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NCERT Solutions for Class 11 Physics Chapter 6 System Of Particles And Rotational Motion

NCERT Solutions for Class 11 Physics Chapter 6 System Of Particles And Rotational Motion

Edited By Vishal kumar | Updated on Jun 22, 2025 11:57 AM IST

When a rubber ball is squeezed it takes a new shape and a metal rod could be bent when pressure is exerted. With very small deformations that do not produce significant changes in the way the object behaves, however, we very frequently assume that the object remains a rigid body in order to simplify calculations.

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This Story also Contains
  1. NCERT Solutions for Class 11 Physics Chapter 6: Download Solution PDF
  2. System of Particles and Rotational Motion Class 11 NCERT Solutions: Exercise Questions
  3. Class 11 physics NCERT Chapter 6: Higher Order Thinking Skills (HOTS) Questions
  4. NCERT Chapter 6 System of Particles and Rotational Motion Topics
  5. NCERT Solutions for Class 11 Physics Chapter 6: Key Formulas
  6. Approach to Solve Questions of NCERT Solutions for Class 11 Physics Chapter 6
  7. What Extra Should Students Study Beyond NCERT for JEE/NEET?
  8. Importance of NCERT Solutions for Class 11 Physics Chapter 6: System of Particles and Rotational Motion
  9. NCERT Solutions for Class 11 Physics Chapter-Wise


An important yet a tricky concept in Class 11 in Physics is system of particles and rotational motion. Below, you will also find the NCERT solutions to Chapter 6 that were developed by experts. The steps are depicted clearly in these NCERT solutions and the topic becomes easier to learn. To do well in your Class 11 exams and competitive exams like JEE and NEET, it is essential to practice solving numerical problems from this chapter and follow a strategic study plan.

NCERT Solutions for Class 11 Physics Chapter 6: Download Solution PDF

With several questions and answers based on these ideas, the chapter concentrates on important subjects like the centre of mass and the moment of inertia. These solutions will help students do well on tests and are especially helpful for self-study.

Download Solution PDF

System of Particles and Rotational Motion Class 11 NCERT Solutions: Exercise Questions

Q6.1 Give the location of the centre of mass of a (i) sphere, (ii) cylinder, (iii) ring, and (iv) cube, each of uniform mass density. Does the centre of mass of a body necessarily lie inside the body?

Answer:

For uniform mass density, the location of the centre of mass is the same as that of the geometric centre.

No, it is not necessary that the centre of mass lies inside the body. For example in the case of a ring the centre of mass outside the body.

Q6.2 In the HCl molecule, the separation between the nuclei of the two atoms is about 1.27AA(1AA=1010m) . Find the approximate location of the CM of the molecule, given that a chlorine atom is about 35.5 times as massive as a hydrogen atom and nearly all the mass of an atom is concentrated in its nucleus.

Answer:

Let us assume that the centre of mass of the molecule is x cm away from the chlorine atom.

So, we can write

m(1.27  x) + 35.5mxm + 35.5 = 0

or x = 1.27(35.51) =  0.037AA

Here negative sign indicates that the centre of mass lies leftward of the chlorine atom.

Q6.3 A child sits stationary at one end of a long trolley moving uniformly with a speed V on a smooth horizontal floor. If the child gets up and runs about on the trolley in any manner, what is the speed of the CM of the (trolley + child) system?

Answer:

Since child and trolley are considered in a system thus speed of the centre of mass change only when an external force will act on the system. When the child starts to move on the trolley the force produced is the internal force of the system so this will produce no change in the velocity of the CM.

Q6.4 Show that the area of the triangle contained between the vectors aandb is one half of the magnitude of a×b .

Answer:

Let a and b be two vectors having Θ angle between them.

Consider Δ MON,

sinΘ = MNMO

or sinΘ = MN|b|

or MN = bsinΘ

|a×b| = |a||b|sinΘ= 2 (Area of Δ MOK)

Therefore the area of Δ MOK = One half of |a×b| .

