Careers360 Logo
NCERT Solutions for Exercise 10.2 Class 12 Maths Chapter 10 - Vector Algebra

NCERT Solutions for Exercise 10.2 Class 12 Maths Chapter 10 - Vector Algebra

Edited By Ramraj Saini | Updated on Dec 04, 2023 08:51 AM IST | #CBSE Class 12th

NCERT Solutions For Class 12 Maths Chapter 10 Exercise 10.2

NCERT Solutions for Exercise 10.2 Class 12 Maths Chapter 10 Vector Algebra are discussed here. These NCERT solutions are created by subject matter expert at Careers360 considering the latest syllabus and pattern of CBSE 2023-24. NCERT solutions for exercise 10.2 Class 12 Maths chapter 10 introduces a few more concepts of vectors. The questions of exercise 10.2 Class 12 Maths deals with the concepts like addition of vectors, magnitude of vectors, unit vectors etc. Also in NCERT solutions for Class 12 Maths chapter 10 exercise 10.2 questions related to collinear vectors and components of vectors. As the Class 12 Maths syllabus is concerned, it is important to practice Class 12 Maths chapter 10 exercise 10.2. Questions related to the Class 12th Maths chapter 10 exercise 10.2 can be expected for CBSE Class 12 Board Exams as well as competitive exams like JEE Mains.

The chapter Vector Algebra exercises will be helpful for both Mathematics and Physics. To solve more problems of NCERT Class 12 Maths, students can also make use of NCERT Exemplar Solutions for Class 12 Maths. 12th class Maths exercise 10.2 answers are designed as per the students demand covering comprehensive, step by step solutions of every problem. Practice these questions and answers to command the concepts, boost confidence and in depth understanding of concepts. Students can find all exercise together using the link provided below.

Also practice-

PTE Registrations 2024

Register now for PTE & Unlock 10% OFF : Use promo code: 'C360SPL10'. Limited Period Offer!

Access NCERT Solutions for Class 12 Maths Chapter 10 Exercise 10.2

Download PDF

Vector Algebra-Exercise: 10.2

Question:1 Compute the magnitude of the following vectors:

(1) \vec a = \hat i + \hat j + \hat k

Answer:

Here

\vec a = \hat i + \hat j + \hat k

Magnitude of \vec a

\vec a=\sqrt{1^2+1^2+1^2}=\sqrt{3}

Question:1 Compute the magnitude of the following vectors:

(2) \vec b = 2 \hat i - 7 \hat j - 3 \hat k

Answer:

Here,

\vec b = 2 \hat i - 7 \hat j - 3 \hat k

Magnitude of \vec b

\left | \vec b \right |=\sqrt{2^2+(-7)^2+(-3)^2}=\sqrt{62}

Question:1 Compute the magnitude of the following vectors:

(3) \vec c = \frac{1}{\sqrt 3 }\hat i + \frac{1}{\sqrt 3 }\hat j -\frac{1}{\sqrt 3 }\hat k

Answer:

Here,

\vec c = \frac{1}{\sqrt 3 }\hat i + \frac{1}{\sqrt 3 }\hat j -\frac{1}{\sqrt 3 }\hat k

Magnitude of \vec c

\left |\vec c \right |=\sqrt{\left ( \frac{1}{\sqrt{3}} \right )^2+\left ( \frac{1}{\sqrt{3}} \right )^2+\left ( \frac{1}{\sqrt{3}} \right )^2}=1

Question:2 Write two different vectors having same magnitude

Answer:

Two different Vectors having the same magnitude are

\vec a= 3\hat i+6\hat j+9\hat k

\vec b= 9\hat i+6\hat j+3\hat k

The magnitude of both vector

\left | \vec a \right |=\left | \vec b \right | = \sqrt{9^2+6^2+3^2}=\sqrt{126}

Question:3 Write two different vectors having same direction.

Answer:

Two different vectors having the same direction are:

\vec a=\hat i+2\hat j+3\hat k

\vec b=2\hat i+4\hat j+6\hat k

Question:4 Find the values of x and y so that the vectors 2 \hat i + 3 \hat j and x \hat i + y \hat j are equal.

