Messages posted by Chirag...I rock.. ..

Ease yourself by preparing for AIEEE without much bothering about the distribution pattern...

If I say Mechanics has a greater weightage, then that doesn't means that you will leave out Calculus part since Kinematics requires lot of Calculus...

So, I say please don't get tensed about the distribution....

Hope this will ease your tension

Thanx...

The answer is 36.

Soln:-

2004-24=1980

1980=2^2*3^2*5*11

therefore, no of factors=(2+1)(2+1)(1+1)(1+1)

=3*3*2*2

=36

hope it solves your problem...If there is any doubt please post it..

please rate for what I have done...

The statement is absolutely wrong..

v

=constant

that constant turns out to be 'c' i.e. speed of light.

v=c/

therefore frequncy is inversely proportinal and not directly..

Please rate for my efforts.......

thanx

I have checked my soln, still I am getting 630m/s.

Pl tell me how did you get 444m/s

The mass on the right will be nearer because velocity 'v'(i am assuming it for convinience) is given to the right mass

therefore, the tension will develop in the string with a effect of Tsin

=m(v-u)/t

but initial velo.=0

Tsin

=m(v)/t.

but v is linear proportional with distance....

Therefore, the length of string increases on the right part since tension is developed and the string is same.

This is the soln, if you have any doubt please free to ask me..

Please rate for what I have put in....

Thanx..

Maths:-

1)M.L. Khanna...IIT Maths(Subjective)

2) Pradeep Publication..IIT Maths

3)Higher Algebra...Hall and Knight

4)+2 Textbooks

Physics:-

1)Halliday and Resnick

2) H.C. Verma

3)Pradeep Publication..IIT Physics

4) Gupta and Gupta....IIT Physics(JPNB Publication)

Chemistry:-

1) O.P Agrawal....IIT Chemistry(JPNB Publication)

2)R.C. Mukherjee

3)Arihant publication's Organic Chemistry..

If you want more help you can join a correspondence course of a reputed institute like FIITJEE or Brilliant Tutorials which are of good use...

tan(C/2)={

/s(s-c)}

tan(C/2)=c^2-(b-a)^2/s(s-c)

tan(C/2)=(c+b-a)(c-b+a)/s(s-c)

tan(C/2)=2(s-a)2(s-b)/s(s-c)

tan(C/2)=4(s-a)(s-b)/s(s-c)

4{tan(C/2)}^2=tan(C/2)

tan(C/2)=1/4

Please rate me...........

In

ABD and

ACD

using sine rule,

sinBAD/BD=sinABD/AD

therefore, AD=sinABD*BD/sinBAD.......(i)

similarly,in

triangle ACD, sinCAD/CD=sinACD/AD........(ii)

Equating (i) and (ii) for AD

sinBAD*2/

3BD=sinCAD*CD

2

therefore we get sinCAD/sinBAD=1/

6

^{Please rate me...............}

a^4+b^4+c^4=2c^2(a^2+b^2)

a^4+b^4+c^4-2c^2(a^2+b^2)=0

adding 2*a^2*b^2 both sides,

a^4+b^4+c^4-2c^2(a^2+b^2)+2*a^2*b^2=2*a^2*b^2

(a^2+b^2-c^2)^2=2*a^2*b^2

taking sqrt both sides,

(a^2+b^2-c^2)=+

2*a*b.............(i)

or

(a^2+b^2-c^2)=-

2*a*b............(ii)

^{using cosine rule,}

^{cos(c)= b^2+a^2-c^2/2ab...................(iii)}

^{Substituting (i) and (ii)}

^{we get,}

^{cos c= 45 or 135}

^{therefore ans(B)}

The answer is irrrational because whenever you take P less than 2, there is exists a number B which adds up to it...........

Please rate me.....

Q1)

x³+27=0

(x+3)(x²-9x+9)=0

x=-3 or x=9

sqrt(45)/2

(c/a)²=126-18*sqrt(45)/36

(b/a)²=126+18*sqrt(45)/36

(c/a)²+(b/a)²=7

(c/a)²*(b/a)²=418.5

therefore eqn is x²-7x+418.5=0

Q2)

By completing the square method, you get the condition, k<16

I think answer may be 4...

Please rate me.........................

Can you explain in detail me what is hudograph?

If the gas is monoatomic, then answer is 2R, if diatomic then 3R....

formula:- c=R/(cp/cv)-1- R/n-1, where n is the co-efficient of V on the question....

Ans:- 5.61 Hours

Soln:-

v(avg)=sqrt(8RT/piM)

sustituting the values, we get,

v(avg)=630m/s

diamter of earth=12732km=12732000m

therfore, time=12732000/(630*60*60)=5.61 hours

according to ist kinematic eqn,

v=u+at

But IInd says that

s=ut+.5at^2

here taking out 't' common and dividing we get,

v=u+.5at

What causes the diff of .5 at in both cases

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