GATE Physics Mock Test Series - 1

60 Questions MCQ Test GATE Physics Mock Test Series | GATE Physics Mock Test Series - 1

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Water from a tap emerges vertically down-wards with an initial speed of 1.0 m/ sec. The cross-sectional area of tap is 10-4m2. Assume that the pressure is constant throughout the stream of water and that the flow is steady, the cross-sectional area of stream 0.15 m below the tap is : 


From conservation of energy

From Continuity equation
A1V1 =A2,V2

A2= 5x10-5m2


A vessel contains a mixture of one mole of oxygen and two moles of nitrogen at 300K. The ratio of the average rotational kinetic energy per O2 molecule to Per N2 molecule is :


Average kinetic energy per molecule per degree of freedom = 1/2 KT. Since both the gases are diatomic and at the same temperature (300K). Both will have the same number of rotational degree of freedom i.e. two Therefore both the gases will have the same average rotational kinetic energy per molecule

So Ratio will be 1 :1


Two particles each of mass m and charge q, are attached to the two ends of a light rigid rod of length 2R. The rod is rotated at constant angular speed about a perpendicular axis passing through its centre. The ratio of the magnitude of the magnetic moment of the system and its angular momentum about the centre of the rod is


Current i = (frequency) (charge)

magnetic moment M = (i) (A)

M = qωR2
L = 2lω = 2(mR2)ω Angular momentum


Two identical containers A and B with frictionless pistons contain the same ideal gas at the same temperature and the same volume V. The mass of the gas in A is MA and that in B is MB. The gas in each cylinder is now allowed to expand isothermally to the same final volume 2V. The changes in the pressure in A and B are found to be MAΔ P and 1.5 MBΔ P respectively. Then  


Process is isothermal .
Therefore, T = constant P α 1/V volume is increasing, therefore, pressure will decrease.

In chamber B


A particle executing S.H.M. in a straight line has velocities 8, 7, 4 at three points distant one foot from each other. The maximum velocity of the particle will be


We know that 

64 - 49 = 15 = ω2 (2 x + 1) 
64 - 16 = 48 = ω2 (4 x + 4)

ω = 3

Maximum velocity = ωa = 3 x [√65/3] = √65 Ans


If A denotes the area of free surface of a liquid and h is the depth of an orifice of area of cross-section a below the liquid surface then the velocity u of flow through the orifice is given by


Applying Bernoulli’s theorem , the rate of flow from surface to orifice is given by
where V = velocity of all surface of liquid.
If u is velocity of efflux, then
AV = au or V = (au/A).....(2)
From eqs. (1) and (2, we get




A particle of mass 500 gm is at rest in the stationary reference frame. The motion of the particle with respect to reference frame rotating with an angular velocity 10π rad/sec2 is.


Mass of particle m = 500 gm = 0.5 kg angular velocity of reference frame ω = 10π rad/sec2 Distance of particle from the axis of rotation r= 15 cm = 0.15 m
We know that the velocity of particle in rotating reference frame

Since the particle is at rest in the stationary reference frame 

so, the particle will appear to be moving in a circular path with respect to rotating reference frame.


P-V diagram of n moles of ideal gas is as shown in figure. The maximum temperature between A and B is.


From the graph the P - V equation can be written as, P =  V + 3 P0 (y = - mx + c) 
or PV =  V2 + 3 P0V or nRT = 3 P 0 V2  (as PV = nRT)

This is the required T - V equation. This is quadratic in V.
T - V graph is parabola. Now, to find maximum or minimum value of T we can substitute. dT/dV = 0


Hence, T is maximum at V = 3/2 V0 and this maximum value is,

Thus, T - V graph is as shown in figure.


The half-life period of a radioactive element X is same as the mean lifetime of another radioactive element Y. Initially, both of them have the same number of atoms. Then:


(t1/2)x = (tmean)y

λx = 0.693 λy
Rate of decay = λN
Initially number of atoms (N) of both are equal but since λxTherefore y will decay at a faster rate than x.


Liquid oxygen at 50 K is heated to 300 K at constant pressure of 1 atm. The rate of heating in constant which of the following graphs represents the variation of temperature with time?


Temperature of liquid oxygen will first increase in the same phase. Then phase change will take place. During which temperature will remain constant After that temperature of oxygen in gaseous state will further increase.


