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as most of the reaction of benzene are electrophilic substitution reaction and aniline has a higher reactivity than benzene it will start reacting.
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Let The point below P in plane of eyes of prisioner at A and B is Q
Let PQ=h
and
AQ=b
From first
tan(31)=h/b
By pythagorus theorm
BQ=sqrt(b^2+15^2)
Therefore
tan(25)=h/sqrt(b^2+15^2)
Using the two equations solve for h and b.
Height of point P=h+1.7 meters
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Go through Ashish's derivation when it comes to
Ey = lamda/(4*pi*epsilon_not*a)*integral of cos(theta)d(theta)
Take limit from
-pi/4 to pi/4 for quater ring
-pi/2 to pi/2 for half ring
-3pi/4 to 3pi/4 for three quater ring
-pi to pi for full ring = 0
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Right answer. Its COOH only
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Let L1 is length of 35 and L2 is that of 37. So number of molecules of 35 (n1) and 37(n2) are proportional to L1/L2
As average molar mass is 35.45
So 35.45= (35*L1+37*l2)/(L1+L2)
Solved this to get L1/L2
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There is a book by P Prakash in the market exclusively for DCE. There you will find graphs for various years given marks to rank correlation. It contains all the three subject and you get previous year's question chapter wise. You can have a look at it.
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If a body is kept at a place where apparant gravity is g' then weight is mass*g'. This weight might be balanced by normal force in most of the cases but there may be cases when weight is not equal to normal force. Suppose you are pulling water from a well using pully. In this case the string will be pulling you upwards and hence normal will be les than the weight.
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Right point indrajeet. While gauss law is valid for such a charge distribution, it is not useful because we cannot convert integral of E.ds into E times some kind of area. We can't find field due to cone by gauss's law because of absence of symetry.
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Only one question at a time.
For the first question. The magnet will always have a magnetic field.
It can also create electric field if you move a loop of wire relative to it.
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It does not cancel each other. The jockey is so placed that the potential between the one end and jockey is exactly equall to potential difference of the secondry circuit
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Pick up NCERT. And go through inorganic chapter by chapter (Do it by giving half hr daily for a longer period of time, like a month or so).
You will find that after this exercise you will only remember 25 percent things. But don't worry and repeat the whole exercise at least 3-4 times and at the end of it (which may be as long as 6 months) you will be good with inorganic chemistry. The idea is to allow it to settle it into your brain.
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k=.314cm^-1=2*pi()/lamda
So lamda = 20 cm
So minimum length = lamda/2 = 10 cm
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Right answer allamraju. Well done.
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Right answer Rolisha. Well done
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If d is distance between them then, assume angle between strings is theta
theta= d/l
Tcos(theta/2)=mg
Tsin(theta/2)= kq^2/d^2
For small angle cos(theta/2)=1
sin(theta/2)=d/2l
So mgd/2l=kq^2/d^2
So
d=(2kq^2l/mg)^(1/3)
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