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Normal forced never perform work on rigid bodies. Here the sphere is rigid. So, even after the collision, KE is conserved, provided no non-conservative force acts on the body..

The basic definition of Elastic collision is that the energy is conserved. If u review the definition of Coefficient of restitution, u will find that it has nothing to do with energy.

But mathematically, it happens that coefficient of restitution is 1 for perfectly elastic collisions.
So, both the definitions are true for perfectly elastic collisions.
Now coming to the problem, I would like to stress on the exact definition of Coefficient of restitution.

e = (rel. vel. of seperation)/(rel. vel. of approach)
In this definition we talk about two relative velocities. Note that both the relative velocities MUST be taken in the direction of common normal. So, inthis particular proble, when u use the definition of Coefficient of restitution, u need to take the relative velocities along the common normal. ie. normal to the sphere at the point of contact. If u go by this method, u will get the same answer as u get by conservation of energy method.

The min no of tosses required for B's cash to go to zero is 20.
So, he will go bankrupt on 21st toss (if he looses on 21st toss)

It's a factor of 3.
The easiest way to confirm is by checking the power.
Let Vline-line = 1 Volts.
(This means Vline-neutral = 1/sqrt(3) V)
Let Zcap,wye = 1 ohm
Choose freq = 1/(2pi) Hz, so w = 1 radian/second
C,wye = 1/(w*Zcap,wye) = 1 F
Total Power = (Vline-line)^2/Zcap,wye = 1 VAR

Let's guess that impedance in delta is 3x impedance in wye.
Zcap,delta = 3 ohm
C,delta = 1/(w*Zcap,delta) = 1/3 F
Check Power:
Power in each line-line branch = (Vline-line)^2/ Zcap,delta = 1/3 VAR
Total Power = 3 * 1/3 VAR = 1 VAR

So, to convert from wye to delta, multiply impedance by 3.
But, in a capacitor, impedance is inversely proportional to capacitance, so be sure to invert the 3!

source: yahoo answers

Do ur cbse stuf fully. That will fill70% of JEE prepn.
The remaining 30% is practice. Make sure that u solve a particular number of problems every day. During your exams, solve atleast 1 problem a day. That willmake sure, that u dont loose contact with problem solving.

The light remains as parallel beam after the first refraction...
So, this parallel beam converges at the focal point on the other side of the prism. So, using refraction at curved surface formula find focal length. The beam converges at the focal point of this prism.

Once double check your question...
Because, the x^3 in the numerator and x in the denominator cancel....

The min no of tosses required for B to go bankrupt is 21.
There is no max no of tosses, as the game can go for ever.
The next min no of tosses is 23. Then 25, 27, 29... so on.....
Find probability for the occurence of each of these cases. and also find probability for B to loose in each of theses cases. Multiply these two probabilities in each case....
Now u will get an infinite series (coresponding to 21,23,25,27,29,......)
Add this series to get answer.

Well done Rohith....

Once again the same answer:

U can leave this problem, as it is out of JEE syllabus.
This problem needs a concept called continious probability to solve. If u r so desperate about solving this problem, then google for the concept and try solving. Its an easy problem, once u know the concept.

U can leave this problem, as it is out of JEE syllabus.
This problem needs a concept called continious probability to solve. If u r so desperate about solving this problem, then google for the concept and try solving. Its an easy problem, once u know the concept.

I menat there are no NRIs.
There are students who lived abroad for quite some time and then came back to India for studies....

Here we need to consider only the expansion of block. (Forget abt the expansion of liquid.)
So, as the temp increases, the density of the block decreases but the weight remains the same.
The weight of the liquid displaced must be equal to the weight of the block. So, the same volume of liquid must be displaced even now. So, the volume of the block inside liquid remains same. Since, the area of cross section of the block increases due to heating, the depth (though with the block dips in water) must decrease to maintain constant volume of liquid displaced. So, x decreases. The height of liquid remain same. So, y must increase.

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