Q6.5 Show that a.(b×c) is equal in magnitude to the volume of the parallelepiped formed on the three vectors , a , b and c.

Answer:

A parallelepiped is shown in the figure given below:-

Volume is given by : abc

We can write :

|b×c| = |b||c|sinΘ n^ (The direction of n^ is in the direction of vector \vec{a})

|b×c|\= |b||c|sin90 n^

|b×c|\= |b||c| n^

Now,

a.(b×c) = a.(bc) n^

= abccosΘ

= abccos0

= abc

This is equal to volume of parallelepiped.

Q6.6 Find the components along the x,y,z axes of the angular momentum 1 of a particle, whose position vector is r with components x,y,z and momentum is p with components px , py and pz . Show that if the particle moves only in the xy plane the angular momentum has only a z-component.

Answer:

Linear momentum of particle is given by :

p = pxi^ + pyj^ + pzk^

And the angular momentum is :

l =r×p

l =(xi^ + yj^ + zk^)×(pxi^ + pyj^ + pzk^)

l =|ijkxyzPxPyPz|

l = i^(ypzzpy)  j^(xpzzpx) + k^(xpyypx)

When particle is confined to x-y plane then z = 0 and p z = 0.

When we put the value of z and p z in the equation of linear momentum then we observe that only the z component is non-zero.

Q6.7 Two particles, each of mass m and speed v , travel in opposite directions along parallel lines separated by a distance d . Show that the angular momentum vector of the two particle system is the same whatever be the point about which the angular momentum is taken.

Answer:

Assume two points (say A, B) separated by distance d.

So the angular momentum of a point about point A is given by : = mv×d = mvd

And about point B : = mv×d = mvd

Now assume a point between A and B as C which is at y distance from point B.

Now the angular momentum becomes : = mv×(dy) + mv×y= mvd

Thus it can be seen that angular momentum is independent of the point about which it is measured.

Q6.8 A non-uniform bar of weight W is suspended at rest by two strings of negligible weight as shown in Fig. The angles made by the strings with the vertical are 36.90 and 53.10 respectively. The bar is 2m long. Calculate the distance d of the centre of gravity of the bar from its left end.

Answer:

The FBD of the given bar is shown below :

Since the bar is in equilibrium, we can write :

T1sin36.9 = T2sin53.1

or T1T2 = 0.8000.600 = 43

or T1 = 43T2 .....................................................(i)

For the rotational equilibrium :

T1cos36.9×d = T2cos53.1×(2d) (Use equation (i) to solve this equation)

or d = 1.21.67 = 0.72 m

Thus the center of gravity is at 0.72 m from the left.

Q6.9 A car weighs 1800kg . The distance between its front and back axles is 1.8m . Its centre of gravity is 1.05m behind the front axle. Determine the force exerted by the level ground on each front wheel and each back wheel.

Answer:

The FBD of the car is shown below :

We will use conditions of equilibrium here :

Rf + Rb = mg

 1800×9.8 = 17640 N ....................................(i)

For rotational equilibrium :

Rf(1.05) = Rb(1.81.05)

or 1.05Rf = 0.75Rb

Rb = 1.4Rf ..............................(ii)

From (i) and (ii) we get :

Rf = 176402.4 = 7350 N

and Rb = 176407350 = 10290 N

Thus force exerted by the front wheel is = 3675 N

and force exerted by back wheel = 5145 N.

Answer:

The moment of inertia of hollow cylinder is given by = mr2

And the moment of inertia of solid cylinder is given by : = 25mr2

We know that : τ = Iα

Let the torque for hollow cylinder be τ1 and for solid cylinder let it be τ2 .

According to the question : τ1 = τ2

So we can write the ratio of the angular acceleration of both the objects.

α2α1 = I1I2 = mr225mr2 = 25

Now for angular velocity,

ω = ωo + αt

Clearly, the angular velocity of the solid sphere is more than the angular velocity of the hollow sphere. (As the angular acceleration of solid sphere is greater).