Consider three identical infinite straight wires A, B and C arranged in parallel on a plane as shown in the figure. The wires carry currents as shown in figure and have mass per unit length m. If the wires Aand C are held fixed and the wire B is displaced by a small distance x from its position, then it (B) will execute simple harmonic motion with a time period -


Let us first calculate the force on wire B due to wire A magnetic field at the position of wire B

force per unit length on wire B due to wire A
  towards right
Similarly force per unit length on wire B due to magnetic field produced by wire C

If wire B is displaced towards right, then net restoring force will be equal to


A thin conducting wire is bent into a circular loop of radius r and placed in a time dependent magnetic field of magnetic induction

Such that, the plane of the loop is perpendicular to  Then the induced emf in the loop is



Which one of the following curves gives the solution of the different at equation  where k1, k2 and k3, are positive constant with initial conditions x = 0 and t = 0 ?


Differentiating w.r.t. to t, we get

Hence, solution is given by

By given condition x = 0 , when t = 0
0 = c1 + c2 => c2 = -c,


A particle is in normalized state |ψ > which is a super position of the energy eigen states |E0 = 10eV > and |E1 = 30eV >. The average value of energy of the particle in the state | ψ > is 20eV. The state | ψ > is given by


<ψ* |E| ψ > = 20ev
Now ψ0* |E0| ψ0
and < ψ| E1 | ψ> = 30ev
only option (d) satisfies this condition as


Two beams A and B of plane polarized light with mutually perpendicular planes of polarization are seen 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 equals


By the law of malus I = I0 cos2- θ
IA '=IA cos2 60º
IA '= IB '
IA 'cos 2 60º= IB 'cos30º 


There are three sources of sound of equal intensity with frequencies 400, 401, and 402 vib/sec. The number of beats per second is


Let n -1 = 400, n = 401, n + 1 = 402
y1 = a sin 2π (n - 1 ) t ; y2 = a sin 2π nt ; y3 = a sin 2π (n + 1)t 
= a sin2π nt + a sin 2π (n-1) +a sin 2π (n+1)t
= a [ 2 cos 2π t + 1] sin 2π nt 
y = a’ sin 2π nt 
a ’ = a[1 + 2 cos 2π t ]
Iα(a1)2α a[1 + 2cos2πt]2
for I to be max or min
⇒ 2πt = 0 or 1 + 2 cos 2πt = 0 
⇒if 1 + 2 cos 2πt = 0 

n = 0 , 1 , 2 , ............

2nπ= nπ ⇒ n  = 0, 1 , 2 , 3 ...


A wire of 9.8 *103 kg n-1 passes over a frictionless light pulley fixed on the top of a frictionless inclined plane which makes an angle of 30° with the horizontal. Masses m and M are tied at the two ends of wire such that m rests on the plane and M hangs freely vertically downwards. The entire system is in equilibrium and a transverse wave propagates along the wire with a velocity of 100 m/sec. Then the mass


For equilibrium Mg = mgsin 30°

According to wave formula, velocity of wave 
where is mass per unit length M = 10kg, m = 20kg


In a test experiment on an aeroplane model in a wind tunnel flow, the speeds the air on the upper and lower surfaces of the wings are 70 m/sec and 63 m/sec respectively. If the area of the wings is 2.5 m2. then find the up thrust on the wings .


By Bernoulli theorem

V, v1 = 70m /sec, v2 = 63 m /sec and p = 1 .Skg/m3

= 605.15 N/m2
Up (thrust) = (P2 - P1 ) A = 605.15x 2.5 
Up (thrust) = 1512.9 N


If the wave function ψ = X e -mωx2 /2ℏ and potential is defined as v = 1/2mω2x2 . Find out the eigen energy value of the particle.



If the op-amp in the figure ; is ideal then Vis :-


Output can be calculated as :-

V0 = -(V1 + V2) sinωt


For a BJT, the current amplification factor “ a = 0.9” . This transistor is connected in CE configuration. When the base current changes by 0.4 mA, the change in collector current will be :-



The DC load line of an amplifier circuit :-


The DC load line of an amplifier circuit has a negative slope.




Assuming human pupil to have a radius of 0.25 cm and a comfortable viewing distance of 25 cm , the minimum separation between two objects that human eye can resolve at 500 nm wavelength is


Resolving angle of naked eye is given by

y = 30x10-6 m ⇒ y = 30μm


In the diagram of the space charge across a p-n junction as shown ; the region X has :-



The region X has uncovered acceptor impurities & mobile holes.


For the three - particle system as shown is figure. Find the rotational interia about an axis perpendicular to the x - y plane and passing through the centre of mass of the system.