Answer:

2 \hat i + 3 \hat j will be equal to x \hat i + y \hat j when their corresponding components are equal.

Hence when,

x=2 and

y=3

Question:5 Find the scalar and vector components of the vector with initial point (2, 1) and terminal point (– 5, 7).

Answer:

Let point P = (2, 1) and Q = (– 5, 7).

Now,

\vec {PQ}=(-5-2)\hat i+(7-1)\hat j=-7\hat i +6\hat j

Hence scalar components are (-7,6) and the vector is -7\hat i +6\hat j

Question:6 Find the sum of the vectors \vec a = \hat i - 2 \hat j + \hat k , \vec b = -2 \hat i + 4 \hat j + 5 \hat k \: \: and\: \: \: \vec c = \hat i - 6 \hat j - 7 \hat k

Answer:

Given,

\\ \vec a = \hat i - 2 \hat j + \hat k ,\\ \vec b = -2 \hat i + 4 \hat j + 5 \hat k \: \: and\: \: \: \\\vec c = \hat i - 6 \hat j - 7 \hat k

Now, The sum of the vectors:

\vec a +\vec b+\vec c = \hat i - 2 \hat j + \hat k + -2 \hat i + 4 \hat j + 5 \hat k + \hat i - 6 \hat j - 7 \hat k

\vec a +\vec b+\vec c = (1-2+1)\hat i +(-2+4-6) \hat j + (1+5-7)\hat k

\vec a +\vec b+\vec c =-4\hat j-\hat k


Question:7 Find the unit vector in the direction of the vector \vec a = \hat i + \hat j + 2 \hat k

Answer:

Given

\vec a = \hat i + \hat j + 2 \hat k

Magnitude of \vec a

\left |\vec a \right |=\sqrt{1^2+1^2+2^2}=\sqrt{6}

A unit vector in the direction of \vec a

\vec u = \frac{\hat i}{\left | a \right |} + \frac{\hat j}{\left | a \right |} +\frac{2\hat k}{\left | a \right |} =\frac{\hat i}{\sqrt{6}}+\frac{\hat j}{\sqrt{6}}+\frac{2\hat k}{\sqrt{6}}

Question:8 Find the unit vector in the direction of vector \vec { PQ} , where P and Q are the points (1, 2, 3) and (4, 5, 6), respectively.

Answer:

Given P = (1, 2, 3) and Q = (4, 5, 6)

A vector in direction of PQ

\vec {PQ}=(4-1)\hat i+(5-2)\hat j +(6-3)\hat k

\vec {PQ}=3\hat i+3\hat j +3\hat k

Magnitude of PQ

\left | \vec {PQ} \right |=\sqrt{3^2+3^2+3^2}=3\sqrt{3}

Now, unit vector in direction of PQ

\hat u=\frac{\vec {PQ}}{\left | \vec {PQ} \right |}=\frac{3\hat i+3\hat j+3\hat k}{3\sqrt{3}}

\hat u=\frac{\hat i}{\sqrt{3}}+\frac{\hat j}{\sqrt{3}}+\frac{\hat k}{\sqrt{3}}

Question:9 For given vectors, \vec a = 2 \hat i - \hat j + 2 \hat k and \vec b = - \hat i + \hat j - \hat k , find the unit vector in the direction of the vector \vec a + \vec b .

Answer:

Given

\vec a = 2 \hat i - \hat j + 2 \hat k

\vec b = - \hat i + \hat j - \hat k

Now,

\vec a + \vec b=(2-1)\hat i+(-1+1)\hat j+ (2-1)\hat k

\vec a + \vec b=\hat i+\hat k

Now a unit vector in the direction of \vec a + \vec b

\vec u= \frac{\vec a + \vec b}{\left |\vec a + \vec b \right |}=\frac{\hat i+\hat j}{\sqrt{1^2+1^2}}

\vec u= \frac{\hat i}{\sqrt{2}}+\frac{\hat j}{\sqrt{2}}

Question:10 Find a vector in the direction of vector 5 \hat i - \hat j + 2 \hat k which has magnitude 8 units.

