NCERT Solutions for Class 8 Maths Chapter 4 Practical Geometry

# NCERT Solutions for Class 8 Maths Chapter 4 Practical Geometry

Edited By Ramraj Saini | Updated on Mar 01, 2024 06:20 PM IST

Practical Geometry Class 8 Questions And Answers provided here. These NCERT Solutions are provided by Careers360 team freely and created considering latest syllabus and pattern of CBSE 2023-24. In this chapter, you will learn some advanced-level constructions like- quadrilaterals. In this, you are going to learn how to construct a unique quadrilateral when different conditions are given like when the lengths of four sides with a diagonal are given, when three sides and two diagonals are given, when two adjacent sides with three angles are given, etc. Geometry is a subset of Mathematics in which you learn about the shape, size & relative position of geometric figures like- Angle, triangles, rectangles and etc.

In this chapter, there are a total of 4 exercises and 33 questions. Students must complete the NCERT Class 8 Maths Syllabus as soon as possible. They must also read the NCERT Class 8 Maths Books and complete all the topics.

## Practical Geometry Class 8 Solutions - Important Points

• A quadrilateral can be uniquely constructed if you know the lengths of its four sides and one diagonal.

• A quadrilateral can be uniquely constructed if you have information about both its diagonals and three of its sides.

• A quadrilateral can be uniquely constructed if you are aware of the lengths of two adjacent sides and the measures of three of its angles.

• A quadrilateral can be uniquely constructed if you are provided with the lengths of its three sides and the measures of two included angles

Free download NCERT Solutions for Class 8 Maths Chapter 4 Practical Geometry for CBSE Exam.

## Practical Geometry Class 8 NCERT Solutions (Intext Questions and Exercise)

Class 8 practical geometry NCERT Solutions - Chapter: Introduction

Answer: No, he cannot construct a unique quadrilateral. The given measurements consist of 2 adjacent sides, the angle between these sides and 2 diagonals. It is not one of the specific combinations to construct a quadrilateral.

Moreover, on trying to construct using these, we find that the vertex C is not fixed and can be varied. Hence, the quadrilateral is not unique.

NCERT Solutions for Class 8 Maths Chapter 4 Practical Geometry - Topic: Constructing A Quadrilateral

Q (i) We saw that 5 measurements of a quadrilateral can determine a quadrilateral uniquely. Do you think any five measurements of the quadrilateral can do thisAnswer: No, any five measurements of the quadrilateral cannot determine a quadrilateral uniquely. For example, consider a quadrilateral ABCD where AB, BC, CD are known and $\angle$ A and $\angle$ C are known. But this cannot uniquely determine the quadrilateral.

Answer: Yes, we can draw a parallelogram BATS where BA = 5 cm, AT = 6 cm and
AS = 6.5 cm (diagonal). We know, diagonals of a parallelogram are equal. Hence, BT = AS =6.5 cm
And opposite sides are equal. Hence, BS =AT = 6 cm;
Therefore, we have 2 diagonals and 3 sides to construct the parallelogram uniquely.
Answer: Yes, we can draw a rhombus ZEAL where ZE = 3.5 cm, diagonal EL = 5 cm.
Because all sides of a rhombus are equal.
Therefore we have four sides and a diagonal to construct the rhombus uniquely.

Q (iv) A student attempted to draw a quadrilateral PLAY where $PL = 3 cm, LA = 4 cm, AY = 4.5 cm, PY = 2 cm$ and $LY= 6 cm$ , but could not draw it. What is the reason?
[Hint: Discuss it using a rough sketch].
Answer: Given, a quadrilateral PLAY where $PL = 3 cm, LA = 4 cm, AY = 4.5 cm, PY = 2 cm$ and $LY= 6 cm$ ,

Rough sketch:

We know, that sum of two sides of a triangle must be greater than the third side. But PL + PY < YL . Hence this is not a triangle. Therefore, this quadrilateral cannot be made.

