GATE Physics Mock Test Series - 6 - GATE

According to thermodynamic equation 
Tds = du + pdv


 

The expectation value of an operator that does not depend on the time and commutes with the Hamiltonian is constant in time. So, (E) will remain constant with time.

If P₁= Pthen P₂= P + 10% of P = 1.10 P
From boyle’s lawPV = constant

∴ Percentage change in volume

The auxiliary equation is 
m²- 3m + 2 = 0 
(m - 1)(m - 2) = 0   ⇒ m = 1, 2
CF = C₁e^x + C₂e^2x

So, (a) is AND gate

So (B) is OR GATE

For a circular motion of small element dx

Here, m₁= m₂ = m, u₁ = u, u₂ = 0. Let v₁, v₂ be their velocities after collision. 
According to principle of conservation of linear momentum, 
mu + 0 = m (v₁ + v₂) or v₁ + v₂ = u ... (i)
By definition, ....(ii)
Add (i) and (ii),  
Subtract (ii) from (i)

ΔS = 0

Let two equal charges Q each, be held at A and B, where AB = 2x. C is the centre of AB. where charge q is held.

For the three charges to be in equilibrium, net force on each charge must be zero.
Now, total force on Q at B is

which was to be proved.

We know intensity I = Es

Activity of S₁ =  (Activity of S₂)

Given N₁ = 2N₂

When a plane-polarised light is incident on half wave plate with angle θ = 30° then output emerging light is also plane polarised and making angle 2θ = 60°

When plane polarised light is incident on quarter wave plate at θ = 45° then out light is circularly polarised. Because the phase difference provided by Q-W plate is and path difference is  so, output light is circularly polarised

Voltage at OP-AMP output
V₁ = - 1 - 5 sin cot
At   
At   
During ‘+’ ve cycle of the waveform, when V₁ < -2V, diode D₂ conducts and V₀ = -2 V
During ‘-’ ve cycle, when V₁ > 2V, D₁ conducts and V₀ = 2V.

Heat given by 1 gm water when it freezes into ice 
Q = mL= 1 gm x 80 cal / gm = 80 cal.
= 80 x 4.2 Joule = 336 Joule
The external work done against the atmosphere when its volume increases by 0.091 cm³

By the first law of thermodynamics, the change in internal energy

Here Q is taken negative because heat is given by the system.
Thus internal energy decreases by 336.0093 Joule.

In I and III region direction of E₁ and E₂ are opposite to each other and In magnitude

So , in I and III region direction of E₁ and E₂ are opposite to each other and in magnitude

So, in I and III region resultant electric field is zero. But in region II^nd i.e. region between plates

Given E = E₀ cosωt 
Displacement vector

So displacement current density

Conduction current density

When S₃ is closed, due to attraction with opposite charge, no flow of charge takes place through S₃. Therefore, potential difference across capacitor plates remains unchanged or V₁= 25 V and V₂ = 15 V

When loss in internal energy = work done by the system, dQ = 0, process must be adiabatic
Further, loss in internal energy indicates decrease in temp. of the gas.

Applying KCL at various nodes, we have

or V_in + V₂ = 2V₁ = 0    (as inverting terminal is at virtual ground)
So V_in = -V₂

or V₃ = 3V₂ = -3V_in    .....(2)

Putting values from equation (1) and (2), we get 

Let p, q and r be the intercepts on X, Y, and Z axes respectively, then

where a, b and c are primitives and h, k and l Miller indices.
Given that a= 1.2 Å, b = 1.8 Å and c = 2.0Å and h = 2 , k = 3 and l = 1
∴       
= 0.6 : 0.6 : 2.0 
As       p = 1.2 Å
∴      1.2 : q = 0.6 : 0.6
or      
Similarly,     1.2 : r = 0.6 : 2.0
or    
Thus the intercepts along Y and Z axes are 1.2 Å and 4 Å respectively.

Here Q is stay at point (2a, 0) which is very for in x-axis (2a).

Motion is simple harmonic only if Q is released from a point not very far from the origin on x-axis. (for small oscillations) other wise motion is periodic but not simple harmonic. So here such type of motion is not simple harmonic motion.

When the rod rotates through an angle θ, the centre of gravity falls through a distance h. from ΔBG’ C,


..(1)
The decrease in P.E. is equal to the kinetic energy of rotation 
  ..(2)
From equation (1) and (2), we get

Clearly, the force will be minimum when upper end of the rod lies on the round support, we are excluding the case when rod is vertical In Figure, sinθ = 

  ...(1)
Now, at P (first minima) path difference between the rays reaching from two edges (A and B) will be

Corresponding phase difference (φ) will be

At junction a potential barrier/depletion layer is formed, with n-side at higher potential and p-side at lower potential. Therefore, there is an electric field at the junction directed from the n-side to p-side.

for stable point (orbit) 

Therefore radius of stable circular orbit is  Angular frequency of oscillation 
about stable paint

We know, in a simple cubic lattice all atoms are at eight comers of cubic lattice of length a.
So, In a simple cubic lattice, the distance of nearest neighbour d₁ = a 
and the distance of next nearest neighbour d₂ = a√2

The situation is shown in Fig. Let R be the radius of the sphere that can just fit into the void.
From figure, 4 + R = a/2
∴     R = (a/2) - r ..(1)

We know that for fee structure
a = 4r/√2 ..(2)
Substituting the value of a from eq.(2) in eq. (1), we get

Ball moves under the effect of grauity.

