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NCERT Solutions for Class 6 Maths Chapter 2 - Lines and Angles

NCERT Solutions for Class 6 Maths Chapter 2 - Lines and Angles

Edited By Komal Miglani | Updated on Jul 07, 2025 09:41 AM IST

Lines and Angles are fundamental topics in the concept of Geometry. In this chapter, we will explore some of the basic definitions and concepts that are included in plane geometry and form the basic building blocks for further classes. In the NCERT class 6 chapter 2, Lines and Angles, we will learn about concepts such as point, ray, line segment, line, angles, etc. This article about NCERT Solutions for Class 6 Maths contains solutions to all the questions in this chapter of the NCERT Class 6 Maths Book.

This Story also Contains
  1. NCERT Solutions for Class 6 Maths Chapter 2 Lines and Angles (Exercise)
  2. Lines and Angles Class 6 Maths Chapter 2 - Topics
  3. NCERT Solutions for Class 6 Maths Chapter 2 Lines and Angles - Notes
  4. NCERT Solutions for Class 6 Maths Chapter 2 Lines and Angles - Points to Remember
  5. NCERT Solutions for Class 6 Maths Chapter Wise
NCERT Solutions for Class 6 Maths Chapter 2 - Lines and Angles
NCERT Solutions for Class 6 Maths Chapter 2 - Lines and Angles

The NCERT Solutions give the step-by-step solutions to the questions that help students build a strong foundation on the topics with conceptual clarity. These solutions are highly reliable as they are created by the subject matter experts of Careers360. NCERT Solutions for class 6 are very helpful for the students during exam preparation to practice more questions and help them face the questions with confidence. For a complete syllabus, detailed notes, and downloadable PDFs, check out this link: NCERT.

NCERT Solutions for Class 6 Maths Chapter 2 Lines and Angles (Exercise)

Page number: 15-17

Question: 6

Question 1:

Can you help Rihan and Sheetal find their answers?

Answer: Rihan can draw an infinite number of lines that pass through the point, whereas Sheetal can draw only one line that passes through both of the points.

Question 2: Name the line segments in Fig. 2.4. Which of the five marked points are on exactly one of the line segments? Which are on two of the line segments?

1748882709021

Answer: In the given figure, there are four line segments, i.e., LM,MP,PQ,QR
Points L and R are on one of the line segments.
Points M, P, and Q are on two of the line segments.

Question 3: Name the rays shown in Fig. 2.5. Is T the starting point of each of these rays?

Answer:

In the given figure, there are two rays:

  • Ray TA: This ray starts at point T and passes through point A, extending infinitely beyond A.

  • Ray TB: This ray also starts at point T and passes through point B, extending infinitely beyond B.

So yes, T is the starting point of both rays.

Question 4: Draw a rough figure and write labels appropriately to illustrate each of the following: a. OP and OQ meet at O. b. XY and PQ intersect at point M. c. Line l contains points E and F but not point D. d. Point P lies on AB.

Answer:

a. OP and OQ meet at O

1748882709076

b. XY and PQ intersect at point M

1748882709108

c. Line l contains points E and F but not point D

1748882709133

d. Point P lies on AB.

1748882709160

Question 5: In Fig. 2.6, name:

a. Five points

b. A line

c. Four rays

d. Five line segments

1748882709184

Answer:

(a) Five points are: B, C, D, E, and O

(b) A line is: DB (or BD)

(c) Four rays are: OB, OC, OD, and OE

(d) Five line segments are: OB, OC, OD, DE, and OE.

Question 6: Here is a ray OA (Fig. 2.7). It starts at O and passes through the point A. It also passes through point B.
a. Can you also name it as OB? Why?
b. Can we write OA as AO? Why or why not?

1748882709230

Answer:

a) Yes, the ray can also be named OB because the ray OA passes through point B as well. Rays are named starting from the initial point and passing through any other point on the ray. Since the ray starts at O and passes through both B and A, it can be named OB.

b) No, we cannot write OA as AO because rays are directional. The ray starts at point 0 and extends through A, so OA indicates the direction from 0 to A. Writing it as AO would imply the ray starts at A and goes towards 0, which is incorrect in this context because 0 is the starting point.

