15 Min Objective Test series for Mechanics for IITJEE,AIEEE,PMT

Multiple choice question with one or more answer

1. Which of these is true of a conservative force?
a. Workdone between two points is independent of the path
b. Workdone in a closed loop is zero
c. if the workdone by the conservative is positive,its potential energy increases
d. None of the these

2.A particle is acted upon by a force of costant magnitude which is always perpendicular to the velocity of the particle.The motion of the particle takes place in a plane.it follows that
a. velocity is constant
b. accelerattion is constant
c. KE is constant
d. Particle moves in a circular path

3. A package is dropped out of airplane in a level flight.If air resistance could be neglected.which of these is true
a. Package will follow a parabolic path with respect to a observer on the groud
b Package will follow a vertically straight line path with respect to the observer in the plane.
c.Package will follow a vertically straight line path with respect to a observer on the groud
b Package will follow a parabolic path with respect to the observer in the plane.


4.A small mass m is suspended from one end of a vertical string. and then whirled in a horizontal circle at a constant speed v.
Which of the followings is true
a.The strings stays vertical
b.The string becomes inclined to the vertical.
c.There is no force on mass m except its weight
d.The angle of inclination of the string does not depend on the v
e.The centripetal force on m is mg

5.A IITJEE text book of mass M rests flat on a horizontal table of mass m placed on the ground.Let RX->Y be the constant force exerted by the body x on body Y.Which of the following is true for the forces Rground->table and Rtable->ground
a. It is action and reaction pair
b. have equal magnitude
c. Opposite direction
d.have resultant zero


6.The tension in cable supporting an elevator is equal to th weight of the elevetor.The elevator may be
a. going up with incresing speed
b. going down with increasing speed
c. going up with uniform speed
d. going down with uniform speed.


7 A simple pendulum consists of a mass attached to a light string l. if the system is oscillating through small angles which of the following is true
a.The freqiency is independent of the acceleration due to gravity g
b.The period depends on the amplitude of the ocsillation
c.the period is independent of mass m
d. the period is independent of lenght l


8.A ball is thrown into the air with an intial speed u.The time interval taken for the ball to rise to its maximum height is t1.The time interval taken for the ball to fall back from this maximum height is t2.Under real life condition,which of the following is satisfied
a.t2 > t2
b t2 < t2
c. t2=t2
d. t1 > t2 if u is great enough


9. Let v and a denote the velolcity and acceleratio respectivly of the particle in one dimension motion.
a.the speed of the particle decreases when v.a <0
b. the speed of the particle decreases when v.a >0
c.the speed of the particle increases when v.a=0
d. the speed of the particle decreases when |v|<|a|

10. A refrence frame attached to the earth
a.is an inertial frame by defination
b.cannot be inertial frame as earth is resolving around the sun
c.is an intertial frame because Newton's law are applicable in this frame
d.cannot be intertial frame because earth is rotating about its own axis


Solutions
1.a,b
2.c,d
3.a,b
4. b
5. a,b,c,d
6. c,d
7. c
8 a
9 a
10 b,d

Challenging Problems in Mechanics to sharpen your mind

1.A particle of mass M is acted upon by a force that has a intial magnitude F0 when t=0 and decrease at a uniform rate until when t=t1,its magnitude is zero.It is moving accross the X-axis.Find the velocity and coordinate of the particle when t=t2 assuming that t2 > t1. Assume xt=0=0 and (dx/dt)t=0=0

Ans
Velcity=F0t1/2M
Displacement=(F0t1/2M)(t2-t1/3)


2.A wheel of radius r rolls without slip along the x-axis. with constant speed. v0.Find out the motion(velocity and acceleration ) of the point A on the rim of the wheel which starts from the origin O.Take Y axis as perpendicular at X axis at origin

Ans

dx/dt=v0[1-cos(v0t/r)]
dy/dt=v0sin(v0t/r)

d2x/dt2=(v02/r)sin(v0t/r)
d2y/dt2=(v02/r)cos(v0t/r)

3.A body moves under the action of a constant force F through fluid that opposes the motion with a force proportional to the square of the velocity that is Ax2.Show that the limiting velocity is VL=(F/A)1/2.

