Doubt from tests


4th ) BCD
A is correct , notes mein se hain
@Navjeet_Kohli for 3rd
I am getting theta = tan^-1( 1-u/1+u) where u is friction coefficient . Trying 6th RN

@Navjeet_Kohli for 6th , doing option D wait.
For option D the problem i am finding is to approximate the path diff from S1 and S2 towards S3.
How do you find intensity at S3 when its neither at maxima nor at minima without phase diff?
The path diff for D is [(d/2+kD/4d)^2+D^2]^1\2 -[(d/2-kD/4d)^2+D^2]^1\2

@Abhinav.08 @LaveshGupta @manan_uppadhyay Please try option D for 6th one no

@LaveshGupta did you got this question?

didn’t try. abhi free nai hu sorry. I am logging out.


@Navjeet_Kohli please confirm answers no

1 Like

Answer for 3rd and 4th is correct! (how did you solve them)
6th one answer is ABD…
But in Option A, B they have asked ratio of intensities on “S2”… What does that mean?
Sorry for the delay…

1 Like

I assumed it to be Screen 2 coz it doesnt make sense if its slit 2

I got AB , not able to solve D

4th one is from notes , theory of beta decays . You need to remember the values of Q for various decays. Should i send pic of my notes?
3rd one mein the container is moving down with a= gsin45-ugcos45 and hence on particles of container a pseudo force will be applied up the incline and a force mg due to their weight , now when we resolve the comp of ma to masin45 and macos45 the net downward force will be m(g-asin45) and net horizontal force will be macos45 , the resultant force will be the net force on liq ( where the liquid is pushed) so angle between resultant force and vertical will be the needed angle coz the force is pushing the liquid and that will be tan^-1( macos45/m(g-acos45)) , should i send written sol for this ? In RRB a similar kind of sum was there and process was a little similar like this too.

1 Like

Yeah right! Thanks!

Yes please…

Nope… I got it…Thank you so much!

Sorry if its a little congested in my notes.

1 Like

plz help

Is the answer 44%

yes bro…can you send the solution?

plz help

The question says that the alpha particle is approaching the nucleus ffrom a far distance. So, the kinetic energy changes into electrostatic potential energy as it approaches the nucleus. Use kqQ/R = 1/2mv^2 for both cases, you will get two radii(the first one is the true radius and the second will be bigger as the alpha particle can’t get to the same position as the first). Use percentage error formula to get the answer.