Here are some moderately difficult problems for your revision:
1)
Consider dD(t)/dt=k1*L(t)-k2*D(t). Also, L(t)=Lo*e^(-k1*t). Find an expression for the critical time Tc: the time when D(t) is maximum. Take D(0)=Do. Note that k1,k2, Lo and Do are constants and k2>k1.
--3 marks
What is this time (Tc) when k1*Lo is less than Do*(k2-k1)?
--3 marks
2)What is the magnetic moment of a disc sector (angle=pi/3) of radius R and charge Q (uniformly spread) rotated with angular velocity w about its natural axis??
--2 marks
3)A point charge is kept at the center of a cylinder of length L and radius R. What is the ratio of electric flux thru the curved surface to the flux thru the end caps?
--2 marks
4)The volume charge density varies with the distance from origin 'r' as rho=a*r+b*r*r.
What is the variation of electric field with r?
--2 marks
5)A 3:1 (by moles) mixture of oxygen and nitrogen effuses through a tiny orifice. What is the molecular mass of the mixture effusing out? (Do NOT consult your class notes!).
--1 mark
Note: Part 2 of Q1 is my favorite. Please try it! It looks (and is) difficult but it is solvable by you!
Note: You're through with JEE if you can solve 4 problems!
ans 1
ReplyDeleteTc= (1/k2-k1) log[(k2/k1){k1L. - D.(k2-k1)}/k1L.]
ans1 (b)
above expression become imaginary when k1L.is less than D.(k2-k1)
ans 2
qr^2/24
ans 3
L/[sqrt(L^2 +4R^2) + L]
ANS 4
r^2(a+br)/E.
ans 5
[96sqrt 7 + 28sqrt8]/[3sqrt7 + sqrt8]
i just missed writing w in the answer of ques 2 so the answer is qr*r*w/24
ReplyDeleteI think these problems are too easy.
DeleteQS-1
ReplyDeleteIts chemical kinetics,elementry reactions in series.[L--k1-->D---k2-->E]Iska toh formula dekha tha,
Tc= Ln(k2/k1)/k2-k1
Will give it a mathematical try.
QS-2
qr^2w/24
Qs-4
E=r^2(5a+rb)/20
Qs-5 {gives a motivation to revise gas laws :P )
(48sqrt14 + 96)/(3sqrt14+4)
1-Tc= Ln(k2/k1)/k2-k1
ReplyDelete2- 0
3- Qwr*r/4
4- r^2(5a+rb)/20
5- 30.94
well goofed up!!
Delete1-Tc= Ln(k2/k1)/k2-k1 , 0
2-Qwr*r/4
3-[root(5)+1]/24
4-r^2(5a+rb)/20
5- 30.94
"I think these problems are too easy" was not written by me. I can't delete it.
ReplyDeleteShivam's answer to problem 1 is close. To verify it, please refer to the wikipedia page on "Streeter Phelps model" (see the expression for T(crit)).
Does no one have the answer to 1(b)? "Above expression become imaginary when k1.L is less than D.(k2-k1)" is hardly an answer! Infact, that's the question.
As I am not in the mood for solving the other problems (seriously, I don't remember the definition of magnetic moment!) , so you can fight the answers out among yourself. Extra points for posting an explanation which cannot be contradicted. (I think this is called crowd-sourcing).
2-
ReplyDeleteanswer is qwr*r/4 since m/L=q/2m
L= (6mr*r/2)/6 w
moi is same as that of a whole disc
@Shivam: do you have anything to say?
Delete4- typical integration and gauss law
ReplyDeletesome mistake in typing by me
it should be r^2(5a+4rb)/20
could someone validate the 3rd one???
lalit u are correct for the second and fourth one maine disc ka moment of inertia mr^2/12man liya tha silly mistake. and yes 4 th main tumhara last waala reply sahi hai maine integrate karke denominator main 4 and 5 likhna bhool gaya tha i hope u got it. and i dont think ki tumhara 3 rd ka ans sahi hai becausethe flux through the ends should depends upon the radius of the cylinders u can check this by taking r to be very very large and in second case r to be very very small.smae is for length of cylinder.
ReplyDeletehere are my solutions
ReplyDeletethe diff equation was
dD(t)/dt=k1L(t)-k2D(t)
L(t)=L.e^-k1t
now D(t) ko y and t ko x rakg ke dekho
ekdam se lagega ki its first order differential equation :D
now u can solve it (no need to search on wiki)
after solving this we got
y.e^k2x=[k1L./(k2-k1)]e^k2-k1)x + c
lalit u must have missed c.
put x=0 c=D.-k1L./(k2-k1)
diff it w.r.t. x and put it equals to 0.
and u can get the value of x or Tc.
the expression for Tc was coming out to be
Tc= (1/k2-k1) log[(k2/k1){k1L. - D.(k2-k1)}/k1L.]
wiki pe jo expression hai usme k2/k1 ki jagah k1/k2 likha hai i dont know where i went wrong.
ans of 1(b)
since the dy/dx is -ve when k1L. is less than (k2-k1)D. for all x so y will be max at x=0 or t=0 actually i should take x=-infinity but since domain 0f x is[0,infinity) (because it is time)
so max value of D(t) will be at t=0.
for question 2 the trick is M=L*q/2m.
for question 3 u need to calculate flux through the circle (no shortcuts sorry) suppose it comes out to be $ now flux through the cylinderical surface=q/E. - 2$(there is another end as well).
for ans of ques no 4 u can see your module of electrostatics there is a similar sse (no i did not see that i just remember)
i think 5 th ka answer sabka same hi aa raha hai so no need to post that.
for flux
Yes, t=0 is the correct answer.
Delete@anuj bahiya i commented in the playlists waala post can u just see that.
ReplyDeleteHi! I've posted a comment on that thread.
Delete@shivam
ReplyDeletei am quiet sure of what i have written about 3rd one
why not you just try once again
flux is just coming in terms of q/epsilon. in both the cases
i did and getting my same previous answer,can u post your integration
ReplyDeleteyep your right!
ReplyDeletei have put the limits wrong
may be took R=L