4.A Bungee Jumper is attached to one end of a long elastic rope.The other end of the elastic rope is fixed to a high bridge.The Jumper steps off the bridge and falls from rest towards the river below..he does not hit the river below.The mass of the jumper is M and length of unstretched rope is L.Force constant of the rope is K. and gravitional field strength is g.Mass of rope is negligible ,air resitance is negligible.
1.Find out the distance y dropped by the jumper before coming instantaneously to rest for the first time
2.Maximum speed attained by the jumper during this drop
3.The time taken during the drop before coming to rest for the first time


Answer
y=[KL+mg+√(2mgKL+m2g2)]/k
v=√(2gL+mg2/k)
t=√(2L/g) + √(m/k)tan-1{-√(2KL/mg)}

Download Ebooks related to physics

1.Fundamentals of Physics by Resnick Haliday:
It is a nice book having good problems.A good no of checkpoint are there to make you aware of the concept.

Download

2. Feynmann Lectures on Physics:
It is a nice book to prepare for olymiads Physics.The lectures in the book are very good.
Download

3. Irodov Problems in Physics

Download

4. Irodov Solutions I and II

Download
Download

5. Benjamin Crowell Books for Physics

Newtonian Mechanics
Conservation Laws
Vibration and Waves
Electricity and Magnetism

IITJEE Papers ,Pattern Analysis And old Physics paper

Question Papers

There will be two question papers of three hours duration, each consisting of separate sections in Physics, Chemistry and Mathematics. Questions in these papers will be of objective type, which are to be answered on a specially designed machine-gradable sheet using HB pencils only. Incorrect answers will be awarded negative marks.

Language of Question Papers

Candidates can opt for Question Papers either in English or in Hindi. The option should be exercised while filling the application form. The option cannot be changed at any later stage.



Formats of papers
The two papers had three different parts each, for physics, chemistry and mathematics respectively.
Each part had four sections. They had the following types of questions:

Straight objective type with One answer
These are multiple choice question pattern with one correct option. The questions are elementary and basic concepts are required to solve the question.. find the answer and pick that from among the choices.


Straight objective type with one or more answer
These are multiple choice question pattern with one or more correct option. The questions are elementary and basic concepts are required to solve the question.. find the answer and pick that from among the choices.

Assertion -- reasoning type.
These are bit tough question.This test the real links between the concepts.
Basically Two statements will be there.One statement will be assertion and One will the reason.You will have to tell the link between them ,followings options will be there
a) Statement I is true ,statement II is true ,statement II is correct explanation for statement I
b) Statement I is true ,statement II is true ,statement II is not a correct explanation for statement I
c) Statement I is true,Statement II is false
d) Statement I is False,Statement II is True

You have to chosse the correct one.These questions test the ability to link two diffrent statement.

Linked comprehension type.
The three questions that followed a small problem description had to be solved on the basis of the given information. It is basically similar long question asked earliar.Basically they have made it in objective pattern.
The student will need to read the paragraph and apply the knowlegde in the subject to solve the question

Matrix -- match type.
These are column based question.You will need to read the item in column I and related it to the items in Column II.There may be two answer in Column II.


Questions in Physics come from all the modules.But there are some module which are easiar to attempt.

Topics which are scoring and should be cover first in Physics
Thermodynamics
Conduction and Convection of Heat
Hydrostatics and Bernoulli's Principle
Waves in Elastic Media
Interference Beats and Doppler's Effect
Electrostatics (full)
Electromagnetic Induction
Lorentz' Forces
Circuits with Capacitors
Modern Physics (full)
Collisions
Rotational Motion
Gravitation
Elasticity.


IITJEE 2008 Physics analysis
Mechanics
Error analysis:1 Straight objective type with One answer,
Gravitation:1 Straight objective type with One answer,2 Assertion Reason questions
Momentum:1 Straight objective type with One or more answer
Collision: Linked Comphrehension Type
Rotational Mechanics: 2 Assertion Reason questions
Fliud Mechanics:1 Assertion Reason questions

Electricty And Magnetism
Electricstatics:1 Linked Comphrehension Type
Magnetic Field:1 Straight objective type with One or more answer
Current Electricity:1 Straight objective type,1 Assertion Reason questions

Modern Physics
X rays:1 Straight objective type.
Binding Energy of Nuclie:2 Straight objective type with One or more answer
Bohr Model:1 Linked Comphrehension Type