Therefore magnitude of acceleration of ball during upward or downward movement remain constant
for first bounce
y = H, u_y = 0, a_y = g 

Therefore time of fall before bounce 
Under gravity, time for going up a certain distance is equal to time for going down.
Therefore time interval between first bounce and second bounce is

Therefore, total time elapsed before ball stops is 
T = t₁+ t₂+13 +.......

Let v₁ is velocity of first mass after collision and v₂ is velocity of second mass after collision.
By the conservation law of momentum
m₁v = m₁v₁ + m₂v₂ 
m₁ = m₂ = m
⇒  mv = mv₁ + mv₂ 
v = v₁ + v₂
By conservation law of energy

Velocity of mass becomes zero whereas velocity of second mass attain the velocity of first. Now, the velocity of centre of mass is given 

Hence the velocity of first w.r.t. centre of mass 
That of second = 
Total kinetic energy as seen from centre of mass = 

Process A → B (isothermal)

Process B → C (isochoric process)

Process C → D (Isothermal process)

Process D → A (Isochoric process)

Here particle is confined to an infinite potential symmetric well of width a as The eigenfunction is given by


and energy eigenvalues are given by 

Coordination number is the number of nearest neighbours to a lattice point.

packing fraction
BCC :-   
simple Cubic structure :-

In the presence of dray force, only the distance covered in upward and downware motion is common, for upward motion, initial speed = V₀ final speed = 0, let height reaced = h
Equation of motion 

on integrating we get,    ________(1)
for downward motion, initial speed = 0 final speed = v (let), height descended = h 
⇒ equation of motion

According to given problem v = o if x = 0.25

at equilibrium position U = o , x = o

(A) Translational KE = 
(C) Rotational KE =  

Hence A is orthogonal matrix

We know that
x = A sin ωt i.e. x = 0 at t = 0 

SO T₁ < T ₂

Given that E(R) = 
for equilibrium separation

The given differential equation is a linear differential equation

The genral solution of the given differential equation is

It is given that y(0) = 0 
o. sec o = o + c⇒ c = 0
Thus the solution satisfying the given condition is

Net work done W = W_AB + W_BC + W_CD + W_DA = 1152 J

Applying continuity equation at upper end and at hole 
A₁V₁ =A₂V₂
Further applying Bernoulli’s equation at two points

In this situation, the source is moving towards the observer. Hence he will hear the sound of frequency n’ as given by

= 259.9 Hz
The same frequency is received by the wall. This frequency is reflected back towards observer. The wall with act as a source of frequency n’. Now source and observer both are at rest. The apparent frequency received by observer
n” = n’ = 259.9 Hz
∴  Beat frequency = n” - n’ = n’ ~ n’ = 0

As in case of diffraction at single slit second order maximum is approximately midway between 2nd and 3rd diffraction minima,
.....(1)
and as for small q,   sin q = q = (y/D)....(2)
from equ.(1) and (2), we have

So    

Fringe width 

When the air is compressed slowly, the heat generated conducts away to the surroundings. Thus it is an isothermal process, temperatures remains constant throughout, i.e., there is no change in temperature.
When the air is compressed suddenly, the process is adiabatic.
Hence

If T₁, V₁ be the initial and T₂, V₂ the final temperature and volume respectively, then

or   

Thus       
or     T₂ = 300 (3)⁰⁴ = 465.5 K = (465.5 - 273) = 192.5°C 
∴    Change in temperature T₂ - T₁ = 192°C - 27°C = 165°C.

Maximum angle of diffraction can be 90°, i.e., θ = 90°.
λ = 5000 x 10^-8 cm
∴   Grating element (a + b) =  For nth maxima , we find 
(a + b) sin θ = nλ

Hence, 6 order will be visible.

let the total No. of non-interacting distinguishable particles are N

Population of particle in the level with energy ∈ is


So, the population of particles is the level with energy ∈ is 500.

Here; we have 

or   

From first law of thermodynamics 
dQ = dU+ dW
where all quantities are measured in same units, say kilo cal.
Here air is heated at constant volume and hence no external work is done
i.e., dW = p. dV = 0.
Let m be the mass, c_v the specific heat at constant volume and dT the rise in temperature of the air, the heat taken in by it is 
dQ = m x c_v x dT kilo cal.
Then from first law of thermodynamics 
dU = dQ - dW = dQ
= m x c_v x dT =0.004 x 0.172 x 2 = 1.376 x 10^-3 kilo cal.

where a = interatomic distance
d = interplanar distance 
h,k,l = Miller indices.
According to the problem h = k = I = 1 (being d₁₁₁ plane) 
q = 30°, I = 1.75 x 10^-10m, n = 1 
from bragg’s equation 2d sin q = nl, we have

Solving we have a = 3.031Å

m = 100 gm = 0.1kg.

angular velocity of frame 
Coriolis force  

Given     ω₀ = 100 
₀ = 5
Phase velocity 
dω = 110.25 - 90.25
dω = 20
dK = 6.25 - 4.25 = 2

According to clausius clapeyron equation 


dP = 1.73 x 10⁸ Pa 
dP = 0.173 x 10⁹ Pa 
dP = 0.173 GPa

From Moseley’s law for K_α -line, we have