IIT-JEE problems

[pauldmps]

Questions from class XI syllabus:

Chemistry: For the reaction : CO(g) + H2O(g) <------> CO2(g) + H2(g) at a given temperature, the equilibrium amount of CO2(g) can be increased by :

A. Adding a suitable catalyst
B. Adding an inert gas
C. Decreasing the volume of the container
D. Increasing the amount of CO(g)
[One or more than one option may be correct]


Physics: Two blocks of masses 3kg & 6kg respectively are placed on a smooth horizontal surface. They are connected by a light spring of spring constant k= 200 N\m. Initially the spring is unstretched. A velocity of 1 m\s is imparted to the 3kg block & 2 m\s to the 6 kg block in the opposite directions which is also opposite to the side on which the spring is attached on both blocks.

The maximum extension in the spring would be:

A. 30cm B. 25cm C. 20cm D. 15cm [Only one option correct]


Maths: If x,y,z are in H.P. ,then ln(x+z) + ln(x-2y+z), (where "ln" is natural logarithm) is equal to :

A. ln(x-z) B. 2ln(x-z) C. 3ln(x-z) D. 4ln(x-z) [Only one option correct]

[nims11]

^^

Spoiler:

Chemistry : seems like 'D' is the only correct answer
Physics : 15cm
Maths : ln(x-z)
are these correct?

I will also try post some good questions i come across.

[Jaskanwar Singh]

ok nims11 and pauldmps. i got that.

thanks asingh. yes it tends to forget if you are not in touch. so not a problem for you now. and we all will keep this up and ready with questions.

i am solving those questions. but please all of you from now on put answers in spoiler tags. it spoils the fun.

[nims11]

see if you are good at Maths and Physics then do not study for CBSE exams now. Study for exam only in March. I was also good at PCM and mostly at P and M. I prepared for exam in March and got 97 in P and 95 in M. For chem you have to study hard. Because ques. In chem are more theory oriented. So, prepare for chem only. Also at present do Organic only. Those reaction based questions are frequent in JEE. Good luck!!

Thanx for the suggestions!! i am currently putting my all into organic chem and i have decided to fill in all my gaps and holes in it by 15th.
i am not taking too much worry about boards until 20th. And the CBSE preparations will be along with my JEE preparation. CBSE Chemistry and Physics syllabus is easy and i have 11 days gap for maths which is more than enough. Also my 5th subjct is C.Sc in which i dont have to give much efforts.

ok nims11 and pauldmps. i got that.

thanks asingh. yes it tends to forget if you are not in touch. so not a problem for you now. and we all will keep this up and ready with questions.

i am solving those questions. but please all of you from now on put answers in spoiler tags. it spoils the fun.

Sorry. i will take care of that.

[Jaskanwar Singh]

Spoiler:

Chemistry D
solving physics now
i have to do log still in mathematics.

[pauldmps]

Solutions:

Spoiler:

Physics: A. 30cm
At max. extension, the velocity of both the blocks are same.

Applying conservation of linear momentum,

(3+6)v = {(6X2) - (3x1)} = 9
=> v = 1 m/s
Let 'x' be the max. extension in the spring:
Applying conservation of Energy,

(1/2)(3)(1)^2 + (1/2)(6)(2)^2 = (1/2)(200)(x)^2 + (1/2)(9)(1)^2
=> x= 0.3m = 30cm




Maths: B. 2ln(x+z)

x,y,z are in H.P => 1/x,1/y,1/z are in A.P
=> 2/y = 1/x + 1/z
=> 2xz = y(x+z)

Now, ln(x+z) + ln(x-2y+z) = ln{ (x+z)^2 - 2y(x+z)}
= ln{(x+z)^2 - 4xz}
= 2ln(x-z)

Chemistry: Option D. only
Since Δn = 0, addition of inert gas does not change the equilibrium.

