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NCERT Solutions for Class 10 Maths Chapter 9 Some Applications of Trigonometry

NCERT Solutions for Class 10 Maths Chapter 9 Some Applications of Trigonometry

Edited By Dinesh Goyal | Updated on Sep 06, 2023 09:12 PM IST | #CBSE Class 10th

Applications Of Trigonometry Class 10 NCERT Solution

NCERT Solutions for Class 10 Maths Chapter 9 - Some Applications of Trigonometry is the 9th chapter of Maths which deals with the applications of trigonometry in our daily life. Chapter 9 class 10 Maths NCERT solutions are provided here. These NCERT solutions are created by subjects experts at Careers360 keeping in mind latest syllabus of CBSE 2023-24. NCERT Class 10 maths solutions for chapter 9 consist of the answers to each question given in the 9th chapter of NCERT class 10 maths book.

There is only one exercise in NCERT chapter 9 Class 10 Maths which contain 16 questions. In these NCERT solutions for Class 10 Maths chapter 9, students will learn how trigonometry is used for finding the distances and heights and various objects, without actually measuring them. Also, check NCERT solutions for Class 6 to 12.

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NCERT Solutions for Class 10 Maths Chapter 9 Some Applications of Trigonometry PDF Free Download

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Applications Of Trigonometry Class 10 NCERT Solution - Important Points

Line of Sight - The Line of Sight is the line formed by our vision as it passes through an item when we look at it.

Horizontal Line - The distance between the observer and the object is measured by a horizontal line.

The angle of Elevation:

  • The angle formed by the line of sight to the top of the item and the horizontal line is called an angle of elevation.

  • It is above the horizontal line, i.e., when we gaze up at the item, we make an angle of elevation.

The angle of Depression:

  • When the spectator must look down to perceive the item, an angle of depression is formed.

  • When the horizontal line is above the angle, the angle of depression is formed between it and the line of sight.

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Free download NCERT Solutions for Class 10 Maths Chapter 9 Some Applications of Trigonometry PDF for CBSE Exam.

NCERT Solutions for Class 10 Maths Chapter 9 Some Applications of Trigonometry (Intext Questions and Exercise)

NCERT Solutions for ch 9 maths class 10 Excercise: 9.1

Q1 A circus artist is climbing a 20 m long rope, which is tightly stretched and tied from the top of a vertical pole to the ground. Find the height of the pole, if the angle made by the rope with the ground level is 30° (see Fig. 9.11).

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Answer:

Given that,
The length of the rope (AC) = 20 m. and \angle ACB =30^o
Let the height of the pole (AB) be h

So, in the right triangle \Delta ABC

sdgfhgjhjfdfg4

By using the Sin rule
\sin \theta = \frac{P}{H} =\frac{AB}{AC}
\sin 30^o =\frac{h}{20}
h =10 m.
Hence the height of the pole is 10 m.

Q2 A tree breaks due to storm and the broken part bends so that the top of the tree touches the ground making an angle 30° with it. The distance between the foot of the tree to the point where the top touches the ground is 8 m. Find the height of the tree.

Answer:

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Suppose DB is a tree and the AD is the broken height of the tree which touches the ground at C.
GIven that,
\angle ACB = 30^o , BC = 8 m
let AB = x m and AD = y m
So, AD+AB = DB = x+y

In right angle triangle \Delta ABC ,
\tan \theta = \frac{P}{B}=\frac{x}{8}
\tan 30^o =\frac{x}{8}=\frac{1}{\sqrt{3}}
So, the value of x = 8/\sqrt{3}

Similarily,
\cos 30^o = \frac{BC}{AC} = \frac{8}{y}
the value of y is 16/\sqrt{3}

So, the total height of the tree is-

x+y=\frac{24}{\sqrt{3}}=8\sqrt{3}

= 8 (1.732) = 13.856 m (approx)

Q3 A contractor plans to install two slides for the children to play in a park. For the children below the age of 5 years, she prefers to have a slide whose top is at a height of 1.5 m, and is inclined at an angle of 30° to the ground, whereas for elder children, she wants to have a steep slide at a height of 3m, and inclined at an angle of 60° to the ground. What should be the length of the slide in each case?

Answer:

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Suppose x m is the length of slides for children below 5 years and the length of slides for elders children be y m.

Given that,
AF = 1.5 m, BC = 3 m, \angle AEF = 30^o and \angle BDC = 60^o

In triangle \Delta EAF,
\sin \theta = \frac{AF}{EF} = \frac{1.5}{x}
\sin 30^o = \frac{1.5}{x}
The value of x is 3 m.

