Hi everyone in a previous video, I looked through this exam here, which is the
2018 JEE main exam and I just had a little look through and showed you
I guess what the syllabus would involve if you were studying for an exam like this today
I'm going to take the same exam and solve three of the problems in-depth
so it's going to be
the first three problems from the physics section and this is for anyone who was still a bit more curious and wanted to see what
Solving these problems is actually like rather than just seeing what the topics are
I'm recording this commentary after I've already solved the problems and am watching back the footage of how I solved it
This is because I only have one tripod and also I think this way will help me to stay kind of concise in my explanations
The first question here is about collisions and I'll read it out
It is found that if a neutron suffers an elastic collinear collision with deuterium at rest
Fractional loss of its energy is Pd while for it's similar
Collision with carbon nucleus at rest the fractional loss of energy is Pc
So we're being asked to find the values of Pd and Pc
So we need to know how much energy is being lost by this neutron through these two different collisions
Now I'm trying to solve these problems with minimal memorization
So sometimes when you look at these solutions to these JEE exams
It seems like if you had memorized a certain step it's kind of a shortcut to solving the problem like maybe you know
Stuff already about elastic collisions and you've done so many of these problems, you know how they sort of go
But without trying to memorize too much at all let's just go back to the basics. What we will need to know
is that the mass of deuterium will be twice as much as the mass of just the single Neutron because
Deuterium has one proton and one Neutron
and then when we get to working with carbon its mass will be 12 times the mass of the neutron because
Carbon's atomic number is six so it's got six protons six neutrons. We will need to know that in an elastic collision
That kinetic energy is being conserved. And that means that after the collision our two masses are not sticking together
What we can do is write down what the kinetic energy is going to be on both sides
So before and after the collision and we know it's conserved so we can put an equal sign in the middle there
If you cancel out the common terms
Which is 1/2 m in this case you get v1 squared is equal to v2 squared plus 2 v3
squared
So I'm rewriting this equation in terms of v3 and then I'm recognizing
that
v1 squared minus v2 squared can be expanded out into these two brackets (v1-v2) and
(v1+v2) that's going to help us combine it with an equation that we'll find soon
That was all from conserving kinetic energy
now the other thing that's being conserved here is momentum so we can write out that in mv1 is equal to
mv2 plus 2mv3 that's just conservation there then again
Canceling common terms we can get v1 is equal to v2
plus 2v3
We've got two different equations at this point, but three unknowns, so I'm going to do
Simultaneous equations and try and sub in what we found from considering momentum into what we were dealing with before
So I'm basically just rearranging the formulas so that I can sub them in nicely
So doing that allows us to cancel some terms
And we end up with v1 plus v2 is equal to v3
but if we go back and rearrange what we had before
v3 is nothing but v1 minus v2
Over 2 so we can get rid of v3 entirely and just have our equations in terms of v1 and v2
this is good because
being able to relate v1 to v2 will help us find what we really want to know which is the
fractional loss of energy
Finding that fractional loss of energy remember this is just for the neutron we will take the neutrons kinetic energy before
subtract off the neutrons kinetic energy after and divide it all by the neutrons kinetic energy before so
Kinetic energy 1/2 m v squared and actually writing that out and canceling those common terms of 1/2 m
We get v1 squared minus v2 squared
Over v1
squared giving us what we want to find now just sub in our relation between v1 and v2
Do a little bit of arithmetic here and we get that the final answer is 8 over 9. That's pretty good
So now we actually need to do all that again
for a collision with
Carbon the only thing that's going to change is that the carbon is now 12 times the mass of the neutron
So we can might be able to go a bit faster because they're doing it again in this example in my diagram. I'm assuming that
The neutron and the carbon move off together in the same direction
I guess v2 could also go backwards. The neutron could have bounced off and been going in the opposite direction
I worked through the math was that with that as well?
And it doesn't matter whether it bounces off or goes forward because you end up squaring the velocity and the final terms
Really the absolute value of that velocity is all that matters there
So again looking at kinetic energy first
and we can jump some of the steps that we were doing on the previous example to get
V 1 squared equals V 2 squared plus 12 V 3 squared this time and just rearranging it in a similar way
so that we can
Cancel some terms again
Conserving momentum we end up that v1 is equal to v2
Plus 12 v3
rearranged that sub it into our kinetic energy equation we
End up with v1 plus v2 is equal to v3
But the 3 is nothing but v1 minus v2 over 12
So we can get rid of v3 entirely get an expression that just relates v1 and v2
Then our frictional loss of energy
Very similar thing, but now we're subbing in - 13 11
V - as being our value of v1, and this algebra gets a little bit more tricky
But we end up doing some fractions
We get 48 over 169 as our fraction there now because the question is asking
For this in terms of decimals, which makes it harder. We have to work that out as a decimal and that will be
0.28 to two decimal places
So we need to go and find the other one decimals - and that would be zero point eight nine
So then we just need to look at our answers
we have zero point eight nine for the first one point two eight for the second one and that gives us
the third option as being correct there our second question is about
Molecules bouncing off a wall and what pressure that's going to exert on the wall?