Page number: 19-21

Question: 6

Question 1: Can you find the angles in the given pictures? Draw the rays forming any one of the angles and name the vertex of the angle.

1748882709327

Answer:

The name of the vertex of ∠BDC is D.

1748882709356

The name of the vertex of ∠POR is O.

1748882709389

The name of the vertex of ∠LKM is K.

1748882709421

The name of the vertex of ∠AYC is Y.

1748882709450

Question 2: Draw and label an angle with arms ST and SR.

Answer: We can draw an angle ∠RST of any measure with vertex S and arms ST and SR.

1748882709476

Question 3: Explain why ∠APC cannot be labelled as ∠P

Answer: Using a single letter name like ∠P is insufficient in case of two or more angles having the same vertex. Here, ∠P can be considered for any of the three angles ∠APB,∠BPC or ∠APC. So, here it is necessary to use three letters in specifying an angle.

Question 4: Name the angles marked in the given figure.

Answer: In the given figure, two angles have been marked with a common arm TR. We can name these angles as ∠RTQ and ∠RTP or ∠PTR and ∠QTR.

Question 5: Mark any three points on your paper that are not on one line. Label them A, B, and C. Draw all possible lines going through pairs of these points. How many lines do you get? Name them. How many angles can you name using A, B, C? Write them down, and mark each of them with a curve as in Fig. 2.9.

Answer: We get three lines AB, BC, and CA by joining pairs of these points. There are three angles, ∠ABC, ∠ACB, and ∠BAC, formed using these points.

1748882709573

Question 6: Now mark any four points on your paper so that no three of them are on one line. Label them A, B, C, D. Draw all possible lines going through pairs of these points. How many lines do you get? Name them. How many angles can you name using A, B, C, D? Write them all down, and mark each of them with a curve as in Fig. 2.9.

Answer: We can draw the figure as shown below. We get 6 lines in all. These are AB, BC, CD, CD, AC, BD.

1748882709602

We get 12 angles in all. These are ∠ABC, ∠ABD, ∠GBD; ∠BCD, ∠BCA, ∠ACD; ∠CDA, ∠CDB, ∠BDA; ∠DAB, ∠DAC, and ∠CAB.

Page number: 23

Question: 3

Question 1: Fold a rectangular sheet of paper, then draw a line along the fold created. Name and compare the angles formed between the fold and the sides of the paper. Make different angles by folding a rectangular sheet of paper and compare the angles. Which is the largest and smallest angle you made?

Answer:

1748882709658

The angles formed with a line along the fold created are ∠AEF, ∠BEF,∠DFE, and ∠CFE, which are marked with 1, 2, 3, and 4. Out of these angles ∠AEF and ∠CFE are the larger ones, whereas ∠BEF and ∠DFE are the smaller ones.

Question 2: In each case, determine which angle is greater and why. a. ∠AOB or ∠XOY b. ∠AOB or ∠XOB c. ∠XOB or ∠XOC Discuss with your friends on how you decided which one is greater.

Answer:

(a) ∠AOB>∠XOY because ∠XOY lies in the interior of ∠AOB.

(b) ∠AOB>∠XOB because both angles have one arm, OB common, but the uncommon arm OX of ∠XOB lies in the interior of ∠AOB.

(c) ∠XOB =∠XOC because OB and OC both denote the same ray, so both angles have the common arms and the same vertex. That is, they superimpose on each other.

Question 3: Which angle is greater: ∠XOY or ∠AOB? Give reasons.

Answer:

∠1 > ∠3 (∠1 has more spread than ∠3)

∠1 + ∠2 > ∠3 + ∠2 (adding ∠2 to both sides)

Hence, ∠XOY > ∠AOB

Page number: 29-31

Question: 4

Question 1: How many right angles do the windows of your classroom contain? Do you see other right angles in your classroom?

Answer: In my classroom, the windows typically contain several right angles, as the corners where the frames meet are usually 90 degrees. Every rectangular or square-shaped window has four right angles.

Besides windows, other objects in the classroom with right angles include the corners of the doors, the edges of the blackboard or whiteboard, the desks or tables, and tiles on the floor or walls may also contain right angles as they are square and rectangular.

