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.

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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.

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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:

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

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 and are proportional,
Therefore, (c) is not true.

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

Therefore, (d) is not true.

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

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.

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.

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NCERT Solutions Subject Wise

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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 C lass 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.

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Questions related to CBSE Class 12th

Have a question related to CBSE Class 12th ?

i couldn't clear 1st compartment (which held in August 2022) cbse class 12th so can I give improvement in all subjects which to be held on March 2023. tell me a solution i need more than 75% in class 12th

hello,

Yes you can appear for the compartment paper again since CBSE gives three chances to a candidate to clear his/her exams so you still have two more attempts. However, you can appear for your improvement paper for all subjects but you cannot appear for the ones in which you have failed.

I hope this was helpful!

Good Luck

Read Complete Answer

Upasana Barman 24 August,2022

i was not able to clear 1st compartment and now I have to give 2nd compartment so can I give improvement exam next year? plz tell me very much tensed(cbse class 12th)

Hello dear,

If you was not able to clear 1st compartment and now you giving second compartment so YES, you can go for your improvement exam next year but if a student receives an improvement, they are given the opportunity to retake the boards as a private candidate the following year, but there are some requirements. First, the student must pass all of their subjects; if they received a compartment in any subject, they must then pass the compartment exam before being eligible for the improvement.

As you can registered yourself as private candidate for giving your improvement exam of 12 standard CBSE(Central Board of Secondary Education).For that you have to wait for a whole year which is bit difficult for you.

Positive side of waiting for whole year is you have a whole year to preparing yourself for your examination. You have no distraction or something which may causes your failure in the exams. In whole year you have to stay focused on your 12 standard examination for doing well in it. By this you get a highest marks as a comparison of others.

Believe in Yourself! You can make anything happen

All the very best.

Read Complete Answer

Mayankjha.student2007 23 August,2022

if i am giving improvement in one subject this year, will i be able to give improvement in more subjects next year?

Hello Student,

I appreciate your Interest in education. See the improvement is not restricted to one subject or multiple subjects and we cannot say if improvement in one subject in one year leads to improvement in more subjects in coming year.

You just need to have a revision of all subjects what you have completed in the school. have a revision and practice of subjects and concepts helps you better.

All the best.

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Nuthalapati Vyshnavi 21 August,2022

I will have my boards in Feb 2023 (Class 12) and I still haven't completed 1 chapter in any subject (Physics, chemistry and maths) , My 11th concepts are not that great but its okish, I wanted to know that If I do start studying from 15th of August Can I get above 85?

If you'll do hard work then by hard work of 6 months you can achieve your goal but you have to start studying for it dont waste your time its a very important year so please dont waste it otherwise you'll regret.

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khushi.thakurbbk 07 September,2022

I want to take a admission in class 12 privately.. so can I???

Yes, you can take admission in class 12th privately there are many colleges in which you can give 12th privately.

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Parvati Solanki 02 June,2022

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Popular CBSE Class 12th Questions

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 3.8×10⁷J 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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