NCERT Solutions for Class 8 Maths Chapter 4 Practical Geometry - Exercise: 4.1

Q1 (i) Construct the following quadrilaterals

AB = 4.5 cm
BC = 5.5 cm
CD = 4 cm
AC = 7 cm

AB = 4.5 cm
BC = 5.5 cm
CD = 4 cm
AC = 7 cm

Step 1. Using SSS condition ,draw $\Delta ABC$

Step 2. Now, CD= 4 cm. Using C as centre, draw an arc of radius = 4 cm

Step 3. Also, AD = 6 cm. Using A as centre, draw an arc of radius = 6 cm.

Step 4. Point D will be the point of intersection of the two drawn arcs. Join AD and CD to complete the quadrilateral.

JU = 3.5 cm
UM = 4 cm
MP = 5 cm
PJ = 4.5 cm
PU = 6.5 cm

JU = 3.5 cm
UM = 4 cm
MP = 5 cm
PJ = 4.5 cm
PU = 6.5 cm

Step 1. Using SSS condition ,draw $\Delta JUP$

Step 2. Now, PM = MP = 5 cm. Using P as centre, draw an arc of radius = 5 cm

Step 3. Also, UM = 4 cm. Using U as centre, draw an arc of radius = 4 cm.

Step 4. Point M will be the point of intersection of the two drawn arcs. Join PM and UM to complete the quadrilateral.

Parallelogram MORE
OR = 6 cm
RE = 4.5 cm
EO = 7.5 cm

Parallelogram MORE

OR = 6 cm
RE = 4.5 cm
EO = 7.5 cm

We know, Opposite sides of a parallelogram are equal in length and also parallel to each other

Therefore,

EM = OR = 6 cm,

OM = RE = 4.5 cm

Step 1. Using SSS condition ,draw $\Delta ORE$

Step 2. Now, EM = OR = 6cm . Using E as centre, draw an arc of radius = 6 cm

Step 3. Also, OM = RE = 4.5 cm . Using O as centre, draw an arc of radius = 4.5 cm.

Step 4. Point M will be the point of intersection of the two drawn arcs. Join OM and EM to complete the parallelogram.

MORE is the required parallelogram.

(iv) Rhombus BEST
BE = 4.5 cm
ET = 6 cm

BEST is a rhombus where:

BE = 4.5 cm
ET = 6 cm (diagonal)

We know that the sides of a rhombus are equal in length and opposite sides are parallel to each other.

Therefore, the sides of Rohmbus BE = ES = ST = TB = 4.5 cm

Step 1. Using SSS condition ,draw $\Delta BET$

Step 2. Now, ES = 4.5 cm . Using E as centre, draw an arc of radius = 4.5 cm

Step 3. Also, TS = 4.5 cm . Using T as centre, draw an arc of radius = 4.5 cm.

Step 4. Point S will be the point of intersection of the two drawn arcs. Join ES and TS to complete the rhombus.

BEST is the required rhombus.

Practical geometry class 8 questions and answers - Topic: When Two Diagonals And Three Sides Are Given

Answer: No, it is not possible. We cannot draw $\Delta$ ABD first, because neither SSS nor SAS condition is there.

Answer: Yes, we can construct a quadrilateral with the given conditions. Here, two diagonals and three sides are given.

(Hint: Construct $\Delta PRS$ using SSS condition. Next, find point Q by drawing arcs from P and R. PQRS is the required quadrilateral.)

Class 8 maths chapter 4 question answer - Exercise: 4.2

LI = 4 cm

IF = 3 cm

TL = 2.5 cm

LF = 4.5 cm

IT = 4 cm

LI = 4 cm

IF = 3 cm

TL = 2.5 cm

LF = 4.5 cm (Diagonal)

IT = 4 cm (Diagonal)

Steps of construction:

Step 1. Using SSS condition ,draw $\Delta LIF$

Step 2. Now, LT = TL = 2.5 cm . Using L as centre, draw an arc of radius = 2.5 cm

Step 3. Also, IT = 4 cm . Using I as centre, draw an arc of radius = 4 cm.

