CSIR NET Physical Science Exam  >  CSIR NET Physical Science Tests  >  CSIR NET Physical Science Mock Test Series 2024  >  CSIR NET Physical Science Mock Test - 5 - CSIR NET Physical Science MCQ

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


Test Description

30 Questions MCQ Test CSIR NET Physical Science Mock Test Series 2024 - CSIR NET Physical Science Mock Test - 5

CSIR NET Physical Science Mock Test - 5 for CSIR NET Physical Science 2024 is part of CSIR NET Physical Science Mock Test Series 2024 preparation. The CSIR NET Physical Science Mock Test - 5 questions and answers have been prepared according to the CSIR NET Physical Science exam syllabus.The CSIR NET Physical Science Mock Test - 5 MCQs are made for CSIR NET Physical Science 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for CSIR NET Physical Science Mock Test - 5 below.
Solutions of CSIR NET Physical Science Mock Test - 5 questions in English are available as part of our CSIR NET Physical Science Mock Test Series 2024 for CSIR NET Physical Science & CSIR NET Physical Science Mock Test - 5 solutions in Hindi for CSIR NET Physical Science Mock Test Series 2024 course. Download more important topics, notes, lectures and mock test series for CSIR NET Physical Science Exam by signing up for free. Attempt CSIR NET Physical Science Mock Test - 5 | 55 questions in 180 minutes | Mock test for CSIR NET Physical Science preparation | Free important questions MCQ to study CSIR NET Physical Science Mock Test Series 2024 for CSIR NET Physical Science Exam | Download free PDF with solutions
CSIR NET Physical Science Mock Test - 5 - Question 1

By folding the given paper, which of the following cube cannot be made?

Detailed Solution for CSIR NET Physical Science Mock Test - 5 - Question 1

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.

CSIR NET Physical Science Mock Test - 5 - Question 2

T is the daughter of P, who is the son of H. H is married to C. F is the daughter of Q, who is married to P. C is the brother of E. How is C related to T?

Detailed Solution for CSIR NET Physical Science Mock Test - 5 - Question 2

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".

1 Crore+ students have signed up on EduRev. Have you? Download the App
CSIR NET Physical Science Mock Test - 5 - Question 3

Seven persons – P, Q, R, S, T, U and V are sitting in a straight line facing north, but not necessarily in the same order. P is sitting at the centre of the row. Between P and Q only one person sits. S sits second from the left corner. Between U and T only one person sits and T at the right corner. R does not sit in any corner. Who is sitting between P and Q?

Detailed Solution for CSIR NET Physical Science Mock Test - 5 - Question 3

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.

CSIR NET Physical Science Mock Test - 5 - Question 4

A house has a number which needs to be identified. The following three statements are given that can help in identifying the house number.

  1. If the house number is a multiple of 3, then it is a number from 50 to 59.
  2. If the house number is NOT a multiple of 4, then it is a number from 60 to 69.
  3. If the house number is NOT a multiple of 6, then it is a number from 70 to 79.

What is the house number?

Detailed Solution for CSIR NET Physical Science Mock Test - 5 - Question 4
  • 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.

CSIR NET Physical Science Mock Test - 5 - Question 5

Select the figure from the options that can replace the question mark(?) and complete the given pattern.

Detailed Solution for CSIR NET Physical Science Mock Test - 5 - Question 5

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

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

CSIR NET Physical Science Mock Test - 5 - Question 6

The police arrested four criminals - P, Q, R and S. The criminals knew each other. They made the following statements:
P says “Q committed the crime.”
Q says “S committed the crime.
R says “I did not do it.”
S says “What Q said about me is false.”

Assume only one of the arrested four committed the crime and only one of the statements made above is true. Who committed the crime?

Detailed Solution for CSIR NET Physical Science Mock Test - 5 - Question 6

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.

CSIR NET Physical Science Mock Test - 5 - Question 7

The probability of having 53 Tuesdays in an ordinary year is:

Detailed Solution for CSIR NET Physical Science Mock Test - 5 - Question 7

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.