[Jaskanwar Singh]

i too got 15m in physx

ok tell me whats wrong with my method in physx -

consider extension produced be x. then force that pulls back the bodies will be Kx. let retardation of 3kg mass be r and 6kg mass be r'. and let extension produced by 3kg be a and by 6kg be b. then 2ra = 1^2 and 2r'b = 2^2. (v^2 - u^2 = 2as) and final velocity after reaching max extension = 0 and moreover Kx = 3r and Kx = 6r'. x=a+b solve these equations. i got 15m.

[nims11]

OH NO....I had interchanged the initial velocities of both the blocks, its now coming to 30m. Reading the question properly is very important.

[pauldmps]

i too got 15m in physx

ok tell me whats wrong with my method in physx -

consider extension produced be x. then force that pulls back the bodies will be Kx. let retardation of 3kg mass be r and 6kg mass be r'. and let extension produced by 3kg be a and by 6kg be b. then 2ra = 1^2 and 2r'b = 2^2. (v^2 - u^2 = 2as) and final velocity after reaching max extension = 0 and moreover Kx = 3r and Kx = 6r'. x=a+b solve these equations. i got 15m.

You cannot use v^2 - u^2 = 2as as retardation is not constant.

[nims11]

consider extension produced be x. then force that pulls back the bodies will be Kx. let retardation of 3kg mass be r and 6kg mass be r'. and let extension produced by 3kg be a and by 6kg be b. then 2ra = 1^2 and 2r'b = 2^2. (v^2 - u^2 = 2as) and final velocity after reaching max extension = 0 and moreover Kx = 3r and Kx = 6r'. x=a+b solve these equations. i got 15m.

you are ignoring the fact that the retardation is not constant and it varies with x.
Also the the final velocity of the two blocks wont be zero.
At max extension, the velocity of the blocks will be same and equal to the velocity of the center of mass(1 m/s for the given question) of the system. velocity of center of mass remains the same throughout the motion, so calculating it is easy.
in other words, the relative velocity of the center of mass and the two blocks is zero with respect to each other.
using conservation of momentum will also lead to the final velocity of the blocks as the velocity of the center of mass. energy and momentum conservation is more handy in such cases. if you want proceed your way, you will need to do some calculus which is undesirable in objective-questions based exams.

BTW do you guyz know a shortcut(Cutting the spring method) for solving these kinds of questions? its really handy and allows us to observe each block independently and then combine the result. i use it frequently and is really helpful in saving time.

[pauldmps]

^^ It is called "reduced mass concept" in which one side of the spring can be fixed while the other side is attached to a single block of resultant mass of the two blocks (not equal to sum of masses) & the velocity of the blocks is replaced by relative velocity between the blocks. The answer comes out to be the same.

[abhijangda]

Oh so many post in such a short time.
Ok solve this now

Q. In triangle ABC, the equation of the perpendicular bisector of AC is 3x-2y+8=0. If A=(1,-1) and B=(3,1), find the equation of BC and the coordinates of the circumcentre??

[nims11]

^^ It is called "reduced mass concept" in which one side of the spring can be fixed while the other side is attached to a single block of resultant mass of the two blocks (not equal to sum of masses) & the velocity of the blocks is replaced by relative velocity between the blocks. The answer comes out to be the same.

its not the reduced mass concept i am talking about. it is a different one.
in the "cutting the spring" method, the spring is divided in two parts across the center of mass. the blocks are given the velocities relative to the the center of mass.
if K is the spring constant of the original spring, then the K of the new spring is = K*(ratio of the original and new length of the spring).
the centre of mass behaves as a fixed wall and the two blocks can be observed independently. It might be unclear what i said. But i will be happy to explain if you guyz want to know it.

[pauldmps]

@abhijangda

Coordinate is my weak point. Is the equation for BC

Spoiler:

4x -5y - 7 =0

? Sorry I couldn't find the circumcentre.

@nims11

Got it.

[Jaskanwar Singh]

OH I SEE. Thanks guys. Nims i will love that shortcut. Tell me.

Abhijanga-

Spoiler:

Circumcenter (-4/5,14/5)

[nims11]

Spoiler:

circumcentre - -4/5,14/5
BC - 29y-16x+18

my answers are strange so they might be wrong...