Q6.11 A solid cylinder of mass 20kg rotates about its axis with angular speed 100rads1 . The radius of the cylinder is 0.25m . What is the kinetic energy associated with the rotation of the cylinder? What is the magnitude of angular momentum of the cylinder about its axis?

Answer:

Firstly we will calculate moment of inertia of the solid cyliner :

Ic= 12mr2

Ic= 12(20)(0.25)2 = 0.625 Kg m2

So the kinetic energy is given by :

Ek=12Iω2=12×(6.25)×(100)2=3125J

And the angular momentum is given by : =Iω=0.625×100=62.5Js

Q6.12 (a) A child stands at the centre of a turntable with his two arms outstretched. The turntable is set rotating with an angular speed of 40rev/min . How much is the angular speed of the child if he folds his hands back and thereby reduces his moment of inertia to 2/5times the initial value? Assume that the turntable rotates without friction.

Answer:

We are given with the initial angular speed and the relation between the moment of inertia of both the cases.

Here we can use conservation of angular momentum as no external force is acting the system.

So we can write :

I1w1 = I2w2

w2=I1w1I2=I(40)25I

w2= 100 rev/min

Q6.12(b) Show that the child’s new kinetic energy of rotation is more than the initial kinetic energy of rotation. How do you account for this increase in kinetic energy?

Answer:

The final and initial velocities are given below :

Ef = 12I2w22 and Ei = 12I1w12

Taking the ratio of both we get,

EfEi = 12I2w2212I1w12

or EfEi= 25×100×10040×40

or EfEi= 2.5

Thus the final energy is 2.5 times the initial energy.

The increase in energy is due to the internal energy of the boy.

Q6.13 A rope of negligible mass is wound round a hollow cylinder of mass 3kg and radius 40cm . What is the angular acceleration of the cylinder if the rope is pulled with a force of 30N ? What is the linear acceleration of the rope ? Assume that there is no slipping.

Answer:

The moment of inertia is given by :

I = mr2

or I= 3×(0.4)2 = 0.48 Kg m2

And the torque is given by :

τ = r×F= 0.4×30 = 12 Nm

Also, τ = Iα

So α = τI = 120.48 = 25 rad/s2

And the linear acceleration is a = αr = 0.4×25 = 10 m/s2

Q6.15 From a uniform disk of radius R , a circular hole of radius R/2 is cut out. The centre of the hole is at R/2 from the centre of the original disc. Locate the centre of gravity of the resulting flat body.

Answer:

Let the mass per unit area of the disc be σ .

So total mass = Πr2σ = m

Mass of the smaller disc = Π(r2)2σ = m4

Now since the disc is removed, we can assume that part to have negative mass with respect to the initial condition.

So, the centre of mass of the disc is given by the formula :

x =m1r1 + m2r2m1 + m2

Or x= m×0  m4×r2m  m4

Or, x= r6

Hence the centre of mass is shifted r6 leftward from point O.

Q6.16 A metre stick is balanced on a knife edge at its centre. When two coins, each of mass 5g are put one on top of the other at the 12.0cm mark, the stick is found to be balanced at 45.0cm . What is the mass of the metre stick?

Answer:

The centre of mass of meter stick is at 50 cm.

Let the mass of meter stick be m.

Now according to the situation given in the question, we will use the rotational equilibrium condition at the centre point of the meter stick.

10g(4512)  mg(5045) = 0

or m = 10×335 = 66 g

Thus the mass of the meter stick is 66 g.

Q6.17 The oxygen molecule has a mass of 5.30×1026kg and a moment of inertia of 1.94×1046kgm2 about an axis through its centre perpendicular to the lines joining the two atoms. Suppose the mean speed of such a molecule in a gas is 500m/s and that its kinetic energy of rotation is two thirds of its kinetic energy of translation. Find the average angular velocity of the molecule.

Answer:

We are given the moment of inertia and the velocity of the molecule.

Let the mass of oxygen molecule be m.