CSIR NET Physical Science Mock Test - 5 - Question 8

Amulya and Amar visited two places A and B respectively in Kashmir and recorded the minimum temperatures on a particular day as –4°C at A and –1°C at B. Which of the following statement is true? 

Detailed Solution for CSIR NET Physical Science Mock Test - 5 - Question 8

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.

CSIR NET Physical Science Mock Test - 5 - Question 9

If then is equal to-

Detailed Solution for CSIR NET Physical Science Mock Test - 5 - Question 9

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.

CSIR NET Physical Science Mock Test - 5 - Question 10

If the compound interest on a certain sum of money for 2 years at 4% per annum is Rs. 3264, then what would be the simple interest on the same sum for 2 years at the same rate?

Detailed Solution for CSIR NET Physical Science Mock Test - 5 - Question 10

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.

CSIR NET Physical Science Mock Test - 5 - Question 11

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. If pipe A and pipe B are opened together and after 10 hours pipe B is closed and pipe C is opened. Find the total time required to fill the tank completely. (in hours)

Detailed Solution for CSIR NET Physical Science Mock Test - 5 - Question 11

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.

CSIR NET Physical Science Mock Test - 5 - Question 12

In the given figure O is the center of the circle.∠PQO = 36° and ∠PRO = 32° 

Find the value of ∠QOR.

Detailed Solution for CSIR NET Physical Science Mock Test - 5 - Question 12


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°  

CSIR NET Physical Science Mock Test - 5 - Question 13

If x is real then find the minimum value of (x + 4)(x + 3)

Detailed Solution for CSIR NET Physical Science Mock Test - 5 - Question 13

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.

CSIR NET Physical Science Mock Test - 5 - Question 14

The following line graph shows the number of bikes sold by Bikash Motors in the last month.

The number of Yamaha bikes sold is what percent of the total sales?

Detailed Solution for CSIR NET Physical Science Mock Test - 5 - Question 14

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.

CSIR NET Physical Science Mock Test - 5 - Question 15

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

If the total population of all the villages is 60000, what is the population of village G?

Detailed Solution for CSIR NET Physical Science Mock Test - 5 - Question 15

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.

CSIR NET Physical Science Mock Test - 5 - Question 16

A Counter Consists of Four Flip-Flops

It the counter is initialized as A0A1A2A3=0110 , the state after the next clock pulse is 

Detailed Solution for CSIR NET Physical Science Mock Test - 5 - Question 16

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,

CSIR NET Physical Science Mock Test - 5 - Question 17

The Van der Waals' equation of state for a gas is given by,represent the pressure, volume V and temperature respectively, and a and b are constant parameters. At the critical point, where all the roots of the above cubic equation are degenerate, the volume is given by-

Detailed Solution for CSIR NET Physical Science Mock Test - 5 - Question 17

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

At critical point (T = TC)

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),

CSIR NET Physical Science Mock Test - 5 - Question 18

The equation of motion of a system described by the time dependent Lagrangian

Detailed Solution for CSIR NET Physical Science Mock Test - 5 - Question 18

Here, time dependent Lagrangian is given as

Now, equation of motion is given as,

Where

From equation (2)

CSIR NET Physical Science Mock Test - 5 - Question 19

In the rest frame S1 of a point particle with electric charge q1, another point particle with electric
charge q2 moves with a speed v parallel to the x -axis at a perpendicular distance l. The magnitude of
the electromagnetic force felt by q1 due to q2 when the distance between them is minimum, is
[In the following

Detailed Solution for CSIR NET Physical Science Mock Test - 5 - Question 19


The electric field at O due to point charge q2 is

Therefore, the force felt by the charge q1 is

CSIR NET Physical Science Mock Test - 5 - Question 20

The Hamiltonian of a two-level quantum system isA possible initial state in which the probability of the system being in that quantum state does not change with time, is

Detailed Solution for CSIR NET Physical Science Mock Test - 5 - Question 20

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

CSIR NET Physical Science Mock Test - 5 - Question 21

The magnetization M of a ferro-magnet, as a function of the temperature T and the magnetic field H, is described by the equation In these units, the zero − field magnetic susceptibility in terms of is given by,

Detailed Solution for CSIR NET Physical Science Mock Test - 5 - Question 21

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,

CSIR NET Physical Science Mock Test - 5 - Question 22

Consider black body radiation contained in a cavity whose walls are at temperature T. The radiation is in equilibrium with the walls of the cavity. If the temperature of the walls is increased to 2T and the radiation is allowed to come to equilibrium at the new temperature, the entropy of the radiation increases by a factor of:

Detailed Solution for CSIR NET Physical Science Mock Test - 5 - Question 22

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),

CSIR NET Physical Science Mock Test - 5 - Question 23

A matter wave is represented by the wave function: where A is a constant. The direction of wave propagation is given by?

