GATE Physics Mock Test Series - 2 - Physics MCQ

GATE Physics Mock Test Series - 2 - Physics MCQ

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60 Questions MCQ Test GATE Physics Mock Test Series 2025 - GATE Physics Mock Test Series - 2

GATE Physics Mock Test Series - 2 for Physics 2024 is part of GATE Physics Mock Test Series 2025 preparation. The GATE Physics Mock Test Series - 2 questions and answers have been prepared according to the Physics exam syllabus.The GATE Physics Mock Test Series - 2 MCQs are made for Physics 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for GATE Physics Mock Test Series - 2 below.
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GATE Physics Mock Test Series - 2 - Question 1

Consider a beam of plane polarised light of wavelength λ is incident on a optical component making angle 45° to the optical axis and output light is circularly polarised then the optical component is

Detailed Solution for GATE Physics Mock Test Series - 2 - Question 1

For quarter wave plate

n = 0,1,2, 3.......
When a plane polarised light is incident on quarter wave plate with angle θ= 45° then output light is circularly polarised.

GATE Physics Mock Test Series - 2 - Question 2

A complex matrix is given as  then

Detailed Solution for GATE Physics Mock Test Series - 2 - Question 2

So , A+ = A
This implies that A is Hermitian matrix and all  eigen values of Hermittian matrix are real.

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GATE Physics Mock Test Series - 2 - Question 3

A particle of mass m, moves under the action of a central force whose potential is V(r) = kmr3 (k > 0 ) , then the angular frequency is _________ .

Detailed Solution for GATE Physics Mock Test Series - 2 - Question 3

For angular frequency, we have to find out equation of motion. We use Taylor's series expansion

V(r) = Kmr3
Ve =

So at r = r0

T =

ω = √ 5Ka

GATE Physics Mock Test Series - 2 - Question 4

A particle of mass m moves in a central force field defined by if E is the total energy supplied to the particle, then its speed is given by .

Detailed Solution for GATE Physics Mock Test Series - 2 - Question 4

Given F =
so , V(r) =
Now energy, E =

GATE Physics Mock Test Series - 2 - Question 5

The space - time coordinates of two events as measured by O are x1= 6 * 104 m ,y1 = z1= Om, t1’ = 2 * 104sec and X2,= 12 x 104m, t2 = 1 x 10-4sec. What must be the velocity of O’ with respect to O if O’ measures the two events to occur simultaneously ?

Detailed Solution for GATE Physics Mock Test Series - 2 - Question 5

Subtracting two Lorentz transformation.

On solving , we obtain,

Therefore , v is in the negative x - direction.

GATE Physics Mock Test Series - 2 - Question 6

The value of the surface integral where s is the surface of the sphere x2+ y2 + z2 = 4, n is the unit outward normal and

Detailed Solution for GATE Physics Mock Test Series - 2 - Question 6

On putting x = x = r sinθ cosφ
y = r sinθ sinφ
z = r cosθ, we get

= π x 8

GATE Physics Mock Test Series - 2 - Question 7

Fourier series which will represents f(x) = x sinx in the interval -π< x< π, then

will have value

Detailed Solution for GATE Physics Mock Test Series - 2 - Question 7

The given f(x) Function f(x) = xsin x is an even function. Hence series will be
f(x) = x sin x

If n =1

f(x) = x sin x

Divided each side by 2

GATE Physics Mock Test Series - 2 - Question 8

A particle moving in a central force located at r = 0 describes the spiral r = e, the magnitude of force is inversely proportional to

Detailed Solution for GATE Physics Mock Test Series - 2 - Question 8

r = e

Equation of orbit

GATE Physics Mock Test Series - 2 - Question 9

Rest mass energy of an electron is 0.51 MeV. A moving electron has a kinetic energy of 9.69 MeV. The ratio of the mass of the moving electron to its rest mass is

Detailed Solution for GATE Physics Mock Test Series - 2 - Question 9

E = mc2 - m0c2 ⇒ E =
9.69 =
10.2 =
m =

GATE Physics Mock Test Series - 2 - Question 10

A system of four particles is x - y plane. Of these, two particles each of mass m are located at (-1, 1) and (1, -1). The remaining two particles each of mass 2m are located at (1, 1) and (-1, 1). The xy component of the moment of inertia tensor of this system of particles is -