Similarily in \Delta CDB,
\sin 60^o = \frac{3}{y}
\frac{\sqrt{3}}{2}= \frac{3}{y}
the value of y is 2\sqrt{3} = 2(1.732) = 3.468

Hence the length of the slide for children below 5 yrs. is 3 m and for the elder children is 3.468 m.

Q4 The angle of elevation of the top of a tower from a point on the ground, which is 30 m away from the foot of the tower, is 30°. Find the height of the tower.

Answer:

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Let the height of the tower AB is h and the angle of elevation from the ground at point C is \angle ACB = 30^o
According to question,
In the right triangle \Delta ABC ,
\tan \theta = \frac{AB}{BC} = \frac{h}{30}
\tan 30^o =\frac{1}{\sqrt{3}}=\frac{h}{30}
the value of h is 10\sqrt{3} = 10(1.732) = 17.32 m
Thus the height of the tower is 17.32 m

Q5 A kite is flying at a height of 60 m above the ground. The string attached to the kite is temporarily tied to a point on the ground. The inclination of the string with the ground is 60°. Find the length of the string, assuming that there is no slack in the string.

Answer:

A
15942903087441594290307251
Given that,
The length of AB = 60 m and the inclination of the string with the ground at point C is \angle ACB = 60^o .
Let the length of the string AC be l .
According to question,
In right triangle \Delta CBA,
\sin 60^{o} = \frac{AB}{AC} = \frac{60}{l}
\frac{\sqrt{3}}{2} = \frac{60}{l}
The value of length of the string ( l ) is 40\sqrt{3} = 40(1.732) = 69.28 m
Hence the length of the string is 69.28 m.

Q6 A 1.5 m tall boy is standing at some distance from a 30 m tall building. The angle of elevation from his eyes to the top of the building increases from 30° to 60° as he walks towards the building. Find the distance he walked towards the building.

Answer:

15942903319101594290329650
Given that,
The height of the tallboy (DC) is 1.5 m and the height of the building (AB) is 30 m.
\angle ADF = 30^o and \angle AEF = 60^o

According to question,
In right triangle AFD,
\\\Rightarrow \tan 30^o=\frac{AF}{DF} = \frac{28.5}{DF}\\\\\Rightarrow \frac{1}{\sqrt{3}} = \frac{28.5}{DF}
So, DF = (28.5)\sqrt{3}

In right angle triangle \Delta AFE ,
\tan 60^o =\frac{AF}{FE}=\frac{28.5}{EF}
\sqrt{3}=\frac{28.5}{EF}
EF = 9.5\sqrt{3}

So, distance walked by the boy towards the building = DF - EF = 19\sqrt{3}

Q7 From a point on the ground, the angles of elevation of the bottom and the top of a transmission tower fixed at the top of a 20 m high building are 45° and 60° respectively. Find the height of the tower.

Answer:

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Suppose BC = h is the height of transmission tower and the AB be the height of the building and AD is the distance between the building and the observer point (D).
We have,
AB = 20 m, BC = h m and AD = x m
\angle CDA = 60^o and \angle BDA = 45^o

According to question,
In triangle \Delta BDA,
\tan 45^o = \frac{AB}{AD}=\frac{20}{x}
So, x = 20 m

Again,
In triangle \Delta CAD,

\\\Rightarrow \tan 60^o = \frac{AB+BC}{AD}=\frac{20+h}{20}\\\\\Rightarrow \sqrt{3}= 1+\frac{h}{20}\\\\\Rightarrow h=20(\sqrt{3}-1)\\\\\Rightarrow 20(0.732) = 14.64 m

Answer- the height of the tower is 14.64 m

Q8 A statue, 1.6 m tall, stands on the top of a pedestal. From a point on the ground, the angle of elevation of the top of the statue is 60° and from the same point, the angle of elevation of the top of the pedestal is 45°. Find the height of the pedestal.

Answer:

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Let the height of the pedestal be h m. and the height of the statue is 1.6 m.
the angle of elevation of the top of the statue and top of the pedestal is( \angle DCB= 60^o )and( \angle ACB= 45^o ) respectively.

Now,
In triangle \Delta ABC ,
\tan 45^o =1 =\frac{AB}{BC}=\frac{h}{BC}
therefore, BC = h m

In triangle \Delta CBD ,
\\\Rightarrow \tan 60^o = \frac{BD}{BC}=\frac{h+1.6}{h}\\\\\Rightarrow \sqrt{3}= 1+\frac{1.6}{h}
the value of h is 0.8(\sqrt{3}+1) m
Hence the height of the pedestal is 0.8(\sqrt{3}+1) m

Q9 The angle of elevation of the top of a building from the foot of the tower is 30° and the angle of elevation of the top of the tower from the foot of the building is 60°. If the tower is 50 m high, find the height of the building.

