CSIR NET Physical Science Mock Test - 5 - CSIR NET Physical Science MCQ

From the given open cube we will find the opposite pairs as shown-


Here, A is opposite to C, B is opposite to H, 9 is opposite to D. 
Now, these opposite pairs can NOT be together by seeing from one side if they are together → that cube can not be made.
Here, Image (II)has opposite pair B and H together → This cube can not be made.
Image (IV)has opposite pair A and C together → This cube can not be made.
Hence, Option (b) is correct.

According to the given information:

1) T is the daughter of P, who is the son of H.

• T is the daughter of P,
• P is the son of H.

2) F is the daughter of Q, who is married to P.
3) C is the brother of E.

• F is the daughter of Q,
• Q is married to P.
• Thus C is the husband of H.

This is the final diagram that can be deduced,
As we can see C is the father of P, who is the father of T.
Thus, C is T's father's father.
​Hence, the correct answer is "Father's father".

Persons: P, Q, R, S, T, U and V
1) P is sitting at the centre of the row.

2) Between P and Q only one person sits.
3) S sits second from the left corner.

4) Between U and T only one person sits and T at the right corner.

5) R do not sit in any corner.

Hence, “U” is sitting between P and Q.

• Condition 1: If the house number is a multiple of 3, then it is a number from 50 to 59.
  • Possibilities: 51, 54, 57
  • But if one of these is the right answer, then it should be a multiple of 4 and 6 which is not the case here.
• Condition 2: If the house number is NOT a multiple of 4, then it is a number from 60 to 69.
  • Possibilities: 61, 63, 65, 66, 67, 69
  • But if one of these is the right answer, it should be multiple of six (i.e. 66) and not a multiple of 3 (but 66 is multiple of 3). So it is not a required sample space.
• Condition 3: If the house number is NOT a multiple of 6, then it is a number from 70 to 79.
  • Possibilities: 70, 71, 73, 74, 75, 76, 77, 79
  • But if one of these is the right answer, it should be multiple of 4 (i.e. 76) and not a multiple of 3 (i.e. 70, 71, 73, 74, 76, 77, 79).

So 76 is the correct answer.

The figure that will replace the question mark (?) in the following figure series is:

Hence, the correct answer is "Option (4)".

Let S1: P says “Q committed the crime.”
S2: Q says “S committed the crime.”
S3: R says “I did not do it.”
S4: S says “What Q said about me is false.”
1. Assume that P committed the crime:
S1 is false, S2 is false, S3 is true, S4 is true
Since S3 and S4 are true, R and S says true but only one of the statements among S1, S2, S3 and S4 is true. Therefore, assuming that P committed the crime is false.
2. Assume that R committed the crime:
S1 is false, S2 is false, S3 is false, S4 is true
Since S4 is true, S says true and only one of the statements among S1, S2, S3 and S4 is true. Therefore, assuming that R committed the crime is true.
3. Assume that S committed the crime:
S1 is false, S2 is true, S3 is true, S4 is false
Since S2 and S3 are true, Q and R both says true but only one of the statements among S1, S2, S3 and S4 is true. Therefore, assuming that S committed the crime is false.
4. Assume that Q committed the crime:
S1 is true, S2 is false, S3 is true, S4 is true
Since S1, S3 and S4 are true, P, R and S says true but only one of the statements among S1, S2, S3 and S4 is true. Therefore, assuming that Q committed the crime is false.

Given:
The probability of having 53 Tuesdays in an ordinary year is:
Concept used:
Probability(event) = Number of favourable events/total number of events
Calculation:
⇒ 1 year = 365 days
A year has 52 weeks,
∴ There will be 52 Tuesday for sure.
⇒ 52 weeks = 52 × 7 = 364 days
In an ordinary year, there will be 52 Tuesdays and 1 day will be left.
This 1 day can be:
Sunday
Monday
Tuesday
Wednesday
Thursday
Friday 
Saturday
Of these total 7 outcomes, the favourable outcomes are 1.
∴ The probability of getting 53 Tuesday in ordinary year = 1/7.

Given:
A = -4°C
B = -1°C
Concept used:
°C in higher negative values is considered as more cooler.
Solution:
-4 < -1
A is cooler than B
The difference = -1 - (-4) = 3°C
The temperature of B is 3°C higher than A.
Hence, the correct option is A.

Given:
x + 1/x = 2

Concept used:
If x + 1/x = p
Then,
x² + 1/x² = p² - 2
Calculation:
⇒ x + 1/x = 2
⇒ x² + 1/x² = 4 - 2
⇒ 2
∴ The correct answer is 2.