The centre of mass

ycm = 1.37m
The squared distances from the centre of mass to each of the particles are

= 11.74 m2
The rotational inertia then follows directly from
= m1r1+ m2r22 + m3r32
2.3 x 2.62 + 3.2 x 3.4 + 1.5 x 11.74
= 34.516 kg m2


Figure shown a horizontal solenoid connected to a battery and a switch. A copper ring is placed on a frictionless, the axis of the ring being along the axis of the solenoid. As the switch is closed, the ring will :-


as ;  (because of lenz law).


In the circuit shown in figure initially the switch is in position ‘1’ for a long time, then suddenly at t = 0, the switch is shifted to position ‘2’. It is required that a constant current should flow in the circuit, the value of Resistance ‘R’ in the circuit :-


With the given circuit & data the constant current scenario cannot be determined


In the circuit shown in figure L = 20 H, R = 15Ω, E = 15V, the switch ‘s’ is closed at t = 0. At t = 4 sec. The current in the circuit is : -



Two waves travel in the same direction along a string and interface. The waves have the same wavelength and travel with the same speed. The amplitude of each wave is 9.7 mm , and there is a phase difference of 110° between them. What is the amplitude of the combined wave resulting from the interface of the two waves ?


The amplitude of the combined wave 
2ym | cos(ΔØ/)/2)= 2(9.7mm) cos(110°/2)
= 11.1 mm

*Multiple options can be correct

Consider the following statement associated with phase transition. Which of the statement(s) is/are correct ?


There is two type of phase transition's
(i) First order phase transition
(ii) Second order phase transition 
In first order phase transition
(i) T and P are constant
(ii) dQ changes
⇒ G, the Gibb's function remain constant.
In second order phase change
(i) density and entropy changes slowly
 are continuous.
(iii) G becomes discontinuous.

*Multiple options can be correct

Consider the following statement associated with quantum mechanical Simple harmonic oscillator. Which of the following statement(s) is/are correct ?


The energy of a linear harmonic oscillator is given as

Where n = quantum number 
v = frequency
⇒ E increases with quantum number n 
Energy corresponding to n = 0 is
 (non - zero)

*Multiple options can be correct

Consider the following statement about quantization of angular momentum, then select the correct statement(s):


According to Bohr electrons revolve around the nucleus in only those orbits for which the angular momentum is an integer multiple of 
i.e., angular momentum is =  ; n=1,2,3, ...
⇒ angular momentum is quantized.
Since, motion of electron is on a plane so, only one component of the angular quantum number is quantized.

*Multiple options can be correct

Consider the following statements when a steel ball hits a clay chunk in air and gets embedded in it, then which of the following statement(s) is/are correct ?


Momentum is always conserved whether collision is elastic. In non elastic collision energy is used to deform the colliding objects, hence in this case energy is not conserved.

*Multiple options can be correct

The torque on a body about a given point is found to be equal to  where  is a constant vector and is the angular momentum of the body about the point from this it follow that.


This relation implies that  is perpendicular to both  .Therefore option A is correct.

Therefore  = 0 or magnitude of  does not change with time. 
(B) So far we are confirm about two points
and L is not changing with time, therefore it is a case when direction of  is changing but its magnitude is constant and is perpendicular to at all points.
This can be written as

Now  is a constant vector and it is always perpendicular to  Thus  can be written as 

so, we can say that component of  along  is zero or component of   along  is zero or component of  along  is always constant. Finally we conclude   are always mutually perpendicular

*Multiple options can be correct

Two blocks A and B each of mass m are connected by massless spring of natural length L and spring constant K. The blocks are initially resting on a smooth horizontal floor with the spring at its natural length as shown in figure. A third identical block C also of mass m moves on the floor with a speed V along the line joining A and B and collides with A. Then


Let the velocity required by A and B be v, then mv = mV + mV   where x is the maximum compression  • At maximum compression , kinetic energy o f the AB system = 

*Multiple options can be correct

Consider the circuit shown in figure with If transistor is operating in the active region, then which of the following statement  is/are correct :- 


Applying KVL, in EC loop we pet ; 
In EB loop ; 

putting R1 in equation we get ;

*Multiple options can be correct

Consider the shunt regulator circuit shown in  figure with Rs. = 0.75 kΩ . Rl = 1 KΩ, Vs= 15V zener breakdown voltage (Vz) = 9V, rated power of zener diode  Pzen = 25 mW, then