Question 2: Join A to other grid points in the figure by a straight line to get a straight angle. What are all the different ways of doing it?

1748882709784

Answer: We can produce BA beyond A to join other points of the grid. Thus, we make a straight angle as shown in the figure below. Clearly, ∠CAB is a straight angle. There is only one way of doing it.

1748882709819

Question 3: Now join A to the other grid points in the figure by a straight line to get a right angle. What are all the different ways of doing it?

Hint: Extend the line further as shown in the figure below. To get a right angle at A, we need to draw a line through it that divides the straight angle CAB into two equal parts.

1748882709883

Answer: We can produce BA beyond A to make a straight angle, then through A, draw the line DE, which is perpendicular to BC as shown in the figure below. Clearly, ∠DAB or ∠EAB is a right angle. There is only one way of doing it.

1748882709916

Question 4: Get a slanting crease on the paper. Now, try to get another crease that is perpendicular to the slanting crease.

a. How many right angles do you have now? Justify why the angles are exact right angles.

b. Describe how you folded the paper so that any other person who doesn’t know the process can simply follow your description to get the right angle.

Answer: We can fold the paper as shown in the figure.

1748882709948

(a) We get 4 right angles. The angle formed at any point and one side of a line is always a straight angle. The perpendicular drawn by the second crease through the vertex of a straight angle bisects it into two right angles.
(b) We first fold the paper to get a slanting crease and then fold the paper again in such a way that a part of the slanting crease falls on the other part of itself. Doing it carefully, we get another slanting line which is perpendicular to the first one. Thus, we get 4 right angles at the meeting point of the two lines as shown in the figure.

Page number: 31-32

Question: 4

Question 1: Identify acute, right, obtuse, and straight angles in the previous figures.

Answer:

Acute angle

Right angle

Obtuse angle

Question 2: Make a few acute angles and a few obtuse angles. Draw them in different orientations.

Answer: Acute angles and obtuse angles in different orientations:

Acute angles

Obtuse angles

1748882710134

1748882710181

Question 3: Do you know what the words acute and obtuse mean? Acute means sharp, and obtuse means blunt. Why do you think these words have been chosen?

Answer: Yes, the words acute and obtuse in geometry have meanings that relate to their everyday usage:

Acute: The word acute means "sharp" in general English, which is why an acute angle is described this way. Acute angles are smaller than 90 degrees and visually appear "sharp" or narrow, like the tip of a knife or a point, which gives a sharp impression.

Obtuse: The word obtuse means "blunt" or "dull." An obtuse angle is greater than 90 degrees and appears wider or broader, which makes it look blunt or less sharp, much like a blunt object or a broad, dull edge.

Question 4: Find out the number of acute angles in each of the figures below

What will be the next figure,, and how many acute angles will it have? Do you notice any pattern in the numbers?

Answer: We find the number of acute angles in the given figures are:

1748882710271

The next figure will be

1748882710316

So, the number pattern is 3, 3 + 9 = 12, 12 + 9 = 21, 21 + 9 = 30,…

Page number: 35

Question: 1

Question 1: Write the measures of the following angles: a. ∠ KAL Notice that the vertex of this angle coincides with the centre of the protractor. So the number of units of 1 degree angle between KA and AL gives the measure of ∠KAL. By counting, we get— ∠KAL = 30°. Making use of the medium-sized and large-sized marks, is it possible to count the number of units in 5s or 10s?
b. ∠WAL
c. ∠TAK

Answer:

(a) Yes, it is possible to count the number of units in 5s or 10s because there are also marking for each 5 and 10 on the protractor and the arms AK and AL pass through the 10 markings.
(b) In the given figure, the arms AL and AW pass through the 10 markings. There are five 10 markings of scale where AL and AW pass. So, WAL=50.
(c) In the given figure, the arms AK and AT pass through the 10 markings. There are twelve 10 markings of scale where AK and AT pass. So, TAK=120.

Page number: 40-43

Question: 9

Question 1: Find the degree measures of the following angles using your protractor.

Answer:

1748882710413

Question 2: Find the degree measures of different angles in your classroom using your protractor.

Answer: Take different angles and measure them.

Question 3: Find the degree measures for the angles given below. Check if your paper protractor can be used here!

