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)

( CSIR NET 2018 December: Part A ) See the detailed explanations and solutions to part A

The article aims to make the best attempt at finding the answers for the recently concluded 2018 CSIR NET. Detailed explanatory answers for

physical sciencessection ( 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.

## CSIR NET 2018 December physical sciences

### Part B

Q - 21. Consider the decayA → B + Cof a relativistic spin1/2particleA. Which of the following statements is true in the rest frame of the particleA? 1. The spin of bothBandCmay be1/2. 2. The sum of the masses ofBandCis greater than the mass ofA. 3. The energy ofBis uniquely determined by the masses of the particles. 4. The spin of bothBandCmay be integral.

**click to see or hide answer to Q – 21**

— the answer is: **option 3**.

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: `E _{B} = ( M_{A}^{2} – M_{C}^{2} + M_{B}^{2} ) / 2 M_{A}`

Also the other options talk about the spin of the particles but we need not bother since option 3 is the correct option.

Hence the correct answer is **option 3**.

Q - 22. Two current-carrying circular loops, each of radiusR, are placed perpendicular to each other, as shown in the figure below. the loop in the xy-plane carries a currentIwhile that in the xz-plane carries a current_{0}2I. The resulting magnetic field at the origin is_{0}

1. 2. 3. 4.

**click to see or hide answer to Q – 22**

— the answer is: **option 3**.

We only need to find the direction of the net magnetic field of the two circular loops to be able to select the correct answer. Thats because all the options have the same common magnitude of the field: ( `μ _{0 }I_{0} ) / 2 R`. For the vertical loop with current

`2I`if we curl our palm along the shown direction for current our thumb points in the – y axis. Thus we should have a vector: . Similarly for the horizontal loop if we curl our palm along the shown direction for the current our thumb points in the +ve z axis and we must have a vector: . As a result the total field created by the two loops of currents is: .

_{0}Hence the correct answer is **option 3**.

Q - 23. An electric dipole of dipole moment is placed at the origin in the vicinity of two charges+qand-qat(L, b)and(L, -b)respectively, as shown in the figure below. The electrostatic potential at the point(L/2, 0)is

1. 2. 3. 4.

**click to see or hide answer to Q – 23**

— the answer is: **option **.

Explanation:

Hence the correct answer is **option **.

Q - 24. A monochromatic and linearly polarised light is used in a Young's double slit experiment. A linear polarizer, whose pass axis is at an angle45to the polarisation of the incident wave, is placed in front of one of the slits. If^{0}Iand_{max}I, respectively denote, the maximum and minimum intensities of the interference pattern on the screen, the visibility, defined as the ratio (_{min}I-_{max}I)/(_{min}I+_{max}I), is 1._{min}√2/32.2/33.2√2/34.√2/√3

**click to see or hide answer to Q – 24**

— the answer is: **option 3**.

Explanation: (updated 10:45 pm, 14-12-2019) For the time being I am uploading a scanned solution sheet. I will update this later. To understand the expressions you need to go through the linked lecture: ( lecture VII ) read about interference of light waves and remember how intensity is dropped by a square of cosine of angle between source polarization and polarizer pass axis. Remember that both intensity (slit 1 and slit 2) are same to begin with (I_zero). Then one of the intensity (either slit) is changed due to polarization through the polarizer.

Hence the correct answer is **option 3.**

Q - 25. An electromagnetic wave propagates in a non-magnetic medium with relative permittivityε = 4. The magnetic field for this wave is whereHis a constant. The corresponding electric field is 1. 2. 3. 4._{0}

**click to see or hide answer to Q – 25**

— the answer is: **option 1**.

Explanation: the idea is to determine the direction of the field as the magnitude is same in all 4 given options. The first thing to calculate is the direction of the em wave. The direction of the em wave ( same as wave propagation vector and the direction of Poynting vector ) in a `cos ( kx – ωt ) = cos ( ωt – kx )` variation is in the `+ x` direction. This means our given wave travels in the direction of the vector . But we know that . The option 1 gives us: . We used the fact that: direction of the magnetic field vector `H/B` is .

Hence the correct answer is **option 1**.

Q - 26. The ground state energy of an anisotropic harmonic oscillator described by the potential:V(x,y,z) = (1/2) mωx^{2}^{2}+ 2mωy^{2}^{2}+ 8mωz^{2}^{2}(in units of ) 1.5/22.7/23.3/24.1/2

**click to see or hide answer to Q – 26**

— the answer is: **option **.

Explanation:

Hence the correct answer is **option **.

Q - 27. The productΔxΔpof uncertainties in the position and momentum of a simple harmonic oscillator of mass m and angular frequencyωin the ground state , is . The value of the productΔxΔpin the state ( wherelis a constant and is the momentum operator ) is 1. 2. 3. 4.