Answer:

Given a vector

\vec a=5 \hat i - \hat j + 2 \hat k

the unit vector in the direction of 5 \hat i - \hat j + 2 \hat k

\vec u=\frac{5\hat i - \hat j + 2 \hat k}{\sqrt{5^2+(-1)^2+2^2}}=\frac{5\hat i}{\sqrt{30}}-\frac{\hat j}{\sqrt{30}}+\frac{2\hat k}{\sqrt{30}}

A vector in direction of 5 \hat i - \hat j + 2 \hat k and whose magnitude is 8 =

8\vec u=\frac{40\hat i}{\sqrt{30}}-\frac{8\hat j}{\sqrt{30}}+\frac{16\hat k}{\sqrt{30}}

Question:11 Show that the vectors 2 \hat i -3 \hat j + 4 \hat k and - 4 \hat i + 6 \hat j - 8 \hat k are collinear.

Answer:

Let

\vec a =2 \hat i -3 \hat j + 4 \hat k

\vec b=- 4 \hat i + 6 \hat j - 8 \hat k

It can be seen that

\vec b=- 4 \hat i + 6 \hat j - 8 \hat k=-2(2 \hat i -3 \hat j + 4 \hat k)=-2\vec a

Hence here \vec b=-2\vec a

As we know

Whenever we have \vec b=\lambda \vec a , the vector \vec a and \vec b will be colinear.

Here \lambda =-2

Hence vectors 2 \hat i -3 \hat j + 4 \hat k and - 4 \hat i + 6 \hat j - 8 \hat k are collinear.

Question:12 Find the direction cosines of the vector \hat i + 2 \hat j + 3 \hat k

Answer:

Let

\vec a=\hat i + 2 \hat j + 3 \hat k

\left |\vec a \right |=\sqrt{1^2+2^2+3^2}=\sqrt{14}

Hence direction cosine of \vec a are

\left ( \frac{1}{\sqrt{14}},\frac{2}{\sqrt{14}} ,\frac{3}{\sqrt{14}}\right )

Question:13 Find the direction cosines of the vector joining the points A(1, 2, –3) and B(–1, –2, 1), directed from A to B.

Answer:

Given

point A=(1, 2, –3)

point B=(–1, –2, 1)

Vector joining A and B Directed from A to B

\vec {AB}=(-1-1)\hat i +(-2-2)\hat j+(1-(-3))\hat k

\vec {AB}=-2\hat i +-4\hat j+4\hat k

\left | \vec {AB} \right |=\sqrt{(-2)^2+(-4)^2+4^2}=\sqrt{36}=6

Hence Direction cosines of vector AB are

\left ( \frac{-2}{6},\frac{-4}{6},\frac{4}{6} \right )=\left ( \frac{-1}{3},\frac{-2}{3},\frac{2}{3} \right )

Question:14 Show that the vector \hat i + \hat j + \hat k is equally inclined to the axes OX, OY and OZ.

Answer:

Let

\vec a=\hat i + \hat j + \hat k

\left | \vec a \right |=\sqrt{1^2+1^2+1^2}=\sqrt{3}

Hence direction cosines of this vectors is

\left ( \frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}} \right )

Let \alpha , \beta and \gamma be the angle made by x-axis, y-axis and z- axis respectively

Now as we know,

cos\alpha=\frac{1}{\sqrt{3}} , cos\beta=\frac{1}{\sqrt{3}} and\:cos\gamma=\frac{1}{\sqrt{3}}

Hence Given vector is equally inclined to axis OX,OY and OZ.