Detailed Solution for CSIR NET Physical Science Mock Test - 5 - Question 23

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:

CSIR NET Physical Science Mock Test - 5 - Question 24

A one-dimensional simple harmonic oscillator with Hamiltonianis subjected to small perturbation The first order correction to the ground state energy is dependent on

Detailed Solution for CSIR NET Physical Science Mock Test - 5 - Question 24

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

CSIR NET Physical Science Mock Test - 5 - Question 25

A walker travels along a one dimensional discrete lattice, labeled by points –N, –N + 1, ...0, ..., N – 1, N, by putting random left and/or right steps of length l with equal probability for every step. Suppose the random walker starts from the lattice position 0 and is found at the same lattice position after (i) 10 and (ii) 7 step walks. The corresponding probabilities are respectively given by

Detailed Solution for CSIR NET Physical Science Mock Test - 5 - Question 25

(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.
CSIR NET Physical Science Mock Test - 5 - Question 26

Consider the following probability density function :
P(x) = 0 x < –2 and x > 3
P(x) = a –2 < x < 2

The value of a is :

Detailed Solution for CSIR NET Physical Science Mock Test - 5 - Question 26

Given-

The sum of all the probabilities is equal to 1.

So, the correct answer is a=11/48

CSIR NET Physical Science Mock Test - 5 - Question 27

The mean kinetic energy per atom in a sodium vapor lamp is 0.8 eV.  Given that the ratio of the Doppler width of an optical line to its central frequency is 4 X 10-8. Find the approximate mass of the sodium atom?

Detailed Solution for CSIR NET Physical Science Mock Test - 5 - Question 27

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 ΔvD 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.

CSIR NET Physical Science Mock Test - 5 - Question 28

The electrical conductivity of cooper is approximately 90 % of the electrical conductivity of silver, while the electron density in the silver is approximately 50 % of the electron density in cooper. In the Drude's model, the approximate ratio τCuAg of the mean collision time in cooper to mean collision time in silver is?

Detailed Solution for CSIR NET Physical Science Mock Test - 5 - Question 28

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:

CSIR NET Physical Science Mock Test - 5 - Question 29

A particle in 1-D moves under the influence of a potential of V(x) a x4, where a is a real constant. For large the quantized energy En depends on as:

Detailed Solution for CSIR NET Physical Science Mock Test - 5 - Question 29

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

where, β12=π/4, for V (x) to be finite a boundary and β12=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 β12= π/4 as the potential is finite at the turning points, and we get:

The potential is symmetric hence,
where 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:

CSIR NET Physical Science Mock Test - 5 - Question 30

Diffuse hydrogen gas within a galaxy may be assumed to follow a Maxwell distribution at temperature 106 K, while the temperature appropriate for the H gas in the inter-galactic space, following the same distribution, may be taken to be 104 K. The ratio of thermal broadening ΔvG / ΔvIG of the Lyman-α line from the H- atoms within the galaxy to that from the inter-galactic space is closest to

Detailed Solution for CSIR NET Physical Science Mock Test - 5 - Question 30

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 ω(2kBT ln 2/Mc2)1/2
Thus Δ ω ∝ T1/2
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

View more questions
5 tests
Information about CSIR NET Physical Science Mock Test - 5 Page
In this test you can find the Exam questions for CSIR NET Physical Science Mock Test - 5 solved & explained in the simplest way possible. Besides giving Questions and answers for CSIR NET Physical Science Mock Test - 5, EduRev gives you an ample number of Online tests for practice
Download as PDF