), (1, 6, – 6) and (4, 9, – 6) are the vertices of an isosceles triangle.

Answer:

Given Three points A=(0, 7, –10), B=(1, 6, – 6) and C=(4, 9, – 6)

The distance between two points (x1,y1,z1) and (x2,y2,z2) is given by

d=(x2x1)2+(y2y1)2+(z2z1)2

The distance AB

AB=(10)2+(67)2+(6(10))2

=1+1+16

=18

The distance BC

BC=(41)2+(96)2+(6(6))2

=9+9+0

=18

The distance CA

CA=(40)2+(97)2+(6(10))2

=16+4+16

=36

=6

As we can see AB=BCCA

Hence, we can say that ABC is an isosceles triangle.

Question 3(ii): Verify the following

(0, 7, 10), (- 1, 6, 6) and (- 4, 9, 6) are the vertices of right angled triangle.

Answer:

Given Three points A=(0, 7, 10), B=(-1, 6, 6) and C=(-4, 9, 6)

The distance between two points (x1,y1,z1) and (x2,y2,z2) is given by

d=(x2x1)2+(y2y1)2+(z2z1)2

The distance AB

AB=(10)2+(67)2+(610)2

=1+1+16

=18

The distance BC

BC=(4(1))2+(96)2+(66)2

=9+9+0

=18

The distance CA

CA=(40)2+(97)2+(610)2

=16+4+16

=36

=6

As we can see

(AB)2+(BC)2=18+18=36=(CA)2

Since this follows pythagorus theorem, we can say that ABC is a right angle triangle.

Question 3(iii): Verify the following: (–1, 2, 1), (1, –2, 5), (4, –7, 8) and (2, –3, 4) are the vertices of a parallelogram.

Answer:

Given A=(–1, 2, 1), B=(1, –2, 5), C=(4, –7, 8) and D=(2, –3, 4)

Given Three points A=(0, 7, –10), B=(1, 6, – 6) and C=(4, 9, – 6)

The distance between two points (x1,y1,z1) and (x2,y2,z2) is given by

d=(x2x1)2+(y2y1)2+(z2z1)2

The distance AB

AB=(1(1))2+(22)2+(51)2

=4+16+16

=36

=6

The distance BC

BC=(41)2+(7(2))2+(85)2

=9+25+9

=43

The distance CD

CD=(24)2+(3(7))2+(48)2

=4+16+16

=36

=6

The distance DA

DA=(12)2+(2(3))2+(14)2

=9+25+9

=43

Here, As we can see

AB=6=CD And BC=43=DA

As the opposite sides of the quadrilateral are equal, we can say that ABCD is a parallelogram.

Question 4: Find the equation of the set of points which are equidistant from the points (1, 2, 3) and (3, 2, –1).

Answer:

Given, two points A=(1, 2, 3) and B=(3, 2, –1).

Let the point P= (x,y,z) be a point which is equidistance from the points A and B.

so,

The distance PA= The distance PB

(x1)2+(y2)2+(z3)2=(x3)2+(y2)2+(z(1))2

(x1)2+(y2)2+(z3)2=(x3)2+(y2)2+(z(1))2

[(x1)2(x3)2]+[(y2)2(y2)2]+[(z3)2(z+1)2]=0

Now lets apply the simplification property,

a2b2=(a+b)(ab)

[(2)(2x4)]+0+[(4)(2z2)]=0

4x88z+8=0

4x8z=0

x2z=0

Hence locus of the point which is equidistant from A and B is x2z=0 .

Question 5: Find the equation of the set of points P, the sum of whose distances from A (4, 0, 0) and B (– 4, 0, 0) is equal to 10

Answer:

Given,

Two points A (4, 0, 0) and B (– 4, 0, 0)

let the point P(x,y,z) be a point sum of whose distance from A and B is 10.

