Analogy 1:Repose Theorem

The questions we solve are too analogous..half the questions in mechanics can be solved using analogies..calling these analogies 'tricks' would be a crime! they are observations made by painfully analysing a set of problems based on a single concept...

Let me show you something:
Posted alongside are three different scenarios..encountered in different contexts. All lead to the same answer. Why should they not???

Let us denote the angle or repose of the incline in each case by 'R'..
in each case the block is to be kept at rest wrt. the incline. i want you to solve for each case:
1)the min
2)the max
3)and the optimum (when no friction acts)
value of the physical qty. specified (under the constraint of no relative motion b/w the block and the incline)..and try to observe.

Obviously i have left some thinking to you..this will take some effort!!

take you time..anyways, i won't be online for a week. next post next Sunday!
(in the third diagram the accn. of the wedge is maintained const. at 'a')

Solution 7

The height rise due to surface tension can be usually calculated using the fact that the drop in P on crossing a spherical meniscus of radius R is 2T/R...

But what if you don't know R..or what if the meniscus is not spherical???

You can always calculate the rise using the fact that the weight of water in the capillary is supported entirely by the forces of Surface Tension.

We have:T(2pi*b)+T(2pi*a)=H(pho)(pi(b^2)-pi(a^2))
weight of water

You can calculate H...you can also enlarge the img. by clicking on it!

Solution 6

Not many tried this one..you people seem to be more interested in mechanics!!

"The flux thru system 1 due to current 'i' in system 2 is same as flux thru system 2 due to current 'i' in system 1"...this is the principle of reciprocity (a term i learned in iit) in my own words...

To find the flux thru the inf. plane..you may as well find it thru the inf. loop that i have shown in the dig..and you may assume the current to be flowing thru the inf. loop rather than in the small square!!

The feild at the square's proximity is effectively that of the nearest edge of the loop: the infinite current carrying wire..find the flux thru the square to get the ans...

Nearly all of you got the supplement..unfortunately, i encountered it in a match the column problem..took some guts to mark it B,B,B,B...

Supplement

The linear mass density of a rod AB of length 'L' is proportional to the 'n'th power of the distance from end A....

Find the distance of the COM from the end 'B'

Problem 7

This is a good one:

This is a capillary with a difference : it is two coaxial cylinders of radius 'a' and 'b', with a < b..
The inner cylinder is solid while the outer, obviously, is hollow..

When this capillary is put into water of surface tension 'T' and contact angle 0..predict the height rise in the capillary.

Supplement

Find out the MI of the hollow hemisphere abt the four axes shown:1,2,3 and 4.

Just give me the arrangement in ASCENDING order.

Question 6

Find the flux due to the magnetic feild generated by the loop thru the infinite plane kept to its right.

Assume the side of the square loop to be 'a' and the current in it to be I..and the closest distance of the loop from the plane as I.

MI of hollow cone

Assume the surface mass density to be 'S'
Area of elemental ring is dA=(2pi(xSin(A))dx)..hence its mass is Sda.

The moment of inertia cantribution is dI=dm(xSin(A))^2...

Integrate to get MR^2/2...

Points to note:
1)While taking a surface element..always take the length along the area.

2)The moment of inertia is same as that of a disc of same mass and radius...isn't it quite obvious..

3)For practice, try the COM of a hollow hemisphere..also find its MI.

Solution 5

Try to analyse the cone wrt. the turntable.
The pseude force acting is -ma..which is equal to mw^2R in the direction shown.

The cone begins to topple abt. the point P when:

mw^2L(R/2)>= mg(R)

or w >= root(2g/L)

Once a day fellows!

Hi buddies...

I took it on myself to fine tune some of your physics..(not that it really needs some i think now).

In this blog i'll post some of the best questions i ever solved and more..as you'll see this Saturday!

But yaar..i just realised that you do not really have as much time as i have..and many of you are going online many times a day..
Personally, i think i'm doing as much bad as good.. Internet was not meant for JEE prepn..takes too much time..really no good for your concentration..

So..here's the deal. You can promise me not to come online more than once a day..or maybe lesser.. post a solution right there..or note down the question and post it later.
I don't want it on me that i had 30 serious JEE candidates coming online everyday..
So, once a day fellows..
Maybe at 8 in the evening..i'll surely have posted the problems by then!

Solution

Supplement

Find the moment of inertia of a hollow cone of mass M.. abt the axis shown..

Correct ans. on the first attempt only!

Is that the question vibhav??

Solution 5

Problem No.5

Today, i got a simple one for you.

There's a turntable rotating about the axis shown. A solid cone of radius 'R' and height '2R' is placed at a distance 'L' from the axis.

Assume that toppling of the cone takes place bfore it slips...
Give me the maxm. angular velocity with which the turntable can rotate!

Solution4

So, you got the solution to Problem No.4 on your own! Well done.

I've posted the diagram of the supplementary problem. Here, the choice of axes matters.

Finally, the velocity of the particle alng. x is zero
so:
0=V-gSin(A)t
this gives t=V/gcosec(A)..