Step 4. Vertex T will be the point of intersection of the two drawn arcs. Join LT and TF to complete the quadrilateral.

OL = 7.5 cm

GL = 6 cm

GD = 6 cm

LD = 5 cm

OD = 10 cm

OL = 7.5 cm

GL = 6 cm

GD = 6 cm

LD = 5 cm

OD = 10 cm

Steps of construction:

Step 1. Using SSS condition ,draw $\Delta OLD$

Step 2. Now, DG = GD = 6 cm . Using D as centre, draw an arc of radius = 6 cm

Step 3. Also, LG = GL = 6 cm . Using L as centre, draw an arc of radius = 6 cm.

Step 4. Vertex G will be the point of intersection of the two drawn arcs. Join G to D and G to O to complete the quadrilateral.

Rhombus BEND
BN = 5.6 cm
DE = 6.5 cm

Answer: Given, BEND is a rhombus.

BN = 5.6 cm (Diagonal)
DE = 6.5 cm (Diagonal)

We know that the diagonals of a rhombus bisect (cut in halves) each other at 90 degrees.

Steps of construction:

Step 1. Draw a line segment BN = 6.5 cm. With radius greater than half of BN, draw arcs on both sides of BN with B and N as the center. The line joining these two intersections is the perpendicular bisector of BN. Let it intersect BN at O (Therefore, O is the midpoint of BN. It will also be the midpoint of DE!)

Step 2. With O as the center, draw two arcs on the perpendicular bisector with radius = $\frac{1}{2} DE = \frac{6.5}{2} = 3.25 cm$ (Since, O is the midpoint of DE.)

Step 3. The intersecting points are vertices D and E.

Step 4. Join D to B and N. Also join E to B and N.

BEND is the required rhombus.

NCERT Class 8 maths ch 4 question answer - Topic: When Two Adjacent Sides And Three Angles Are Known

Answer: Yes, we can draw the quadrilateral MIST if we have 100° at M instead of 75°. (It will be a different quadrilateral!)

Moreover, The vertex T will now change. And since the sum of all interior angles of a quadrilateral is 360°, hence angle T will also change.

We know that sum of all the interior angles of a quadrilateral is 360°.

$\therefore$ ∠P + ∠L +∠A + ∠N= 140° + 75° + 150° + ∠N = 360°

$\implies$ ∠N = 360° - (140° + 75° + 150°) = - 5°

But this is not possible, because an angle of a quadrilateral cannot be negative.

Hence, We cannot construct a quadrilateral PLAN.

Answer: Yes, we still need at least the measure of the angle between the two adjacent sides to create a unique parallelogram.

Class 8 maths Chapter 4 NCERT Solutions - Exercise: 4.3

MO = 6 cm
OR = 4.5 cm
∠M = 60°
∠O = 105°
∠R = 105°

MO = 6 cm
OR = 4.5 cm
∠M = 60°
∠O = 105°
∠R = 105°

Rough:

Steps of construction:

Step 1. Draw a line segment MO = 6 cm. Construct an angle of 105° at O. As OR =4.5 cm, draw an arc with O as centre and radius = 4.5 cm. We get point R.

Step 2. Construct an angle of 105° at R.

Step 3. Construct an angle of 60° at M. This ray intersects the previous one at E.

PL = 4 cm
LA = 6.5 cm
∠P = 90°
∠A = 110°
∠N = 85°

PL = 4 cm
LA = 6.5 cm
∠P = 90°
∠A = 110°
∠N = 85°

We know, Sum of all angles of a quadrilateral = 360°

$\therefore$ ∠P + ∠L + ∠A+ ∠N = 360° $\implies$ 90° + ∠L + 110° + 85° = 360°

$\implies$ ∠L = 360° - 285° = 75°

Steps of Construction

Step 1. Draw a line segment PL = 4 cm. Construct an angle of 75° at L. As LA =6.5 cm, draw an arc with L as centre and radius = 6.5 cm. We get point A.

