how many positive integral soutions are there for the equation x+y+z+w=20?

 please do using marker method and explain.

let the complex numbers z₁ z ₂and z₃ be the vertices of equliateral triangle .let be z ₀ the circumcentre of the triangle then pt     

 

   z_1²+ z_2²+ z_3² = z_0²      

                                                      z_{1^{2   means}} z_{1 ^square}

a(n) = 1 + 1/2 + 1/3 + 1/4 + .............+ 1/(2ⁿ-1) then

 

A)a(100)100                           B)a(100)>100

C)a(200)100                            D) a(200)>100

 

                                        please give the full solution of the problem.thank you

hi,

the complex number (sinx + icos2x) and cosx - isin2x are conjugate to each other for:

 A)x = n                         B)x=(n+0.5)

 C)x=0                              D)no value of x

 

 

            please also do mention  how can two complex numbers be conjugate wrt each other.   

THIS is only to members of goiit team .

                                            as previously you have conducted a exam on dec 16 ,are you conducting any more exams like that in future?my request is that please do conduct such exam atleast once more .

                                                                   thanking you,

                                                                                                            kusum

yes you are write so the only change in my solution is in the energy eqution where instead of mv²/2 =mg(R+r)(1-cos)we will have  that

I²/2 + mv²/2 =mg(R+r)(1-cos) where I=mr2      =v/r. 

the question can be understood better and made  much simpler by drawing the appr diagram and writing all the forces acting at the point the small sphere leaves the larger one

                           let  be the angle with the vertical ,so the point to remmeber here is  AT THE POINT CONTACT THE CONTACT FORCE IS ZERO.

let say at the point of leaving v is the velocity(which will be in a direction tangntial to the surface)

 to find v

         gain in K.E=loss in P.E

         mv²/2 =mg(R+r)(1-cos)

         mv² =mg(R+r)(1-cos)*2..........................(i)

 

 at the point of leaving,

N + mgcos=mv²/r .................(ii)

here N=0  substituting (i) in (ii)

 

i wish to answer this .please give me 5 min.is the answer cos =2/3 and

=sqrt(2g/3(r+R))

 

 

this one seems to quite easy.
                                         when u draw the diagram u assume the angle the strng makes with the horizontal to be .let the tension be T (it will be equal in both strings to left & right of midpoint).let m be the mass of body hung.

     

 =(5.1*10^-2/.915)   

2Tsin=mg

  =5.1*10^-2 /.915 .......which is very small for tansin.

2T=mg

from above we get T in terms of m

 

2L is the total elongation of the wire,L initial legth of wire=0.951

considering one half of the wire (take either right or left),we proceed

Y=(T/A)/(L/L).....................................(i)

 now L=(0.951)/2cos-0.951

definitely Y has to be given in the prob

 substituting all values in (i) we get m.

                  i hope i am clear in all aspects.

hi nishant,

                 you could conserve the angular momentum of the system,

                    initial angular momentum=final angular momentum

                    I-mrv=(I+mr²)'

                 '=( I-mrv)/(I+mr²)

              i have used a -ve sign considering that angular momentum direction due to the cockroach is opposite to that of disc.ALL THE BEST.

 

also i have assumed that v is taken wrt ground .if it is v wrt disc

then take v in the exp as v+r.