So the mass of each oxygen atom is given by : m2

Moment of inertia is :

I = m2r2 + m2r2 = mr2

or r = Im

or r = 1.94×10465.36×1026 = 0.60×1010 m

We are given that :

Erot = 23Etra

or 12Iω2 = 23×12mv2

or ω = 23×vr

or ω= 6.80×1012 rad/s

Class 11 physics NCERT Chapter 6: Higher Order Thinking Skills (HOTS) Questions

Question 1) A solid sphere of mass M and radius R is rolling without slipping with a velocity v0 on a horizontal surface. It encounters an inclined plane making an angle θ with the horizontal. Assuming no energy loss due to friction, find the maximum height h the sphere will reach before coming to rest.
1) 5v026q2

2) 3v024g

3) 7v0210g

4) v022g

Answer:

For a rolling solid sphere, the total kinetic energy is the sum of translational and rotational kinetic energy:

Etotal =KEtranslational +KErotational 
KEtotal =12Mv02+12Iω2


For a solid sphere, the moment of inertia about its center is:

I=25MR2


Using the rolling without slipping condition, v0=Rω, so:

ω=v0R


Substituting I and ω :

KErotational =12×25MR2×(v0R)2=15Mv02


So, the total kinetic energy:

KEtotal =12Mv02+15Mv02
KEtotal =510Mv02+210Mv02=710Mv02


At maximum height, all kinetic energy converts into gravitational potential energy:

PE=Mgh

Equating energy:

710Mv02=Mghh=7v0210g

Hence, the answer is the option (3).

Question 2) A rod starts rotating with angular velocity ωo but due to viscous force, its angular speed starts decreasing with variable retardation: α=cω Find out the angle traversed by the rod as a function of time.
1) θ=2ω0c[1ect]

2) θ=ω0c[1ect]

3) θ=ω02c[1ect]

4) θ=ω03c[1ect]

Answer:

We know

α=cωdωdt=cωa0ωdωω=0tcdtlnωω0=ctω=ω0ectdθdt=ω0ect0θdθ=ω00tectdtθ=ω0c[1ect]
Hence, the answer is the option (2).

Question 3) A wedge of mass M=2kg carries two blocks A and B, as shown in figure. Block A lies on a smooth-incline surface of the wedge inclined at angle θ to the horizontal. B lies inside a horizontal groove made in the wedge. The two blocks have m=1kg each and are connected by a light string. Length of the groove is l=2m as shown. The entire system is released from rest. What is the distance travelled by the wedge (in m) on the smooth horizontal surface by the time block B comes out of the groove?

Answer:

Let the displacement of the wedge be x towards the right. Displacement of B=(x1) towards right.
Displacement of A w.r.t wedge (on the incline) is l.
Displacement of A in horizontal direction =lcosθ+x
The COM of the entire system suffers no displacement in horizontal direction.

ΔxCOM=0m1Δx1+m2Δx2+m3Δx3=0Mx+m(xl)+m(lcosθ+x)=0x=ml(1cosθ)M+2m
Now, putting m=1 kg,M=2kgI=2 m and θ=60 in (i), we get

x=1×2(1cos60)2+2×1x=0.25 m
Hence, the answer is 0.25 .

Question 4)

A sphere is released on a smooth inclined plane from the top. When it moves down its angular momentum is:
1) Conserved about every point
2) Conserved about the point of contact only
3) Conserved about the centre of the sphere only
4) Conserved about any point on a line parallel to the inclined plane and passing through the centre of the ball.

Answer:

As the inclined plane is smooth, the sphere can never roll, rather it will just slip down. Hence, the angular momentum remains conserved at any point on a line parallel to the inclined plane and passing through the centre of the ball.

Hence, the answer is the option (4).

Question 5) A square plate of mass m and length l is rotated with angular velocity ' w ' with a vertical axis passing through its centre, and it is kept on a rough surface. It takes ' t0 ' time to stop because of uniform friction. Now if we hinge it at the corner & give same angular velocity, In kt0 time it will take to stop, Find k.