Question:15 (1) Find the position vector of a point R which divides the line joining two points P and Q whose position vectors are i + 2 j - k and - i + j + k respectively, in the ratio 2 : 1 internally

Answer:

As we know

The position vector of the point R which divides the line segment PQ in ratio m:n internally:

\vec r=\frac{m\vec b+n\vec a}{m+n}

Here

position vector os P = \vec a = i + 2 j - k

the position vector of Q = \vec b=- i + j + k

m:n = 2:1

And Hence

\vec r = \frac{2(-\hat i+\hat j +\hat k)+1(\hat i+2\hat j-\hat k)}{2+1}=\frac{-2\hat i+2\hat j +2\hat k+\hat i+2\hat j-\hat k}{3}

\vec r = \frac{-2\hat i+2\hat j +2\hat k+\hat i+2\hat j-\hat k}{3}=\frac{-\hat i+4\hat j+\hat k}{3}

\vec r = \frac{-\hat i}{3}+\frac{4\hat j}{3}+\frac{\hat k}{3}

Question:15 (2) Find the position vector of a point R which divides the line joining two points P and Q whose position vectors are \hat i + 2 \hat j - \hat k and - \hat i + \hat j + \hat k respectively, in the ratio 2 : 1 externally

Answer:

As we know

The position vector of the point R which divides the line segment PQ in ratio m:n externally:

\vec r=\frac{m\vec b-n\vec a}{m-n}

Here

position vector os P = \vec a = i + 2 j - k

the position vector of Q = \vec b=- i + j + k

m:n = 2:1

And Hence

\vec r = \frac{2(-\hat i+\hat j +\hat k)-1(\hat i+2\hat j-\hat k)}{2-1}=\frac{-2\hat i+2\hat j +2\hat k-\hat i-2\hat j+\hat k}{1}

\vec r = -3\hat i +3\hat k

Question:16 Find the position vector of the mid point of the vector joining the points P(2, 3, 4) and Q(4, 1, –2).

Answer:

Given

The position vector of point P = 2\hat i+3\hat j +4\hat k

Position Vector of point Q = 4\hat i+\hat j -2\hat k

The position vector of R which divides PQ in half is given by:

\vec r =\frac{2\hat i+3\hat j +4\hat k+4\hat i+\hat j -2\hat k}{2}

\vec r =\frac{6\hat i+4\hat j +2\hat k}{2}=3\hat i+2\hat j +\hat k

Question:17 Show that the points A, B and C with position vectors, \vec a = 3 \hat i - 4 \hat j - 4 \hat k , \vec b = 2 \hat i - \hat j + \hat k \: \: and \: \: \: \vec c = \hat i - 3 \hat j - 5 \hat k , respectively form the vertices of a right angled triangle.

Answer:

Given

the position vector of A, B, and C are

\\\vec a = 3 \hat i - 4 \hat j - 4 \hat k ,\\\vec b = 2 \hat i - \hat j + \hat k \: \: and \\\vec c = \hat i - 3 \hat j - 5 \hat k

Now,

\vec {AB}=\vec b-\vec a=-\hat i+3\hat j+5\hat k

\vec {BC}=\vec c-\vec b=-\hat i-2\hat j-6\hat k

\vec {CA}=\vec a-\vec c=2\hat i-\hat j+\hat k

\left | \vec {AB} \right |=\sqrt{(-1)^2+3^2+5^2}=\sqrt{35}

\left | \vec {BC} \right |=\sqrt{(-1)^2+(-2)^2+(-6)^2}=\sqrt{41}

\left | \vec {CA} \right |=\sqrt{(2)^2+(-1)^2+(1)^2}=\sqrt{6}

AS we can see

\left | \vec {BC} \right |^2=\left | \vec {CA} \right |^2+\left | \vec {AB} \right |^2

Hence ABC is a right angle triangle.

Question:18 In triangle ABC (Fig 10.18), which of the following is not true:

15948360623461594836059639

A ) \overline{AB}+ \overline{BC}+ \overline{CA} = \vec 0 \\\\ B ) \overline{AB}+ \overline{BC}- \overline{AC} = \vec 0 \\\\ C ) \overline{AB}+ \overline{BC}- \overline{CA} = \vec 0 \\\\ D ) \overline{AB}- \overline{CB}+ \overline{CA} = \vec 0

Answer:

From triangles law of addition we have,

\vec {AB}+\vec {BC}=\vec {AC}

From here

\vec {AB}+\vec {BC}-\vec {AC}=0

also

\vec {AB}+\vec {BC}+\vec {CA}=0

Also

\vec {AB}-\vec {CB}+\vec {CA}=0

Hence options A,B and D are true SO,

Option C is False.

