Messages posted by feynmann

Got the same ans by a diff procedure .

see x^8 & y^ 8 are + ve nos. , so A.M. >= G. M. applies , giving

x^8 + y^ 8 >= 2 (xy) ^4

Now, from the given eqn , putting xy=p , we get

4p - 3 >= p^4 ;

whoch gives

p^4 - 4p + 3 <= 0;

factorising the LHS we get

(p-1)^2 { ( p+1 ) ^2 + 2 } <=0

now { ( p+1 ) ^2 + 2 } > 0 for real p

therefore the only possibility is (p-1 ) ^2 = 0

so p = 1 ;

i. e. xy = 1 ;

Now putting this value in , the eqn . becomes

x^8 + y^8 = 2 ;

but again from A.M . - G. M. theorem , x^8 + y^8 >= 2 , the equality sign holding iff x= y ;(if x = - y then the given eqn has no real soln , check it )

so we get x^2 = 1 ;

x = + or - 1

therefore reqd soln set is ( 1, 1 ) & ( -1 , -1 ) ;

Howzaaaaaaaaattt!!!!!!!!!!!!!!!!!!!!!!!!!!

ans 42 / 143

Nr.being simply the coeff. of x^10 in the expn. of ( x + x^2 + x^3 + ............) ^4

Dr. is coeff. of x^10 in the expn. of ( 1 + x + x^2 + ........... ) ^4

ans is 55 .

its simply the coeff. of x^12 in the expn. of ( x + x.^2 + x^3 + ...........) ^3

got it ?

Karthik has made a mistake in calculating the work done by friction . See , the frictional force is not const . so you have to integrate to find out the total work done by friction .

ans ( b ) n ;

soln

= - k

or, d/ d = -k

giving , = -k

from which the result follows .

it asks you to find the point that divides the line joining the two given points in the ratio 2: 1

Its the 'abstract mathematical fields( E & B )' that oscillate in EM waves . But mind you , this fields are as real as you , in the sense that they carry angular momentum , linear momentum , energy and what not . They are detected in the mobile set that you use . So try to think of fields as real entity .

Break the limit in two parts ( 0 , 1 ) & ( 1 , infinity ) .

Now put z = 1/ x in the second integral . See that after adjusting the limits accordingly , the second integral is nothing but the first integral with a -ve sign .

Hence the answer is zero .

Clear enough ?

no. of soln = 1 .

proof : 1 . First see that it has at least one soln .

Write the eqn. in the form log ( x + 5 ) + ( x -6 ) = 0 ;( x> - 5 )

Form the fn . F( x) = LHS of the above eqn .

first put x = 1/ 32 - 5 ( or whatever you like ) .So F ( x) is -ve at this point .

Again put x = 27 ( again choice is yours ) . For this F( x) is +ve .

Now F( x) being a continuous fn ( x > -5) the above two result suggests that F(x) must have a zero in between these two points .

2 . Now we prove for the uniqueness of the root

differentiating F(x) w.r.t. x we get the derivative is :

log ( e )(base 2 ) / ( x+ 5 ) + 1

see that for x> -5 ( for which the eqn. is valid ) the derivative is always > 0

therefore F(x) is an monotonically increasing fn .

which means that it can't attain a zero value more than once .

hence the root is unique .

Ohhhhhhhhhhh, I have forgotten to put the value of m. i. I . Well, its value is 2/3ml^2 about the axis mentioned . So for the latter case the complete answer would be angular accn. = 3/8 g/l .

Do you need a derivation of the above m.i. ?

ans A..P.

soln.

take the first one , cosA cot A/2 = cos A (s-a)/

= k1 cos A ( b+c -a )

= k2 { b+c - a ( cos A + cos B + cos C ) }

( k1 & k2 are consts that I dont bother to write )

{ apply the formula b= c cos A + a cos C

& similarly for c }

similarly write the other two given terms .

now from each of the terms cancel the term -k2 ( a+ b+ c) { since the given terms are in A. P . we are permitted to do so and the reulting terms are also in A.P.}

which means that the terms like

-k2 a ( 1 + cos A + cos B + cos C ) are in A.P.

which means terms like a i.e. a , b , c are in A. P. ( proved )

satisfied ? rate if u like it .

See, ideal gas obeys maxwellian velocity distribution. Which means that the probability of a gas molecule to lie between speeds v & v + dv is given by F(v)dv,

where F(v) is the probability disn. fn. whose expression (rather cumbersome ) has been worked out by Maxwell.

Now avg speed =

_{[ ]}

^{[ ]} v F(v) dv ( a simple statistical avg or mean )

rms speed = sqrt of

_{( [ ]}

^{[ ]} v.^2 F(v) dv )

m.p. speed is that speed for which F(v) is maximum , i.e. d F(v) / dv =0

satisfied ?

note : for all integrations limits are from 0 to infinity

Feynmann

Suppose that total resistance ( electrical) of the wire be R ;

Now emf inuced in the wire E = Bvl;

therefore current i= Bvl/R;

Force necessary = frictional forces + B i l;

= frictional forces + B^2 l^2 v/R;

got it ?

Rate if you like.

Use the formula for E in polar co- odinates for dipole .

Er = 2 p cos

/ r^3 ;

E

= p sin

/r^3;

here r= y;

= 135 degrees ;

got it ? rate if you like .

let the minimum force reqd. to tilt the cube be Fm;

then we get

Fm * 3l/4 = mg * l/2 ;

which gives , Fm = (2/3) *mg;

a> therefore for the first case , the cube will remain at rest

angular accn. =0

b> In this case total unbalanced torque = mgl/4;

Now calculate the moment of inertia of the cube about one of its edges .

Let it be I ;

therefore angular accn .= mgl/4I

a> appy the formula for entropy

dS = dQ/ T

but dQ = mcdT ( c= sp. heat of water )

therefore , dS = mc dT/T

integrating,

S= mc ln ( T2/ T1);

putting the values in , we get

S = 1310.8 J/K

b> heat released by the reservoir is

Q = 1 * 4200 * ( 373- 273) J

= 42000 J

therefore

S of reservoir is = - 42000/ 373 J/K

= - 112.6 J/K

c> total change in entropy = 1310.8 + ( - 112.6) J/K

= 1198.2 J/K

note that the process is irreversible ,so total change in entropy > 0

Methink that the final result does not depend on the nature of the spherical surface . It seems that the point source referred to is a source of light emitting photons of frequency nu .

Is it zehra ?

If so ,then I will post the solution. Also tell the intensity of the source .

Don' t need to short .

Leave the collector as it is (i. e. isolated) and use base as one terminal and emitter as the other terminal of the diode .

Those are very very simple sums .

find cos 15 from sqrt( ( 1+ cos 30 ) /2 );

The next one being tan 60 = tan ( 85 - 25)

are you sure that you require these two solns ?

I wonder , these being so simple !!!!!!!!!!!!!!!!!!

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