Step 2. Construct an angle of 110° at A.

Step 3. Construct an angle of 90° at P. This ray intersects the previous one at N.

Parallelogram HEAR
HE = 5 cm
EA = 6 cm
∠R = 85°

Answer: Given, HEAR is a parallelogram.
HE = 5 cm
EA = 6 cm
∠R = 85°

We know, Diagonally opposite angles of a parallelogram are equal.

Therefore, ∠E = ∠R = 85°

And, Opposite sides of a parallelogram are equal and parallel to each other.

Therefore, AR = HE = 5 cm
RH = EA = 6 cm

Steps of construction:

Step 1. Draw a line segment HE = 5 cm. Construct an angle of 85° at E. As EA = 6 cm, draw an arc with O as center and radius = 6 cm. We get to point A.

Step 2. With A as centre, draw an arc of radius = AR = HE = 5 cm.

Step 3. With H as centre , draw an arc of radius = HR = EA = 6 cm. The intersection of this arc with the previous arc is R.

Step 4. Join R to H and R to A.

HEAR is the required parallelogram.

Alternatively,

Steps of construction

Step 1. Draw a line segment HE = 5 cm. Construct an angle of 85° at E. As EA = 6 cm, draw an arc with O as center and radius = 6 cm. We get to point A.

Step 2. Construct an angle of (180°-85° = 95°) at A. (Adjacent angles in a parallelogram are supplementary)

Step 3. Construct an angle of (180°-85° = 95°) at H. This ray intersects the previous one at R.

HEAR is the required parallelogram.

Q1 (iv) Construct the following quadrilaterals.

Rectangle OKAY
OK = 7 cm
KA = 5 cm

Answer: Given, OKAY is a rectangle.
OK = 7 cm
KA = 5 cm

We know that all four angles of a rectangle are right angles ( $90^{\circ}$ ) and opposite sides are equal

Therefore, OK = AY = 7 cm. KA = YO = 5 cm

Steps of construction:

Step 1. Draw a line segment OK = 7 cm. Construct a right angle at K. With K a center, draw an arc of radius = 5 cm. We get to point A.

Step 2. Construct another right angle at A. With A as the center, draw an arc of radius = AY =OK = 7cm. We get the point Y.

Step 3. Join Y to O and Y to A

OKAY is the required rectangle.

NCERT Solutions for Class 8 Maths Chapter 4 Practical Geometry Exercise: When Three Sides And Two Included Angles Are Given

Answer: Other than BC, we could have drawn AB or CD first.

(i) Quadrilateral ABCD with AB = 5 cm, BC = 5.5 cm, CD = 4 cm, AD = 6 cm and $\angle B = 80\degree$ .

(ii) Quadrilateral PQRS with PQ = 4.5 cm, $\angle P = 70\degree$ , $\angle Q = 100\degree$ , $\angle R = 80\degree$
and $\angle S = 110\degree$ .

Construct a few more examples of your own to find sufficiency/insufficiency of the
data for the construction of a quadrilateral.

Answer: Yes, there can be other sets of five measurements to draw a quadrilateral.

Examples:

(i) Quadrilateral ABCD with AB = 5 cm, BC = 5.5 cm, CD = 4 cm, AD = 6 cm and $\angle B = 80\degree$ . (GIVEN)

(All 4 sides and 1 angle)

(ii) Quadrilateral PQRS having PQ = 4.5 cm, $\angle P = 70\degree$ , $\angle Q = 100\degree$ , $\angle R = 80\degree$
and $\angle S = 110\degree$ .