Thermodyanamics And Properties of matter
Ideal gas laws:1 Straight objective type.
Properties of matter:1 Linked Comphrehension Type

Optics:1 Straight objective type with One answer,1 Straight objective type with One or more answer


IITJEE solutions for Physics Previous Papers

IITJEE 2008 Physics Part 1
IITJEE 2008 Physics Part 2
IITJEE 2007 Physics Part 1
IITJEE 2007 Physics Part 2
IITJEE 2006 Physics
IITJEE 2005 Physics
IITJEE 2004 Physics
IITJEE 2003 Physics
IITJEE 2002 Physics
IITJEE 2001 Physics

AIEEE solutions for Physics Previous paper


AIEEE 2008 Physics
AIEEE 2007 Physics
AIEEE 2006 Physics
AIEEE 2005 Physics
AIEEE 2004 Physics
AIEEE 2003 Physics

Problem Solving tips for IITJEE Mechanics

Vector Mathematics

You will come across vectors in physics problem very frequently.So it is must to know to solve the vector mathematics in short time.And Making sure you have done in correctly.You will find vectors in every module be it mechanics ,electrostatics,magentics

Addition or Substraction
When you need to add or substract two or more vectors, you can use following procedure

1. Select a coordinate system that is convenient. (Try to reduce the number of
components you need to find by choosing axes that line up with as many
vectors as possible.)

2. Draw a labeled sketch of the vectors described in the problem.

3. Find the x and y components of all vectors

4. Compute the resultants in x and y direction

5. If necessary, use the Pythagorean theorem to find the magnitude of the resultant vector and select a suitable trigonometric function to find the angle
that the resultant vector makes with the x axis.

Products
When u need to perform scalar or vector product.It is nice to following procedure

1. Select a coordinate system(x,y,z) that is convenient. (Try to reduce the number of
components you need to find by choosing axes that line up with as many
vectors as possible.).Use i,j,k as the unit vector across x,y,z axis.

2. Draw a labeled sketch of the vectors described in the problem.

3. Express all the vector in form in i,j,k forms

4. Perform the multipication




Motion in a Two dimensional Plane

- Select a coordinate system and resolve the initial velocity vector into x and y
components.

-Find out acceleration in each direction and solve in each direction according to one rectilinear motion equation.

-if the acceleration is in vertical direction only.Follow the techniques for solving constant-velocity problems to analyze the horizontal motion. Follow the techniques for solving constant-acceleration problems to analyze the vertical motion. The x and y motions share the same time of flight t.

- There might be question about trajortory in the Problem ,find out the motion in x and y direction with respect to time from previous point.And then find the value of t from one equation and then put that value in another equation to find out the equation of trajactory

Motion in a Three dimensional Plane

- Select a coordinate system and resolve the initial velocity vector into x , y and z components

-Find out acceleration in each direction and solve in each direction according to one rectilinear motion equation.



Uniform Circular Motion

-Draw a simple, neat diagram of the system.

-Firstly consider the origin of the forces acting on the each object.To do this find out the field forces acting on the each object.Wherever contact in available account the contact force carefully

-Find out the force acting on the body.The resultant force should provide the required centrepatal required for Circular motion

-Centrepatal force=mv2/R will give the velotiy accordingly.


Newtons Law Problem
-Draw a neat diagram of the system.

-Firstly consider the origin of the forces acting on the each object.To do this find out the field forces acting on the each object.Wherever contact in available account the contact force carefully

-Isolate the body whose motion is to be analyzed. Draw a free-body diagram for this body. For systems containing more than one objects, draw separate free-body diagrams for each objects This way there will not any confusion about the force acting on each object.Newtons Third law will help in obtaining the action reaction pair. Do not include in the free-body diagram forces exerted by the object on its surroundings. Establish convenient coordinate axes for each objects and find the components of the forces along these axes.

- Use pseudo force if viewing from the non intertial frame of refrence

-Apply Newton's second law, F= ma, in component form.

-Solve the component equations for the unknowns. Remember that you must have as many independent equations as you have unknowns to obtain acomplete solution.

-Make sure your results are consistent with the free-body diagram.

- If the object are attached to each others by the strings,then make use of constraints theory to find out the acceleration equation between the objects


Work And Energy

- Choose your frame of refrence.KE will differ in each refrence frame while PE will remains constant

-Define your system, which may include two or more interacting particles, as well as springs or other systems in which elastic potential energy can bestored. Choose the initial and final points.

