The results of Kerala Engineering/Medical Entrance Examinations conducted during the last week of April 2007 by the Commissioner for Entrance Examinations, Govt. of Kerala are announced. The results are available at www.keralaresults.nic.in and at www.results.kerala.nic.in
Friday, May 25, 2007
Wednesday, May 23, 2007
IIT-JEE 2007 Assertion-Reason Type MCQ on Friction
The following Assertion-Reason Type MCQ which appeared in IIT-JEE 2007 question paper is meant for checking your understanding of the conservative nature of gravitational force and the non-conservative nature of the frictional force:
STATEMENT-1
A block of mass m starts moving on a rough horizontal surface with a velocity v. It stops due to friction between the block and the surface after moving through a certain distance. The surface is now tilted to an angle of 30º with the horizontal and the same block is made to go up on the surface with the same initial velocity v. The decrease in the mechanical energy in the second situation is smaller than that in the first situation
because
STATEMENT-2
The coefficient of friction between the block and the surface decreases with the increase in the angle of inclination.
(A) Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1.
(B) Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1.
(C) Statement-1 is True, Statement-2 is False
(D) Statement-1 is False, Statement-2 is True
In both situations the initial kinetic energy of the block is reduced to zero. In the first situation, the initial K.E. is entirely used up in doing work against friction and is irrecoverably lost since friction is a non-conservative force. In the second situation, part of the initial K.E. is used in doing work against gravitational force and is stored as gravitational potential energy in the block since gravitational force is a conservative force. The decrease in mechanical energy in the second situation is therefore smaller than that in the first situation.
Statement-1 is therefore true.
Statement-2 is false since the coefficient of friction is independent of the inclination and is dependent only on the nature of the two surfaces in contact.
The correct option therefore is (C).
Sunday, May 20, 2007
MCQ on Wien’s Law
The following question on Wien’s law is a typical one and is of the type often found in Medical and Engineering Entrance test papers:
A black body emits radiations of maximum intensity for the wave length of 7000Ǻ, when the temperature of the body is 3927ºC. If the temperature of the body is increased by 1000ºC, the maximum intensity would be observed at nearly
(a) 3650Ǻ (b) 4650Ǻ (c) 5650Ǻ
(d) 6000Ǻ (e) 6650Ǻ
The wave length λm corresponding to the maximum intensity emission from a black body at a temperature T is given by Wien’s law,
λmT = constant [= 0.29 cm K, if you express λm in cm].
In other words, λm1T1 = λm2T2
[Remember, you have to substitute the absolute (Kelvin) temperatures in this equation].
Therefore, 7000×4200 = λm2×5200, from which λm2 =5650Ǻ, nearly.
[For question setters: When a question is set using Wien’s law, it should be remembered that the product λmT satisfies the condition, λmT = 0.29 cmK, if λm is expressed in cm. I write this because I have seen questions in which this point is ignored]
Monday, May 14, 2007
Two Kerala Engineering Entrance 2007 Questions from Nuclear Physics
The following MCQ (on radioactivity) which appeared in Kerala Engineering Entrance 2007 question paper is simple, but it differs slightly from the conventional type:
Radium has half life of 5 years. The probability of decay of a radium nucleus in 10 years is
(a) 50% (b) 75% (c) 100% (d) 60% (e) 25%
Since the half life is 5 years, the amount getting decayed in 10 years will be 75%.
[This can be found very easily: After 5 years half the initial amount will be decayed; after another 5 years, half of the remaining will be decayed. If the half life and the time period given are not so simply related, you will have to calculate the number of nuclei undecayed (N) using the equation, N = N0/2n where N0 is the initial number and ‘n’ is the number of half lives in the given time. The percentage decayed in the given time is then calculated].
The probability for decay of any given nucleus in 10 years is therefore 75%.
The following question involving the relative abundance of isotopes is a popular one and it has found place in Kerala Engineering Entrance 2007 question paper:
The natural boron of atomic weight 10.81 is found to have two isotopes B10 and B11. The ratio of abundance of isotopes in natural boron should be
(a)
This is a simple arithmetical problem. If there are n1 boron atoms of atomic weight 10 and n2 boron atom of atomic weight 11 in a sample of natural boron, the mean atomic weight (which is given as 10.81) is related to n1 and n2 as
(10n1 + 11n2)/ (n1 + n2) = 10.81.
Rearranging, 0.81n1 = 0.19n2, from which n1/n2 = 19/81.
Sunday, May 13, 2007
AP Physics Exams.
Many among the regular visitors of this site might have already noted that the posts here are useful for preparing for the AP (Advanced Placement) Physics Exams. Multiple choice questions discussed here have five choices as in the AP Physics Exams and are designed and expected to be worked out without using calculators and physical and mathematical tables. Even though multiple choice questions are discussed here, you will find the posts useful for answering the free-response section of the AP Physics Exams also, since the questions are answered with all necessary theoretical details.
You will find more questions (with solution) covering the content of AP Physics at physicsplus
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