Answer:

15942904215031594290420227
It is given that, the height of the tower (AB) is 50 m. \angle AQB = 30^o and \angle PBQ = 60^o
Let the height of the building be h m

According to question,
In triangle PBQ,
\tan 60^o = \frac{PQ}{BQ} = \frac{50}{BQ}
\\\sqrt{3} = \frac{50}{BQ}\\ BQ = \frac{50}{\sqrt{3}} .......................(i)

In triangle ABQ,

\tan 30^o = \frac{h}{BQ}
{BQ}=h\sqrt{3} .........................(ii)
On equating the eq(i) and (ii) we get,

\frac{50}{\sqrt{3}}=h\sqrt{3}
therefore, h = 50/3 = 16.66 m = height of the building.

Q10 Two poles of equal heights are standing opposite each other on either side of the road, which is 80 m wide. From a point between them on the road, the angles of elevation of the top of the poles are 60° and 30°, respectively. Find the height of the poles and the distances of the point from the poles.

Answer:

15942904402531594290438964
Given that,
The height of both poles are equal DC = AB. The angle of elevation of of the top of the poles are \angle DEC=30^o and \angle AEB=60^o resp.
Let the height of the poles be h m and CE = x and BE = 80 - x

According to question,
In triangle DEC,

\\\Rightarrow \tan 30^o = \frac{DC}{CE} = \frac{h}{x}\\\\\Rightarrow \frac{1}{\sqrt{3}}= \frac{h}{x}\\\\\Rightarrow x=h\sqrt{3} ..............(i)

In triangle AEB,
\\\Rightarrow \tan 60^o = \frac{AB}{BE}=\frac{h}{80-x}\\\\\Rightarrow \sqrt{3}=\frac{h}{80-x}\\\\\Rightarrow x=80 - \frac{h}{\sqrt{3}} ..................(ii)
On equating eq (i) and eq (ii), we get

\sqrt{3}h=80 - \frac{h}{\sqrt{3}}

\frac{h}{\sqrt{3}}=20 So, x = 60 m

Hence the height of both poles is ( h=20\sqrt{3} )m and the position of the point is at 60 m from the pole CD and 20 m from the pole AB.

Q11 A TV tower stands vertically on a bank of a canal. From a point on the other bank directly opposite the tower, the angle of elevation of the top of the tower is 60°. From another point 20 m away from this point on the line joining this point to the foot of the tower, the angle of elevation of the top of the tower is 30° (see Fig. 9.12). Find the height of the tower and the width of the canal.

15942904616351594290459973

Answer:

15942904764911594290475054
Suppose the h is the height of the tower AB and BC = x m
It is given that, the width of CD is 20 m,
According to question,

In triangle \Delta ADB ,
\\\Rightarrow \tan 30^o = \frac{AB}{20+x}=\frac{h}{20+x}\\\\\Rightarrow \frac{1}{\sqrt{3}}=\frac{h}{20+x}\\\\\Rightarrow 20+x = h\sqrt{3}\\\\\Rightarrow x = h\sqrt{3}-20 ............(i)

In triangle ACB,
\\\Rightarrow \tan 60^o = \frac{h}{x}=\sqrt{3}\\\\\Rightarrow x= \frac{h}{\sqrt{3}} .............(ii)

On equating eq (i) and (ii) we get:

h\sqrt{3}-20= \frac{h}{\sqrt{3}}
from here we can calculate the value of h=10\sqrt{3}= 10 (1.732) = 17.32\: m and the width of the canal is 10 m.

Q12 From the top of a 7 m high building, the angle of elevation of the top of a cable tower is 60° and the angle of depression of its foot is 45°. Determine the height of the tower.

Answer:

15942904947031594290493105
Let the height of the cable tower be (AB = h+7 )m
Given,
The height of the building is 7 m and angle of elevation of the top of the tower \angle ACE = 60^o , angle of depression of its foot \angle BCE = 45^o .

