Messages posted by agnit_thebest

Two stars P and Q have slightly different surface temperatures T_pand Tq respectively with T_p>T_q.Both stars are receding from the earth with speed v_pand v_qrelative to the earth. The wavelength of light at which they radiate the maximum energy is found to be the same for both-

a) v_p>v_qb) v_p<v_q c) v_p=v_qand the size of Q>size of P

d) Nothing can be said regarding v_pand v_qfrom the given data

Two stars P and Q have slightly different surface temperatures T_pand Tq respectively with T_p>T_q.Both stars are receding from the earth with speed v_pand v_qrelative to the earth. The wavelength of light at which they radiate the maximum energy is found to be the same for both-

a) v_p>v_qb) v_p<v_q c) v_p=v_qand the size of Q>size of P

d) Nothing can be said regarding v_pand v_qfrom the given data

A cyclotron is accelerating proton where the applied magnetic field is 2T, the potential gap is 100 kV then how much turn the proton has to moive between the dees to acquire a kinetic energy 20 meV

a) 200 b) 300 c) 150 d)100

Please show the steps!!!

Hey good question!!!Experts pl help!!!

Ya I have

lim(n->

) [

1+

2+

3+..............+

n] / [n

n]

What is the product formed due to ozonolysis of C₆H₅CH=CH₂?

Thanks Puneet Bhaiya for your confirmation. This question was given in the Narayan study material and I had my work cut out!!!!

OPTION (d)
The molecules of 0 degree water have greater kinetic energy than that of 0 degree ice and thus owing to conservation of energy the potential energy of molecules of 0 degree water has lower k.e than that of 0 degree ice. Potential energy does not depend on the temperature of the system.!!!

Ya u r right.i am getting option (b)

Normal force acting on the insect N= ( m+ M )g
frictional force = kN = k( m+ M )g
Therefore acceleration a =[ k( m+ M )g ]/m
We know that S=ut + 1/2 at^2
Here initial vel. is zero
Therefore putting S = L we get:-
L=1/2 [ k( m+ M )g ]/m * t^2
or t^2 = 2Lm/ {k( m+M)g}

Hence option (b) is correct

Pls help experts!!!

Ya I know!!! Thanks anyway

Let 'a' and 'b' be non collinear vectors of which 'a' is a unit vector. The angles of the triangle whose two sides are represented by

3( a x b ) and b-(a.b)a are

A) pie/2,pie/3,pie/3 B) pie/2,pie/4,pie/4 C) pie/3,pie/3,pie/3 D) data insufficient

This has been done in another topic with same heading!!! Pls refer to thet

For all positive integers k, prove that:

sin 2kx

--------- = 2 [ cos x + cos 3x + ................+ cos( 2k-1 )x]

sin k

Hence prove that

_[0]

^[pie/2]sin(2kx)cotx dx = pie/2

Congrats Di!!!

If arg(z)<0 then arg(-z) - arg(z) =

a) pie b) -(pie) c) -(pie)/2 d) (pie)/2

Pls show the steps!!!

Ya I wanted to know the last thing i.e.

x4+x2+1=(x4+2x2+1)-x2 =(x2+1)2-x2 =(x2+x+1)(x2-x+1)...

Thanks a lot

I have understood everything but my only querry is that in the examination how will u immediately apply

x^4+x^2+1=(x^2+x+1)(x^2-x+1)

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