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Draw the figure.
The circles meet at one point only (4,0).
At this point the tangent is parallel to Y axis.
Hence y intercept is infinity
The options seem incorrect.
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A great book to accompany the problem book of Irodov. It does require an elementary knowledge of vector calculus. But the concepts are explained in short and very elegant ways.
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A great book to accompany the problem book of Irodov. It does require an elementary knowledge of vector calculus. But the concepts are explained in short and very elegant ways.
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The minimum length of the wire between any two corners is L/n, where L is the length of the wire.
Its resistance is
R' =  (L/n)/A =[  L/A]/n = R/n
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First of all ask your friend to stop acting like a sissy. Ask him to hold his head high and say onto himself "I have got what it takes". Thats all. No second opinions needed, no help from anybody required. A man is his own best mentor.
Noone can tell whether or not he will make it to other institutes after dropping a year. Only he can say that. If he feels he is upto it, he should give it a shot. Otherwise if he isn't sure, he should ask his dad to cough up some money and go to a college by donation. These are the only two options. So ask him to take his pick.
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The correct answer is the left hand side bottom one.
At the position of the bob shown, it has two acceleration components:
The tangential acceleration gsin along the tangent and oriented toward left
And the centripetal accln. v2/L directed inwards.
Their vector sum will be something as shown in the left hand side bottom one.
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You can solve both by LH rule ... but lets try a different tune.
First problem
x a (sin-1x - sin-1a) / (x - a)
=   ( - )/(sin - sin ) where sin-1x =  and sin-1a = 
= ( - )/2cos[( +)/2]sin[( -)/2]
=(( - )/2)/sin[( -)/2] x (2/cos[( +)/2])
=1 x 2
=2
Second problem
x /6 sin (x - /6) /  3 - 2cos x
=x/6 sin (x - /6) / 2( 3/2 - cos x)
=x/6 sin (x - /6) / 2(cos/6 - cos x)
= x/6 2sin [(x - /6)/2] cos[(x - /6)/2]/ 4sin [(x + /6)/2]sin [(x - /6)/2]
=x/6 (1/2)[cos[(x - /6)/2]/sin [(x + /6)/2]]
=(1/2) x 1 / (1/2)
= 1
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Consider the function v(t), where v is the velocity of the particle.
v(0) = v(1) = 0
Then by Rolle's theorem, there exists at least one value of t at which v'(t) = 0, i.e, = 0.
Hence, changes sign in the interval (0,1).
Hence (a).
The reason for (c) is provided by akku
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See, the vector relation is
r1 - r2 = (v2 - v1)T
which is similar to
A =  B
where  is a scalar.
Then A is parallel to B
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Well, being an IITian myself, I might be a bit biased. But I'll try to be as neutral as possible.
You see, its not about the infrastructure, its not about the teachers.... its about the kind of individual you become after 4 or 5 years in an IIT. The sheer number of extra cirricular activities possible here groom your persona to make you a balanced individual in all respects.
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HCV HCV HCV HCV
Do it first.
Then move on to Irodov and more flashy stuff.
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Telecommn. is a very good and happening field to be in. Go ahead bro
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CONGRATULATIONS to the new fellow IITian.
I agree with Bipin. It will be a tough call for CSE in IITM. Try at IIT Kgp
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