Detailed Solution for GATE Physics Mock Test Series - 2 - Question 10

The xy component of the moment of inertia tensor

Two particles of mass m located at (-1,1) and (1, -1) other particles of mass 2m are located at (1, 1) and (-1, 1)
W = - [m1x1y1 + m2x2y2 + m3X3y3 + m4x4y4]
= - [(mx -1 x 1 ) + (m x 1 x -1 ) + (2m x 1x1) + (2m x 1x-1)]
= - [-m -m + 2m - 2m]
= [2m] ⇒ Ixy = 2m

GATE Physics Mock Test Series - 2 - Question 11

What is the energy of a uniformly charged spherical shell of total charge q and radius R.

Detailed Solution for GATE Physics Mock Test Series - 2 - Question 11

The self energy stored in the system

Now the potential at the surface of the sphere is

GATE Physics Mock Test Series - 2 - Question 12

A current I is uniformly distributed over a wire of circular cross section, with radius a, suppose the current density in the wire is proportional to the distance from the axis J = Kx (K is constant)
Find the total current in the wire

Detailed Solution for GATE Physics Mock Test Series - 2 - Question 12

The area - perpendicular to flow is πa2, so
J =
J varies with x. Then the current I =

GATE Physics Mock Test Series - 2 - Question 13

For a Gaussian wave packet described by The expectations value of the momentum operator is

Detailed Solution for GATE Physics Mock Test Series - 2 - Question 13

Expectation value of the momentum operator is

< p > = 0

GATE Physics Mock Test Series - 2 - Question 14

The wave function of a particle is given by ψ= c exp (-x2α2 ) ,  co where c and a are constants. The probability of finding the particle in the region

Detailed Solution for GATE Physics Mock Test Series - 2 - Question 14

The probability of finding the particle in the region 0 < x <  is

Now using the normalization condition

GATE Physics Mock Test Series - 2 - Question 15

A particle is moving with one component of constant velocity parallel to the axis of y and another component of velocity parallel to the axis of x proportional to y. It will describe a

Detailed Solution for GATE Physics Mock Test Series - 2 - Question 15

When a body is moving parallel to the axis of y and moving parallel to x-axis then it's intersecting point is like a parabola .

GATE Physics Mock Test Series - 2 - Question 16

The solution of the differential equation for subject to the initial condition y(O) = 0 and ,is

Detailed Solution for GATE Physics Mock Test Series - 2 - Question 16

The differential equation is ,
complementary function (C.F) =
(D2 - 1 ) y = 0 , so D = ±1
C.F. =
Now particular integral PI =

Complete solution of given differential equation is
Apply boundary conditions, y(0) and
At t = 0 , y = 0 , we have y(0) = A + B
At t = 0 , we have

from those equations A = 0 , B = 0
y(t) = t sin ht

GATE Physics Mock Test Series - 2 - Question 17

A nozzle in the shape of a trun cated cone, as shown in the figure. The area at the wide end is 25 cm2 and and the narrow end has an area of 1 cm2, water enters the wider end at a rate of 500 gm/sec the height of the nozzle is 50 cm and it is kept vertical with the wider end at the bottom. The magnitude of the pressure difference in k Pa between the two ends of the nozzle is

Detailed Solution for GATE Physics Mock Test Series - 2 - Question 17

According to Bernoulli’s equation

Now given that θ = 500 gm / sec

According to equation of continuity

GATE Physics Mock Test Series - 2 - Question 18

What is the magnitude of the linear momentum of a photon of radiation having electric field described by e =

Detailed Solution for GATE Physics Mock Test Series - 2 - Question 18

The equation for the electric field of an electro- magnetic wave can be written as

GATE Physics Mock Test Series - 2 - Question 19

Given that the magnetic flux through the closed loop PQRSP is  along PQR, the value of  along PSR is

Detailed Solution for GATE Physics Mock Test Series - 2 - Question 19

Consider the closed loop PQRS as shown in the figure

It is given that

GATE Physics Mock Test Series - 2 - Question 20

Two infinitely extended homogeneous isotropic dielectric media (medium - 1 and medium - 2 with dielectric constant  respectively) meet at the z = 0 plane as shown in the figure. A uniform electric field is given by  The interface separating the two media is charge free. The electric displacement vector in the medium 2 is given by

Detailed Solution for GATE Physics Mock Test Series - 2 - Question 20

Given that

And the interface separating of the two media is charge free.
Since, there is no free charge at the interface. So, normal component of  is continuous at the boundary

GATE Physics Mock Test Series - 2 - Question 21

Which of the following truth table for the given logic is true ?