1748882710444

Answer:

The degree measures for the angles using a paper protractor are given below:

1748882710474 1748882710512

Question 4: How can you find the degree measure of the angle given below using a protractor?

Answer: We can draw a ray opposite to one of the arms and split the angle into two parts - a straight angle and an acute angle. Measure the acute angle (∠1 = 80°) using a protractor and add it to 180°. It would be the measure of the required angle (180° + 80° = 260°).

1748882710575

Question 5: Measure and write the degree measures for each of the following angles:

1748882710643

Answer:

(a) 80°

(b) 120°

(c) 60°

(d) 130°

(e) 125°

(f) 62°

Question 6: Find the degree measures of ∠BXE, ∠CXE, ∠AXB, and ∠BXC.

Answer: The degree measures of ∠BXE, ∠CXE, ∠AXB, and ∠BXC are as follows:

∠BXE = 115°

∠CXE = 85°

∠AXB = 65°

∠BXC = 30°

Question 7: Find the degree measures of ∠PQR, ∠PQS, and ∠PQT

Answer: The degree measures of ∠PQR, ∠PQS, and ∠PQT are as follows:

∠PQR=45°

∠PQS=100°

∠PQT=152°

Question 8: Make the paper craft as per the given instructions. Then, unfold and open the paper fully. Draw lines on the creases made and measure the angles formed.

Answer:

Step 1: Fold the square diagonally to form a triangle.

Step 2: Fold the bottom tip of the triangle upwards.

Step 3: Fold the right and left corners upwards to make ears.

Step 4: Fold the side edges inward to form the sides of the face.

Step 5: Fold the ear tips outward slightly.

Step 6: Fold the top corner behind the model to round the head.

Step 7: Fold the bottom point backward.

Step 8: Draw eyes, nose, and whiskers to complete the bunny!

In the complete bunny face, we will have multiple angles:

The ears form acute angles (around 30° to 45°). Carefully unfold the entire model to return to a flat sheet.

The face near the chin forms an obtuse angle, close to 120°.

The sides of the face are around 90° to 120°, depending on the precision of the folds.

Question 9: Measure all three angles of the triangle shown in Fig. 2.21 (a), and write the measures down near the respective angles. Now add up the three measures. What do you get? Do the same for the triangles in Fig. 2.21 (b) and (c). Try it for other triangles as well, and then make a conjecture for what happens in general! We will come back to why this happens in a later year.

Answer:

a) The sum of the three angles of the triangle: 45° + 65° + 70° = 180°

1748882710844

b) The sum of the three angles of the triangle: 56° + 62° + 62° = 180°
msup

1748882710880

c) The sum of the three angles of the triangle: 30° + 52° + 98° = 180°
msup

1748882710916

Page number: 45-46

Question: 5

Question 1: Angles in a clock:

a. The hands of a clock make different angles at different times. At 1 o’clock, the angle between the hands is 30°. Why?

b. What will be the angle at 2 o’clock? And at 4 o’clock? 6 o’clock?

c. Explore other angles made by the hands of a clock.

Answer:

a) The clock dial has been divided into 12 equal parts. And the dial of the clock represents the angle of 360°.

So, the angle measure of each part is 360°12 = 30°.

Thus, at 1 o'clock, the angle between the hands is 30°.

b) From 12 to 2, we have 2 parts at 2 o'clock. So, the angle between the hands is 60°.

Similarly, between 12 and 4, we have four parts.

So, at 4 o'clock, the angle between the hands is 120°. Similarly, at 6 o'clock, the angle between the hands is 180°.

c) The other angles made by the hands of a clock:

Angle between hands at 5o′ clock = 5 × 30° = 150°

Angle between hands at 7 o' clock = 7 × 30° = 210°

Angle between hands at 8o′ clock = 8 × 30° = 240°

Question 2: The angle of a door: Is it possible to express the amount by which a door is opened using an angle? What will be the vertex of the angle, and what will be the arms of the angle?

Answer: Yes, it is possible to express the amount by which a door is opened using an angle. The vertex of the angle will be the hinge of the door, and the arms of the angle will be the wall and the door itself.