-Identify zero points for potential energy (both gravitational and spring). If there is more than one conservative force, write an expression for the potential energy associated with each force.

-Determine whether any nonconservative forces are present. Remember that if friction or air resistance is present, mechanical energy is not conserved.

-Determine whether any external forces are present. Remember that external forces are present, mechanical energy is not conserved.

-If mechanical energy is conserved, you can write the total initial energy at some point Ei=PEi+KEi. Then, write an expression for the total final energy at the final point that is of interest Ef=PEf+KEf. Because mechanical
energy is conserved, you can equate the two total energies and solve for the quantity that is unknown.

-If frictional forces are present (and thus mechanical energy is not conserved), first write expressions for the total initial and total final energies. In this case, the difference between the total final mechanical energy and the total initial mechanical energy equals the change in mechanical energy in the system
due to friction.

-If external forces are present (and thus mechanical energy is not conserved), first write expressions for the total initial and total final energies. In this case, the difference between the total final mechanical energy and the total initial mechanical energy equals the change in mechanical energy in the system
due to external force.

-If both external forces and frictional forces are present (and thus mechanical energy is not conserved), first write expressions for the total initial and total final energies. In this case, the difference between the total final mechanical energy and the total initial mechanical energy equals the change in mechanical energy in the system due to external force and friction


Momentum and Collisions

- Choose your frame of refrence

-Set up a coordinate system and define your velocities with respect to that system.It is usually convenient to have the x axis coincide with one of the initial velocities.

-In your sketch of the coordinate system, draw and label all velocity vectors and include all the given information.

-Write expressions for the x and y components of the momentum of each object before and after the collision. Remember to include the appropriate signs for the components of the velocity vectors.

-Write expressions for the total momentum in the x direction before and after
the collision and equate the two. Repeat this procedure for the total momentum
in the y direction. These steps follow from the fact that, because the momentum of the system is conserved in any collision(law on conservation of linear momentum), the total momentum along any direction must also be constant. Remember, it is the momentum of the system that is constant, not the momentan of the individual objects.

-If the collision is inelastic, kinetic energy is not conserved, and additional information is probably required. If the collision is perfectly inelastic, the final velocities of the two objects are equal. Solve the momentum equations for the unknown quantities.

-If the collision is elastic, kinetic energy is conserved, and you can equate the
total kinetic energy before the collision to the total kinetic energy after the
collision to get an additional relationship between the velocities.And you can solve the energy and momentum equation to find out the find velocities.

- Centre of mass can be useful feature in solving the momentum problems


Body in Static equilibrium

-Draw a neat diagram of the system.

-Firstly consider the origin of the forces acting on the each object.To do this find out the field forces acting on the each object.Wherever contact in available account the contact force carefully


-Isolate the object being analyzed. Draw a free-body diagram and then show
and label all external forces acting on the object, indicating where those forces are applied.This way there will not any confusion about the force acting on each object.Newtons Third law will help in obtaining the action reaction pair. Do not include forces exerted by the object on its surroundings.(For systems that contain more than one object, draw a separate free-body diagram for each one.) Try to guess the correct direction for each force. If the direction you select leads to a negative force, do not be alarmed; this merely means that the direction of the force is the opposite of what you guessed.

-Establish a convenient coordinate system for the object and find the components of the forces along the two axes. Then apply the first condition for equilibrium. Remember to keep track of the signs of all force components.

-Choose a convenient axis for calculating the net torque on the object. Remember that the choice of origin for the torque equation is arbitrary; therefore, choose an origin that simplifies your calculation as much as possible.
Note that a force that acts along a line passing through the point chosen as
the origin gives zero contribution to the torque and thus can be ignored.

Thermodynamics Numerical

1.An ideal diatomic gas is enclosed in an insulated chamber at temperature 300K. The chamber is closed by a freely movable massless piston, whose initial height from the base is 1m. Now thegas is heated such that its temperature becomes 400 K at constant pressure. Find the new height of the piston from the base.If the gas is compressed to initial position such that no exchangeof heat takes place, find the final temperature of the gas.


