Science based numerology for this year
Science based numerology for this year Just for jest.
Science based numerology for this year Just for jest.
The CSIR NET 2018 held on December 16: Indian
Assistant Professor and PhD scholarship exam
solution, prepared by me. The answers and
detailed explanations are available for 18
out of 25 questions of Part-B. The detailed
explanations and answers to Part-A is also
available, see link below.
Please point out any inadvertent errors.
(this is entirely free stuff: help spread the word)
The article aims to make the best attempt at finding the answers for the recently concluded 2018 CSIR NET. Detailed explanatory answers for physical sciences section ( part-B ) is available ( for 18 out of 25 questions at the moment ). Also full explanation based solutions to part-A is available, check link above.
Q – 21. Consider the decay A → B + C of a relativistic
spin 1/2 particle A. Which of the following statements
is true in the rest frame of the particle A?
1. The spin of both B and C may be 1/2.
2. The sum of the masses of B and C is greater than
the mass of A.
3. The energy of B is uniquely determined by the
masses of the particles.
4. The spin of both B and C may be integral.
Explanation: obviously the second option is incorrect as it violates conservation of energy in relativistic kinematics, rest-masses of the product particles can not be more than that of the parent particle. Option 3 is explained with a diagram. The value of the energy of one of the daughter particle ( B ) is determined uniquely as evinced by the given formula for the same.
In a two body relativistic decay in the parent rest frame: EB = ( MA2 – MC2 + MB2 ) / 2 MA
Also the other options talk about the spin of the particles but we need not bother since option 3 is the correct option.
The article aims to make the best attempt at finding the answers for the recently concluded 2018 CSIR NET. Detailed explanatory answers are available to 19 out of 20 questions in Part A, at the moment. Only physical sciences paper part B and C will be added.
CSIR NET 2018 December physical sciences
Part A
Q – 1. A tourist drives 20 km towards east, turns
right and drives 6 km, then drives 6 km towards west.
He then turns to his left and drives 4 km and finally
turns right and drives 14 km. Where is he from his
starting point?
1. 6 km towards east
2. 20 km towards west
3. 14 km towards north
4. 10 km towards south
Q – 2. In an examination 100 questions of 1 mark
each are given. After the examination, 20 questions
are deleted from evaluation, leaving 80 questions
with a total of 100 marks. Student A had answered
4 of the deleted questions correctly and got 40
marks, whereas student B had answered 10 of the
deleted questions correctly and got 35 marks.
In this situation
1. A and B are equally benefited.
2. A and B lost equally.
3. B lost more than A.
4. A lost more than B.
Q – 3. A circular running track has six lanes,
each 1 m wide. How far ahead ( in meters )
should the runner in the outermost lane
start from, so as to cover the same distance
in one lap as the runner in the innermost lane?
1. 6 π 2. 10 π
3. 12 π 4. 36 π
i. We assume a dilute gas which is enclosed by a thermally insulated container on all sides.
Dilute gas in a thermally insulated container: Dilute means concentration of gas molecules is low. Insulated implies there is no reasonable flow of heat energy across the walls of the container.
ii. Each molecule is assumed to be a hard sphere which moves randomly in all directions such that its velocity vary from – infinity to + infinity.
Hard sphere: Remember a hard sphere is a classical analogy of a rigid sphere whose surfaces do not deform when an external object comes into contact. This essentially means the incoming object is scattered elastically that is without loss of kinetic energy, only momenta magnitude and directions are changed in accordance with the conservation of linear momentum.
iii. When molecules collide they do not lose energy or time. They bounce off each other so that ‘energy’ and ‘momenta’ are conserved.
We would like to obtain an expression for velocity distribution function. That is we would like to know the fraction of molecules having velocity between v to v+dv for all possible values of velocity.
For this we assume an ensemble of molecules in equilibrium. The ratio of number of molecules in a velocity range to the total number of molecules N gives the probability of finding a molecule in that velocity range.
v. The “phase space” of the ensemble of molecules is defined by a 6 N dimensional space, which constitutes of 3 N spatial components and 3 N velocity components of the N molecules in equilibrium. For a more advanced concept of phase space check the following statistical mechanics lecture.
A. structure of the nucleus
Every atom consists of a dense positive central core of mass, known as nucleus. Its size is much smaller compared to the size of the atom, nut nonetheless it contains almost all of the mass of the atom.
The nucleus is made of only neutrons and protons. — These are collectively known as nucleons.
Nucleons are not the smallest constituents of matter. — In-fact nucleons are made of different combinations of 3 quarks, of which only two of the quarks can be of the same type.
Combinations of 3 quarks which form into a bound state of material system are known as baryons. We will study about baryons in the last part of this lecture series.
Thus baryons are a naturally occurring collective matter, built from 3 quarks, where the 3 quarks interact among each other, but can’t escape this bound state of formation, even under the impact of tremendous force.
This fact is known as asymptotic freedom, such freedom is only a dream for them, and for us. This is possible in principle, when the distance of separation between them can be made infinite, in order to weaken the existing attractive force between them, to zero.