Messages posted by Yagyadutt Mishra

during the motion force is acting on the particle in opposite direction due to which its velocity starts decreasing and at some instant it become zero or can say stop.After that particle again start in the motion in same direction of electric feild.

Now the force acting on the particle by EF is q.E.

work done will be= -q.E.x where x is the distance travelled before v=0.

energy changed will be = 0-1/2mv^2 ...{final - intial}

now, -1/2mv^2 = -q.E.x..{ x is to be calculated}

put all the values you will get x = 0.2.

Really a good and conceptual question.....

Suppose the body is compressed to a distance y..for the maximum work done.

Then force at this instant will be k.y(y-l).

so work done will be F.x = k.y(y-l).y = ky^3 - ky^2l.

differential it with respect to y because we have to find the y for max.work done.

d(w)/dy = 3ky^2 - 2kyl = 0

y = 0 0r y = (2l/3).....this the compression for maximum work done.

See the easiest way to solve such problem that i know is following.

~~ Consider the Point P as (h,k).

~~Suppose line cuts the X-axis at A and Y-axis at B with co-ordinant as (a,0) and (b,0)..

~~ One relation you will hve is a^2 + b^2 = (158)^2...-------------(1).

~~ ratio in that point P divides the line is 3:155 { because p is at 3 cmm from X-axis so it will be 155 cm from Y-axis}

~~ apply simple geometry... P divides A and B in 3:155..so get the co-ordinant (h,k).

~~ h = (155a + 3x0)/158

~~k = (3b + 155x0)/158.

~~ from here a= 158.h/155

~~ and b=158k/3

~~ put in equation first you will have..

h^2/155^2 + k^2/9 = 1

~~ means locus will be ellipse.

NOTE: if the point is just middle point then it will give locus as CIRCLE..you can analise this.

EMF will be induced only when any geometrical figure cuts the magnetic line of feild perpendicular to its motion.I think this much we all know..Now it does not mean that the figure is close or open..just emf is induced..and if the figure is syymetric then total emf is zero..{here we are talking of total EMF}.As in case a wire of certain.will hane emf equal to BLV...same wire is bent into Curve with the two end joined together the also emf will induced but in this net become zero.

Also emf induced in any figure does not there will be current...NO,current will flow only when figure is close.

It depends on the structure of Alkene..

1> H2C == CH2....in this Grinard reagent shows no reaction because there is no acidic hydrogen.

2> H2C == CH --- CH3... in this case it shows reaction with Grinard reagent because it has acidic hydrogen......

H2C == CH --- CH3 ----------------> H2C == CH----CH2--Cl ----------------> H2C==CH-------CH--MgCl----------->.H2C==CH----CH---CH3 + MgCl2.

( Cl2/hv) (Mg/Dry ether) ( CH3Cl)

So,we can't generalise that Grinard reagent will not show any reaction with Alkene..

Aromaticity------->Resonance -------------->Hyper Conjugation ---------------->Inductive Effect.

This the trend to Check the stability of a Carbocation.{Because carbocation shows all the effect}

for free radical you Go through resonance and Hyper conjugation.

For Carbo anion You check only by Aromaticity,Resonance,Inductive effect.{Because hyper conjugation is not the property of Carbo Anion}

Now in your question Intermediate will be two compound (Normal pentene) with + on the third carbon and and second one will be with positive on second carbon. and.

Now the Major product will be by the intermediate having stable carbocation.The intermediate compound do not show Resonance,and Aromaticity.So,now you will check its stability By Hyper Conjugation Or by Inductive Effect..

Hence in the previous case we have the same situation so considering I.E we have MAJOR product 3-bromopentane

Note: In the case when you are finding the stability of a compound and if the compound shows both effect Hyper Conjugation and I.E then I.E will dominant if the -i.e group is attached otherwise...hyper conjugation.

X^1/2 = a

Y ^1/2 = b

a + b = 1.

a = 1- b

a^2 = 1 + b^2 - 2b

2b = 1+b^2 - a^2

again square it..

4b^2 = Ka^4 + Zb^4 + C(a.b)^2 + P.......where Z,K,C,P are constants..--------------------- (1)

put all the corresponding variable a and b to get the parabola.. now the parabola you will have after subsititution will be perfect parabola...X>0 and Y > 0..is the condition because it is in the squarroot.

So,the part of parabola in the first quardrant will be the only consideration.Hence your equation is apart of parabola of equation ({1)

1> Reimer Tiemann reaction...

2> Reduction of Benzo nitrile with SnCl2.

Delhi ke Bear bina hair ke hote hain hain,,,so they have colour spectrum...can't determine their colour..HI..HI..Hi.//HIii..HIIII

A will be structure of Phenol with CH3 at META position.And B will be same structure with Br at all ortho p[osition.The hint is that there are Three Br group attached in the structure (B) which is possible only when all the ortho position is vacant,since both are ortho directing group.

Also compound A is soluble in not soluble in NaHCO3 then remeber the compound must be phenol.....After considering Phenol we have CH3 left which can be attached any where..according to question.

OK mohit...

A2X3 --------> 2A + 3X.

y 0 0

0 2y 3y.

solubility product will be [A]^2.[X]^3 = 108y^5.

I don't know why you all are fighting..Just leave ,,this there is nothing happen to the colour of bear..whether it is in the forest of ASSAM or it in the POLE..black remain black and white as white.

IIT is not going to ask such slum question in paper so,plz no more discussion..

When the ecentricity is same then the b/a ratio will be same.But area is pi(a.b) as posted by "KABI"..

so, it is clear that area will be same.

x^2/18 + y^2/9 = 1.

x^2/4 + y^2/2 = 1.

both have eccentricity equal to (1/2)^1/2....but if you sketch the curve then the first one will occupy more space in the plane as compared to second one.I think it is clear now.

yes mr. decoder is right...@pink you have used the wrong identitity of Log..

Hi dude I don't think any Child is so much less serious about such Tuff exam..But let me tell u that I am in IIIt allahabad and after my first semster am preparing for it...And I have also One month,,,so It does not mean that u can't prapare but the fact is that how you prepare.

Mr. Rghav has written the table is necessary but alon with that there is some concept which is used for any function.

F(x) has period equal to T.

F( c.x ) has period equal to T/|c|{ |c| mod c}

cF(x) has same period T.

Like Sin 2x has perid equal to pi. because Sin x has period 2pie...

F(x) + G(x) + H(x) will have period equal to lcm of the individual Period.

I think you are getting these.

Ex. Sin2x + Sinx Solve this,

Friction Is the force which comes in the picture when the relative motion between the two surface start.Now this force will try to oppose with its maximum effort to stop the relative Motion.{ let maximum friction is 50 }

Suppose you are applying force 20 N then friction force will be 20..{ this is the actual friction force}

you apply 30 the friction force will be 30.

you apply 49.9 then friction will be 49.9..

if you apply 50.1 then friction will be 50.{ because the friction has maximum effort equal to 50 }...

So. the 50 N will be the limiting friction and the other values are the actual friction experienced by the particle in motion..I think now it is clear.

plz clearify your question what is the [ ] notify..If u not post the modified question we can't solve.So plz.

Sorry i got confused that 124.7 fractional point is 0.7 not 7.....so its 5,,i think you agree with my solution.

When H is in first excited state then it will realese energy equal to 10.2 which is absorbed by He+ ....{which is in its first exc,state}.After absorbing 10.2 energy He+ will shift in 3rd excited state.

why? Because E in 3rd excited state is -3.4 and in first excited is -13.6..So to jump He+ from 1st to 3rd requires absorption of 10.2 energy which is given by the exc.H.

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