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If it returns back to the same position after striking then it must have striked the parabola normally at (-1,1). Hence it must have been projected normally from (1,1).
slope of normal at (1,1) = -1/(dy/dx) = -1/2x (at x=1) = -1/2
Hence the angle of projection with horizontal is tan -1(1/2).
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Radical alkyl intermediates are stabilized by similar criteria as carbocations, the more substituted the radical center is, the more stable it is. This will direct their reactions: formation of a tertiary radical (R3C·) is favored over secondary (R2HC·) or primary (RH2C·). However, radicals next to functional groups, such as carbonyl, nitrile, and ether are even more stable than tertiary alkyl radicals. Radicals attack double bonds, but unlike similar ions, they are slightly less directed by electrostatic interactions. For example, the reactivity of nucleophilic ions with ?,?-unsaturated compounds (C=C-C=O) is directed by the electron-withdrawing effect of the oxygen, resulting in a partial positive charge on the carbonyl carbon. There are two reactions that are observed in the ionic case: the carbonyl is attacked in a direct addition to carbonyl, or the vinyl is attacked in conjugate addition, and in either case, the charge on the nucleophile is taken by the oxygen. Radicals add rapidly to the double bond, and the resulting ?-radical carbonyl is relatively stable.
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I think the question is :
I =  {1+x+  (x+x2)}/{(x+1) + x} dx
Now rationalize it :
[{1+x+  (x+x2)}/{(x+1) + x}].[{(x+1) - x}/{(x+1) - x}]
= (1+x)1/2 - x1/2 + x(1+x)1/2 - x3/2 + x1/2 + x3/2 - x(1+x)1/2
= (1+x)1/2
Hence I =  (1+x)1/2 dx = (2/3)(1+x)3/2 + c
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With mild oxidizing agent it will form CH3CHO.
With strong oxidizing agent it will form CH3COOH.
The difference is when it oxidized by stronger agent first it oxidizes to aldehyde and it further oxidizes to acid while with mild agent further oxidation will not take place after formation of aldehyde.
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Use a concentric Gaussian sphere of radius r outside the shell.
E.dA = Qin/ 0
E(4 r2) = Q/0
E = (1/40)(Q/r2)
V = infinityr (-Edr) = infinityr (-1/40)(Q/r2).dr
V = (1/40)(Q/r)
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The term chiral is used to describe an object which is non-superimposable on its mirror image.
A molecule is chiral if it cannot be superimposed on its mirror image the two mirror images of such a molecule are referred to as enantiomers. A mixture of equal amounts of the two enantiomers is said to be a racemic mixture.
Human hands are perhaps the most universally recognized example of chirality.
 The two enantiomers of bromochlorofluoromethane
Here carbon atom is the centre of chirality.
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Nobody knows the exact weightage of the topics asked in JEE. Instead try to revise your strong topics and do your best in the exam.
Himanshu had given some links , refer them for shortcuts.
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r = (area)/(semi-perimeter)
area = (1/2)(3)(4) = 6
semi-perimeter = (3+4+5)/2 = 6
Hence r = 6/6 = 1
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I think the that the eligibility criteria will remain as it is in JEE 2007. But nobody knows what plans the have. Even if there is a change in plan you will be provided with one time exception as it was the case last year. If your percentage is above 60 and you are confident , i would suggest you to take a drop.
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The shape of a molecule is determined by the number of groups of electrons around the central atom. The 'groups' might be a non-bonding single electron, a non-bonding or bonding pair of electrons, a double pair of bonding electrons or triple pair of bonding electrons etc. The electron 'groupings' repel to minimise the potential energy of the system i.e. to make the A-B-C angle as wide as possible.
Any lone pairs of non-bonding electrons on the central atom X, are closer to X than bond pairs because there is no Q atom attracting/sharing the lone pair electron charge.This will increase the repulsion between a lone pair of electrons on X and any other bonding/non-bonding on X.
In terms of electron pair repulsion: lone pair - lone pair > lone pair - bond pair > bond pair - bond pair.
As the lone pair - 'other pair' repulsion increases, the angle between these pairs increases, so the Q-X-Q angle will be slightly reduced compared to what might be expected from the 'simple' geometry of the shape.
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In calculus, an antiderivative, primitive or indefinite integral of a function f is a function F whose derivative is equal to f, i.e., F ? = f. The process of solving for antiderivatives is antidifferentiation (or indefinite integration).
Antiderivatives are related to definite integrals through the fundamental theorem of calculus, and provide a convenient means for calculating the definite integrals of many functions.
They can be used to compute definite integrals, using the fundamental theorem of calculus: if F is an antiderivative of the integrable function f, then:
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1 man in 1 day can perform 1/(24x16) work.
1 woman in 1 day can perform 1/(32x24) work.
Hence in 12 days 16 men and 16 women perform 12{16/(24x16) + 16/(32x24)} = 3/4 work
Let additional men required be x.
In remaining 2 days 1/4 amount of work is to be done.
Hence 2{(16+x)/(24x16) + 16/(32x24)} = 1/4
x = 24 additional men required.
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y = (x2+3x-40)/(x2+5x-14)
y = {(x-5)(x+8)}/{(x+7)(x-2)}
y < 0 when x  (-8,-7)  (2,5)
Hence the least integral value of x for which y < 0 is 3.
Hence the greatest integral value of x for which y < 0 is 4.
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Time period is given by T = 2(m/k)
Let mass of chair be M.
1 = 2(M/600)
M = 600/42
When a man of mass m sits on chair :
2.5 = 2(M+m/k)
M+m = 600(2.52/42)
Hence m = 600(2.52/42) - 600/42
Approx. m = 78.75 kg
Weight = 787.5 N
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KE when it is striked = (1/2)m(52)
Suppose its velocity is u when it makes and angle  with vertical.
Gain in KE = Loss in PE
(1/2)mu2 - (1/2)m(52) = mglcos
u2 = 25 + 40 cos
Force equation at maximum tension :
Tmax - mgcos = mu2/l
2mg - mgcos = mu2/l (since Tmax = 2mg)
2mg - mgcos = m(25+40cos)/2
4 - 2cos = 2.5 + 4cos
6cos = 1.5
cos = 1.5/6 = 1/4
Hence = cos-1(1/4)
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