Enjoyed your TS??

Some of you had your TS today. Bet you got loads of interesting Physics problems.
Anyone wanna discuss some? If yes, post them as comments to this post.

I'll be happy to have good new problems in my stock!

Supplement:

Just for fun:
A particle is projected with a velocity V at an angle Q with the horizontal..
After what time will its velocity become perpendicular to its initial velocity??

Solution3

Solution3:
So guys.. not much response to QNo3.. i guess busy with TS??

Anyways..
The line integral of the magnetic feild is synoymous with the Ampere's law..
Coupled with a situation where one can actually evaluate the integral.. it leads to a dangerous wastage of 5 precious exam minutes...

In the diagram, i've shown the top view of the loops and the wire. The line integral along the common axis is equivalent to that along the infinite loop containing the axis (the other sides of the loop contribute nothing as the feild reduces to zero there)
This leads to a closed integral whose value is that of a line integral!!

Apply amperes law to get the value of the closed integral to be 3U*.

Problem No.4

Just an hour ago, my seniors here were making this paper for a JEE mock test. I, like a good junior, was happily analyzing it for its complexity.

There was a question which i couldn't solve at first sight..(though maybe i could have with pen n paper)

It went like this:

There is a triangle ABC and a point P in space. Specify the point P if PA(square)+PB(square)+PC(square) is to be minimized.

The options were the usual: circumcenter, orthocenter, incenter or centroid??
(And this Digvijay, has a truly 'fantastic', solution)

Would you people care to post your methods too?? in brief?

The Origins

The Origins2:

There comes a time during JEE prepn. when one is all crazy abt integration..and moment of inertia..center of mass etc. One feels powerful with this ability! I got this question at that time.

But one may get stumped when the integration has to be replaced by a summation..as in problem no. 2..

The question i originally solved was to find out the moment of inertia of the rod abt an axis perpendicular to it and passing thru its COM.

Not easy??? anyone wanna give a shot?

So guys.. Saturday-Sunday you get your cute problems. Other days..when i got some less fun tasks and lesser time at hand..you get some good problems for practice..related to these big ones...

Problem No3

Problem No3.

Ready for some different action??

Give me the value of the quantity mentioned in the diagram...

i.e the line integral of the magnetic feild (due to the two coaxoial loops) along the common axis of the two current carrying loops.

Assume the radius of both the loops to be some 'R';
Represent the permeability by U*

Solution 2

Solution2:
One of the first definitions of center of gravity that a student learns is that it is the point at which the object can be balanced using a pinhead...
The point of hinging is the center of mass (equivalent to center of gravity here) of the Rod-balls system. The Rod being massless contributes nothing.

COM=∑(Mi*Xi)/∑Mi

The center of mass of the balls are at the positions of the rod where they are placed...at least the x-coordinates are.

→COM=(1.1+2.2+3.3+...+100.100)/(1+2+3+...+100)

=(100*101*201/6)/(100*101/2)....applying sum of 1st n naturals...and of their squares.

→COM=67...which is the point of hinging...

The origins1
JEE 2008 featured a paragraph on collision of a particle with an inclined plane...
The catch was eliminating the component of velocity perpendicular to the plane of collision..
(plz see it. you will be able to solve it)

The concept became so popular that every 2nd test paper i saw featured a question. I will post some of them soon...
But this pretty one came from one of my friends... never asked where he found it. The motion of the ball was worth thought. I really enjoyed solving it, as proof i can say that even after more than a year, i remember the question and solution vividly.

Solution1:
The ball collides with the ground with a velocity v=sqrt{2g(L/2)}. Due to inelastic nature of the collision v2 is lost and only v1 is left.

v1=v(1/2)....the angle is 30*.
Now the ball continues to move left till the string becomes taut again. The angle is again 30*. Only the tangential component of velocity ie. v3 survives the jerk given by the string.

v3=v(1/2)
Now you can apply conservation of eng. of h=v3*v3/2g;
(v3 comes out to be sqrt(gL)/4.)

So:h=L/32.

Problem No.2

I got this rod with balls on it and i want to hinge it at some point so that it does not... you know wobble like a see-saw.. or fall down...

The balls and the rod have the following property:
At '1', mass of the ball is 1kg;
At '2', the mass is 2kg..at '100' the mass is 100 kg.

Assume the balls to be really small and the rod to be massless.
Tell me the point of hinging.

Buck up fellows!!
Seems like ye all got it wrong..or is it me??
Let me give you a few hints...

In the inelastic collision, the ball loses all the velocity normal to the surface. (in the diagram v2 is lost) Coz thats the defn. Then it moves left horizontally (because of v1) till...Well i'll leave the 'till' to you...

Anyways the answer was L/32. Giving you a day more bfore the solution...

Problem No. 1:

The ball shown in the figure is released from rest... It undergoes a completely inelastic collision with the floor.
To what height does it rise when it leaves contact with the surface for the 1st time...... (obviously in terms of L)

You may want to visit this blog for your daily dose of powerful JEE problems.
You ready?