**click to see or hide answer to Q – 27**

— the answer is: **option **.

Explanation:

Hence the correct answer is **option **.

Q - 28. Let the wavefunction of the electron in a hydrogen atom be , where are the eigenstates of the Hamiltonian in the standard notation. The expectation value of the energy in this state is 1.-10.8 eV2.-6.2 eV3.-9.5 eV4.-5.1 eV

**click to see or hide answer to Q – 28**

— the answer is: **option 4**.

Explanation: the given state is already normalized ( amplitude squares add up to 1 ). All we need to do is compute `( amplitude ) ^{2 }`

`/ n`and multiply to

^{2 }`-13.6 eV`. The

`( amplitude )`and corresponding

^{2}`n`are given as:

^{2 }`( 1/6, 4 )`,

`( 2/3, 4 )`,

`( 1/6, 1)`. This gives

`-5.1 eV`, answer that is given in option 4.

Hence the correct answer is **option 4**.

Q - 29. Three identical spin1/2particles of massmare confined to a one-dimensional box of lengthL, but are otherwise free. Assuming that they are non-interacting, the energy of the lowest two energy eigenstates, in units of , are 1.3and62.6and93.6and114.3and9

**click to see or hide answer to Q – 29**

— the answer is: **option 1**.

Explanation: the energy eigenstates of the 1 dimensional box are given as: and the eigenvalues are given as . Accordingly the lowest two energy eigenstates correspond to the quantum numbers` ( 1, 1, 1 )` where all the `3 `particles are in the ground state and `( 2, 1, 1 )` where at-most `1` particle is in the first excited state and the other two in the ground state. Accordingly the lowest two energy states correspond to `1 ^{2}`

`+1`

^{2}`+1`

^{2 }`= 3`and

`2`

^{2}`+`

`1`

^{2}`+`

`1`

^{2}`= 6`in units of .

Hence the correct answer is **option 1**.

Q - 30. The heat capacityCat constant volume of a metal, as a function of temperature, is_{V}αT+βT, where^{3}αandβare constants. The temperature dependence of the entropy at constant volume is 1.αT+(1/3)βT2.^{3}αT+βT3.^{3}(1/2)αT+(1/3)βT4.^{3}(1/2)αT+(1/4)βT^{3}

**click to see or hide answer to Q – 30**

— the answer is: **option 1**.

Explanation: The specific heat at constant volume is given by the expression: . An integration of the same yields the temperature dependence of the entropy: `S = [∫ ( C_{V}/T ) dT ]_{N, V }= ∫α dT + ∫βT^{2 }dT = αT + (1/3)βT^{3}.`

Hence the correct answer is **option 1**.

Q - 31. The rotational energy levels of a molecule are wherel = 0, 1, 2, ...andIis its moment of inertia. The contribution of the rotational motion to the Helmholtz free energy per molecule, at low temperatures in a dilute gas of these molecules, is approximately 1. 2. 3._{0}-k_{B}T4.

**click to see or hide answer to Q – 31**

— the answer is: **option **.

Explanation:

Hence the correct answer is **option **.

Q - 32. The vibrational motions of a diatomic molecule may be considered to be that of a simple harmonic oscillator with angular frequencyω. If a gas of these molecules is at a temperature T, what is the probability that a randomly picked molecule will be found in its lowest vibrational state? 1. 2. 3. 4.

**click to see or hide answer to Q – 32**

— the answer is: **option 2**.

Explanation: when a system is in thermal equilibrium with a heat reservoir at temperature `T`, the probability `P _{r }`that the system be found in a state ( with index

`r`) with energy

`E`is given by this temperature

_{r}`T`and energy of the state

`E`in the following manner ( its called canonical distribution ) :

_{r}`P`. Here

_{r }= C e^{-βEr}`β = (`

`k`T)

_{B}^{-1}and

`C`is the constant which is the inverse of

`Z`the sum of states or partition function (

`C = Z`) given by:

^{-1}`Z = ∑`

_{r}`e`. We can take

^{-βEr}`C`to be

`1`. What remains is the probability or the Boltzmann factor:

`e`. So the probability of finding the randomly picked molecule in an energy state

^{-βEr}`E`is

_{r}`e`. But energy of the diatomic molecules is given by: . In the ground state (

^{-(1/(kBT))Er}`n = 0`) the energy is: . Thus the probability of finding a randomly picked molecule in its lowest vibrational state is: .

Hence the correct answer is **option 2**.

Q - 33. Consider an ideal Fermi gas in a grand canonical ensemble at a constant chemical potential. The variance of the occupation number of the single particle energy level with mean occupation number is 1. 2. 3. 4.

**click to see or hide answer to Q – 33**

— the answer is: **option 1**.

