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GATE Physics Mock Test Series - 1 - Physics MCQ


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30 Questions MCQ Test GATE Physics Mock Test Series 2025 - GATE Physics Mock Test Series - 1

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GATE Physics Mock Test Series - 1 - Question 1

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 : 

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

From conservation of energy


From Continuity equation
A1V1 =A2,V2




A2= 5x10-5m2

GATE Physics Mock Test Series - 1 - Question 2

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 :

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

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

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GATE Physics Mock Test Series - 1 - Question 3

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

Detailed Solution for GATE Physics Mock Test Series - 1 - Question 3

Current i = (frequency) (charge)

magnetic moment M = (i) (A)

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

GATE Physics Mock Test Series - 1 - Question 4

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  

Detailed Solution for GATE Physics Mock Test Series - 1 - Question 4

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

In chamber B


GATE Physics Mock Test Series - 1 - Question 5

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

Detailed Solution for GATE Physics Mock Test Series - 1 - Question 5

We know that 
∴ 

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

ω = 3

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

GATE Physics Mock Test Series - 1 - Question 6

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

Detailed Solution for GATE Physics Mock Test Series - 1 - Question 6

Applying Bernoulli’s theorem , the rate of flow from surface to orifice is given by
 .....(1)
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

or

∴ 

GATE Physics Mock Test Series - 1 - Question 7

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.

Detailed Solution for GATE Physics Mock Test Series - 1 - Question 7

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.

GATE Physics Mock Test Series - 1 - Question 8

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

Detailed Solution for GATE Physics Mock Test Series - 1 - Question 8

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)
or

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

or 



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


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

GATE Physics Mock Test Series - 1 - Question 9

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:

Detailed Solution for GATE Physics Mock Test Series - 1 - Question 9

(t1/2)x = (tmean)y

λx = 0.693 λy
λxy
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.

GATE Physics Mock Test Series - 1 - Question 10

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?

Detailed Solution for GATE Physics Mock Test Series - 1 - Question 10

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.

GATE Physics Mock Test Series - 1 - Question 11

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 -

Detailed Solution for GATE Physics Mock Test Series - 1 - Question 11

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



GATE Physics Mock Test Series - 1 - Question 12

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

Detailed Solution for GATE Physics Mock Test Series - 1 - Question 12


GATE Physics Mock Test Series - 1 - Question 13

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 ?

Detailed Solution for GATE Physics Mock Test Series - 1 - Question 13


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,

GATE Physics Mock Test Series - 1 - Question 14

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

Detailed Solution for GATE Physics Mock Test Series - 1 - Question 14

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

GATE Physics Mock Test Series - 1 - Question 15

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

Detailed Solution for GATE Physics Mock Test Series - 1 - Question 15

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

IA * 1/4 = IB * 3/4

IA/IB = 3
 

GATE Physics Mock Test Series - 1 - Question 16

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

Detailed Solution for GATE Physics Mock Test Series - 1 - Question 16

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 
y=Y1+y2+y3
= 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 ...

GATE Physics Mock Test Series - 1 - Question 17

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

Detailed Solution for GATE Physics Mock Test Series - 1 - Question 17

For equilibrium Mg = mgsin 30°

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

GATE Physics Mock Test Series - 1 - Question 18

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 .

Detailed Solution for GATE Physics Mock Test Series - 1 - Question 18

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

GATE Physics Mock Test Series - 1 - Question 19

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.

Detailed Solution for GATE Physics Mock Test Series - 1 - Question 19







GATE Physics Mock Test Series - 1 - Question 20

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

Detailed Solution for GATE Physics Mock Test Series - 1 - Question 20

Output can be calculated as :-

V0 = -(V1 + V2) sinωt

GATE Physics Mock Test Series - 1 - Question 21

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 :-

Detailed Solution for GATE Physics Mock Test Series - 1 - Question 21


GATE Physics Mock Test Series - 1 - Question 22

The DC load line of an amplifier circuit :-

Detailed Solution for GATE Physics Mock Test Series - 1 - Question 22

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

GATE Physics Mock Test Series - 1 - Question 23

Detailed Solution for GATE Physics Mock Test Series - 1 - Question 23

GATE Physics Mock Test Series - 1 - Question 24

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

Detailed Solution for GATE Physics Mock Test Series - 1 - Question 24

Resolving angle of naked eye is given by



y = 30x10-6 m ⇒ y = 30μm

GATE Physics Mock Test Series - 1 - Question 25

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

 

Detailed Solution for GATE Physics Mock Test Series - 1 - Question 25

The region X has uncovered acceptor impurities & mobile holes.

GATE Physics Mock Test Series - 1 - Question 26

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.

Detailed Solution for GATE Physics Mock Test Series - 1 - Question 26

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

GATE Physics Mock Test Series - 1 - Question 27

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 :-

Detailed Solution for GATE Physics Mock Test Series - 1 - Question 27

as ;  (because of lenz law).

GATE Physics Mock Test Series - 1 - Question 28

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 :-

Detailed Solution for GATE Physics Mock Test Series - 1 - Question 28

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

GATE Physics Mock Test Series - 1 - Question 29

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 : -

Detailed Solution for GATE Physics Mock Test Series - 1 - Question 29


GATE Physics Mock Test Series - 1 - Question 30

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 ?

Detailed Solution for GATE Physics Mock Test Series - 1 - Question 30

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

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