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The energy density and pressure of aphoton gas are given by and P = where T is the temperature and σ is the radiation constant. The entropy per unit volume is given by
According to thermodynamic equation
Tds = du + pdv
Let ψ_{0} and ψ_{2}. denote respectively the ground state and second excited state energy eigenfunction of a particle moving in a harmonic oscillator potential with frequency ψ. If at time t the particle has the wavefunction,
The expectation value of the energy as a function of time
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.
The wave function of a certain particle is than the value of A is
At constant temperature on increasing the pressure of a gas by 10% will decrease its volume by
If P_{1}= Pthen P_{2}= P + 10% of P = 1.10 P
From boyle’s lawPV = constant
∴ Percentage change in volume
The independent solutions of the equation = cos hx are —
The auxiliary equation is
m^{2} 3m + 2 = 0
(m  1)(m  2) = 0 ⇒ m = 1, 2
CF = C_{1}e^{x} + C_{2}e^{2x}
So, (a) is AND gate
So (B) is OR GATE
A Ushaped tube contains a liquid of density ρ and it is rotated about the line as shown in the figure. The difference in the levels of liquid column is
For a circular motion of small element dx
The first ball of mass m moving with the velocity v collides head on with the second ball of mass m at rest. If the coefficient of restitution is e, then the ratio of the velocities of the first and the second ball after the collision is
Here, m_{1}= m_{2} = m, u_{1} = u, u_{2} = 0. Let v_{1}, v_{2} be their velocities after collision.
According to principle of conservation of linear momentum,
mu + 0 = m (v_{1} + v_{2}) or v_{1} + v_{2} = u ... (i)
By definition, ....(ii)
Add (i) and (ii),
Subtract (ii) from (i)
One mole of an ideal monoatomic gas is kept in a volume V and is at temperature T. If the volume and temperature of the gas are respectively changed to V’ and T’ in such a way that the entropy of the gas is unchanged, then which of the following is correct.
ΔS = 0
A charge q is placed at the centre of the line joining two equal charges Q. The system of three charges will be in equilibrium if q =
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.
At a given point in space the total light wave is composed of three phases
The intensity of light at this point is
We know intensity I = Es
A radioactive sample S_{1} having an activity of 5μCi has twice the number of nuclei as another sample S_{2} which has an activity of 10μCi. The half lives of S_{1} and S_{2} can be
Activity of S_{1} = (Activity of S_{2})
Given N_{1} = 2N_{2}
When a plane polarised light is normally incident on half wave plate with an angle 30° to optical axis the output light is
When a planepolarised 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 polarized light wave is incident on a quarter wave plate at θ= 45° angle then what is the nature of output wave.
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 QW plate is and path difference is so, output light is circularly polarised
For the circuit shown in the given figure, assuming ideal diodes, the output waveform V_{0} will be
Voltage at OPAMP output
V_{1 }=  1  5 sin cot
At
At
During ‘+’ ve cycle of the waveform, when V_{1} < 2V, diode D_{2} conducts and V_{0} = 2 V
During ‘’ ve cycle, when V_{1} > 2V, D_{1} conducts and V_{0} = 2V.
At normal temperature (0°C) and normal pressure (1.013 x 10^{5} N/m^{2}) when 1 gm. of water freezes, its volume increases by 0.091 cm^{3}. What is the change in internal energy.
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 cm^{3}
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.
Two infinite parallel plates are uniformly charged. And the charge density on these plates are +σ and  σ respectively. The intensity of electric field between these plates will be
In I and III region direction of E_{1} and E_{2} are opposite to each other and In magnitude
So , in I and III region direction of E_{1} and E_{2} 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
A conductor is placed in a variable electric field . The ratio of displacement current density to conduction current density is (σ is conductivity)
Given E = E_{0} cosωt
Displacement vector
So displacement current density
Conduction current density
For the circuit shown, which one of the following statement is true?
When S_{3} is closed, due to attraction with opposite charge, no flow of charge takes place through S_{3}. Therefore, potential difference across capacitor plates remains unchanged or V_{1}= 25 V and V_{2} = 15 V
The internal energy of an ideal gas decreases by the same amount as the work done by the system
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.
