Messages posted by 3774answerer

I think you should probably consider repeating if you are sure of getting in the Merit List next year or else you would have been selected in other exams, haven't you?

I don't know about substitution but I can prove it like this:

Let a^2 b = m , b^2 c = n and c^2 a = o; and a, b and c.

Without any loss of generality, let's assume that a <= b <= c ( as the expression is symmetrical )

Obviously, m <= n <= o.

Now the value of ma + nb + nc is greater than or equal to that of mb + nc + oa ( From the rearrangement inequality )

So a^3 b + b^3 c + c^3 a >= a^2 b^2 + b^2 c^2 + c^2 a^2, or a^3 b + b^3 c + c^3 a - a^2 b^2 - b^2 c^2 - c^2 a^2 >=0.

This is nothing but the expanded form of the given expression.

Hence Proved.

18 = 3^2 . 2

For the factorials greater than or equal to 6!, they are divisible by 18. So they leave no remainder.

For those before,

1! + 2! + 3! + 4! + 5! = 1 + 2 + 6 + 24 + 120 = 153.

153 / 18 leaves the remainder 9.

PV = nRT

delta P V = nR delta T.

divide the two

delta P / P = delta T / T.

So the percentage increase in pressure is the same as that in temperature = 1 / 250 * 100 = 0.4 %

No Vishnu {1/x} is not equal to 1/{x}.

10 Volume H2O2 means that 1 Litre of this H2O2 liberates 10 Litres of O2 at STP.

So 500 cm^3 of O2 are liberated by 50 cm^3 = 50 mL of this H2O2.

Sorry Kapil I don't think square root of a rational number is always rational. Take 3 for example. root(3) is irrational.

Let the Radii of curvature be R1(air side) and R2(water side)

Then for rays coming from infinity, u = -infinity.

Apply the refraction formula

u2 / v - u1 / u = (u2 - u1) / R1

4/3v - 0 = 1/(3R1)

v = 4R1

Then for the second surface

3/2v - 4/(3(4R1)) = -(4/3 - 3/2)/R2

3 / 2v - 1/3R1 = 1/6R2

v = focal length

3 / 2f = 1/6R2 + 1/3R1

Depends on the equation. What is the polynomial equation?

It's a unitary method question. If he can feed 4 cows for three weeks, he can feed a cow for 4*3 = 12 weeks.

So he can feed 6 cows for 12 / 6 = 2 weeks.

I don't know this one, but wikipedia says Nathan Astle.

I think it is Sir Don Bradman(officially approaching 100, unofficially > 100)

If the drop is falling down with constant speed, its linear momentum is not changing, and so no force is acting on it.

The gravitational force cancels the viscous and buoyant forces exactly and no force acts on the drop.

Magnitude - 0, direction - not defined.

This is a recent one.... Rahul Dravid with 183 catches I think.

Great answer Madhusudan sir. Only an expert with thorough knowledge of chemistry would have been able to answer the question.

Although I'll have to say that the question was out of JEE syllabus.

I think 2, one of them was Anil Kumble.

The process will probably take place at constant volume, if it happens in a closed container.

So the molar heat capacities will be 3R for CO2 and 2.5R for O2

Let the final temperature be T.

Heat recieved by CO2 = Heat given by O2

3R(T-27)(22/44) = 2.5R(37-T)(16/32)

Solve it to get T = 31.55 degrees approx.

On the other hand, if it is given that the process takes place at constant pressure, then the heat capacities will be 4R for CO2 and 3.5R for O2.

4R(T-27)(22/44) = 3.5R(37-T)(16/32)

Solve to get T = 31.66 degrees approx.

Viv Richards, off 56 balls.

Shahid Afridi in 37 balls in 1996 against Sri Lanka.

No, if n = 10, the answer is 2^10.

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