(1 side, all 4 angles)

(iii) Quadrilateral ABCD with AB = 5 cm, BC = 5.5 cm, CD = 4 cm, BD = 8 cm (Included angle diagonal) and $\angle B = 80\degree$ .

(3 sides, 1 included angle, and included angle diagonal)

(iv) Quadrilateral ABCD with AB = 5 cm, BC = 5.5 cm, BD = 8 cm , $\angle B = 80\degree$ and $\angle C = 95\degree$

(2 sides, 2 angles (1 included), and included angle diagonal).

There can be plenty of other examples.

NCERT Solutions for Class 8 Maths Chapter 4 Practical Geometry Exercise: 4.4

DE = 4 cm
EA = 5 cm
AR = 4.5 cm
∠E = 60°
∠A = 90°

DE = 4 cm
EA = 5 cm
AR = 4.5 cm
∠E = 60°
∠A = 90°

Steps of construction:

Step 1. Draw a line segment DE = 4 cm. Construct an angle of 60° at E.

Step 2. With E as centre, draw an arc of radius = EA = 5 cm. We get point A.

Step 3. Construct an angle of 90° at A. With A as centre, draw an arc of radius = AR = 4.5 cm. We get the point R.

Step 4. Join R to D.

TR = 3.5 cm
RU = 3 cm
UE = 4 cm
$\angle R = 75\degree$
$\angle U = 120\degree$

TR = 3.5 cm
RU = 3 cm
UE = 4 cm
$\angle R = 75\degree$
$\angle U = 120\degree$

Steps of construction:

Step 1. Draw a line segment TR = 3.5 cm. Construct an angle of 75° at R.

Step 2. With R as a center, draw an arc of radius = OR = 3 cm. We get point U.

Step 3. Construct an angle of 120° at U. With U as a center, draw an arc of radius = UE = 4 cm. We get the point E.

Step 4. Join E to T

NCERT Solutions for Class 8 Maths Chapter 4 Practical Geometry Exercise: Some Special Cases

Answer: A rectangle is a special case of a parallelogram whose opposite sides are equal and all the angles are $90^{\circ}$ .

Therefore, we know all 4 sides and all 4 angles.

(Steps:

1. Draw PQ. Draw right angle at R and Q. Draw an arc of radius = QR = PS to get the points S and R.

3. Join S to R. PQRS is the required rectangle.)

Answer: We know, in a Kite one diagonal perpendicularly bisects the other diagonal.

Steps of construction:

Step 1. Draw a line segment AY= 8 cm. Draw the perpendicular bisector of AY.

Step 2. With Y as a center and radius = EY = 4 cm, draw an arc on one side of the perpendicular bisector. The intersection is point E.

Step 3. Again, With Y as a center and radius = SY = 6 cm, draw an arc on the other side of the perpendicular bisector. The intersection is point S.

Step 4. Join E to A and Y. And also join S to A and Y.

EASY is the required kite.

NCERT Solutions for Class 8 Maths Chapter 4 Practical Geometry Exercise: 4.5

We know, All the sides of a square are equal and all angles are right angles.

Therefore, RE = EA = AD = DR = 5.1 cm

Steps of construction:

Step 1. Draw a line segment RE = 5.1 cm. Construct a right angle at R and E.

Step 2. With R and E as centre, draw an arc of radius = EA = RE = 5.1 cm. We get the point D and A respectively.

Step 3. Join D to A.

Answer: Let ABCD be the rhombus such that:

AC = 6.4 cm (longer diagonal) and BD = 5.2 cm (shorter diagonal)

We know that the diagonals of a rhombus bisect (cut in half) each other perpendicularly, i.e at 60°

In other words, the midpoint of the diagonals coincide.

Steps of construction:

Step 1. Draw a line segment AC =6.4 cm. Now, construct the perpendicular bisector of AC. Let it intersect AC at O.