According to question,

In triangle \Delta DBC ,
\\\tan 45^o = \frac{CD}{BD} = \frac{7}{BD} = 1\\ BD =7\ m
since DB = CE = 7 m

In triangle \Delta ACE ,

\\\tan 60^o = \frac{h}{CE}=\frac{h}{7}=\sqrt{3}\\ \therefore h = 7\sqrt{3}\ m

Thus, the total height of the tower equal to h+7 =7(1+\sqrt{3}\) m

Q13 As observed from the top of a 75 m high lighthouse from the sea-level, the angles of depression of two ships are 30° and 45°. If one ship is exactly behind the other on the same side of the lighthouse, find the distance between the two ships.

Answer:

Given that,
The height of the lighthouse (AB) is 75 m from the sea level. And the angle of depression of two different ships are \angle ADB = 30^0 and \angle ACB = 45^0 respectively
15942905447991594290543400
Let the distance between both the ships be x m.
According to question,

In triangle \Delta ADB ,

\tan 30^0 = \frac{AB}{BD}=\frac{75}{x+y} = \frac{1}{\sqrt{3}}
\therefore x+y = 75 \sqrt{3} .............(i)

In triangle \Delta ACB ,

\tan 45^0 = 1 =\frac{75}{BC}=\frac{75}{y}
\therefore y =75\ m .............(ii)

From equation (i) and (ii) we get;
x = 75(\sqrt{3}-1)=75(0.732)
x = 54.9\simeq 55\ m

Hence, the distance between the two ships is approx 55 m.

Q14 A 1.2 m tall girl spots a balloon moving with the wind in a horizontal line at a height of 88.2 m from the ground. The angle of elevation of the balloon from the eyes of the girl at any instant is 60°. After some time, the angle of elevation reduces to 30° (see Fig. 9.13). Find the distance traveled by the balloon during the interval.

15942906011101594290599656

Answer:

Given that,
The height of the girl is 1.2 m. The height of the balloon from the ground is 88.2 m and the angle of elevation of the balloon from the eye of the girl at any instant is ( \angle ACB =60^0 ) and after some time \angle DCE =30^0 .
15942906424181594290640376
Let the x distance travelled by the balloon from position A to position D during the interval.
AB = ED = 88.2 - 1.2 =87 m

Now, In triangle \Delta BCA ,

\\\tan 60^0 = \sqrt{3}=\frac{AB}{BC}=\frac{87}{BC}\\ \therefore BC = 29\sqrt{3}

In triangle \Delta DCE ,

\\\tan 30^0 = \frac{1}{\sqrt{3}}=\frac{DE}{CE}=\frac{87}{CE}\\ \therefore CE = 87\sqrt{3}

Thus, distance traveled by the balloon from position A to D

= CE - BC =87\sqrt{3}-29\sqrt{3} = 58\sqrt{3} m

Q15 A straight highway leads to the foot of a tower. A man standing at the top of the tower observes a car at an angle of depression of 30°, which is approaching the foot of the tower with a uniform speed. Six seconds later, the angle of depression of the car is found to be 60°. Find the time taken by the car to reach the foot of the tower from this point.

Answer:

15942906686221594290666652
Let h be the height of the tower (DC) and the speed of the car be x\ ms^{-1} . Therefore, the distance (AB)covered by the car in 6 seconds is 6 x m. Let t time required to reach the foot of the tower. So, BC = x t

According to question,
In triangle \Delta DAC ,
\\\tan 30^0 = \frac{1}{\sqrt{3}}=\frac{h}{6x+xt}\\ x(6+t) = h\sqrt{3} ..........................(i)

In triangle \Delta BCD ,

\\\tan 60^0 = \sqrt{3} = \frac{h}{xt}\\ \therefore h = 3.xt ...................(ii)

Put the value of h in equation (i) we get,
\\x(6+t) = (\sqrt{3}.\sqrt{3})xt\\ 6x +xt = 3xt\\ 6x = 2xt
t = 3

Hence, from point B car take 3 sec to reach the foot of the tower.

Q16 The angles of elevation of the top of a tower from two points at a distance of 4 m and 9 m from the base of the tower and in the same straight line with it are complementary. Prove that the height of the tower is 6 m.

Answer:

15942906874451594290685895
Let the height of the tower be h m.
we have PB = 4m and QB = 9 m
Suppose \angle BQA = \theta , so \angle APB =90- \theta

According to question,

In triangle \Delta ABQ ,

\\\tan \theta = \frac{h}{9}\\ \therefore h = 9 \tan \theta ..............(i)

In triangle \Delta ABP ,

\\\tan (90-\theta)=\cot \theta = \frac{h}{4}\\ \therefore h = 4\cot \theta .....................(ii)

multiply the equation (i) and (ii), we get

\\h^2 = 36\\ \Rightarrow h = 6 m

Hence the height of the tower is 6 m.