Given:
the compound interest on a certain sum of money for 2 years at 4% per annum is Rs. 3264

Formula used:
Compound interest = Principal
Simple interest = 
Calculation:
let P be Principal,

51P = 2040000
Principal = 2040000/51
= 40000
the simple interest on the same sum for 2 years at the same rate

= 3200
Answer is 3200.

Given: 
Pipe A and pipe B can fill the tank completely in 20 hours and 30 hours respectively.
Pipe C can empty the completely filled tank in 24 hours.

Concept used:
The time is taken by pipe A to fill the tank = a hours
The time is taken by pipe B to fill the tank = b hours
The time is taken by pipe C to empty the completely filled tank = c hours
The capacity of the tank = L.C.M of (a, b and c) L

Calculation:
According to the question,
The time is taken by pipe A to fill the tank = 20 hours
The time is taken by pipe B to fill the tank = 30 hours
The time is taken by pipe C to empty the completely filled tank = 24 hours
The capacity of the tank = L.C.M of (20, 30 and 24) = 120 L
Now,
Efficiency of pipe A = 120/20 = 6 L per hour
Efficiency of pipe B = 120/30 = 4 L per hour
Efficiency of pipe C = 120/24 = -5 L per hour
Again according to the question,
The combined efficiency of pipe A and pipe B = 6 + 4 = 10 L per hour
The amount of tank filled by pipe A and pipe B in 10 hours = 10 × 10 = 100 L
The remaining quantity of tank to be filled = 120 - 100 = 20 L
Now,
The combined efficiency of pipe A and pipe C = 6 + (-5) = 1 L per hour
The time is taken by pipe A and pipe C to fill 20 L of water = 20/1 = 20 hours
The total time required = 10 + 20 = 30 hours
Therefore, '30 hours' is the required answer.

Given:
O is the center of the circle.
∠PQO = 36° and ∠PRO = 32° 

Concept used:
Join PO.
Then PO = OR =OQ [ Radius of the circle]
The angle made in the center is twice than the angle made in the arc

Calculation:
∠PQO = ∠QPO = 36°
Again, ∠OPR =  ∠ORP = 32°
So,  ∠QPR = ∠QPO + ∠OPR 
⇒ 36 + 32 = 68°
So, ∠QOR = 2 × ∠QPR
⇒ 2 × 68° = 136°  
The value of ∠QOR is 136°  

Given:
(x + 4)(x + 3)

Concept used:
dy/dx = 0, the value of x gives the minimum value when d2y/dx2 is greater than 0, and gives the maximum value when d2y/dx2 is less than 0

Calculation:
(x + 4)(x + 3)
⇒ x² + 3x + 4x + 12
⇒ x² + 7x + 12      ----(i)
By differentiating equation (i),
dy/dx = 0
⇒ 2x + 7 = 0      ----(ii)
⇒ x = - 7/2
Now by double differentiating equation (ii),
d²y/dx² = 2 > 0
that means, the minimum value of equation (i) is at x = - 7/2
Minimum value of equation (i)
⇒ 49/4 - 49/2 + 12
⇒ (49 - 98 + 48)/4
⇒ - 1/4
∴ The minimum value is - 1/4.

Calculation:
Total number of bike sold last month is  35 + 25 + 45 + 20 + 10 = 135
The Yamaha bike sold last month was 25
So the required percentage is (25 /135) × 100 = 18.51 %
The number of Yamaha bikes sold is 18.50 % of the total sales.

Given:
The population of 7 villages is shown in the following pie chart.

If the total population of all the villages is 60000
Calculations:
According to the question,
Population of village G = 15% of total population of all villages
Population of G = 15% × 60000
Population of G = 0.15 × 60000 = 9000
∴ The population of village G is 9000.

Truth Table of JK Flips Flop:

Truth Table of JK Flips Flop
Initially counter is,

Therefore from figure,

Also, from figure inputs of the flips-flops are,

Therefore for initially inputs of flip-flip from results (1) are;

Now, first clock is given to flip-flops, from truth table of JJ flipsflop,

Therefore after one cycle the counter is,

Here, the Van der Waals' equation of state for a gas is given as,

At critical point (T = T_C)

Now, from equation (2), first derivative of the equation (1) is given as,

Now, from equation (2), first derivative of the equation (1) is given as,

From equation (3), second derivation of equation (1) is zero.