So,

The distance PA+The distance PB=10

(x4)2+(y0)2+(z0)2+(x(4))2+(y)2+(z)2=10

(x4)2+(y)2+(z)2+(x+4)2+(y)2+(z)2=10

(x4)2+(y)2+(z)2=10(x+4)2+(y)2+(z)2

Squaring on both side :

(x4)2+(y)2+(z)2=10020(x+4)2+(y)2+(z)2+(x+4)2+(y)2+(z)2

⇒ ${(x-4)^2-(x+4)^2=100-20\sqrt{(x+4)^2+(y)^2+(z)^2}$

16x=10020(x+4)2+(y)2+(z)2

20(x+4)2+(y)2+(z)2=100+16x

5(x+4)2+(y)2+(z)2=25+4x

Now again squaring both sides,

25((x+4)2+(y)2+(z)2)=625+200x+16x2

25x2+200x+400+25y2+25z2=625+200x+16x2

9x2+25y2+25z2225=0

Hence, the equation of the set of points P, the sum of whose distances from A and B is equal to 10 is 9x2+25y2+25z2225=0.

NCERT Three-Dimensional Geometry Class 11 Solutions: Miscellaneous Exercise
Page Number: 215
Total Questions: 6

Question 1: Three vertices of a parallelogram ABCD are A(3, – 1, 2), B (1, 2, – 4) and C (– 1, 1, 2). Find the coordinates of the fourth vertex.

Answer:

Given

Three vertices of the parallelogram, ABCD are:

A(3, – 1, 2), B (1, 2, – 4) and C (– 1, 1, 2).

Let the fourth vertice be (a, b, c )

Now, As we know the concept that the diagonal of the parallelogram bisect each other,

Hence here in parallelogram ABCD, the midpoint of the line segment AC will be equal to the midpoint of the line segment BD.

So, (312,1+12,2+22)=(1+a2,2+b2,4+c2)

On comparing both points, we get

a=1,b=2andc=8

Hence the Fourth vertice of the is (1,-2,8).

Question 2: Find the lengths of the medians of the triangle with vertices A (0, 0, 6), B (0,4, 0) and (6, 0, 0).

Answer:

Given,

Three vertices of the triangle,A (0, 0, 6), B (0,4, 0)and C(6, 0, 0).

Now, let D be the midpoint of the AB, E be the midpoint of the BC and F be the midpoint of the AC.

Vertice of the D =

(0+02,0+42,6+02)=(0,2,3)

Vertices of E =

(0+62,4+02,0+02)=(3,2,0)

Vertices of the F =

(0+62,0+02,6+02)=(3,0,3)

Now, Medians of the triangle are CD, AE, and BF

So the lengths of the medians are:

CD=CD=(60)2+(02)2+(03)2=36+4+9=49=7

AE=(03)2+(02)2+(60)2=9+4+36=49=7

BF=(30)2+(04)2+(30)2=9+16+9=34

Hence lengths of the median are 7,7and34 .

Question 3: If the origin is the centroid of the triangle PQR with vertices P (2a, 2, 6), Q (– 4, 3b, –10) and R(8, 14, 2c), then find the values of a, b and c.

Answer:

Given,

Triangle PQR with vertices P (2a, 2, 6), Q (– 4, 3b, –10) and R(8, 14, 2c),

Now, As we know,

The centroid of a triangle is given by

(x1+x2+x33,y1+y2+y33,z1+z2+z33)

Where coordinates of vertices of the triangle are (x1,x2,x3),(y1,y2,y3)and(z1,z2,z3)

Since the Centroid of the triangle, PQR is origin =(0,0,0)

(2a4+83,2+3b+143,610+2c3)=(0,0,0)

On equating both coordinates, we get

a=2,b=163andc=2

Question 4: Find the coordinates of a point on y-axis which are at a distance of 52 from the point P (3, –2, 5).

Answer:

Let the point Q be (0,y,0)

Now Given

The distance of point Q From point P = 52

So,

(30)2+(2y)2+(50)2=52

(30)2+(2y)2+(50)2=25×2

9+(2+y)2+25=50

(2+y)2=16

(2+y)=4or2+y=4

y=2ory=6

The coordinates of the required point is (0,2,0) or (0,-6,0).

Question 5: A point R with x-coordinate 4 lies on the line segment joining the points P(2, –3, 4) and Q (8, 0, 10). Find the coordinates of the point R. [Hint Suppose R divides PQ in the ratio k : 1. The coordinates of the point R are given by (8k+2k+1,3k+1,10k+4k+1)

Answer:

Let R divides PQ in the ratio k: 1.