(2) & if RL = 1.25 KΩ , V $= 15 - 9 = 6v

I z = 8 - 7.2 = 0.8 mA 
Pz = 0.8 mA x 9V = 7.2mW

*Multiple options can be correct

A particle of mass ‘m’ & charge - q is projected with a horizontal speed v into an electric field of intensity ‘E’ directed downward. The following statements are given ; Mark the one which is correct: -


In + y - direction : - y = 
In + x - direction :- x = vt .............. (2)
From ( 1 ) & ( 2 ) :- 

*Multiple options can be correct

Two protons move parallel to the x-axis in opposite direction with same speed v (<c) At the instant, then which of the following is/are not the ratio of magnetic force to electric force on the upper proton ?


From Coulomb’s law, the magnitude of the electric force on the upper proton is

The forces are repulsive, and the force on the upper proton is vertically upward  (in the + y-direction).

Fig. Electric magnetic forces between two moving protons.
From the right-hand rule for the cross product the  due to the lower proton at the position of the upper proton is in the +z-direction. The magnitude of  is


The velocity of the upper proton is  and the magnetic force on it is . combining this with the expressions for  we find 

The magnetic interaction in this equation is also repulsive. The ratio of the magnitudes of the two forces is

Using the relationship  we can express our result very simply as

when  is small in comparison to c, the speed of light, the magnetic force is much smaller than the electric force.


A uniform solid cylinder is released on a horizontal surface with speed 15 m/ sec without any rotation (slipping without rolling). The cylinder eventually starts rolling without slipping. If the mass and radius of the cylinder are 100 gm and 10 cm respectively, the final linear velocity of the cylinder is _____ m/sec.


Given that
The speed of solid cylinder = 15 m/sec 
mass of cylinder = 100 gm 
radius of cylinder = 10 cm 
By Applying the angular momentum conservation 
mvr = mV cm r + lcm w

mvr = 
mvr =  
V = 


Let If   is a vector such that and the angle between 


So we need 



A hydrogen like atom of atomic number z is in an excited state of quantum number 2n. It can emit a maximum energy photon of 204 eV. If it makes a transition to quantum state n, a photon of energy 40.8 is emitted. The minimum energy (in eV) that can be emitted by this atom during de - excitation i s _____ .


Let ground state energy (in eV) be E1
Then from the given condition

From equation 

E1 = -(13.6 )Z2 ⇒ z = 4

Emin = 10.58 eV


Two moles of an ideal monoatomic gas is taken through a cycle ABCA as shown in the P-T diagram. During the process AB, pressure and temperature of gas vary such that PT=constant. If T1 = 300 K. The work done on the gas in the process Ab i s ____


9950 - 9980

PdV + VdP = nRdT

Pt = constant
PdT + TdP = 0 
Pdv = 2nRdT

= 2nR(TB- TA)
= 2nR(T1 - 2T1)
= 2 x 2 x R(300 - 600) 
 = 1200R= 9972


A body of mass 2 k.g. moves under the influence of a central force, with a potential energy function V(r) =  Joule, where r is the radius of circular orbit, and r = 5 m. (Given e3 ≈ 20). Find its total energy E. (in the unit of mJ)



Given r = 5m.,m = 2}

Then angular momentum = mvr

Total energy E = K.E. + P.E.
  (Given e3 ≈ 20)

= (0.25 - .1) x 10-2 
= 15 x 10-2


The value of   where  and s is the surface of the cube 


(2x + 2y + 2z)dv 
2(x + y + z)dx dy dz
= 3


An ice cube of mass 0.1 kg at 0°C is placed in an isolated container which is at 227° C. The specific heat S of the container varies with temperature T according to the container to the empirical relation S = A + BT, where A = 100 cal/kg.k and B = 2 x 10 2 cal/kg.k2. If the final temperature of the container is 27° C. Then the mass of the container is ______ kg.
[Latent heat of fusion for water = 8 x 104 cal/kg. specific heat of water = 103 cal/ kg.k ]


0.49 or 0.5
Let m be the mass of the container Initial temperature of container 
T1 = (227 + 273) = 500kFinal temperature of container TF = 27 + 273 = 300 k 
Now Heat gained by the ice cube = Heat lost by the container.
⇒ (mass of ice) (Latent heat of fusion of ice) + (mass of ice) (specific heat of water) (300-273)

⇒ (0.1)(8x104)+ (0.1)(103)27
(A + BT)dT
10700 = 
m = 0.495 kg


A 100 Ω resistor, a 100 μF capacitor and a .01 H inductor are connected in series to an a.c. source operating at an angular frequency ω. The a.c. source is operated at a peak voltage of 200 V and a frequency equal to half the resonance frequency of the circuit. Find the peak value of the current in the circuit.


