GATE Physics Mock Test Series - 6 - Physics MCQ

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