2.A cylinder of mass 1 kg is given heat of 20000J at atmospheric pressure. If initially temperature of
cylinder is 200C, find
(a) final temperature of the cylinder.
(b) work done by the cylinder.
(c) change in internal energy of the cylinder.
(Given that Specific heat of cylinder= 400 J kg-1 0C–1, Coefficient of volume expansion
= 9 × 10-5 °C-1, Atmospheric pressure = 105 N/m2 and Density of cylinder = 9000 kg/m3)


3.A gas container has a walls that can bear maximum pressure of 2.2*106 pa.The container contains gas at 4*105 pa and 350K.If the container is steadly heated then neglecting change in volume .Calculate the temperarture at which container will break?

4.A vessel of volume V=20L contains a mixture of hydrogen and helium at T=20° C and Pressue P=2.0 atm.The mass of the mixture is equal to m=5g.Find the ratio of the mass of the hydrgen to that mass of the helium in the mixture Given R=.082 L.atm.mol-1.K-1

5.The first law of thermodynamics dU=dQ-dW,indicates that when a system goes from its intial state to a final state
a. dU is same for every path
b. dQ is same for evry path
c dQ+dW is same for every path
d dQ-2dW is same for every path

Physics Champion Contest-2

Welcome to the Physics Champion Contest!!!!!!!!!!!

We present the First Physics Champion Contest


Physics Champion Contest-2

Winner will get stationary gift pack worth Rs 200.!!!!!!!!!!!!!!!!!!!!!

Procedure
1 This contest is for all the students studying in 10th to 12th ,students preparing Engineering and Medical examination
2.There are 5 multiple choice questions.You have pick the correct choice ,send to us in Format given below
3.You will be choosen for lucky draw only when all the question are correct.
4. Winner will be choosen from the lucky draw from All those candidate who give all write answer
5. In the format given below,you have to also write the name of the friends who reffred you this site
There is lot of importance of this field.Let me explain that
Based on the data provided by you all,we will make a list like this

Candidate:No of times it is refrered

If that candidate has sent the entry for the contest and also if he qualifies for the lucky draw,then his name will be

entered in the lucky draw as many times as he was refrred.
So students you good chance of winning,if you reffred this site to more and more people

6 You have to send the answer and the information to the following mail address in following format


TO address: contest2win_phygoeasy@rediffmail.com
Subject:Physics Champion Contest-I

Body:

About Yourself
Name:
Father's Name:
ADDRESS :
Refrred by:
Mail address of Refrred by:


Answers
1.
2.
3.
4.
5.

Thanks


7 This contest is valid from 23 July 2008 to 15 June 2008.Winner will be announced on the site on 30 Sep 2008.
8.I will cover the cost of shipping. There is no cost to you.


Contest Questions

Multiple choice question with one or more answer

1. A block of mass M is pulled along a horizontal friction surface by a rope of mass m. If a force P is applied at the free end of the rope, the force exerted by the rope on the block is
a)Pm/m+M
b)P
c)PM/m+M
d)Pm/M-m

2.A spring balance is attached to the ceiling of a lift. A man hangs his bag on the spring and the spring reads 49 N, when the lift is stationary. If the lift moves downward with an acceleration of 5 m/s2, the reading of the spring balance will be
(A) 24 N
(B) 74 N
(C) 15 N
(D) 49 N

3. Which of the following functions represent a travelling wave
a. y=pcos(qx)sin(rt)
b. y=psin(qx+rt)
c. y=psin(qx-rt)
d. none of the above

4.There are two statement about Ideal gases
A. The Vrms of gas molecules depends on the mass of the gas molecule and the temperature
B. The Vrms is same for all the gases at the same temperature
which one of the following is correct
a. A and B both
b. A only
c B only
d. A and B both are incorrect

5. A particle moves in a straight line according to
x=t3-4t2+3t

Find the acceleration of the particle at displacement equal to zero
a.(-8,-2,10)
b. (-1,-2,10)
c. (8,2,10)
d. (1,2,10)

Thermodynamics IITJEE test series -2

Mulitiple Choice questions with Single answer

1. Two rigid boxes containing different ideal gases are placed on a table. Box A contains one mole of nitrogen at temperature T0, while Box B contains one mole of helium at temperature (7/3) T0.The boxes are then put into thermal contact with each other and heat flows between them until
the gases reach a common final temperature. (Ignore the heat capacity of boxes). Then, the finaltemperature of the gases, T, in terms of T0 is
a)T=5/2T0
b.T=3/7T0
c.T=7/3T0
d.T=3/2T0

