isme kya hai

cos40= sqrt(1-m^2)

tan40=    m/(sqrt(1-m^2)

thanks to all.

abey yaar aisi faltu baatoin ka time milta hai tujhe.

ab to waise bhi usne apna decision wapas le liya.

i have a conceptual doubt in this question

TWO CHARGED PARTICLES OF MASSES m AND 2m ARE PLACED AT A DISTANCE d APART ON A SMOOTH HORIZONTAL TABLE. BECAUSE OF THEIR MUTUAL ATTRACTION THEY MOVE TOWARDS EACH OTHER AND COLLIDE. WHERE WILL THE COLLISSION OCCOUR WITH RESPECT TO THE INITIAL POSITIONS

my doubt is that as the particles are charged there is a coulomb force of attraction.

1.  then is this not an external force?

       the answer explained and given is as the centre of mass(i.e. 2d/3 FROM m)

also centre of mass of system should have an aceleration.

please explain................. rates assured

ex. 9.3 HCV I

let theta=@

tan (pi cos @)=cot (pi sin@)

sin(pi cos@) / cos( pi cos@) = cos(pi sin@) / sin ( pi sin @)

cross multiplying and taking lhs to rhs,

cos( pi cos@ + pi sin@) = 0

=>  pi (cos @ + sin@)= [ (+-) pi /2]

=> cos@ +sin @= (+-) 1/2

dividing both sides by sqrt(2)

we get cos(@- pi/4)= (+-)1/ 2*(sqrt(2)) ................................ans.

Sorry but this is the wrong answer.

the answer given at the back is (-2*mu*g*L /9)

but i don't know how did we reach there

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PLEASE EXPLAIN HOW DO WE SOLVE  Q 39, ...............RATES ASSURED .

PLEASE PROVIDE DETAILED SOLN.

Good question

now according me answer is as follows

consider the whole function in the fourth quadrant just near 3pi/2

then tan x tends to (-infinity)

cot x is very nearly zero.

cosecx is nearly (-1)

and cos x can be neglected.

so ans should be  (-infinity.

experts please verify

good yaar,me too understood the ques. as this was a mcq and i did not have detailed soln.

thanks both of you

so is ans. given at the back wrong . you got 3 values while ans. says 2.

yes qakit your answer is correct.

sorry taran you got it wrong this time.

still i rated you bcoz you were quite close

why/what  are you considering fractions

plz.explain

rates assured..............

Find the total number of integral values of N,so that

sin x(sin x+cos x)=N

has atleast one solution

PROVE THAT:

[ cos(PI/8) ]^4+[ cos(3 PI/8) ]^4+[ cos(5 PI/8) ]^4+[ cos(7 PI/8) ]^4= 3/2

DETAILED SOLN. PLZ.

RATES ASSURED........

IF A+B+C=pi

PROVE THAT :

SIN(A/2)+SIN(B/2)+SIN(C/2)= 1 +4*SIN[ (PI-A)/4]*SIN[ (PI-B)/4]*SIN[ (PI-B)/4)

PLZ. PROVIDE DETAILED SOLN.

RATES ASSURED..........

I AM GIVING SOLN ONLY FOR THE LINE IN WHICH THE PARTICLE WOULD BE IN SECOND QUADRNT

LET THETA BE THE ANGLE MADE BY THE LINE PASSING THROUGH (UNDER ROOT 3,1),

THEN TAN THETA=[1/SQRT(3)]

SO WE GET ANGLE IS 30 DEGREE.

ANGLE MADE BY THE NEW POSITION OF THE POINT IS 30+90=120 DEGREES.

ALSO SINCE THE THE LINE JOINING THE NEW POINT AND THE ORIGIN IS OF THE FORM y=mx,

WE GET THAT THE EQUATION OF NEW LINE IS y=(tan 120)x    Y=(-SQRT(3))X...................[1]

SOLVING FOR THE EQUATION OF LINE AND CIRCLE SIMULTANEOUSLY WE GET,

x^2+3*x^2=4

x^2=1

x=[-1].......................................................IN SECOND QUADRANT

but slope of the line passing through the new point IS[1/SQRT(3)]

AS @=180-60-90!!!!!!!!!

BUT THERE WILL BE 2 LINES ALONG WHICH PARTICLE MAY MOVE.!!!!!!!!!!

YOU MAY SOLVE TO GET THE ANSWER.