« first day (1513 days earlier)   

6:06 AM
The main doubt is, Is the equation of wavefront same as that of the equation of wave i.e pulse C2?
 
 
2 hours later…
8:03 AM
@JohnRennie Hii sir
Here, they show how to derive the value of 2 order determinant
Now , after solving for x and y . I got these values

What I noticed is that they put green colour value in 2nd equation and not 1st one. Why is that ?
 
@SrijanM.T hi :-)
 
This is my Q sir
 
Are you asking how to solve the two simultaneous equations?
 
@JohnRennie It is that why to choose the green value of x
@JohnRennie Equation I solved sir. I got values of x and y after that but only certain x and y was chosen . My Q is why is that
 
If you have two equations in x and y the approach is to find x from on equation then substitute it into the other equation.
 
8:12 AM
Yes.
But I got 2 v
 
You could find x from equation 1 and substitute into equation 2, or you could find x from equation 2 and substitute into equation 1. Both will work.
In this case they found x from equation 1 (the green value) and substituted into equation 2.
@SrijanM.T what do you mean by 2v ?
 
Nothing. Typo. 2 values of x I meant
 
I must admit I'm not sure what you are asking ...
 
When we put green value in equation 1
Or we put red value in equation 1 at place of x
The value of determinant would be different since we have different variable
 
The green value is derived from equation 1. It's just taking equation 1 and rearranging it. So putting the green equation for x back into equation 1 won't do anything useful.
 
8:17 AM
Ok. Can I put red in equation 1 value of x
Form the image of textbook , values of x appear to be same.
 
Yes. The red value is the expression from x derived from eqn 2, so the next step is to substitute this into eqn 1 and that gives you an equation to find y.
Although I would approach this in a slightly different way ...
 
Ohk. Sir . The thing is this. When we put red value in equation 1. We get the value of determinant as shown in the textbook. Now , if we put green value in equation 1. Then , can I do that since that step would give different variable as the value of determinant
 
Do you mean:
> if we put green value in equation 2
 
You can't put the green value in equation 1 as you derived it from equation 1.
 
8:22 AM
@JohnRennie Ohk. Why is that wrong ? Even if I derived it from equation 1
@JohnRennie yes. Why ? Thati
 
Well let's try it. Equation 1 is:
 
@JohnRennie K
 
a₁ x + b₁ y = d₁
And from that we get the green equation:
 
Yes
 
x = (d₁ - b₁y)/a₁
OK so far?
 
8:24 AM
Yes
 
OK, so now substitute back for x in equation 1 and we get:
a₁ [ (d₁ - b₁y)/a₁ ] + b₁ y = d₁
 
And a₁/a₁ = 1 so this simplifies to:
d₁ - b₁y + b₁ y = d₁
 
d1 = d1
Yes.
 
Right, and that's not very helpful is it?
 
8:26 AM
I should have had solved the equation fully.
@JohnRennie Ofc not
Thanks a lot sir.
Now , I understood for each equation
And my Q
 
:-)
 
@JohnRennie Sir , I got different answer from textbook.
Instead of + a1b2 , I got -.
 
You got the same equation, but you've multiplied by -1 on both sides. Just multiply both sides of your equation by -1 and you'll see it's the same as the book.
 
 
1 hour later…
9:38 AM
@JohnRennie Ohk. Thanks sir
 
10:17 AM
@JohnRennie, Sir I have a question, are you available?
 
@Satwik hi :-)
Yes I'm free.
 
Sir, in my textbook it is written that if we bring a positively charged body near an electroscope which is already positively charged then due to induction a negative charge appears on the top part of rod of electroscope hence the net positive charge between the two gold foils increase and they diverge even more.
I am having the question that instead of the top part of rod getting negatively induced, why doesn’t the charged object gets polarized instead of the rod and then we get a wrong conclusion using the book’s method.
 
Both objects get polarised. Suppose we have a ball with a positive charge and the electroscope rod has a positive charge, then when I bring the ball near the top of the electroscope the positive charges repel.
So on the ball the positive charge flows to the side farthest from the electroscope, and on the electroscope the charge flows to the part farthest from the ball.
 
Then wont the negative charge now on the top of rod and on nearer side of object repel again?
 
Ah, OK, I see what you mean.
 
10:23 AM
I think maybe the charged object we bring is an insulator.
 
Yes, I think the book assumes the positive charge is greater on the ball than on the electroscope.
Or, as you say, that the ball isan insulator so the charge can't move.
 
@JohnRennie So sir if two positive charges are there, one of them greater, then the smaller charge one will be polarised, right?
 
Yes
 
Thank You very much sir :-)
and @JohnRennie if we again have a smaller and bigger positive charge but now the smaller charge is an insulator, then the bigger charge will be polarized ?
 
Yes
 
10:28 AM
ok sir :)
 
 
5 hours later…
3:02 PM
@RishiNandhaVanchi So , what are you doing now bro ?
 
@JohnRennie Chemistry D:
like?
 
Ohk
@RishiNandhaVanchi Prep for jee adv ?
Did you decide any colleges , stream for now ?
 
thats the thing, I still havent spent enough time thinking about what to do if advanced fails
I could get some with mains, but yeah, havent spent enough time on it
but i think stream's gonna be electrical mostly, or physics, or music technology if abroad
 
Ohk. Cool that’s nice.
I’m sure you’ll get the nicest college.
 
3:12 PM
@RishiNandhaVanchi I know you can . Btw , there is a site online where there is an online tutorial by Amit Trivedi ( The musician who made the beat for andhadhun and many other great songs) is going to teach how he did it.
I suggest you do check it. You will like it.
 

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