look ill give u the soln. to a general case....

the free body diagram is as posted below.......let the mass be 'm' the length be 'l'........let 'N' be the normal force which the rod experiences from the bottom..............." " be the coeff of friction..............

analysing rotational motion abt the pt rotation....ie the pt of contact with ground............

mgcos *l/2 =ml²/3 * ...........................1)     Torque abt pt P

so we have =3gcos /2l.......................2)

analysing accleration of centre of mass in horizontal direction.....

ma_x=N..............3)

analysing acceleration in vertical dirn....

mg-N= ma_y........................4)

but we know that ay is nothing but component of accn of c.m in vertical dirn........and as no slipping.....

accn of c.m= *l/2........

thus a_x=*l/2cos................................5) 

and a_y=*l/2sin ..................................6)

so using 1, 2, 3, 4, 5 and 6 and solving for

we get....=

here u can substitute theta as pi/3 to get the reqd. result....

from the above eqn.s solve for  and the 2 accns wud be l/2 and l respectively...(interms of g)

ill try to post the free body diagram...

hope it helps...

rate if useful....

look by symettry i can say that they collide at the midpt.....proof: there is no shift in cm of the system(ext F=0)

so evn towards the end the centre of mass will remain at the midpt only...assume the midpt to be the origine....and the two particles at (D/2,0) and (-D/2,0) respctly....the force eqn for particle at origine will be sumwat like this....

- mVdV/dx = Gm²/(2x)²....if it is at a dist. 2x frm the other mass...it will be at a dist of x frm origine...

integrate it frm D to 'x' and frm 0 to V....write V as dx/dt and u know wat 2 do(integrate frm 0 to 'T'..and frm D to 0) and T is ur time taken....jus. giving the brief mtd...

hope it helped

rate if useful......

if it is falling due to its own weight then how can u conserve L??

u cannot conserve L becoz of the simple reason that there is a net external torque....arising due to the gravitional force....acting on the centre of mass towards the centre of the earth which is responsible for its rotation....hmm one thing that u can do is to conserve mechanical energy....ie mgL/2 (1-cos@) =1/2 I w².....and put cos @ to be 4/5 to get the reqd. answer....(pardon ne silly mistakes...) this is the rough method....

rate if useful....

yaar summore replies plz....

frnz ...my AIR is 4322...i have my councelling tomorrow...

plz tell me what happens in the counclling.....what shud i apply for and what can i hope of getting....this is especially for those that have experienced the councelling......replies reqd. urgently.....

applying bernouli's theoram for water flow thru' the 2 surfaces S₁ and S₂.......

P₁+ hdg  + 1/2dv₁² = P₂+ hdg  + 1/2dv₂²............where h= the height from the, the two surfaces are located....

P₁=P + H₁dg........and P₂=P + H₂dg.......P=atmospheric pressure...

applying we have dH g = 1/2(v₂²-v₁²)

but we have from eqn. of continuity S₁v₁=S₂v₂

applying this we have SV =

rate if useful......

thanks a lot mate....thats just about what i required to know.....thanks a lot.....lets get sum expert views on this one...

Sirs... i have a problem with my altitude metre....ive got many got many salutes coming in today and were not accounted for in my altitude...ive tried contacting the admin and a few moderators in this matter...but none of them have shown interest in it yet....so plz help me out in sorting out this issue......

Ok....when it comes to motivation...a lot of us need it despirately.....when it comes to inspiring lines there are a few that i always make......

" what ever u do....always maintain a CALM MIND.........for a calm mind can even overcome a storm"

"U wanna see a miracle??....BE THE MIRACLE....."

"He who has achieved success without facing defeat even once is the bigest goofer.....failure is the first and the most important step to success"

" fruit of hardwork and efforts is the sweetest of all times....itna hard work karo ki sabko phodke rakhde!!"