Dividing equation (4) by equation (5),

Here, time dependent Lagrangian is given as

Now, equation of motion is given as,

Where

From equation (2)

The electric field at O due to point charge q₂ is

Therefore, the force felt by the charge q₁ is

The possible initial state in which probability of the system being in that quantum state does not change with time, will be an eigenstate of the Hamiltonian

Here, the magnetization M of a ferro - magnet, as a function of T and the magnetic field H is given as,

Partially differentiate above equation by H

Therefore magnetic susceptibility in terms of M(0)=M(H=0) is,

For black body radiation energy is given as,

Where, σ is the Stefan's constant.
The entropy of the radiation is given as,

Now, the temperature of the walls is increased to 2T, from equation (1) entropy is,

From equation (1) and (2),

The wave propagation can be found out from the direction of wave vector.
So, from the wave function we have:

we know that
Hence the unit vector along wave vector is given by:

We have
The first order correction is given by:

Now in one-dimensional case the odd power of x's expectation value is 0.
Therefore

(i) 10 Steps:

• In this case, after 10 steps, the walker must take exactly 5 steps to the left (L) and 5 steps to the right (R), but the order in which these steps occur can vary. 
• The total number of possible paths is 2^10 because each of the 10 steps can be either left or right, so there are 2 choices for each step. Now, we need to calculate the number of valid paths that return to the starting position. The number of ways to choose 5 out of 10 steps to be leftward (L) is given by the binomial coefficient :
   
• Each of these combinations corresponds to a valid path that returns to the starting position. Therefore, there are 252 valid paths. Now, let's calculate the probability.

So, the correct probability for the walker to be at the same lattice position after 10 steps is approximately 0.2461, which can be rounded to 0.25.

(ii) 7 Steps:

• To find the probability that the walker is at the same lattice position after 7 steps, we need to consider the number of ways the walker can take steps to the left and right in such a way that the net displacement is zero. This is equivalent to counting the number of ways to arrange 3 steps to the left (L) and 4 steps to the right (R), or vice versa.
• The total number of arrangements is given by the binomial coefficient 
• However, for the walker to end up in the same position after 7 steps, there must be an equal number of steps to the left and right.
• Since 3 is not equal to 4, there are no valid arrangements that satisfy this condition. Therefore, the probability of the walker ending up at the same lattice position after 7 steps is indeed 0.

Given-

The sum of all the probabilities is equal to 1.

So, the correct answer is a=11/48

Concept:
The Doppler effect or Doppler shift is the apparent change in frequency of a wave in relation to an observer moving relative to the wave source. It is named after the Austrian physicist Christian Doppler, who described the phenomenon in 1842.

Explanation:
The atoms in motion actually obeys a gaussian distribution with a mean and a standard deviation value.
This is given as:

where Δv_D is the deviation and v0 as the mean value of the distribution formed by the atoms in motion.
We have 

Now ln 2 =0.693

mean kinetic energy= 3/2 kT is given to be 0.8 eV.

We are given with the electrical conductivity of copper is approximately 90 % of the electrical conductivity of silver.
That is:

Again, we have electron density in silver is approximately 50 % of the electron density in copper.

According to Drudes model, the conductivity is given as:

Therefore:
So:
Plugging in all the numbers we get:

Concept:
According to WKB approximation:
We have to use the relation that:

where, β₁=β₂=π/4, for V (x) to be finite a boundary and β₁=β₂=0 , for V(x) to be infinite at the boundary.
Explanation:
We have a particle in 1-D that moves under the influence of a potential of V(x) a x⁴.
We use the above formula with β₁=β₂= π/4 as the potential is finite at the turning points, and we get:

The potential is symmetric hence,
where A is some constant
Taking the substitutionwe get:

Now pulling out all the E terms, out and doing the integration we get:
This constant carries a constant after doing the integration.
Now solving it a bit we get:

Concept:
The radiation damping broadening is negligible, so that, for all practical purposes the spread of the frequencies emitted by a collection of atoms in a gas is infinitesimally narrow. The observer, however, will not see an infinitesimally thin line. This is because of the motion of the particles in hot gas. Some atoms are moving hither, and the wavelength will be blue-shifted; others are moving on, and the wavelength will be red-shifted. The result will be a broadening of the lines, known as thermal broadening.

Calculation:
The expression for thermal broadening is given by
Δ ω = 2 ω₀ (2k_BT ln 2/Mc²)^1/2
Thus Δ ω ∝ T^1/2
Diffuse hydrogen gas within a galaxy may be assumed to follow a Maxwell distribution at temperature 106 K
The temperature appropriate for the H gas in the inter-galactic space is taken to be 104 K
Thus 
Δ ω₁ : Δ ω₂ = (10⁶ : 10⁴)^1/2
∴ Δ ω1 : Δ ω2  = 10