Detailed Solution for GATE Physics Mock Test Series - 2 - Question 21

At point X1
At X2 output is X2 = X + Y (OR GATE)

So, Z = X1 +X2(OR GATE)

GATE Physics Mock Test Series - 2 - Question 22

Magnetic flux linked with a stationary loop of resistance R varies with respect to time during the time period T as follows:

Find the amount of heat generated in the loop during that time. Inductance of the coil is negligible

Detailed Solution for GATE Physics Mock Test Series - 2 - Question 22

Given that =  at(T - 1)
Induced e.m.f.e =
= dt (0 - 1) + a (T - 1)
= a (T - 2t)
So induced e.m.f. is also a function of time.
∴  Heat enerated in time T is

GATE Physics Mock Test Series - 2 - Question 23

At t = 0, a particle of mass m having v0 starts moving through a liquid kept in a horizontal tube and experiences a drag force  It covers a distance L before coming to rest. If the times taken to cover the distances  are t2 and t4, respectively, then

Detailed Solution for GATE Physics Mock Test Series - 2 - Question 23

By Applying the second law of motion

The characterstic equation is

General solution

using the condition
C+ c2 = 0

using the condition V(0) = V0

using equations → x(t)

from the question

GATE Physics Mock Test Series - 2 - Question 24

The simplified logic expression for the following logic diagram is

Detailed Solution for GATE Physics Mock Test Series - 2 - Question 24

GATE Physics Mock Test Series - 2 - Question 25

For the logic circuit shown in the figure, the output y is given by

Detailed Solution for GATE Physics Mock Test Series - 2 - Question 25

So, y =
By dinorgen theorem

GATE Physics Mock Test Series - 2 - Question 26

An open pipe is suddenly closed at one end with the result that the frequency of third harmonic of the closed pipe is found to be higher by 200 H2 then the fundamental frequency of the open pipe - The fundamental frequency of the open pipe is

Detailed Solution for GATE Physics Mock Test Series - 2 - Question 26

Length of the organ pipe is same in both the cases, fundamental frequency of open pipe is  and frequency of third harmonic of closed pipe will be
Given that f2 = f1+ 200

GATE Physics Mock Test Series - 2 - Question 27

A paint particle of mass M attached to one end of a massless rigid non - conducting rod of length L. Another paint particle of the same mass is attached to the other end of the rod. The two particles carry changes + q and -q respectively. This arrangement is held in a region of a uniform electric field E such that the rod makes a small angle  with the field direction. Find the minimum time needed for the rod to become parallel to the field after it is set free

Detailed Solution for GATE Physics Mock Test Series - 2 - Question 27

A torque will act on the rod, which tries to align the rod in the direction of electric field. This torque will be of restoring nature and has a magnitude PE sinθ. Therefore

since θ is very small so that sin θ Sin θ ≈ θ

As α is proportional to - θ, motion of the rod is s impleharmonic in natural time period of which is given by

the desired time will be

GATE Physics Mock Test Series - 2 - Question 28

Which of the following statement regarding the electric fields  is correct

Detailed Solution for GATE Physics Mock Test Series - 2 - Question 28

Only   field can represent electrostatic field..