[Jaskanwar Singh]

Sorry couldnt find bc eqn.

Nims our centres match.

[nims11]

Then perhaps our centers are correct.
Is the equn. for BC is

Spoiler:

x+y=4

[abhijangda]

@abhijangda

Coordinate is my weak point. Is the equation for BC

Spoiler:

4x -5y - 7 =0

? Sorry I couldn't find the circumcentre.

@nims11

Got it.

calculus, co-ordinate and algebra are important units for maths. They should be practiced very much and specially calculus.
Also answer to previous question is
circumcentre
(-4/5,14/5)
and
equation of BC is
x + 4y = 7. This was not a difficult one i would say. Solution can be done in this way. Assume C(x1,y1) then find midpt of AC by midpt formula then this midpt should lie on perpendicular bisector of AC. Also slope of AC multiplied by its perpendicular bisector is -1. Solve these equations to get pt C. I guess finding circumcentre easy.
I will post next question shortly.

Here's another one,
Q. The extremities of a diagonal of a square are (1,1) and (-2,-1). Find the other vertices and the equation of the other diagonal.
Solve it.

[nims11]

Spoiler:

eq of the diagonal - 4y+6x+3=0;
other points - (1/2,-3/2) and (-3/2,3/2)

[Jaskanwar Singh]

Abhijangda -

Spoiler:

(-7,39/4)
(6,-39/4)
6x+4y+3=0

[nims11]

OK... here's a physics question which i came across yesterday.

A string is wrapped around a cylinder of mass M and radius R. The string is pulled vertically upward to prevent the center of mass from falling as the cylinder unwinds the string.If the length of the string unwound when the cylinder has reached the angular speed 'w' is (Rw)^2/(Z*g). what is the value of Z?(answer in 0-9).

[Jaskanwar Singh]

Nims-

Spoiler:

Z=2

[pauldmps]

@nims11

Spoiler:

z=4

[nims11]

@paul right!!
@Jaskanwar

Spoiler:

Let the force on the cylinder applied through the string be T.
so according to the question, T=mg
torque=Iα=TR
α=TR/I
given that final angular speed=w
angular distance=(w^2)/2α=(w^2)I/2TR
putting I=(MR^2)/2 and T=mg
angular distance=(w^2)R/4g
thus, distance=angular distance * r=(Rw)^2/(4g)

I think you took I=MR^2 as i didnt specify the cylinder as a "solid cylinder". But if not specified in any question, it is to be taken as a solid one.

[Jaskanwar Singh]

oh another mistake.

tell me whats wrong with this now -

S=1/2gt^2 and v=gt. v=wr. s=(Rw)^2/(Z*g). t is same at that instant..

[nims11]

from what i can make out, you are applying kinematics on a string which is massless, which is against the rules....

Actually i am confused now after seeing your solution, lets see what paul has to say about it.

[Jaskanwar Singh]

that could be a reason then i suppose. lets wait for paul and abhijangda.

[pauldmps]

Ok so this is how I did it:

Spoiler:

Since the C.M of the cylinder is at rest, net force acting on the cylinder is zero.

For vertical equilibrium:

T (upwards due to string) = Mg (downwards)

However since T does not pass through the C.M of the cylinder, there is a net Torque due to T. The net torque due to mg is zero.

Torque = Iɑ = TR
=> ɑ = TR/I = MgR/I
=> ɑ = 2MgR/MR^2
=> ɑ = 2g/R

Using w = w(initial) + ɑt (t is time)
=> w = 0 + 2gt/R = 2gt/R
=> t = Rw/2g

Using θ = θ(initial) + 1/2ɑt^2
=> θ = 1/2 * 2g/R * (Rw/2g)^2
=> θ = R*(w)^2/(4*g)

reqd expr. = s = θR
=> (R*w)^2/4g

Hence Z = 4

[Jaskanwar Singh]

^but whats wrong with my method?