What is the gain V_{0}/V_{i} of the circuit of figure
Applying KCL at various nodes, we have
or V_{in} + V_{2 }= 2V_{1} = 0 (as inverting terminal is at virtual ground)
So V_{in} = V_{2}
or V_{3} = 3V_{2} = 3V_{in} .....(2)
Putting values from equation (1) and (2), we get
In a crystal whose primitives are 1.2A° 1.8A° and 2A°, along whose miller indices are (2, 3, 1) cuts intercept 1.2 A° along X  axis. What will be the length of intercepts along y and z  axes.
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.
Two equal negative charges q are fixed at points (0, a) and (0, a) on yaxis. A positive charge Q is released from rest at the point (2a, 0) on the xaxis. The charge Q will be
Here Q is stay at point (2a, 0) which is very for in xaxis (2a).
Motion is simple harmonic only if Q is released from a point not very far from the origin on xaxis. (for small oscillations) other wise motion is periodic but not simple harmonic. So here such type of motion is not simple harmonic motion.
A uniform rod of length I is free to rotate in a vertical plane about a fixed horizontal axis through B. The rod begins rotating from rest from its unstable equilibrium position . When it has turned through an angle θ , its angular velocity ω is given as
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
Consider the situation shown in the figure. All the surfaces are smooth and the rod has mass M what is the minimum value of F required to keep the rod in equilibrium ?
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 =
A parallel monochromatic beam of light is incident normally on a narrow slit. A diffraction pattern is formed on a screen placed perpendicular to the direction of the incident beam. At the first minimum of the diffraction pattern, the phase difference between the rays coming from the two edges of the slit is
...(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
In a pn junction diode not connected to any circuit
At junction a potential barrier/depletion layer is formed, with nside at higher potential and pside at lower potential. Therefore, there is an electric field at the junction directed from the nside to pside.
A planet of mass m and angular momentum L moves in a circular orbit in a potential v(r)= where k is a positive constant If the planet is slighthy perturbed, find angular frequency of radial oscillation.
for stable point (orbit)
Therefore radius of stable circular orbit is Angular frequency of oscillation
about stable paint
What is the ratio of the nearest neighbour distance to the next nearest neighbour distance in a simple cubic crystal
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 d_{1} = a
and the distance of next nearest neighbour d_{2} = a√2
What is the maximum radius of the sphere that can just fit into the void at the body centre of the fee structure coordinated by the facial atom. Given r is the radius of atom.
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
A ball is dropped from a height H. It bounces back up to a height e times after hitting the ground . If e < 1, after what time the ball will finally come to rest.
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 = t_{1}+ t_{2}+13 +.......
An object of mass m moving with a velocity v is approaching a second object of the same mass but at rest. Which of the following are correct?
Let v_{1} is velocity of first mass after collision and v_{2} is velocity of second mass after collision.
By the conservation law of momentum
m_{1}v = m_{1}v_{1} + m_{2}v_{2}
m_{1} = m_{2} = m
⇒ mv = mv_{1} + mv_{2}
v = v_{1} + v_{2}
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 =
The figure shown PV diagram for one mole of a monoatomic ideal gas. AB and CD are isothermal process at 500 K and 300 K, respectively and V_{2} = 2V_{1}.
Process A → B (isothermal)
Process B → C (isochoric process)
Process C → D (Isothermal process)
Process D → A (Isochoric process)
A particle of mass m is subjected to a potential
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
Which of the following statements are Correct:
Coordination number is the number of nearest neighbours to a lattice point.
packing fraction
BCC :
simple Cubic structure :
A Germanium transistor with α = 0.98 gives a reverse saturation Current of 40μ A, when used in CB configuration when the transistor is connected in the CE mode with a base current of 0.2 ma. Then the following results have been obtained. Choose the correct options .
A particle of unit mass is thrown vertically upward with inifial speed v_{0}. If is acted upon by a drag force bv^{2} in addition to gravity where b is constant and v is instantaneous velocity of particle. Then the speed of the particle when it returns to the point from where it was thrown,
In the presence of dray force, only the distance covered in upward and downware motion is common, for upward motion, initial speed = V_{0} 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
A solid cylinder is attached to a horizontal mass less spring on that it can roll without slipping along a horizontal surface. The spring constant k is 3 N/m. If the system is released from rest at a point in which the spring is stretched by 0.25 m .