so we have the mass of a hydrogen molecule is three point three two times ten to the minus 27 kg if
10 to the 23 hydrogen molecules strike per second a
fixed wall of area two centimeters squared at an angle of 45 degrees to the normal and
Rebound elastically with a speed of 10 to the 3 meters per second
Then the pressure on the wall is merely and were asked to find that value
So the basic ideas here are going to be change in momentum and working out pressure
This is a pretty easy one to remember pressure as force over area
And in terms of like impulse force well force is going to be the change in momentum
over the change in time and
If we're dealing with constant mass
Which I assume these molecules are then that's going to come down to mass times change in velocity over a change in time
Starting off just considering a single molecule. We can draw a diagram
We're coming in at an angle losing all of that momentum and then leaving in the opposite direction
With that momentum again
So because we're being asked about the pressure, that's it exerted on the wall
all we really care about here is the
Horizontal component of what's going on to the horizontal component of the change in momentum or the change in velocity?
To work out that horizontal component, which I'm calling VH we can do V one which has our initial velocity
times the sine of 45
That's the same as doing it
The one divided by the square root of two
Now because we need to work out the final velocity in the horizontal
direction we're actually going to do two times that amount because we're losing at all getting and gaining it all again so our
Change in momentum is going to be two times that value
putting that into our formula for force we get this and remember that
two over the square root of 2 is just equal to square root two and
The time that this is going on over is just one second. So we've got a 1 on the bottom there
Yeah, that was for one molecule. And for n molecules times it by n
That gives us our total force on the wall then define the total pressure on that wall
We want to divide it by our area which was 2 centimeters squared
so just plug in all of our values here most of working this out comes to looking at our like
Exponents, so we have 10 to the 23 a 10 to the 3 that would be 10 to the 26. We've got a minus
27 so we'd have a total of 10 to the minus 1 on the top
We've got a 10 to the minus 4 on the bottom
minus 1
minus negative 4 we're going to end up with 10 to the positive 3 as being the
Exponent on our answer and then working out the rest of it in decimals here. It gives us our final answer
Which corresponds to option number three on our exam?
our last question
Now number three we've got a solid sphere of radius R
made of a soft material of bulk modulus K is
surrounded by a liquid in a cylindrical container a massless piston of area a
floats on the surface of the liquid
covering the entire cross-section of the cylindrical container
When a mass M is placed on the surface of the piston to compress the liquid the fractional decrement in the radius of the sphere
D R over R. Is what?
So essentially we've got a sphere in a container and it's being squashed so a bulk modulus
Which I think is the only thing you'll need to know to get started here
Is that a bulk modulus is the change in pressure?
Over like the fractional change in volume. So I've written the formula here and we're going to want the I guess
Absolute value of it. We also need to know the volume of the sphere here. So
4/3 PI R cubed now
What we want is D R over R, which is relating to our radius and that little fraction there
it looks like we're going to need to differentiate to get there because that little D R is really
Indicating that there's some calculus going on some infinitesimal change
So what we can start with is by differentiating our equation for the volume
And we can do that with respect to the radius so we can find DV
D R as being 4 PI R squared notice that that is the surface
Area of a sphere so actually that makes sense. Now, we've kind of got a DV in our formula for the bulk modulus
So if we have a look at that and take inspiration from there and do a DV over V
we can cancel out quite a few terms actually and we can get left with a formula that has an R and
A d R in it now, that's pretty close to what we want
so if we rearrange for that we end up getting DV over 3v and
If we rearrange our formula for bulk modulus in terms of the change in volume over volume
That would be equal to the change in pressure over the modulus itself
So we have enough we need
To be able to sub things and so pressure is force over area
Our change in pressure is going to be coming from our masse that we've been putting on this
So M times G for our force
Over the area a of that piston and they've given it to us that the bulk modulus is K
Meaning that all this comes to be mg
Over three ka and that is one of our answers. So that's very nice. I
Hope that gave you some more insight into
Solving these problems from my understanding you don't get calculators in this exam
Correct me if I'm wrong, but if you don't get calculators
Actually that makes some of those last steps very difficult those converting fractions to decimals and working out those final decimal answers
So congrats to anyone who can do that and good luck to anyone who's preparing for exams like this. Thanks for watching