Messages posted by feynmann

field inside a conductor is 0 only when no current is flowing (electrostatics ) . But when current is flowing field is non - zero .

j =

E

Now I am going to solve the prob by the substitution u suggested ( going to be a bit longer but the final integration is easy )

Let z = x + sqrt ( x^2 + 2 x - 1 )

so , ( z - x )^2 = x^2 + 2x - 1

giving x = ( z^2 + 1 ) / 2( 1 + z )

dx = ( z^2 + 2z - 1 ) / 2 ( 1 + z ) ^2

sqrt( x^2 + 2 x - 1 ) = z - x = z - (z ^2 + 1 ) / 2 ( 1 + z )

= (z^2 + 2z -1) / 2 ( 1 + z )

Now put the values in terms of z in the integrand , giving

2 dz / (z^ 2 + 1 )

which on integration yields 2 arctan ( z )

= 2 arctan ( x + sqrt(x^2 + 2x - 1 )) (ans )

can I do the prob by a diff substitution ?

If yes , here it is

the qty under the radical sign is ( x + 1 ) ^2 - 2

put( x + 1 ) = sqrt (2 ) tan

so the integral becomes sqrt(2 ) sec

tan

d

/ ( sqrt(2 ) sec

- 1 ) sqrt ( 2 ) tan

which becomes

d

/ sqrt( 2 ) - cos

rest of the integration is trivial

limit is 1

let y = given limit

so, ln y = lim ln ( tan x / x ) / x

Now the R. H. S. is in (0/ 0 ) form as x tends to 0

Now apply L Hospitals rule

this gives , log y = 0

so y = 1

edited

d) down ( since the electronic charge is negative & applying F = q ( v * B ) )

My dear Tka , what you have written is the formula for divergence .

Please note curl is a vector qty given by ( in cartesian coordinates) in determinant form as follows

i j k

d/dx d/dy d/dz

y z x

d stands for partial diff . ( the left & rt bar of determinant is omitted )

which is clearly non zero . Got it ?

ans b) 0.5

sin a + sin b + sin ( a + b )

= 2 sin (a + b ) / 2 { cos ( a- b ) / 2 + cos ( a + b ) / 2 }

=4 sin ( a + b ) / 2 cos a/2 cos b/2 ........ (1 )

sin a + sin b - sin ( a + b )

= 4 sin ( a+ b ) /2 sin a/2 sin b/2 ............ ( 2 )

dividing ( 2 ) by (1 ) we get the reqd. result (proved )

units digit of x is 3

the last expression is not clear

see, you are assuming that the mass per unit length is distributed uniformly ( by taking lambda to be const .) which is not the case here .It is changing with r .

Here the eqn of motion for block A is

Tcos30 = N ( it is at eq.)

T sin 30 = 10g + 0.2 *N

find T & N from these

Now the eqn of motion for B is

20g - 2 * 0.2 * N = 20 aB

find aB

X = ( 2 , inf )

Y = ( 2 , inf )

prob = (she selects chem ) ( she gets A on it )

1/2 * 2/3 = 1/3

Soln to this type of Qn can be given after writing eqns for each of the vector components seperately and the solving the simultaneous diff eqn .

F = q ( E + v * B )

write F = m d2 / dt2 ( r )

where r = x i + yj + z k and seperate each i, j, k comp.

The Curl is non zero here !!!!!

So the answer is going to be different for two paths .

For the 2nd case ( shown by green arrow ) W = 0 dx + a dz = a^2

similarly calculate for the other path.

( 3 / 40 ) ( ans )

first let y = sin inv( x )

dy / dx = ( 1 - x^ 2 ) ^ (- 1/2 )

expanding the binomial , its coeff for x^ 4 is 3/8

now integration of this term gives a2 x^5

so a2 = 3/40

edited

Balls striking per unit area per unit time = n/ 2 ;

Now , the total momentum change of a single ball = 2mv ( mv - ( - mv) )

hence the moment of force applied at the region of plate between x & x + dx from the axis is = x * 2mv * n/2 * a dx

= nmva x dx

So the total moment of force applied due to impact = nmva x dx ( from 0 to b )

= nmvab^2/2

this must balance the torque due to gravity = Mg b/2

so we get , Mg b/2 = nmvab^2 /2

or, v = Mg/ ( nmab ) = 15 m/s

Here is the soln .

Let us consider a point at a distance x from the axis of rotation . Let pressure at x & x + dx be p & p + dp ;

So , applying eqn of motion for the gas confined beyween x & x + dx , we can write ( assuming cross section at x be A )

A dP = (

A dx)

^2 x

but from ideal gas eqn .

= (M/RT) P

substituting this value we get

dP/P = (M/ R T )

^2 x dx

integrating and putting the limits , we get

ln ( P / Po) = ( M/ R T )

^2 r^2 /2

so, P = Po exp { ( M/ R T )

^2 r^2 / 2 ) }

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