d.prop to s⁵.

 S=K(cube root of V)

 S³=K.V

a=3s²v=ks⁵

 

 

 

Yes,I got the ans.substitute T=373K & L=539.9 in the above mentioned formula.

 

 

 

(I did the calculation using calculator)

ans is A 0.08C.

in 10⁴ atoms,1 e is removed.so,5*10¹⁷electrons are totally removed.

so,charge=5*10¹⁷*1.6*10^-19=0.08C

 

 

 

 

rate me,if the ans is correct

K_b=0.002T²/L(whereT is boiling point of water(solvent) &L is latent heat of vapourisation)

 

 

 

 

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Oh,sorry.I didnot read the question correctly.so,the answer is

pir^2/16-1/2r^2sin45/2(where r=10cm)

Q1.(1+cospi/8)(1-cospi/8)(1+cos3pi/8)(1-cos3pi/8)

     =sin^2pi/8.sin^2 3pi/8

     =1/4(cospi/4-cospi/2)²

 

 

 

 

 

Q1.the ans must be r^2/16.If we cut the circle into 16 equal parts,each subtends an angle of 45/2 at the centre

 

 

 

 

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let V(rain)=ai+bj,V(man)=2i in1 case & 4i in 2 case

rel vel=(a-2)i+bj=-Kj so,a=2

similarly in second case,rel vel=(a-4)i+bj=c/2i-croot3/2j

(we assume c to be vel in that direction).From this,c=-4,b=2root3

so,V of rain=2i+2root3j,whose mag is4

 

 

pls rate me if my sol is correct

Let the dis travelled be x by each.then dis bet them is

           

           ^{x^2 +(10-x)^2}

ans is D.calculate net force on one of the four charges.it comes to beKQ^2/2a^2(22+1),diagonally outward.to balance this,central charge must be -ve.balancing force,we get ans asD

 

 

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