in the given equation multiply both sides ndT, u get,

nCdT=nCvdT+3nat^2dT

i.e

dQ=dU+dW

 

so dW=3nat^2dT

i.e PdV=3nat^2dT

or nRTdV/V=3nat^2dT------using PV=nRT

 

seperating the varibles and integrating u get,

 

ln V = (3aT^2)/(2R)

 

this is the reqiured eqution defining the process.

 

plz rate!

ans 1: k= [1/L] meter(inverse)

         c=[ML^3/(A^2)T^4]  Nm^2/c^2

 

ans 2: b( i.e the electric field at x=0 may b greater than 20 V/cm)

 

 soln 2: because the electric field may have a component along any other direction also(i.e the field along x axis may only b a component)

 

ans =40

 

let the ray incident on the concave portion pass trough the C.

after first refraction from the concave portion the ray moving towards the sivered portion should appear to be coming from its C(in that medium only).

 

so R comes out to be 40 cm(considering the combination to b thin) 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

ans 1= 1+5

ans 2= a,b,c

ans 3= all x>=4(d)

ans 4=30

ans 5=0

 

soln:

1)max distanxe of z lying on the given circle from i will be along the line throgh i passing through the centre of the circle. from 2 solutions obtained the one which lies on the other side of centre as i,gives the ans.

2)let z1,z2,z3 be the roots,

so c=1,

a=Iz1+z2+z3I<=Iz1I+Iz2I+Iz3I<=3

similarly b<=3

3)posted above

4)if u check for the maximum power of x that u'll get it comes out to b 625

to obtain 615 u have to omit some powers of x while mutiplying the brackets.

u'll have 2 remove (1,9) and (3,7) powers while actually multiplying the term along them.......

this gives 1*9+3*7=30

5)posted above

 

plz rate.

let v1=xi+yj+zk

v2=ai+bj+ck

 

Iv1I=5

Iv2I=4

 

v1.v2=20 so v1 s parallel to v2

 

let c=i+j+k

 

v2.c=A+b+c

v1.c=x+y+z

 

so (a+b+c)/(x+y+z)=v2/v1 as cos of angle between v1 and c = that between v2 and c

 

so ans is 4/5

 

plz rate!

dV(ba)=0

 

dV(cb)=dv(ca)=E*7

the set of favorable  triplets are (2,6,3), (8,2,4), (2,5,10)

 

and u can select a triplet of 3 distinct digits out of 10 in 10C3 ways

 

fav selection is 3c1

 

there4 the probability is 3C1/10C3= 1/40

 

 

 

 

x+iy=i^(x+iy)

 

ln(x^2+y^2)=-pi*y

 

and

tan(x(pi)/2)=y/x i.e y=xtan(x(pi)/2)

 

so we get on solving,

 

x=-(pi)/2*sin((pi)x)

 

this implies x=0

 

so y=0

 

so x^2+y^2=0 (ans)

 

 

 

 

ans=1/e

 

it converts to e^{[xlnx-x](0 to 1)}

 

 

 

 

 

soln1:

for x+y+z=0

 

we say x^3+y^3+z^3=3xyz

 

applying the same result after taking x=cis(),y=2cis(),z=3cis(),

 

we get 3cis()*2cis()*3cis()=18cis( _+_ +_)

 

to get sin compare the imaginary parts both sides!!

 

hope u get it!!

ans1=

18sin(++)

(x+d)y/(vx)