The coordinates of the point R are given by

(8k+2k+1,3k+1,10k+4k+1)

Now, as given the x coordinate of the R is 4.

So, 8k+2k+1=4

k=12

Hence the coordinates of R are:

(8×12+212+1,312+1,10×12+412+1)=(4,2,6)

Question 6: If A and B be the points (3, 4, 5) and (–1, 3, –7), respectively, find the equation of the set of points P such that PA2+PB2=k2 , where k is a constant.

Answer:

Given Points,

A (3, 4, 5) and B (–1, 3, –7),

Let the coordinates of point P be (x,y,z)

Now,

Given condition :

PB2=(x(1))2+(y(3))2+(z(7))2

PB2=x2+2x+1+y26y+9+z2+14z+49

PB2=x2+y2+z2+2x6y+14z+59

And

PA2=(x3)2+(y4)2+(z5)2

PA2=x2+y2+z26x8y10z+50

Now, Given Condition

PA2+PB2=k2

x2+y2+z2+2x6y+14z+59+x2+y2+z26x8y10z+50=k2

2x2+2y2+2z24x14y+4z+109=k2

Hence Equation of the set of the point P is

2x2+2y2+2z24x14y+4z+109=k2 .

Also, read,

Class 11 Maths NCERT Chapter 11: Extra Question

Question: L is the foot of the perpendicular drawn from a point (3, 4, 5) on the X-axis. The coordinates of L are:

Solution:
We know that, on the X-axis, the y and z coordinates are zero.

⇒ y = z = 0

Thus, the coordinates of L are: (3, 0, 0).

Hence, the correct answer is (3, 0, 0).

Introduction to Three-Dimensional Geometry Class 11 Chapter 11: Topics

Given below are the topics discussed in the NCERT Solutions for class 11, chapter 11, Introduction to Three-Dimensional Geometry:

  • Introduction
  • Coordinate Axes and Coordinate Planes in
  • Three-Dimensional Space
  • Coordinates of a Point in Space
  • Distance Between Two Points

Three-Dimensional Geometry Class 11 Solutions: Important Formulae

Coordinate Axes in Three Dimensions:

In three dimensions, a rectangular Cartesian coordinate system has three mutually perpendicular axes: X, Y, and Z.

Coordinate Planes:

The pair of axes determines three coordinate planes: XY, YZ, and ZX planes. These planes divide space into eight regions called octants.

Coordinates of a Point in Space:

The coordinates of a point P in three-dimensional space are represented as a triplet (x, y, z), which corresponds to the perpendicular distances from P on three mutually perpendicular coordinate planes: YZ-plane, ZX-plane, and XY-plane.

Coordinates of points on the coordinate axes are:

  • X-axis: (x, 0, 0)

  • Y-axis: (0, y, 0)

  • Z-axis: (0, 0, z)

Distance Formula in 3D Space:

Distance between two points P(x1, y1, z1) and Q(x2, y2, z2) isPQ=(x2x1)2+(y2y1)2+(z2z1)2

Distance from Origin in 3D Space:

Distance between origin P(0, 0, 0) and Q(x2, y2, z2) is
OQ=x22+y22+z22

Section Formula for Internal and External Division in 3D Space:

The coordinates of a point R that divides the line segment joining two points P(x1, y1, z1) and Q(x2, y2, z2) internally or externally in the ratio m : n are given by:

  • Internal Division: (mx2+nx1m+n,my2+ny1m+n,mz2+nz1m+n)

  • External Division: (mx2nx1mn,my2ny1mn,mz2nz1mn)

Midpoint Formula in 3D Space:

The coordinates of the mid-point of the line segment joining two points P(x1, y1, z1) and Q(x2, y2, z2) is
(x1+x22,y1+y22,z1+z22)

Coordinates of Centroid of a Triangle in 3D Space:

The coordinates of the centroid of the triangle, whose vertices are (x1,y1,z1),(x2,y2,z2),(x3,y3,z3) is(x1+x2+x33,y1+y2+y33,z1+z2+x33)