Explanation: For a general statistical variable `x` the variance ( `Δ` which is the square of the standard deviation `σ` ) is given by: where `p `gives the probability of occurrence and `q=(1-p)` the probability of non-occurrence. Since we have a single particle we can safely take `N = 1`. This gives us the correct option when we realize that the probability of occurrence `p` is the same as the mean of the occupation number . Since grand canonical ensemble of a single fermion system is only a subsystem of the general statistical system this result follows. Other options can’t follow from this general formula. ( I have checked the answer to be correct, although there is a rigorous proof, but I am unable to plough further at the moment. )

Hence the correct answer is **option 1**.

Q - 34. Consider the following circuit, consisting of an RS flip-flop and two AND gates.

Which of the following connections will allow the entire circuit to act as a JK flip-flop? 1. ConnectQto pin 1 and to pin 2 2. ConnectQto pin 2 and to pin 1 3. ConnectQto K input and to J input 4. ConnectQto J input and to K input

**click to see or hide answer to Q – 34**

— the answer is: **option 2**.

Explanation: let’s follow the figure. Also I have given the truth table for a JK flip-flop, for the enthused. ( One can verify the flip flop works correctly with the truth table. )

C ( clk ) |
J |
K |
Q_{n+1} |
Action |

↑ | 0 |
0 |
Q(last)_{n } |
No change |

↑ | 0 |
1 |
0 |
RESET |

↑ | 1 |
0 |
1 |
SET |

↑ | 1 |
1 |
(toggle) | Toggle |

Hence the correct answer is **option 2**.

Q - 35. The truth table below gives the valueY(A,B,C)whereA,BandCare binary variables.

A |
B |
C |
Y |

0 |
0 |
0 |
1 |

0 |
0 |
1 |
0 |

0 |
1 |
0 |
0 |

0 |
1 |
1 |
1 |

0 |
0 |
0 |
1 |

0 |
0 |
1 |
0 |

0 |
1 |
0 |
0 |

0 |
1 |
1 |
1 |

The outputYcan be represented by 1. 2. 3. 4.

**click to see or hide answer to Q – 35**

— the answer is: **option 2**.

Explanation: Only option 2 satisfies all rows of the table. eg if we take row 1: `A = 0, B = 0, C = 0`. Their inversions are respectively `1, 1, 1` and we get `Y = 1+ 0+0+0=1` which is also shown in the table. Let’s take the first row of table and apply on option 1. We see `Y = 0`. But table says `1`. Hence option 1 can’t be correct. Similarly easily it can be checked that option 3 and 4 are incorrect.

Hence the correct answer is **option 2**.

Q - 36. A sinusoidal signal is an input to the following circuit.

```
Which of the following graphs best describes the
output waveform?
```

**click to see or hide answer to Q – 36**

— the answer is: **option **.

Explanation:

Hence the correct answer is **option **.

Q - 37. A sinusoidal voltage having a peak value of Vp is an input to the following circuit, in which the DC voltage is Vb.

```
assuming an ideal diode, which of the following best
describes the output waveform?
```

**click to see or hide answer to Q – 37**

— the answer is: **option **.

Explanation:

Hence the correct answer is **option **.

Q - 38. One of the eigenvalues of the matrixeis^{A}e, where . The product of the other two eigenvalues of^{a}eis 1.^{A}e2.^{2a}e3.^{-a}e4.^{-2a}1

**click to see or hide answer to Q – 38**

— the answer is: **option 4**.

Explanation: To solve this we need to know 2 identities of the elegant matrix method. one: `det (e ^{A}`

`) = e`two: the product of eigenvalues of

^{Tr (A)}`e`=

^{A}`det (e`

^{A}`)`. Since

`Tr (A) = a`therefore we see that the product of eigenvalues of

`e`=

^{A }`e`. Thus the product of the other two eigenvalues is just

^{a}`1`.

Hence the correct answer is **option 4**.

Q - 39. The polynomialf(x) = 1 + 5x + 3xis written as a linear combination of the Legendre polynomials (^{2}P,_{0}(x) = 1P,_{1}(x) = xP) as_{2}(x) = (1/2)(3x^{2}- 1)f(x) = Σ. The value of_{n}c_{n}P_{n}(x)cis 1._{0}1/42.1/23.24.0

**click to see or hide answer to Q – 39**

— the answer is: **option 3**.

Explanation: Let’s write `f(x) = c _{0}P_{0}(x) + c_{1}P_{1}(x) + c_{2}P_{2}(x) = 1 + 5x + 3x^{2}`. This means

`c`

_{0}`+`

`c`

_{1}x+`c`

_{2}`(`

`3x`–

^{2}`1)/2`

`= 1 + 5x + 3x`and

^{2 }`c`

_{0}`-(`

`c`

_{2}`/2)=1`,

`c`,

_{1}=5`(c`

_{2}`)×(`

`3/2) =3`. Solving for

`c`this gives:

_{2 }`c`

_{2 }`= 2`. Solving for

`c`gives

_{0 }`c`.