Answer:

The torque by friction will be τ = Cμmgl and moment of inertia will be,I=kml2

τ= Iα

Hence, angular acceleration will be

α=Cμgk(1l)

α is inversely proportional to l

Now as it is hinged at one corner, we can assume plate of double side length

Now this is again the central axis, the value of angular acceleration will be half.

Hence, it will take 2t0 time to stop.

Hence, the answer is 2.

NCERT Chapter 6 System of Particles and Rotational Motion Topics

The objects in the world are hardly simply point masses, and objects are composed of a large follow-through of particles. Chapter 6 of NCERT Class 11 Physics System of Particles and Rotational Motion talks about the problem of behavior of these systems.

6.1 Introduction
6.1.1 What Kind Of Motion Can A Rigid Body Have?
6.2 Centre Of Mass
6.3 Motion Of Centre Of Mass
6.4 Linear Momentum Of A System Of Particles
6.5 Vector Product Of Two Vectors
6.6 Angular Velocity And Its Relation With Linear Velocity
6.6.1 Angular Acceleration
6.7 Torque And Angular Momentum
6.7.1 Moment Of Force (Torque)
6.7.2 Angular Momentum Of A Particle
6.8 Equilibrium Of A Rigid Body
6.8.2 Centre Of Gravity
6.9 Moment Of Inertia
6.10 Kinematics Of Rotational Motion About A Fixed Axis
6.11 Dynamics Of Rotational Motion About A Fixed Axis
6.12 Angular Momentum In Case Of Rotation About A Fixed Axis
6.12.1 Conservation Of Angular Momentum

NCERT Solutions for Class 11 Physics Chapter 6: Key Formulas

Get a better grasp of Physics with our class 11 physics chapter 6 solutions. These solutions include important formulas (listed below) and easy-to-follow diagrams. Improve your understanding of the System of Particles and Rotational Motion and perform well in the Chapter 6 physics problems.

1. Centre of mass

For a system of particles: RCOM=mirimi

2. Motion of COM

M=mi
vcm=miviM

pcm=Mvcm

acm=FextM

3. Angular Velocity & Linear Velocity

Average angular velocity =∆θ/∆t

Instantaneous angular velocity, ω=dθ/dt

4. Moment of Inertia (MI)

I=miri2 (Discrete system) I=r2dm (Continuous body) 

5. Moment of Inertia of few useful configurations

6. Torque (Moment of Force)

τ=r×F=rFsinθ

7. Angular Momentum

L=r×p=mr×v
For a rotating body:

L=Iω

8. Torque-Angular Acceleration Relation

τ=Iα

9. Rotational Kinetic Energy

KErot=12Iω2

10. Angular Equations of Motion (Uniform Angular Acceleration)

ω=ω0+αtθ=ω0t+12αt2ω2=ω02+2αθ

Approach to Solve Questions of NCERT Solutions for Class 11 Physics Chapter 6

To solve questions based on NCERT Class 11 Physics Chapter 6: the first step to solving these questions is to gain an understanding of some of the simpler concepts such as the center of mass, the torque, angular velocity, moment of inertia, and the rotational equivalents of the motion. Look keenly to determine whether the problem is a linear motion, a rotational motion or both. Employ the free-body diagrams to depict forces and torque. Use Newton laws, conservation of angular momentum and rotational laws of motion where applicable. In numerical problems make sure to convert the units properly and use of parallel or perpendicular axis theorems to compute moment of inertia. Organization and practice of various problems are important to succeed in this chapter.

What Extra Should Students Study Beyond NCERT for JEE/NEET?