L - C - R combination will behave as a pure resistive load at a resonance i.e. when reactance X = 0

Given L = 0.01 H 
C = 1 00 x 1 0-6 F

= 500S-1
Then Impedance Z =  
R = 100 Ω

= 20-5
X = 15 Ω

Z = 101.12 Ω
Given V= 200 V


Given that the work function of a metal is 5eV and the photons of 7.5eV are incident on it, the maximum velocity of emitted electrons is approximately_________ x 109 m/sec.


937 or 938
K-Emax = 7.5-5 = 2.5 eV 

= 0.937x 106 
= 937x109


If the normalized wavefunction for a particle in a one dimensional box is given by ψ(x) = A(1- x2) for -1 ≤x ≤1 then the value of A2 is _______ .


0.93 to 0.94
Applying normalisation condition


= 0.937


A train moves towards a stationary observer with speed 34m/sec. The train sounds a whistle and its frequency registered by the observer is f1, if the speed of train is reduced to 17m/sec, the frequency registered is f2, if the speed of sound is 340 m/sec then the ratio f1/f2 is ______ .


By using f ’ = 
f1 = 


A hollow pipe of length 0.8 m is closed at one end At its open end a 0.5 m long uniform string is vibrating in its second harmonic and its resonates with the fundamental frequency of the pipe, if the wire is 50 N and the speed of sound is 320 m/sec, the mass (gm) o f the string is _______ .


Fundamental frequency of close organ pipe  second harmonic frequency of string 
m = 40 gm


The circuit diagram of a common emitter amplifier using a silicon transistor with hfe = 100 is shown in the above figure. It is transistor with hfe = 100 is shown in the above figure. It is operated at VCE = 8V, I= 2mA ; RL = 11K Ω . In this circuit R3 (in KΩ) will be ___________ ?



from (1) we get
Vcc = 22 - 8 = 14V
Putting the value of Vcc in equation we get

Thus answer is RB = 700KΩ


Consider the transistor Circuit shown in figure β = 50, (VBE)sat, (VCE)sat = 0.2V; RB = 100 K Ω . Calculate the minimum value o f Rc such that the transistor operates in the saturation region is :-


Applying KVL in both the loop we get

equating these two equation & putting values we get ; 
Rc = 2.084KΩ          & IB = 0.142mA


Assuming VCE,sat = 0.2V &  β = 50, the minimum base current (lB) required to drive the transistor in figure to saturation is ___________ mA.


3V - ICRC =0 


Calculate the no. of lattice points present in a primitive cell?


Primitive cell is the cell where the total no. of atoms is equal to ‘one’ only.


A Ring having Radius ‘R’ carries as charge Q = -1 C uniformly distributed over it. The magnitude of charge that should be placed at the centre of the ring such that the electric field becomes zero at a point distance ‘R’ away from the centre on the axis of the ring (in C)?


E.F on the axis of the ring = 
at a distance ‘R’ → 
Thus ; Q1 =0.35 C
Q1 = 0.35 C


A grating having 15000 lines per inch produces spectra of a mercury arc. The green line of the mercury spectrum has a wavelength of 5461 A. What is the angular separation between the first-order green line and the second-order green line ?


Use the grating formula and solve for 0 for both n = 1 and n =2 , with distances in meters :
nλ = d sin θ

or 0.3225 = sin θ1 
and e1 = 18.8° first order 
Next, for n =2,

or 0.3225 = sin θ1
and θ1 = 40.2° second order
Finally 02 -01 = 40.2° - 18.8° = 21.4° separation.


A resistor of 1KΩ and an inductor of 5mH are connected in series with a battery of emf uv through a switch. The switch is closed at time t = 0 find the current (mA) flowing the circuit at 20 μsec,


At t = 0 , switch s is closed , let current following in the circuit at time t is i.
The induced emf developed is

At t = 0 ,i = 0

i = 3.92 mA


A pulse of light of duration 100 ns is absorbed completely by a small object initially at rest. Power of the pulse is 30 mV and the speed of light is C. The final momentum o f the object is in the unit o f 10 -17 kg - ms-1 is _______.


Final momentum of object = 
Final momentum of object = 
Final momentum of object = 1.0 x 10-17 kg - m / sec