GATE Physics Mock Test Series - 2 - Question 29

A system of four identical distinguishable particles has energy  The single particle states are available at energies   Find the average number n(E) of particles in energy e =

Detailed Solution for GATE Physics Mock Test Series - 2 - Question 29

The possible distribution which can give a total energy are shown in table

GATE Physics Mock Test Series - 2 - Question 30

Consider 10 atoms fixed at lattices sites. Each atom can have magnetic moment   in the z - direction. Let f (μ) denote the probability that the magnetic moment of the system is μ. Assuming statistical mechanics to hold. Find the value of

Detailed Solution for GATE Physics Mock Test Series - 2 - Question 30

The magnetic moment can be  if 6 atoms in state and 4 in  state. This can happen in n

for magnetic moment to be zero , five atoms should be in state and five in state. This can happen in n2 ways where

The probability will be proportional to the no. of microstates. Hence

*Multiple options can be correct
GATE Physics Mock Test Series - 2 - Question 31

A solid sphere is in pure rolling motion on an inclined surface having inclination θ :

Detailed Solution for GATE Physics Mock Test Series - 2 - Question 31

In case of Pure rolling

Therefore, as θ decreases force friction will decrease.

*Multiple options can be correct
GATE Physics Mock Test Series - 2 - Question 32

The linear mass density of a rod of length L varies from one end to the other aswhere x is the distance from one end with tensions T1 and T2 in them (see figure) and λ0 is a constant. The rod is suspended from a celling by two massless strings. Then, which of the following statements is/are correct

Detailed Solution for GATE Physics Mock Test Series - 2 - Question 32

The mass of rod is
The Centre of gravity
Force equation T1 + T2 = torque equation

Putting the value of T2 =  in equation

*Multiple options can be correct
GATE Physics Mock Test Series - 2 - Question 33

Let mbe the mass of proton. mn the mass of neutron. M1 the mass of  nucleus and M2 the mass of  nucleus. Then :

Detailed Solution for GATE Physics Mock Test Series - 2 - Question 33

Due to mass defect (which is responsible for the binding energy of the nucleus), mass of nucleus is always less than the sum of masses of its constituent particles.
is made up to 10 protons + 10 neutrons. Therefore, mass of  nucleus
M1< 10 (mp + mn)
Also, heavier the nucleus, more is the mars defect.
Thus 20 (mn+mp)-M2 > 10 (mp +mn) - M1
10(mp + mn) > M2 - M1
M2<m1 +10(mn + mp)
Now since M1 < 10 (mp + mn)
M2 < 2m1

*Multiple options can be correct
GATE Physics Mock Test Series - 2 - Question 34

Consider the following statement regarding radioactive element emitting β - particle :
Which of the following are correct ?

Detailed Solution for GATE Physics Mock Test Series - 2 - Question 34

We radioactive element emits β- particle then the atomic number of the element increases by one unit where as the mass number of the element remain same : Hence, the equation for β- decay can be written as

Since in β -decay, a neutron is contributed into a proton, the neutron proton ratio decrease.

*Multiple options can be correct
GATE Physics Mock Test Series - 2 - Question 35

The graph between 1/λ and stopping potential (v) of three metals having work function   in an experiment of photoelectric effect is plotted as shown in the figure. Which of the following statement is / are correct ?

Detailed Solution for GATE Physics Mock Test Series - 2 - Question 35

From the relation eV =

This is equation of straight line
slope is tanθ =

Violet colour has wavelength 4000A0.
So, Violet colour can eject photoelectrons from metal -1 and metal - 2.

*Multiple options can be correct
GATE Physics Mock Test Series - 2 - Question 36

A long straight conductor, carrying a current I is bent into the shape shown in the figure.The radius of the circular loop is 'r' .The magnetic field at the centre of the loop is :-

Detailed Solution for GATE Physics Mock Test Series - 2 - Question 36

Field due to straight part(B1) =  out of the page
Field due to circular part (B2) =  into the page.
Not field (B) =B1-B2  into the page.

*Multiple options can be correct
GATE Physics Mock Test Series - 2 - Question 37

The doping of a semiconductor is such that the electron density in the conduction band is given by n = (x) = ni (1 +gx). Find incorrect expression/s for the electric field intensity in the conduction band.

Detailed Solution for GATE Physics Mock Test Series - 2 - Question 37

Due to density gradient electrons would diffuse in conduction band and due to electric field intensity holes drift in a direction opposite to that of electron diffusion. At equilibrium

or
or

*Multiple options can be correct
GATE Physics Mock Test Series - 2 - Question 38

A simple pendulum performs simple harmonic motion about x = 0 with an amplitude A and time period T. The speed of the pendulum at x = A/2 won’t be:

Detailed Solution for GATE Physics Mock Test Series - 2 - Question 38

*Multiple options can be correct
GATE Physics Mock Test Series - 2 - Question 39

A screen is placed 50 m from a single slit, which is illuminated with 6000  light. If distance between the first and third minima in the diffraction pattern is 3,000 mm, which of the following is/are not the width of the slit?