Concept NameJEENCERT
Rotational Motion Of Rigid Body
Center Of Mass
Center Of Mass Of The Uniform Rod
Centre Of Mass Of Semicircular Ring
Centre Of Mass Of Semicircular Disc
Centre Of Mass Of A Triangle
Centre Of Mass Of Hollow Hemisphere
Centre Of Mass Of Solid Hemisphere
Centre Of Mass Of Hollow Cone
Centre Of Mass Of A Solid Cone
Motion Of The Centre Of Mass
Relationship Between Linear And Angular Motion
Torque
Rotational Equilibrium
Work, Energy And Power For Rotating Body
Moment Of Inertia
Moment Of Inertia Of A Rod
Moment Of Inertia Of A Rectangular Plate
Parallel And Perpendicular Axis Theorem
Moment Of Inertia Of A Ring
Moment Of Inertia Of A Disc
Moment Of Inertia Of Hollow Cylinder
Moment Of Inertia Of The Solid Cylinder
Moment Of Inertia Of Hollow Sphere
Moment Of Inertia Of A Solid Sphere
Moment Of Inertia Of Solid Cone
Angular Momentum
Law Of Conservation Of Angular Momentum
Rigid Bodies: Translational Motion And Rotational Motion
Rolling Without Slipping
Rolling Without Slipping On An Inclined Plane

Importance of NCERT Solutions for Class 11 Physics Chapter 6: System of Particles and Rotational Motion

  • Conceptual Clarity: The solutions provide a clear and detailed explanation of important concepts like the centre of mass, moment of inertia, and rotational motion, helping students develop a strong conceptual foundation.
  • Step-by-Step Solutions: Each problem is solved step by step, making it easier for students to understand the methods and approaches used to tackle complex problems.
  • Boosts Problem-Solving Skills: By practising the problems from the solutions, students improve their analytical and problem-solving skills, which are essential for both board exams and competitive tests.
  • Reinforces Core Concepts: The solutions help reinforce key topics such as Newton's laws in rotational motion, torque, and angular momentum, which are essential for mastering the chapter.
  • Effective Exam Preparation: These solutions are designed to help students prepare efficiently for exams by covering all important topics and providing a strong grasp of the subject.
  • Quick Revision: The solutions are concise and well-organized, making them a great resource for quick revision before exams.

In summary, the NCERT solutions for Class 11 Physics help students effectively understand and solve problems in the chapter, strengthening their knowledge and enhancing their performance in exams.

Also, Check NCERT Books and NCERT Syllabus here

NCERT Solutions for Class 11 Physics Chapter-Wise

NCERT solutions for class 11 Subject-wise

Frequently Asked Questions (FAQs)

1. What is the weighage for the Class 11 NCERT chapter 6 for NEET exam?

For NEET exam around 5% of questions are asked from the chapter system of particles and rotational motion. NEET questions can be practiced refering the previous papers of NEET. For more number of  questions refer to NCERT exemplar questions for Rotational Motion.

2. How many questions can be expected from the chapter system of particles and rotational motion for JEE Main exam?

2 questions can be expected from System of Particles and Rotational motion for JEE Main exam. Oscillation and Waves is one of the important topics for exams like KVPY, NSEP and JEE Main

3. How to solve difficult questions related to Chapter 6 of NCERT Solutions system of particles and rotational motion?

To answer difficult questions regarding Chapter 6 of NCERT Solutions for Class 11 Physics, it's essential to carefully read the question, review relevant concepts, break down the question, apply concepts, check your answer, and seek help if needed.

4. What is the Difference Between Work, Energy, and Power in Physics?
  • Work: The transfer of energy when a force acts on an object (W = F × d).
  • Energy: The capacity to do work. It exists in various forms like kinetic and potential energy.
  • Power: The rate at which work is done or energy is transferred (P = W/t).
5. How to Solve Numerical Problems from Class 11 Physics Chapter 6?
  • Identify the given data (mass, velocity, torque, etc.).
  • Apply relevant formulas (e.g., I=mr2 for moment of inertia).
  • Solve step by step, keeping track of units and dimensions.
  • Use the principle of conservation of momentum and energy where applicable.
6. What is the Work-Energy Theorem? Explain with Examples.
  • Work-Energy Theorem states that the work done on an object is equal to the change in its kinetic energy (W = ΔK).
  • Example: When a car accelerates, the work done by the engine increases its kinetic energy, resulting in faster motion.

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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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