" Kabhi kabhi kuch paane ke liye kuch khona padhta hai......jo haari hui baazi ko jeete wahi baazigar kehlata hai"

and finally.....if u feel that u are a looser jus. becoz u have not done well in an exam like JEE or EEE or BITS or CEE or CET or UPTU or VIT (tab toh u shud feel happy) or WBJEE or for that matter any exam....lemme tell u that according to the philosophy of life...a man doesnt prove himself to be a winner just by faring well in a 3hr out of the many decades that he lives.....(im seriously telling it is true)so all the best and take care of urself....aur bohut time hai yaar zindagi main....

~~~~~~~ALL THE BST~~~~~~~~

rate if useful....

good mtd mudit...although a rough procedure...a nice one......

and akku....superb soln. to that integration one....and about that vectors one.....ur soln is 16d²/(a²+4b²+9c²)......the answer is 8/7{options are 8/7 7/8 1 4}.....(the fact that these are unit vectors is perhaps missed out).....but i am jus. thinking...using this identity(cauchy-schwartz if i amnt mistaken) how do we make use of the fact that these vectors are coplanar.....neways i feel like kicking my self for missing out on that(identity)....neways rated u ppl....

assume the resistance to be 'r'

pot. accross internal resistance = 6/(1+r)

so total pot. = 6 - 6/(1+r) = 5.8

or 6r/(r + 1)=5.8

or r= 29 ohms

rate if useful....

here goes the technical solution...

240 = 2⁴3¹ 5¹

we want factors of the form 2(2m+1) so we want odd multiples of 2....

so out of four powers we want only one 2.....rest to be choosen frm 3 and 5....

i have two objects 3 and five...total no of selections that i can make....

so total no of ways = (exp of 3 +1)(exp of 5 +1)= 2*2 =4

rate if useful.....

ppl......i was jus. trying these probs today.....mere palle nahi pad rahe the in many of these....jus try it out....

1) if A and B are the roots of the eqn. tanx = 2x.........find the value of _-1S⁺¹sinAx . sinBx dx  (A,B>0)

2) find _-3S⁺³x⁸{x¹¹} dx (my formula edittor isnt working)

3)the pts represented by vectors a,b,c and d are coplanar and (sinA)a + (2sin2B)b + (3sin3C) -4d =0

then, find the least value of sin²A + sin²2B + sin²3C

4) find the domain of f(x) =log_e(3(pi)x - x² - 2(pi)² - sinx )

try it out and nudge if u get it......

actually it is a ring with inner radius 1 and outer radius 2 and centre at origine....or the intersection of

x2 + y2 >= 1 and x2 + y2 < 2

but if u are looking for integral solutions.....we have (1,1)...(-1,1)...(-1,-1)..and (1,-1)

actually i didnt get a very clear picture of ur Q.

INPhO 2000 if i amnt mistaken......dekho..when i first saw the Q. this is all that i cud figure out.....

assume that there is a sphere and we are depositing charge over it uniformly on its surfare.....

assume there to be a charge q already and we are adding a charge of dq......work done to do so will be

dW = Kqdq/R..........integrating from 0 to Q

and W = KQ²/2R

this is stored as potential energy of the system

so when u splitt into 2 equal spheres.i assume that the total volume is unaltered......so

4/3pi R³ = 2 * 4/3pi r³

or r= R/2^1/3......

and the energy of  first sphere = E = KQ²/R + TA₁........(A1 and A2 are the surface areas of the first and the second 2 spheres)(TA= energy due to surface tension)

where T= surface tension of water...(given in the value list and not used in any other problem)

and energy of the smaller sphere = U = KQ²/4(R/2^1/3) + TA₂

and for this change to be feasible we need E = 2U

equating which and solving we get Q and for 3 bubbles i guess u shud do it using r= R/3^1/3

and E = 3U....

third part main mere palle nahi padh rahe(dielectric break down)

but i have given a very brief method for the first 2 parts.....

correct if wrong....

rate if useful......