Messages posted by feynmann

d>

See the particle can't leave the the xy plane

so it can't reach z= 1

electric forces are in xy plane , magnetic force also ( being a cross product with B ) is in the same plane . Take note of this .

again all the options given dimensionally don't match with that of velocity .

See maxwell's proof does not differentiate between translational , rotational or whatever co- ordinates . It only assumes that the Hamiltonian for motion has n (say ) independent co- ordinates and assuming Boltzmann disn. it gets at the result just by using some standard statistical & analytical calculation . So you

need to learn the general proof . Got it ?

this is a variable mass prob with external force applied = 0

hence the eqn of motion is

m dv/dt = -u dm/dt ( m = instantaneous mass of the body)
but dm/dt = - m ( given)

putting the value of dm/dt and slving the eqn we arrive at the given result .

see that the fn is periodic with period 1

now confine attention only in { - 1 , 1 }

applying x = [ x ] + { x } we see easily that the fn is an even one

2 let us find potential at a distance of D from the centre

take a circular strap of radius r & width dr

so dV =( 1/( 4 pi epsilon naught ) ) * 2 pi r dr / sqrt (r^2 + D^2 )

r runs from 0 to R

integrating , we get reqd. potenial = V = ( q / 2 epsilon naught ) { sqrt( R^2 + D^2 ) - D }

I am giving you hints for some probs .

1> write the binomial expn. for ( 1 + x) ^n & integrate both sides putting the const. value from x= 0 ;

now in the obtained expn pu x= 1 & - 1 & then add

2> just subtract in the above instead of adding

3>put x = 1 & then x= -1 and then add the two ( RHS is not coming , plz check qn )

5> let n be odd

then coeff of x^n comes from

_{[ ]}

^{[ ]} 1/fact(r) * 1/ fact (n-r) = 2^n/fact n

if n be even then an additional term ( 1/fact (n/2))^2 will come

It easily comes from the geometric interpretation of cplx nos

let (a- b) = mod ( a-b ) exp( i )

similarly (c- d) = mod( c- d) exp( i)

so, (a-b)/ (c- d) = mod( a- b) / mod ( c- d) exp( i ( -) )

now if they are , there arguments must differ by + or - pi/2 ;

hence the result

remember its only the rest mass of photon that is zero .

when it moves at the speed of light it does pocess mass = hv/c^2 ;

it is this mass that is converted into energy

remember the eqn .

E^2 = ( pc)^2 + ( m0 c^2 ) ^2

at const volume work done = 0 ;

but at const pressure there is some work done beyond increasing temp.

It is this work for which Cp > Cv

first find the total height ascended(h) .

for this the eqn. of motion is v dv/dx = - ( g+ kv^2 /m)

solve it with the boundary condition given& find h from it .

Now for the downward motion the eqn of motion is

v dv/dx= g - kv^2 /m

now find v at x= h

As for the second Qn. see ( assuming that u know vector calculus)

curl E =- d/ dt ( B) ( partial diff)

If B is const. wrt time then curl E = 0 , which means possible soln of E may be central .

Again curl H = j ( unmodified ampere's law holding only for magnetostatics. )

Now u know that For H field to create statically , j must be non zero . Which means that a possible soln for H (and hence B) can't be a central one .

Satisfied ?

Did u notice that the coins are given to be idetical ? In that case the sample space would not be 4^10 . Looking for a reply .

For A consider the forces acting on rope ;one is F in the forward direction & other is the friction . For equlibrium of rope friction on rope must be backwards .Again from third law friction on wheel must be in the forward direction ( they are action - reaction pairs) . Clear enough ?

see , don' t substitute for x directly

write ( x- 3 )^2 = -16

or , x^2 - 6x + 25 = 0 ;

now take this factor common from the given expn , giving a residue of 5 ;

i. e . given expn. = ( x^2 - 6x + 25 ) ( x^2 + 3x - 4 ) + 5

= 5

clear ?

try to draw the FBD for both the rope and wheel . Soln will be clear

in forward direction at both points . Its friction that is causing motion

There is no scope for confusion at all. The vectorial sum reduces to' algebraic ' sum with proper phase difference when the displacements are along the same axis .For the general case we have to take the vectorial sum .

here we have to consider the field momentum also

plz type the qn in a littlebit clear way so that no confusion arises

Agreed , have not noticed the word 'different' . Thanks for pointing out the mistake .

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