(Therefore, O is the midpoint of both the diagonals AC and BD)

Step 2. With O as center and radius half of BD = $\frac{5.2}{2} = 2.6\ cm$ , draw two arcs on both sides of AC intersecting the perpendicular bisector at B and D.

Step 3. Join B to A and C. Also join D to A and C.

ABCD is the required rhombus.

Answer: Let ABCD be the rectangle such that :

Where AB = 5 cm and BC = 4 cm.

We know that the opposite sides of a rectangle are equal and all the angles are 90°.

Steps of construction:

Step 1. Draw a line segment AB = 5 cm. Construct an angle of 90° at A and B.

Step 2. Construct an arc of radius =AD = BC = 4 cm with A and B as the centre . The intersection points are C and D.

Step 3. Join C to D.

ABCD is the required rectangle.

Answer: Given, OKAY is a parallelogram where OK = 5.5 cm and KA = 4.2 cm.

Steps of construction:

Step 1. Draw a line segment OK = 5.5 cm. Draw a ray KX from point K (With any suitable angle). Extend OK in the direction of KO.

Now we make a ray parallel to KX from O.

Step 2. With K as the center and a suitable radius, draw an arc cutting both OK and KX at P and Q respectively.

Step 3. With the same radius and O as the center, draw an arc cutting extended OK at P'.

Step 4. With P as the center, measure PQ using the compass. Using this as radius, cut the previous arc with P' as the center and mark it Q'. Draw a ray OZ passing through Q'.

Step 5. With radius = KA = 4.2 cm, cut two arcs on OZ and KX with O and K as centers respectively. These intersection points are Y and A respectively. Join Y to A.

OKAY is the required parallelogram. It is not unique as the angle can be varied keeping opposite sides parallel to each other.

## Practical geometry class 8 solutions - Topics

• Some Special Cases

## NCERT Solutions for Class 8 Maths - Chapter Wise

 Chapter -1 Rational Numbers Chapter -2 Linear Equations in One Variable Chapter-3 Understanding Quadrilaterals Chapter-4 Practical Geometry Chapter-5 Data Handling Chapter-6 Squares and Square Roots Chapter-7 Cubes and Cube Roots Chapter-8 Comparing Quantities Chapter-9 Algebraic Expressions and Identities Chapter-10 Visualizing Solid Shapes Chapter-11 Mensuration Chapter-12 Exponents and Powers Chapter-13 Direct and Inverse Proportions Chapter-14 Factorization Chapter-15 Introduction to Graphs Chapter-16 Playing with Numbers

## NCERT Solutions for Class 8 - Subject Wise

### How to use NCERT Solutions for Class 8 Maths Chapter 4 Practical Geometry?

• You must have knowledge about the classification of geometric figures.
• You should go through some examples to understand the pattern of constructions.
• Once you have a clear idea, you can draw the shapes for unsolved problems given in the practice exercises.
• NCERT solutions for Class 8 Maths chapter 4 can be used to help while practicing.
• Keep working hard and happy learning!

Also Check NCERT Books and NCERT Syllabus here:

1. What are the important topics of chapter Practical Geometry ?

Polygons, angle sum property of polygons, quadrilaterals, trapezium, parallelogram, rhombus, rectangle, and square are the important topics of this chapter.

2. Which is best book for CBSE class 8 maths ?

NCERT textbook is the best book for CBSE class 8 maths. You don't need to buy any supplementary book. Be though with the NCERT problems.

3. Where can I find the complete solutions of NCERT for class 8 ?

Here you will get the detailed NCERT solutions for class 8 by clicking on the link.

4. Where can I find the complete solutions of NCERT for class 8 maths ?

Here you will get the detailed NCERT solutions for class 8 maths by clicking on the link.

5. Does CBSE class 8 maths is tough ?

CBSE class 8 maths is damn simple and very basic math. It is a base and foundation for the upcoming classes. Most of the topics in class 8 maths are related to previous classes.

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