_{0 }= 2Hence the correct answer is **option 3**.

Q - 40. The value of the integral whereCis a circle of radiusπ/2, traversed counter-clockwise, with centre atz = 0, is 1.42.4i3.2i4.0

**click to see or hide answer to Q – 40**

— the answer is: **option 4**.

Explanation: According to the residue theorem: . Here: `f(z)=g(z)/zh(z)` where `g(z) = tanh z` and `h(z) = sin πz.` We see that `f(z)` has a simple pole at `z = 0`, `h(0) = 0` and `g(z)` is analytic. So residue is given as: `R(0) = lim _{z→0} [zf(z)]= lim _{z→0} [zg(z)/(zh(z))]`. Then by L’Hospital’s rule:

`R(0) = g(0) lim`. So the integral is

_{z→0}[1/h(z)] = g(0) lim_{z→0}[1/h'(z)]=g(0)/h'(0) = 0/1=0*zero*.

Hence the correct answer is **option 4**.

Q - 41. A particle of mass m, moving along the x-direction, experiences a damping force-γv, where^{2}γis a constant andvis its instantaneous speed. If the speed att = 0isv, the speed at time_{0}tis 1.v_{0}e2.^{-(γv0t)/m}v/{1+ln[_{0}1+(γv3._{0}t)/m]}mv/(m+_{0}γv4._{0}t)2v_{0}/[1+ e^{(γv0t)/m}]

**click to see or hide answer to Q – 41**

— the answer is: **option 3**.

Explanation: from the given information we can write: `m(d ^{2}x/dt^{2}) = -γv^{2}` so

`m(dv/dt) = -γv`and (mdv)/v

^{2 }`=`

^{2}`-γ`dt

`.`Integrating and using given initial conditions (

`v = v`at

_{0}`t = 0`) we get option 3:

`v = mv`.

_{0}/(m+γv_{0}t)Hence the correct answer is **option 3**.

Q - 42. The integralI = ∫is evaluated from the point_{c}e^{z}dz(-1,0)to(1,0)along the contourC, which is an arc of the parabolay = x- 1, as shown in the figure.^{2}

`The value of ``I` is
1. *0* 2. `2 sinh 1`
3. `e`^{2i}` sinh 1` 4. `e + e`^{-1}

**click to see or hide answer to Q – 42**

— the answer is: **option **.

Explanation:

Hence the correct answer is **option **.

Q - 43. In terms of arbitrary constants A and B, the general solution to the differential equationx^{2}(dis 1.^{2}y/dx^{2}) + 5x(dy/dx) + 3y = 0y = A/x + Bx^{3}2.y = Ax + B/x^{3}3.y = Ax + Bx4.^{3}y = A/x + B/x^{3}

**click to see or hide answer to Q – 43**

— the answer is: **option 4**.

Explanation: it can be easily checked that `1/x `and `1/x ^{3}` satisfy the given differential equation hence a linear combination of them viz.

`y = A/x + B/x`also satisfies the differential equation.

^{3}Hence the correct answer is **option 4**.

Q - 44. In the attractive Kepler problem described by the central potentialV(r) = -k/r( where k is a positive constant ), a particle of massmwith a non-zero angular momentum can never reach the center due to the centrifugal barrier. If we modify the potential toV(r) = -k/r - β/one finds that there is a critical value of the angular momentumr^{3}lbelow which there is no centrifugal barrier. This value of_{c}lis 1._{c}[12km^{2}β]2.^{1/2}[12km3.β]^{2}^{-1/2}[12km4.β]^{2}^{1/4}[12kmβ]^{2}^{-1/4}

**click to see or hide answer to Q – 44**

— the answer is tentative: **option 1**.

Explanation: `k `has units of ( energy × distance ) / mass and `β `has units of ( energy × volume ) / mass. So `[12km^{2}β]^{1/2} ` has units of energy × area. Angular momentum has units of energy × time. So option 1 seems to be close to the correct answer ( I am guessing: other options have energy units in the square root or inverse ). Or something is missing.

Hence the correct answer seems to be **option 1**.

Q - 45. The time period of a particle of massm, undergoing small oscillations aroundx = 0, in the potentialV = V, is 1. 2. 3. 4._{0}cosh (x/L)

**click to see or hide answer to Q – 45**

— the answer is tentative: **option 3**.

Explanation: `F = -dV/dx = -V _{0}/L sinh (x/L)`. For small oscillations

`sinh (x/L) ~ x/L`. So,

`F = -(V`. So,

_{0}x/L^{2}) = -kx`k = V`. So time period is given as: .

_{0}/L^{2}Hence the correct answer is **option 3**.

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