Messages posted by KAB

Ans is 24

Eqn is a(x^2+2gx/a+g^2/a^2)+b(y^2+2fy/b+f^2/b^2)=g^2/a+f^2/b-c=t

a(x+g/a)^2+b(y+f/b)^2=t

((x+g/a)^2)/(t/a)+((y+f/b)^2)/(t/b)=1

Eccentricity e=sqrt(1-((t/b)/(t/a)))

e=sqrt(1-(a/b))

e=sqrt((b-a)/b) Assuming b>a

Given=(2a+3/a)+(2b+3/b)+(2c+3/c)

A.M.>=G.M.

So (2a+3/a)/2>=sqrt(6) [sqrt-square root]

or (2a+3/a)>=2sqrt(6)

Similarly (2b+3/b)>=2sqrt(6)

and (2c+3/c)>=2sqrt(6)

Adding them we get

(2a+3/a)+(2b+3/b)+(2c+3/c)>=6sqrt(6)

THEREFORE

(a+3/b+c)+(b+3/c+a)+(c+3a+b)>=6sqrt(6)>=14.69

SO THE MINIMUM VALUE IS 6sqrt(6)=14.69

SO THE ANSWER IS 4)NONE OF THESE.

Sir,

please tell me if i'm wrong anywhere

Sir if we treat fuse wire as a cylinder then shouldn't its volume be

r^2h?and doesnt fuse wire has low melting point?

answer is t+(t^3)/3-t^2+c

I agree with saket.

Thanks for a salute.

a>b

a/c>b/c

Now suppose

(a-c)/(a+c)>(b-c)/(b+c)

(a-c)/(a+c)-(b-c)/(b+c)>0

(ab+ac-bc-c^2-ab-bc+ac+c^2)/(a+c)(b+c)>0

2c(a-b)/((a+c)(b+c))>0

which is true since a>b and a,b and c are all positive.

THEREFORE HERE COMPONENDO - DIVIDENDO CAN BE APPLIED.

[(a-b)/(a+b)]^a>[(b-c)/(b+c)]^b

HENCE MY METHOD IS RIGHT!!!!!!!!!

Solve the practice papers.Revise the important concepts.Be positive.Have confidence.Best of luck!!!!!!

Substitute t=5 to get displacement and distance ie 527

Well Titun,
You have given an example where denominators are not equal.You have taken 4/5>1/2 as eg.Can you find other examples where the denominators are equal and you cant apply componendo-dividendo.
(Keeping in mind a>b>c).If you have found then my method is wrong otherwise my method is SHORT AND CORRECT!!!!!!!!!!

You are right.But 2bcosx=5 therefore b=5/2 since x tends to zero and cosx tends to 1 which is not the reason given by you.

I wasnt sure that they couldnt be evaluated.

Sir,

Plz explain how to solve.

If limit x tending to zero of

(acosx+bxsinx-5)/x⁵ is a finite value then (a,b)=?

Ans is given as (5,5/2)

a > b

implies a/c>b/c

(a-c)/(a+c)>(b-c)/(b+c)

Obviously

[(a-c)/(a+c)]^a>[(b-c)/(b+c)]^b since a > b

Try to remember the formulae(focus,directrices etc) of all the types of parabolas and then you will find it very easy.

Sir,

How to derive,

i²

r³

where,

i is the maximum current that can pass through the fuse wire without melting.

r is the radius of the fuse wire.

Is it applicable to any wire or only fuse wire?

When water is used as the solvent, the dissolving process is called hydration.
Hydration enthalpy of an ion is the amount of energy released when a mole of the ion dissolves in a large amount of water.

Hi,
Better dont leave mechanics and dont keep it pending.Since now you are thorough with its concepts it would be easier for you to finish the topic right now.Remember "for whom mechanics is easy for them the whole physics will be easy".

thanx

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