NCERT Solutions for Class 10 Maths for Other Chapters

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

Trigonometry

Chapter 9

Some Applications of Trigonometry

Chapter 10

Circles

Chapter 11

Constructions

Chapter 12

Areas Related to Circles

Chapter 13

Surface Areas and Volumes

Chapter 14

Statistics

Chapter 15

Probability

NCERT Exemplar solutions - Subject wise

The main topics covered in the application of trigonometry class 10

9.1 Introduction: This section of ch 9 maths class 10 introduces the concept of how trigonometry is used in everyday life, and how it was first developed for use in astronomy. The chapter will also look at how trigonometry can be used to find heights and distances without directly measuring them.

9.2 Heights and Distances: this ch 9 maths class 10 delves into the concepts of line of sight, angle of deviation, angle of elevation, and angle of depression, and demonstrates how to solve problems using trigonometric ratios.

9.3 Summary: This section of chapter 9 class 10 maths provides a summary of the key points covered in the chapter, helping students to focus on the most important concepts.

Let's understand a few terms used in Some Applications of Trigonometry Class 10 Chapter through the diagram.

  • The line of sight- It is an imaginary line drawn from the observer's eye to the object viewed by the observer.

  • The angle of elevation- The angle of elevation of an object viewed by an observer, is the angle formed by the line of sight with the horizontal level when the object is above the horizontal level,


The angle of elevation1594290860577

  • The angle of depression- The angle of depression of an object viewed by an observer, is the angle formed by the line of sight with the horizontal level when the object is below the horizontal level.

The angle of depression

How to use NCERT Solutions for Class 10 Maths Chapter 9?

  • Before coming to this ch 9 class 10 maths, please ensure that you have completed the previous chapter Introduction to Trigonometry.

  • Read the conceptual theory given in the NCERT textbook and have a look over some examples present in the textbook for the deeper understanding of the topics.

  • Once you complete the above points then you can jump to the practice exercises available in the NCERT book.

  • While practicing the questions in practice exercises, you can use ch 9 class 10 maths solutions to know how to solve, what is the final answer and so on.

  • When you have done the practice exercise then the best thing to make your concepts strong is the last 5 years CBSE class 10 question papers and the NCERT exemplar problems.

NCERT Solutions for Class 10 for other subjects

NCERT Books and NCERT Syllabus

Frequently Asked Question (FAQs)

1. Where can one find the applications of trigonometry class 10 ncert solutions?

ch 9 maths class 10 examples solutions can be downloaded for free in PDF format from the Craerrs360 website. These NCERT solutions can also find above in this article. Students can practice them to score well in the exam. Also, students can download the application of trigonometry pdf format to study offline.

2. Are the NCERT Solutions for Class 10 Maths Chapter 9 sufficient for CBSE ex

It is important to thoroughly study and practice all the questions in the NCERT Solutions for trigonometry applications class 10. Once you have completed all the questions, you can supplement your learning by consulting other reference materials and the questions provided in the NCERT Exemplar textbooks.

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

In CBSE board exam most of the questions are directly asked from NCERT textbook, so students must know topics in the NCERT syllabus very well. In NCERT class 10 maths book. , you will get are detailed solutions provided by the experts who knows how best to answer in the board exam in order to get good marks. Students can practice more questions from NCERT Exemplar.

4. What are the benefits of studying the class 10 maths applications of trigonometry solutions?

Some of the benefits of studying NCERT Solutions for application to trigonometry class 10 include the ability to calculate heights and distances of objects without direct measurement and a better understanding of trigonometry-based questions. This knowledge will aid students in answering questions related to trigonometry and performing well in class tests and CBSE exams.

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Dear aspirant !

Hope you are doing well ! The class 10 Hindi mp board sample paper can be found on the given link below . Set your time and give your best in the sample paper it will lead to great results ,many students are already doing so .

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Hope you get it !

Thanking you

Hello aspirant,

The dates of the CBSE Class 10th and Class 12 exams are February 15–March 13, 2024 and February 15–April 2, 2024, respectively. You may obtain the CBSE exam date sheet 2024 PDF from the official CBSE website, cbse.gov.in.

To get the complete datesheet, you can visit our website by clicking on the link given below.

https://school.careers360.com/boards/cbse/cbse-date-sheet

Thank you

Hope this information helps you.

Hello aspirant,

The paper of class 7th is not issued by respective boards so I can not find it on the board's website. You should definitely try to search for it from the website of your school and can also take advise of your seniors for the same.

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Thank you

Hope it helps you.

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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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