Detailed Solution for GATE Physics Mock Test Series - 2 - Question 39

In case of diffraction at single slit, the position of minima is given by
d sin θ= nλ
and for small θ sin θ = θ =
so from eq.(1) and (2), we have

so that,
and hence,

*Multiple options can be correct
GATE Physics Mock Test Series - 2 - Question 40

An unpolarized beam of light is incident on a group of four polarizing sheets which are arranged in such a way that the characteristic direction of each polarizing sheet makes an angle of 30° with the preceding sheet. What fraction of light is/are not transmitted?

Detailed Solution for GATE Physics Mock Test Series - 2 - Question 40

Here

∴
∴

GATE Physics Mock Test Series - 2 - Question 41

Calculate the change in the melting point of wax for a pressure of 50 atmospheres from the following data :
melting point = 64°C, Volume at 0°C = 1 cc; volume of the solid at the melting point = 1.161cc; volume of the liquid at the melting point = 1.166 c.c.; density of the solid at 64° C = 0.96 gm. / c.c latent heat = 97 cal. gm.

Detailed Solution for GATE Physics Mock Test Series - 2 - Question 41

0. 016 - 0.020
Here the melting point (64°C) is given to be at one atmosphere pressure and we are required to find the change in the melting point for a pressure of 50 atmosphere, i.e for a pressure change of (50 - 1) = 49 atmosphere.
dP = 49 x 76 x 13.6 x 981 dynes / cm2
T = 64°C = 64 + 273 = 337 K,
L = 97 cal. = 97 x 4.2 x 107 ergs.
Now mass of the solid at melting point = volume x density
= 1.161 x 0.96 gm.
∴ Specific volume of the liquid at melting point

and specific volume of the liquid at melting point

V2 - V1 = 1.0461 - 1.0417 = 0.0044 c.c. / gm.
Now Clausius Clapeyron equation is

So, that   = 0.018 K = 0.018°C

GATE Physics Mock Test Series - 2 - Question 42

A bead of mass 0.1 kg solids at rest from x (which is 4m above the ground) along a frictionless wire as shown in the figure given above, (Take g = 1Om/s2) when at B, the wire exerts a force on the bead equal to ______ N.

Detailed Solution for GATE Physics Mock Test Series - 2 - Question 42

Applying the conservation law of energy
Total energy at A = Total energy at B
(KE)A + (PE)A = (KE)B + (PE)B

mu2 = 4mg
Now let R' be reaction of ring on the bead directed upwardly as shown in figure. Then centripetal force = mg - R'

R' = mg - mu2 (R= 1m)
R' = mg -4mg
R' = -3m g ⇒ R' = - 3 x (0.1)x 10
R' = - 3N
|R' |= 3N

GATE Physics Mock Test Series - 2 - Question 43

10 gm. of steam at 100°C is blown into 90 gms. of water at 0°C contained in a calorimeter of water equivalent 10 gms. The whole of the steam is condensed. Calculate the increase in the entropy of the system.

Detailed Solution for GATE Physics Mock Test Series - 2 - Question 43

3.6 - 3.7
Given m1 = 10 gm., T1 = 100°C = 373 K
m2 = 90 + 10 = 100 gm., T2 = 0°C = 273 K.
Let the final temperature be T K.
∴   10 x 540 + 10 ( 373 - T) = 100 (T - 273)
T= 331.2 K.
Now (i) Change in entropy when temperature of water and calorimeter increases from 273 K t o 331.2 K.

(ii) Change in entropy when 10 gms. steam at 373 K condenses to water at the same temperature

(iii) Change in entropy when 10 gm. water is cooled from 373 K to 331.2 K.

or ΔS3 = - 1.188 cal/K
The net change (or increase in entropy)
ΔS = ΔS1 + ΔS2 + ΔS3 = + 19.324 - 14.477 - 1.188 = + 3.659 cals./K.