2.According to Newton’s law of cooling, the rate of cooling of a body is proportional to (`T)n, where `T is the difference of the temperature of the body and the surroundings, and n is equal to
(A) two
(B) three
(C) four
(D) one


3.The pressure of the gas in constant volume gas thermometer are 80 cm,90cm and 100cm of mercury at the ice point,the steam point and in a heated wax bath resp.Find the temperature of the wax bath
a)200° C
b)201° C
c)202° C
d) none of the above

4.The pressure P of one mole of gas varies with volume V as
P=4-3V2
Highest absolute temperature to gas can be heated is
a.16/9R
b.9/16R
c 1/16R
d None of the above

5.A sphere,a cube and a circular plate have the same mass and are made of same material.All of them are at the same temperatrure T.Which one will have maximum rate of cooling.
a. Sphere
b Cube
c. Circular plate
d. All will have same rate of cooling

Mulitiple Choice questions with More than one answer

6.For an ideal gas
a) Heat added or Heat taken is not null in adiabatic process
b) Internal energy does not change in the isothermal process
c) Change is internal energy is zero in the cyclic process
d. None of the above

7).In which of the following process,convection does takes place
a. sea and land breeze
b.boiling of the water
c.heating air around the furnace
d.warming of glass of bulb due to filament

8)if a gas has n degrees of freedom.Then the expression (Cp/Cp) +1 is equal to
a) 2(1/n +1)
b) 2(n + 1/2)
c) 2+n/2
d) none of the above

Assertion and Reason
a) Statement I is true ,statement II is true ,statement II is correct explanation for statement I
b) Statement I is true ,statement II is true ,statement II is not a correct explanation for statement I
c) Statement I is true,Statement II is false
d) Statement I is False,Statement II is True


9. STATEMENT 1:The molar specfic heat capacity of the ideal gas in isothermal process is infinity
STATEMENT 2:Heat transfer is non zero in the Isothermal process

10.STATEMENT 1:Equation of state for a real gas is (P+a/V2)(V-b)=nRT

STATEMENT 2: Molecular attraction is not neglible and the size of molecules are not neglibile in comparison to average sepeartion between them

11.STATEMENT 1:Root mean square velocity of the gas does not change in a Isothermal process
STATEMENT 2:Isothermal process may be achieved by immersing the system in large reservior and performing the process slowly


Matrix Match type
12 Three identical cylinder A1,A2,A3 contains equal moles of Helium,Nitrogen,Carbon dioxide at the same temperature

Match the column I to column II

A)Traslational Energy is maximun in the cylinder
B)Total energy is minimum in the cylinder
C)Root mean square velocity is maximum in the cylinder
D)Mean speed will be minimum in the cylinder

P) A1
Q) A2
R) A3
S) No appropiate match given


Linked Comphrehension type

A box Of interior Volumme V1 has a heavy airtight hinged lid of mass M and area A.The box contains n1 moles of gas at Temperature T0.The box is inside a chamber which also contains additional n2 moles of the same gas at the same temperature..The gas in the chamber occupies the volume V2.

13 find the pressure of the gas in the box
a)n1RT0/V1
b)n2RT0/V2
c)(n1+n2)RT/(V1+V2)
d) none of the above.

14 find the pressure of the gas in the chamber
a)n1RT0/V1
b)n2RT0/V2
c)(n1+n2)RT/(V1+V2)
d) none of the above.

15.Intailliy the hinged lid is closed.Then which of these expression is true
a.(n1RT0/V1) -(n2RT0/V2) <=Mg/A
b. (n1RT0/V1) -(n2RT0/V2) => Mg/A
c. (n2RT0/V2) -(n1RT0/V1) => Mg/A
d. (n2RT0/V2) -(n1RT0/V1) <= Mg/A

16.If the whole system is heated,at what temperature T will the gas pressure lift the hinged lid
a)T=Mg/RA(n1/V1 -n2/V2)
b)T=Mg/RA(n2/V2 -n1/V1)
c)T=Mg/RA(n1/V1 +n2/V2)
d) None of the above

Physics Champion Contest -1 Winner

Hi All

We receive lot of entries for the contest.Thanks all for the Response

Answer to the questions are
1.A
2. A
3.A and C
4. A,B,C
5.B

The winner of the contest is Muni Agarwal from Nainital
He will soon recive the stationary Gift worth RS 200

Thanks to all overwhelming response!!!!!

