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This cant be possibly true.
Because:
1. Indians have never used farenhiet in their vocabulary (when he says about the minimum temperature possible)
2. The concept of dualism is not encouraged in indian philosophy courses, most colleges offer vedantic or indian philosophy
3. There is no credit as to where this article is from, cannot even confirm whether true or not
Please check with these facts
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This function is non-integrable under the JEE syllabus.
Similar non-integrable functions are sqrt(Cos x), Sin(x)/x, Cos(x)/x.
Even though they can be integrated by special methods, those methods are neither under 10+2 scope nor under JEE syllabus.
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Let the no. of small shirts be x
Then , the no. of large shirts is x/2 as there are half no. of large shirts as small shirts.
Now, the no. of xtra-large shirts is again x, as the no. of large shirts and xtra-large shirts are the same.
Now, as there are 1/3rd small shirts as medium shirts,
the no. of medium shirts is 3x.
Now, the percentage of medium shirts are:
=[(3x)/(x+ x + x/2 + 3x)]*100
=600/11
which is approx equal to 54.55 %.............
Rate me if I am correct.
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The question is a bit misleading...
The question means that for smaller imparted tangential velocities
,the particle will undergo a parabolic path which involves physical contact with the sphere, hence the particle "slides" on it.
There will be a minimum velocity for which the particle "just" touches the sphere, hence goes undisturbed on it's parabolic path.
All velocities greater than this will involve no contact with the sphere.
Maybe, this is the minimum velocity asked in this question.
Verify me.
Rate me iff correct......
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It is due to q2 only.
My Reason:
The gaussian surface encloses to equal and opposite charges, hece the enclosed charge due to these two charges is zero.
Hence, by gauss' law, the electric field due to q1 and -q1 is zero on the surface.
The remaining is q2 which has an independent feild of it's own which is spherically symmetric about q2, hence, on the given gaussian surface a non-zero electric field exists due to q2.
Hence, the net field on the given surface id due to q2 only.
VERIFY my answer.
Rate me iff correct.
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1 mole can be understood and defined as that quantity of entities that are in .012 kg of carbon 12.
It so happens that the no. of units in 1 mol of a substance is equal
to 6.022 x 10^23.
If you consider the mass of 1 mol in kg then it will remain the same, and in relative atomic mass units, it will become half times the original.
The confusion here is apparently that one has taken the definition of 1 mol on the basis of relative atomic mass units. the definition of both these quantities is independent of each other.....
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Just draw a neat diagram for this question.
Use the formula
F=G(m1)(m2)/r^2 and find the force vectors for each four forces ech corresponding to one pair of centrally located mass and the corner located mass.
Show this in the diagram.
Now, Add the vector-add the forces to get the resultant force.
If you want the vector form of the force, you can assume the centrally located mass at the origin.
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to find the area of the parallelogram encompassed by two vectors, just take the cross product of the two vectors and find it's magnitude. the value of the magnitude will give you the area. This is because: |a x b|=|a| |b| Sin(A) where a is the angle between the two vectors. also |b| Sin(A) is the value of the perpendicular projection of 'b' vector on 'a' vector, therefore |a| |b| Sin(A) is equivalent to the area of the parallelogram area formula (base)*(perpendicular projection or altitude). Hence, the area.
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Since this is a question from kinematics, we must probably assume the value of g to be 9.80 m/s^2. I agree with gautam, that in most problems the value is taken to be 9.8. also if the "great height" is not given or mentioned with a numerical value, we must assume it to be within calculation limits, and do as gautam has done. he has presented the correct answer as far as his assumption is concerned.
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Hi,
Can you get me
Resnick, Halliday and Walker
Vol.1 and 2
John wiley and sons
I couldnt find a hard copy of this book.
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If s denotes the displacement and x the time,
then the displacement s is =56*5, that is 280 units.
Furthermore, if you differentiate this equation w.r.t time you get the velocity v=5 units/time.
Even with this velocity, you can apply the formula s=v*t where s is displacement, v is velocity and t is time, and get the same result.
(Its the same thing whatever you do, as long as it is correct.).
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Hey Akshat, yeah, I agree with magiclko, its the stress thats bugging you, just chill out and try to imagine that you are attempting your exam at your home, then your panic levels will drop down. The B.Mat scores count lesser than what actually you know. If you are sure that you know everything well, then there should be no trouble regarding the real JEE. These exams are only meant to give you the actual atmosphere JEE creates for itself, if you get used to this, then, you can score at the test-centres too.
Best of Luck on part of me too.
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If (2m -1) is a prime, how will you prove that 2m-1(2m-1) is a perfect number, (ie, those numbers whose sum of divisors is equal to the number itself, for example, 6=1+2+3 and 28=1+2+4+7+14, thus 6 and 14 are perfect nos., if we use this formula we get the next perfect number to be 496,). Can you think of other such formulae? How can this be solved using the principles of the binomial theorem?
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if a,b,c are three positive real numbers such that a>b>c. Then, prove that [(a-c)/(a+c)]^a > [(b-c)/(b+c)]^b
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