Approach to Solve Questions of Introduction to Three-Dimensional Geometry Class 11

The steps to approach questions related to three-dimensional geometry are:

  • Understand the basics: You can start by understanding the basic concepts of three-dimensional geometry first, like the coordinates of a point as (x,y,z), also the coordinate of origin as (0, 0, 0).
  • Visualise the 3-d space: While solving any three-dimensional geometry questions, you should always try to visualise the three-dimensional coordinate axes, i.e. the X, Y, and Z axes, to get a clear picture of the problem.
  • Apply the distance formula: Distance between two points P(x1, y1, z1) and Q(x2, y2, z2) isPQ=(x2x1)2+(y2y1)2+(z2z1)2
  • Use the section formula:
    The coordinates of a point R that divides the line segment joining two points P(x1, y1, z1) and Q(x2, y2, z2) internally or externally in the ratio m : n are given by:
    Internal Division: (mx2+nx1m+n,my2+ny1m+n,mz2+nz1m+n)
    External Division: (mx2nx1mn,my2ny1mn,mz2nz1mn)
  • Learn to find the mid-point and the centroid: The coordinates of the mid-point of the line segment joining two points P(x1, y1, z1) and Q(x2, y2, z2) are
    (x1+x22,y1+y22,z1+z22)
    The coordinates of the centroid of the triangle, whose vertices are (x1,y1,z1),(x2,y2,z2),(x3,y3,z3) is(x1+x2+x33,y1+y2+y33,z1+z2+x33)
  • Tips and tricks to Improve Speed & Accuracy: To improve your speed and accuracy, you have to practice many different types of questions from the ncert book, the exemplar book, and the previous year papers.

What Extra Should Students Study Beyond NCERT for JEE?

Here is a comparison list of the concepts in Introduction to Three-Dimensional Geometry that are covered in JEE and NCERT, to help students understand what extra they need to study beyond the NCERT for JEE:

NCERT Solutions for Class 11 Mathematics - Chapter Wise

Given below is the chapter-wise list of the NCERT Class 11 Maths solutions with their respective links:

Also Read,

NCERT solutions for class 11- Subject-wise

Given below are some useful links for NCERT books and the NCERT syllabus for class 10:

NCERT Books and NCERT Syllabus

Here are the subject-wise links for the NCERT solutions of class 10:

Frequently Asked Questions (FAQs)

1. What are the important topics in NCERT Class 11 Maths Chapter 11?

The important topics in NCERT Class 11 Maths Chapter 11 are Coordinate Axes and Coordinate Planes in Three-Dimensional Space, Coordinates of a Point in Space, Distance between Two Points and its formula etc.

2. Why is three-dimensional geometry important in Class 11 Maths?

In the three-dimensional geometry chapter of Class 11 maths, we go beyond our understanding of two-dimensional geometry to the X, Y, and Z coordinate axes. Understanding the key concepts of this chapter will help us find the coordinates of a point in three dimensions, and we can also learn about the distance formula, section formula, midpoint formula in 3D space etc.

3. What is the shortest distance between two skew lines in 3D?

Two skew lines in 3D space are lines that do not intersect each other also they are not parallel to each other.

The shortest distance between two skew lines in 3D space is the length of the perpendicular drawn from one line to the other.

4. How can I download NCERT Solutions for Class 11 Maths Chapter 11 PDF?

You can download the NCERT Solutions for Class 11 Maths Chapter 11 PDF from the link given below:
https://cache.careers360.mobi/media/uploads/froala_editor/files/ncert-solutions-for-class-11-maths-chapter-12-introduction-to-three-dimensional-geometry.pdf
NCERT solutions for other subjects and classes can be downloaded via:
school.careers360.com/ncert/ncert-solutions  

5. What is the Midpoint Formula in three-dimensional geometry?

The coordinates of the mid-point of the line segment joining two points P(x1, y1, z1) and Q(x2, y2, z2) is
$\left(\frac{x_{1}+x_{2}}{2}, \frac{y_{1}+y_{2}}{2}, \frac{z_{1}+z_{2}}{2}\right)$  

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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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