Mechanics Test Series -II

Multiple choice question with one or more answer
1. A block of mAss M is moving with a velocity v on straight surface.What is the shortest distance and shortest time in which the block can be stopped if μ is coefficent of friction
a.v2/2μg,v/μg
b. v2/μg,v/μg
c.v2/2Mg,v/μg
d none of the above


2.A horizontal force of F N is necessary to just hold a block stationary against a wall. The coefficient of friction between the block and the wall is μ. The weight of the block is
a.μF
b. F(1+μ)
c. F/μ
d none of these


3.Consider the following two statements.
STATEMENT 1 Linear momentum of a system of particles is zero.
STATEMENT 2 Kinetic energy of system of particles is zero.
(A) A does not imply B and B does not imply A.
(B) A implies B but B does not imply A
(C) A does not imply B but b implies A’
(D) A implies B and B implies A.


4.The Position vector of the center of mass of uniform semi circular ring of radius R and Mass M whose center coincided with the origin
a.r=(2R/π)j
b.r=(R/π)j
c.r=(4R/π)j
d. none of the these

5.A body is sliding down a rough inclined plane of angle of inclination θ for which coefficent of friction varies with distance y as μ(y)=Ky where K is constant.Here y is the distance moved by the body down the plane.The net force on the body is zero at A.Find the value of constant K
a. tanθ/A
b. Acotθ
c. cotθ/A
d. A tanθ

6.Chosse the correct option
a.if Workdone by the conservative force is positive then Potential energy decreases
b. Rate of change of momentum of many particles system is proportional to net external force on the system
c.The workdone by the conservative force in closed loop is zero
d. None of the above

7.Two electrons(e) are the at the point (-a,0) and (a,0).Their mass is m.They are released from rest.The acceleration of the center of mass of the system when the electron at 4a distance apart
a. e2/64mπε0a2
b. zero
c. e2/16mπε0a2
d. none of the above

8.The potential energy of a certain particle is given by
U=20x2+35z3.Find the vector force on it
a. -40xi-105z2k
b. 40xi-105z2k
c.-10xi-105z2k
d 40xi+105z2k

Matrix Match type

9.
Column I
a. Frictional force
b. Gravitational force
c. Electrical force
d Viscous force

Column II
P. Workdone by the force in closed loop is zero
Q. Workdone by the force in closed loop is not zero

Mechanics Test Series -II

Multiple choice question with one or more answer
1. A block of mAss M is moving with a velocity v on straight surface.What is the shortest distance and shortest time in which the block can be stopped if μ is coefficent of friction
a.v2/2μg,v/μg
b. v2/μg,v/μg
c.v2/2Mg,v/μg
d none of the above


2.A horizontal force of F N is necessary to just hold a block stationary against a wall. The coefficient of friction between the block and the wall is μ. The weight of the block is
a.μF
b. F(1+μ)
c. F/μ
d none of these


3.Consider the following two statements.
STATEMENT 1 Linear momentum of a system of particles is zero.
STATEMENT 2 Kinetic energy of system of particles is zero.
(A) A does not imply B and B does not imply A.
(B) A implies B but B does not imply A
(C) A does not imply B but b implies A’
(D) A implies B and B implies A.


4.The Position vector of the center of mass of uniform semi circular ring of radius R and Mass M whose center coincided with the origin
a.r=(2R/π)j
b.r=(R/π)j
c.r=(4R/π)j
d. none of the these

5.A body is sliding down a rough inclined plane of angle of inclination θ for which coefficent of friction varies with distance y as μ(y)=Ky where K is constant.Here y is the distance moved by the body down the plane.The net force on the body is zero at A.Find the value of constant K
a. tanθ/A
b. Acotθ
c. cottanθ/A
d. Atantanθ

6.Chosse the correct option
a.if Workdone by the conservative force is positive then Potential energy decreases
b. Rate of change of momentum of many particles system is proportional to net external force on the system
c.The workdone by the conservative force in closed loop is zero
d. None of the above