GATE Physics Mock Test Series - 2 - Question 44

The sum of the eigen value of the matrix

Detailed Solution for GATE Physics Mock Test Series - 2 - Question 44

Let A =

cos2θ- 2λcosθ+ λ2 + sin2 θ = 0
λ2 - 2λcosθ+ 1 = 0

at  θ = 60°
λ = cos 60 ± sin60
The sum of the eigen value is
= 2 cos 60 = 1

GATE Physics Mock Test Series - 2 - Question 45

A police car moving at 22 m/s chases a motorcyclist. The police man sounds his horn at 176 Hz, while both of them move towards a stationary siren of frequency 165 Hz. Calculate the speed of the motorcycle. If it is given that the motorcyclist does not observe any beats (speed of sound = 330 m/s)

Detailed Solution for GATE Physics Mock Test Series - 2 - Question 45

The motorcyclist observes no beats. So, the apparent frequency observed by him from the two sources must be equal.

Solving this equation, we get
v = 22m/s

GATE Physics Mock Test Series - 2 - Question 46

The period of a disk of radius 10.2 cm executing small oscillation about a pivot at its rim is measured to be _____ sec. { The value o f the all acceleration due to gravity at that location is 9.83 m/sec2}

Detailed Solution for GATE Physics Mock Test Series - 2 - Question 46

0.6-0.8
The rotational inertia of a disk about an axis through its centre is  where R is the radius and M is the mass of the disk. The rotational inertia about the pivot at the rim is, using the parallel axis theorem

The period of this physical pendulum T =
With d = R, is then

independent of the mass of the disk The simple pendulum having the same period has a length

GATE Physics Mock Test Series - 2 - Question 47

2kg of ice at - 20°C in an insulating vessel having a negligible heat capacity.The final mass of water remaining in the container i s ------kg. It is given that the specific heat of water and ice are 1KCal/kg/°C and 0.5 kcal/kg/°C while the latent heat of fusion of ice is 80 k cal/kg.

Detailed Solution for GATE Physics Mock Test Series - 2 - Question 47

Heat released by 5 kg of water when its temperature falls from 20°C to 0°C is
Q1 = mcΔT = 5x103 (20 - 0) = 105cal
when 2 kg ice at - 20°C comes to a temperature of 0°C, if takes an energy
Q2 = mcΔT = 2 x 500 x 20 = 0.2 x 105cal
The remaining heat
Q = Q1 - Q2 = 0.8 x 10cal will melt a mass m of the ice, where

So, the temperature of the mixture will be 0°C , mass of water in it is 5+1 = 6kg

GATE Physics Mock Test Series - 2 - Question 48

A particle of mass m is placed in the ground state of a one- dimensional harmonic oscillator potential of the form

where the stiffness constant K can be varied externally. The ground state wavefunction has the form

where a is a constant. If suddenly, the parameter k is changed to 4K, the probability that the particle will remain in the ground state of new potential is _____ .

Detailed Solution for GATE Physics Mock Test Series - 2 - Question 48

where A is normalisation constant
when parameter K is suddenly changed to 4K, the wave function associated with ground state of new potential is given by

Probability that particle will be found in ground state p =

Put

From Normalisation equation

Let

Probability P =

GATE Physics Mock Test Series - 2 - Question 49

The given figure shows a silicon transistor connected as a common emitter amplifier. Calculate the approximate value of quiescent collectorvoltage of circuit.

Detailed Solution for GATE Physics Mock Test Series - 2 - Question 49

GATE Physics Mock Test Series - 2 - Question 50

What will be the input independence Zi for the network shown below ?

Detailed Solution for GATE Physics Mock Test Series - 2 - Question 50

∴

GATE Physics Mock Test Series - 2 - Question 51

Find out the minimum number of NAND gates required to implement A + AB + ABC.

Detailed Solution for GATE Physics Mock Test Series - 2 - Question 51

Hence, number of NAND gates required is 0.

GATE Physics Mock Test Series - 2 - Question 52

For the given circuit calculate the output current. Where R1 = 6 kΩ, Rf = 24 kΩ, Vi= 1 V and load register of 6 kΩ.