Question 3: Vidya is enjoying her time on the swing. She notices that the greater the angle with which she starts the swinging, the greater is the speed she achieves on her swing. But where is the angle? Are you able to see any angle?

1748882711023

Answer: Vidya's original position is along the vertical line. When she starts swinging, the rope of the swing is along the slanting line, which makes an angle with the imaginary vertical line. The speed depends on this angle.

1748882711057

Question 4: Here is a toy with slanting slabs attached to its sides; the greater the angles or slopes of the slabs, the faster the balls roll. Can angles be used to describe the slopes of the slabs? What are the arms of each angle? Which arm is visible and which is not?

1748882711092

Answer: Yes, angles can be used to describe the slopes of the slabs. Horizontal imaginary line and slab are the arms of each angle. Here, the slab arm is visible, but the horizontal line is not visible.

1748882711130

Question 5: Observe the images below where there is an insect and its rotated version. Can angles be used to describe the amount of rotation? How? What will be the arms of the angle and the vertex? Hint: Observe the horizontal line touching the insects.

Answer: Observe the vertical and horizontal lines touching the insects. The rotation from vertical to horizontal position makes an angle. We can imagine the meeting point of the two lines as the vertex, and these two lines as arms of the angle, as shown in the picture.

1748882711214

1748882711248

Page number: 49-50

Question: 3

Question 1: In Fig. 2.23, list all the angles possible. Did you find them all? Now, guess the measures of all the angles. Then, measure the angles with a protractor. Record all your numbers in a table. See how close your guesses are to the actual measures.

Answer:

The table to record my guesses and actual measurements:

Angle

Guessed measure

Actual measure

∠PAC

120°

115°

∠APD

45°

50°

∠DPS

60°

60°

∠LPR

110°

110°

∠PLS

90°

80°

∠ARP

360°

360°

∠PRS

110°

115°

∠RSL

80°

75°

∠ALC

45°

50°

Question 2: Use a protractor to draw angles having the following degree measures: a. 110° b. 40° c. 75° d. 112° e. 134°

Answer:

a) 1748882711313

b) 1748882711350

c) 1748882711383

d) 1748882711416

e) 1748882711453

Question 3: Draw an angle whose degree measure is the same as the angle given below:

Also, write down the steps you followed to draw the angle.

Answer: We can draw an angle equal to the given angle by following these steps:

Step I: Draw a ray AB.

1748882711520

Step II: Measure the angle IHJ. It is 118°.

Step III: Place the centre of the protractor at A and the zero edge along AB.

Step IV: Start counting from zero near B. Mark a point C at 118°.

Step V: Join AC.

Thus, ∠BAC = ∠IHJ.

Page number: 51-52

Question: 2

Question 1: In each of the below grids, join A to other grid points in the figure by a straight line to get:

a. An acute angle

b. An obtuse angle

1748882711584

c. A reflex angle

Mark the intended angles with curves to specify the angles. One has been done for you.

Answer:

a) Acute angle

1748882711652

b) Obtuse angle

1748882711683

c) Reflex angle

1748882711716

Question 2: Use a protractor to find the measure of each angle. Then classify each angle as acute, obtuse, right, or reflex.
a. ∠PTR
b. ∠PTQ
c. ∠PTW
d. ∠WTP

1748882711752

Answer:

Using a protractor, we find the measure of each angle and then classify each angle as follows:

(a) ∠PTR = 31°, acute angle

(b) ∠PTQ = 60°, acute angle

(c) ∠PTW = 104°, obtuse angle

(d) ∠WTP = 360° − 104° = 256°, reflex angle.

Page number: 53-54

Question: 7

Question 1: Draw angles with the following degree measures: a. 140° b. 82° c. 195° d. 70° e. 35°

Answer:

a) 1748882711784

b) 1748882711814

c) 1748882711843

d) 1748882711875

e) 1748882711907

Question 2: Estimate the size of each angle and then measure it with a protractor:

Classify these angles as acute, right, obtuse, or reflex angles.

Answer:

(a) 45°, acute angle

(b) 169°, obtuse angle

(c) 120°, obtuse angle

(d) 33°, acute angle

(e) 99°, obtuse angle

(f) 348°, reflex angle

Question 3: Make any figure with three acute angles, one right angle, and two obtuse angles.