Question:19 If are two collinear vectors, then which of the following are incorrect:
(A) \vec b = \lambda \vec a for some saclar \lambda
(B) \vec a = \pm \vec b
(C) the respective components of \vec a \: \:and \: \: \vec b are not proportional
(D) both the vectors \vec a \: \:and \: \: \vec b have same direction, but different magnitudes.

Answer:

If two vectors are collinear then, they have same direction or are parallel or anti-parallel.
Therefore,
They can be expressed in the form \vec{b}= \lambda \vec{a} where a and b are vectors and \lambda is some scalar quantity.

Therefore, (a) is true.
Now,
(b) \lambda is a scalar quantity so its value may be equal to \pm 1

Therefore,
(b) is also true.

C) The vectors 1517995000744796 and 1517995001522472 are proportional,
Therefore, (c) is not true.

D) The vectors 1517995002319114 and 1517995003131589 can have different magnitude as well as different directions.

Therefore, (d) is not true.

Therefore, the correct options are (C) and (D).

More About NCERT Solutions for Class 12 Maths Chapter 10 Exercise 10.2

The NCERT Class 12 Math chapter 10 exercise 10.2 starts with the questions to find the magnitude of vectors. The 5th question of exercise 10.2 Class 12 Maths is related to the concept of position vector. The concepts of unit vectors are explained in questions 7 to 10. NCERT solutions for Class 12 Maths chapter 10 exercise 10.2 also goes through the topics like direction cosines, and section formula.

Also Read | Vector Algebra Class 10 Chapter 10 Notes

Importance of NCERT Solutions for Class 12 Maths Chapter 10 Exercise 10.2

  • The Class 12 Maths NCERT exercises are solved by math experts and can be used for preparations of various examinations.
  • By practising the NCERT Solutions for Class 12 Maths chapter 10 exercise 10.2 students will be able to revise the concepts studied before the exercise and also will be able to clarify doubts.
JEE Main Highest Scoring Chapters & Topics
Just Study 40% Syllabus and Score upto 100%
Download EBook

Key Features Of NCERT Solutions for Exercise 10.2 Class 12 Maths Chapter 10

  • Comprehensive Coverage: The solutions encompass all the topics covered in ex 10.2 class 12, ensuring a thorough understanding of the concepts.
  • Step-by-Step Solutions: In this class 12 maths ex 10.2, each problem is solved systematically, providing a stepwise approach to aid in better comprehension for students.
  • Accuracy and Clarity: Solutions for class 12 ex 10.2 are presented accurately and concisely, using simple language to help students grasp the concepts easily.
  • Conceptual Clarity: In this 12th class maths exercise 10.2 answers, emphasis is placed on conceptual clarity, providing explanations that assist students in understanding the underlying principles behind each problem.
  • Inclusive Approach: Solutions for ex 10.2 class 12 cater to different learning styles and abilities, ensuring that students of various levels can grasp the concepts effectively.
  • Relevance to Curriculum: The solutions for class 12 maths ex 10.2 align closely with the NCERT curriculum, ensuring that students are prepared in line with the prescribed syllabus.

Also see-

NCERT Solutions Subject Wise

Subject Wise NCERT Exemplar Solutions

Frequently Asked Question (FAQs)

1. Can two different vectors have the same magnitude?

Yes, two different vectors can have the same magnitude. These vectors may be pointing in a different direction. An example is shown in question number 2 of NCERT solutions for Class 12 Maths chapter 10 exercise 10.2.

2. Can two different vectors have the same direction?

Yes, two different vectors can have the same direction. It can be vectors with different magnitudes. The third question of exercise 10.2 Class 12 Maths gives an example of this.