Detailed Solution for GATE Physics Mock Test Series - 2 - Question 52

This is the relation for a non inverting amplifier

and load current can be calculated using

We know that current I1 and lL both are passing away from the output terminal
so
but
So l0 = 0.17 mA + 0.83 mA = 1 mA and this is flowing towards the output junction.

GATE Physics Mock Test Series - 2 - Question 53

Calculate the total output voltage for a differential amplifier in which the signal applied to inverting and non inverting terminals are respectively = -0.45 mV and -0.48 mV and  Adiff =106 and CMRR = 80 dB.

Detailed Solution for GATE Physics Mock Test Series - 2 - Question 53

CMRR=

So
The differential input is
Vdiff = V2 - V1 = - 0.48 - (0.45) mV = -.03 mV and common mode input is

Total output voltage

≈ -30 volt

GATE Physics Mock Test Series - 2 - Question 54

Two spherical nuclei have mass number 256 and 4 with their radii R1 and R2 respectively then find out the ratio

Detailed Solution for GATE Physics Mock Test Series - 2 - Question 54

∵
Given A1 = 254
A2 = 4

∵
So,

GATE Physics Mock Test Series - 2 - Question 55

Image of an object approaching a convex mirror of radius 20 m along its optical axis is observed to move from  in 20 sec. The speed (km/hr) of the object is ______

Detailed Solution for GATE Physics Mock Test Series - 2 - Question 55

4.5
Using mirror formula.

and
speed of object =
speed of object =

GATE Physics Mock Test Series - 2 - Question 56

A ball of mass m , initially at rest , is dropped from a height of 5 meters. If the coefficient of restitution is 0.9, the speed of the ball just before it hits the floor the second time is approximately______ m/sec. (take g = 10 m/sec2)

Detailed Solution for GATE Physics Mock Test Series - 2 - Question 56

9
Given that height = 5m
g = 10 m/sec2
Then the velocity of ball before hitting the surface first time =

coefficient of restitution = 0.9
So, after collision its velocity = 0.9 x 10 = 9 m/sec
So, before hits the floor second time velocity in = 9 m/sec
speed of object =

GATE Physics Mock Test Series - 2 - Question 57

AB and CD are two slabs. The medium between the slabs has refractive index 2. Find the minimum angle (In degree) of incidence of Q, so that the ray is totally reflected by both the slabs -

Detailed Solution for GATE Physics Mock Test Series - 2 - Question 57

60
Critical angles at 1 and 2

Therefore, minimum angle of incidence for total internal reflection to take place on both slabs should be 60º
imin = 60º

GATE Physics Mock Test Series - 2 - Question 58

Screen s is illuminated by two point sources A and B. Another source C sends a parallel beam of light towards point P on the screen and the lines AP, BP and C P are in one plane. The distances AP, BP and CP are in are plane. The distances AP and BP are 3m and 1 ,5m respectively. The radiant powers of sources A and B are 90ω and 180 W respectively. The beam From c is of intensity 20 V\//m2. calculate intensity at P on the seen

Detailed Solution for GATE Physics Mock Test Series - 2 - Question 58

13.9 - 14.0
Resultant intensity at P

= 0.97+ 3.18+10
lp = 13.97 W/m2

GATE Physics Mock Test Series - 2 - Question 59

Two beams , A and B, of plane polarised light with through a polaroid, from the position when the beam A has maximum intensity ( and beam B has zero intensity) a rotation of polaroid through 60°makes the two beams appear equally bright. If the initial intensities of the two beams are lA and lB respectively, then lA / lB equals________.

Detailed Solution for GATE Physics Mock Test Series - 2 - Question 59

3
By law of malus I = I0 cos2 θ

GATE Physics Mock Test Series - 2 - Question 60

The change in specific volume when 1 kg of water freezes is 91 x 10-6 m3. The pressure at 273 K i.e. Find out the pressure at which the ice would freeze, when it is given that latent heat of ice = 3.66 * 105 J/kg & one atmosphere=10-5

Detailed Solution for GATE Physics Mock Test Series - 2 - Question 60

136.2
From Maxwell’s relation

As given,

T =273 K

It means the pressure under which the ice would freeze = 1 + 135.2 = 136.2 atm.

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