7.Two electrons(e) are the at the point (-a,0) and (a,0).Their mass is m.They are released from rest.The acceleration of the center of mass of the system when the electron at 4a distance apart
a. e2/64mπε0a2
b. zero
c. e2/16mπε0a2
d. none of the above

8.The potential energy of a certain particle is given by
U=20x2+35z3.Find the vector force on it
a. -40xi-105z2k
b. 40xi-105z2k
c.-10xi-105z2k
d 40xi+105z2k

Matrix Match type

9.
Column I
a. Frictional force
b. Gravitational force
c. Electrical force
d Viscous force

Column II
P. Workdone by the force in closed loop is zero
Q. Workdone by the force in closed loop is not zero

Momentum and Center of Mass

Linear Momentum is defined as
p=mv

where v is the velocity of the particle

-Momentum is a velocity vector
-Momentum of system of particles is the vector sum of individaul momentum of the particle
ptotal=∑viMi



Center Of Mass: Rcm=∑riMi/∑Mi

In co -ordinates system
xcm=∑xiMi/∑Mi

ycm=∑yiMi/∑Mi
zcm=∑ziMi/∑Mi

Velocity of CM=∑viMi/∑Mi

Acceleration of CM=∑Fext/∑Mi


Law Of conservation of Momentum:
if ∑Fext=0

then ∑viMi=constant

Work And Energy

Work:- Work done by the force is defined as dot product of force and displacement vector
- For constant Force
W=F.s

where F is the force vector and s is displacement Vector
-For variable Force

dW=F.ds

or W=∫F.ds

-It is a scalar quantity

Conservative And Non Conservative Forces
-If the workdone by the force in a closed path is zero,Then it is called conservative Force
-If the workdone by the force in a closed path is not zero,Then it is called non conservative Force
-Conservative Forces are gravtitional ,electrical force
-Non Conservative Forces are friction

Kinetic Energy:-It is the energy possesed by the body in motion
-it is defined as
K.E=(1/2)mv2
- Networkdone by the external force is equal to the change in the kinetic energy of the system
W=Kf-Ki


Potential Energy:
-It is the kind of energy possesed due to configuration of the system
-It is due to conservative force
-it is defined as

dU=-F.dr
Uf-Ui=-∫F.dr
Where F is the conservative force

F=-(∂U/∂x)i-(∂U/∂y)j-(∂U/∂z)k

For gravtitional Force
Change in Potential Energy =mgh

where h is the height between the two points


Mechanical Energy is defined as
=K.E+P.E

Law Of conservation of Energy

In absence of external forces,intenal forces being conservative ,total energy of the system remains constant.
K.E1+P.E1=K.E2+P.E2

Uniform Circular Motion,Friction And Frame Of refrence

Uniform Circular Motion
-A body moving with constant speed on the circle
-Acceleration is Directed towards centre
a=v2/r

Where v is the speed and r is the radius of the circle

-Centrepetal force is required to move the body in the circle
F=mv2/r




Friction:

Frictional force acts between the bodies whenever there is a relative motion between them.
When bodies do the slip ,frictional force is called static frictional force and when the bodies does not slip ,it is called kinetic frictional force.

Kinetic Frictional force:-When bodies slip over each other

f=μkN


Where N is the normal contact force between the surface
And μk is the coefficent of kinetic Friction

Direction of frictional force is such that relative slipping is opposed by the friction

Static Frictional force:

-Frictional force can also act even if there is no relative motion.Such force is called static Frictional force
-Maximum Statix friction that a body can exert on other body in contact with it is called limiting Friction.
fmaxsN

Where
N is the normal contact force between the surface
And μs is the coefficent of static Friction
fmax is the maximum possible force of static Friction


Some Imp points
1. μs > μk
2 Angle of friction tanλ=μs


Frame Of refrence:

Inertial Frame Of refrence:

-Inertial frame of refrences are objects which are at rest or moving at constant Velocity
-Exmaple:A person standing in a train moving at constant velocity
-Newtons law are valid


Non Inertial Frame Of refrence:-Inertial frame of refrences are objects which are accelerated
-Exmaple: A person standing in a train moving with increasing speed.
-Newtons law are not valid.

To apply newtons law ,pseudo force has to be introduced in the equation whose value will be
F=-ma

where a is the accelration of the non inertial frame of refrence

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