Answer: Figure with three acute angles, one right angle and two obtuse angles:

1748882711975

Question 4: Draw the letter ‘M’ such that the angles on the sides are 40° each and the angle in the middle is 60°.

Answer: The letter 'M' such that the angles on the sides are 40° each and the angle in the middle is 60°.

1748882712015

Question 5: Draw the letter ‘Y’ such that the three angles formed are 150°, 60°, and 150°.

Answer: The letter 'Y′ such that the three angles formed are 150°, 60°, and 150°.

1748882712049

Question 6: The Ashoka Chakra has 24 spokes. What is the degree measure of the angle between two spokes next to each other? What is the largest acute angle formed between two spokes?

1748882712086

Answer: The Ashoka Chakra resembles a circle. So its degree measure is 360°.

Dividing it into 24 equal parts by spokes, the measure of each part = 360°24 = 15°

Thus, the degree measure of the angle between two spokes next to each other is 15°.

The largest acute angle formed between two spokes is 5 × 15° = 75°.

Question 7: Puzzle: I am an acute angle. If you double my measure, you get an acute angle. If you triple my measure, you will get an acute angle again. If you quadruple (four times) my measure, you will get an acute angle yet again! But if you multiply my measure by 5, you will get an obtuse angle measure. What are the possibilities for my measure?

Answer: As per the given condition, we have 4× acute angle <90 and 5× acute angle >90

acute angle <904=2212 and acute angle >905=18

So, possible measures of the acute angle would be 19,20,21 and 22 in whole numbers.

Lines and Angles Class 6 Maths Chapter 2 - Topics

Given below are the key topics for the NCERT class 6 maths chapter 2, Lines and Angles:

  • Point
  • Line segment
  • Line
  • Ray
  • Angle
  • Types of Angles

NCERT Solutions for Class 6 Maths Chapter 2 Lines and Angles - Notes

All the important concepts, formulas, and definitions of the NCERT class 6 maths chapter 2, Lines and Angles, are given below as notes:

Point: If we mark a dot somewhere on our notebook, we can call it as a point in geometry. It is represented by capital letters like P,Q and doesn't have any fixed length, breadth, or height.

Line Segment: If we mark any two points and join them with a line that has a fixed length, then it is a line segment. It is represented by capital letters like AB

Line: A line is extended in both directions till infinity and does not have a fixed length. It is represented by showing arrows on both sides of a line.

Ray: A ray is defined as having a fixed point, and the other side extends to infinity. Some of the examples include light coming from a lighthouse or sun's rays.

Angle: An angle is formed when two rays have a common starting point and the other two ends extend to infinity. The common point is called the vertex, and the rays are called as the arms of the angle.

We can find angles in our daily life, for example, the hands of a clock. The exact measurement of an angle can be calculated with the help of a protractor.

Right angle: It is an angle that measures exactly 90 degrees. For example, the angle made by the Clock when it turns 3 pm.

Acute angle: It is an angle that measures less than 90 degrees.

Obtuse angle: It is an angle that measures more than 90 degrees, but less than 180 degrees.

Reflex angle: It is an angle that measures more than 180 degrees, but less than 360 degrees.

NCERT Solutions for Class 6 Maths Chapter 2 Lines and Angles - Points to Remember

Given below are some important points about the NCERT class 6 maths chapter 2, Lines and Angles:

Point: A dot representing a location or a position with no length or breadth is a point.

Line: A straight path extending in both directions infinitely.

Line Segment: A line with a definite length is called a line segment.

Ray: A line starting at a fixed point and extending infinitely in the other direction is called a ray.

Angle: An angle consists of two rays with a common starting point.

Types of angles: Angles are of 4 types, namely, acute angle, right angle, obtuse angle and reflex angle.

NCERT Solutions for Class 6 Maths Chapter Wise

Here are all the links for chapter-wise NCERT Solutions of Class 6 Maths:

NCERT Solutions for Class 6 Subject Wise

Practising the exercise questions of each chapter is an important part of exam preparation. Students can refer to the subject-wise solutions for the exercise questions of each chapter using the link below.

Students can also check the NCERT Books and the NCERT Syllabus for Class 6 here:

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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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