3. What type of questions are expected from Class 12th Maths chapter 10 exercise 10.2 for the CBSE board exam?

The question may be to find a unit vector in the direction of a given vector, or it may be related to collinear vectors etc. 

4. How many questions are covered in NCERT Class 12 Maths exercise 10.2?

There are a total of 19 questions. Practising these may help in the CBSE Class 12 Maths Paper

5. What is the use of Vector Algebra in NEET exam?

The concepts in Vector Algebra help to solve NEET physics problems. A good part of Class 11 and Class 12 NCERT Physics Syllabus uses the concepts of vectors. So vectors will be useful for the NEET exam as well as for engineering entrance exams like JEE Main.

6. If 7i+6j=2xi+3yj, then the value of x and y is.

The given vectors are equal. Which implies that 2x=7. Therefore x=3.5. And 3y=6, therefore y=2

7. How to verify whether the given vector forms a right angled triangle?

One method to solve such kinds of problems is to find the magnitudes of given vectors and verify Pythagoras Theorem.

Articles

Explore Top Universities Across Globe

Questions related to CBSE Class 12th

Have a question related to CBSE Class 12th ?

Hi,

The Medhavi National Scholarship Program, under the Human Resources & Development Mission (HRDM), offers financial assistance to meritorious students through a scholarship exam. To be eligible, candidates must be between 16 and 40 years old as of the last date of registration and have at least passed the 10th grade from a recognized board. Higher qualifications, such as 11th/12th grade, graduation, post-graduation, or a diploma, are also acceptable.

To apply, download the Medhavi App from the Google Play Store, sign up, and read the detailed notification about the scholarship exam. Complete the registration within the app, take the exam from home using the app, and receive your results within two days. Following this, upload the necessary documents and bank account details for verification. Upon successful verification, the scholarship amount will be directly transferred to your bank account.

The scholarships are categorized based on the marks obtained in the exam: Type A for those scoring 60% or above, Type B for scores between 50% and 60%, and Type C for scores between 40% and 50%. The cash scholarships range from Rs. 2,000 to Rs. 18,000 per month, depending on the exam and the marks obtained.

Since you already have a 12th-grade qualification with 84%, you meet the eligibility criteria and can apply for the Medhavi Scholarship exam. Preparing well for the exam can increase your chances of receiving a higher scholarship.

Yuvan 01 September,2024

hello mahima,

If you have uploaded screenshot of your 12th board result taken from CBSE official website,there won,t be a problem with that.If the screenshot that you have uploaded is clear and legible. It should display your name, roll number, marks obtained, and any other relevant details in a readable forma.ALSO, the screenshot clearly show it is from the official CBSE results portal.

hope this helps.

Hello Akash,

If you are looking for important questions of class 12th then I would like to suggest you to go with previous year questions of that particular board. You can go with last 5-10 years of PYQs so and after going through all the questions you will have a clear idea about the type and level of questions that are being asked and it will help you to boost your class 12th board preparation.

You can get the Previous Year Questions (PYQs) on the official website of the respective board.

I hope this answer helps you. If you have more queries then feel free to share your questions with us we will be happy to assist you.

Thank you and wishing you all the best for your bright future.

Hello student,

If you are planning to appear again for class 12th board exam with PCMB as a private candidate here is the right information you need:

  • No school admission needed! Register directly with CBSE. (But if you want to attend the school then you can take admission in any private school of your choice but it will be waste of money)
  • You have to appear for the 2025 12th board exams.
  • Registration for class 12th board exam starts around September 2024 (check CBSE website for exact dates).
  • Aim to register before late October to avoid extra fees.
  • Schools might not offer classes for private students, so focus on self-study or coaching.

Remember , these are tentative dates based on last year. Keep an eye on the CBSE website ( https://www.cbse.gov.in/ ) for the accurate and official announcement.

I hope this answer helps you. If you have more queries then feel free to share your questions with us, we will be happy to help you.

Good luck with your studies!

View All

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

Back to top