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
A particle executes simple harmonic motion between x =  A and x = + A. The time taken for it to go from 0 to and to go from to A is T_{2}. Then which of the following is/are incorrect
We know that
x = A sin ωt i.e. x = 0 at t = 0
SO T_{1} < T _{2}
The total energy of an inert  gas crystal is given by where R is the inter  atomic spacing in Angstroms. The equilibrium seperation between the atoms in Angstroms.
Given that E(R) =
for equilibrium separation
Consider the differential equation + y tanx = cosx if y(0) = 0 the is
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
Two moles of helium gas undergoes a cyclic process as shown in figure. Assuming the gas to be ideal, the net work done in this process is _________ (inJ).
Net work done W = W_{AB} + W_{BC} + W_{CD} + W_{DA} = 1152 J
An object travels along the xdirection with velocity in a frame O. An observer in a frame O’ sees the same object travelling with velocity The relative velocity of O ’ with respect to O. is ___________ (in the unit of C)
Water is filled in a cylindrical container to a height of 3m. The ratio of the cross sectional area of the orifice and the beaker is 0.1. The square of the speed of the liquid coming out from the orifice is (g = 10 m/s^{2})______ m^{2}/sec^{2}.
Applying continuity equation at upper end and at hole
A_{1}V_{1} =A_{2}V_{2}
Further applying Bernoulli’s equation at two points
A source of sound of frequency 256 Hz is moving rapidly towards a wall with a velocity of 5m/s. How many beats per sec, will be heard if sound travels at a speed of 330 m/s, and observer is between the source and the wall?
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
The second order bright fringe in a single slit diffraction pattern is 1.4 mm from the center. The screen is 80 cm from the slit of width 0.80 mm. Assuming monochromatic incident light calculate the wavelength of light.
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
Two coherent plane waves of light of equal amplitude and each of wavelength 10π x 10^{8} m, propagating at an angle of radian with respect to each other, fall almost normally on a screen. Find out the fringewidth (in mm) on the screen?
Fringe width
A quantity of air (y = 1.4) at 27°C is suddenly compressed to one third of its volume. Calculate the change in temperature in °C.
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 T_{1}, V_{1} be the initial and T_{2}, V_{2} the final temperature and volume respectively, then
or
Thus
or T_{2} = 300 (3)^{04 }= 465.5 K = (465.5  273) = 192.5°C
∴ Change in temperature T_{2}  T_{1} = 192°C  27°C = 165°C.
How many orders will be visible if the wavelength of incident radiation be 5000 Å and the number of lines on the grating be 7620 to an inch. (Take 1 inch = 2.54cm)
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.
Consider 1000 non  interacting , distinguishable particles in a twolevel system at temp T. The energies of the levels are 0 and ε, whwere ∈> 0. In the high temperature limit (K_{B}T > ε). What is the population of particles in the level with energy ε.
let the total No. of noninteracting 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.
In a silicon transistor, a change of 8.89 mA in the emitter current produces a change of 8.8 mA in the collector current. What change in the base current is necessary to produce an equivalent change in the collector current ?
Here; we have
If for a transistor, then what would be the value of l_{c} (in mA)?
A differential amplifier has a differential gain of 20,000 . CMRR = 80 dB. Find out the common mode gain?
or
What is the change in internal energy (in cal) when 0.004 kg. of air is heated from 0°C to 2°C. The specific heat of air at constant volume being 0.172 kilo cal/kg °C.
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.
The Bragg angle corresponding to the first order reflection from (1,1,1) planes in a crystal is 30° when X  rays of wavelength 1.75_{Å} are used. Calculate the interatomic spacing.
where a = interatomic distance
d = interplanar distance
h,k,l = Miller indices.
According to the problem h = k = I = 1 (being d_{111} plane)
q = 30°, I = 1.75 x 10^{10}m, n = 1
from bragg’s equation 2d sin q = nl, we have
Solving we have a = 3.031Å
A body of 100 gm. mass is moving with velocity m/sec. in an inertial frame which it self (frame) rotating with angular velocity radian/sec. with respect to a stationary reference frame. Then find out the magnitude of Coriolis force on moving body?
m = 100 gm = 0.1kg.
angular velocity of frame
Coriolis force
A wave packet in a certain medium is constructed by superposing wave of frequency ω around ω_{0} = 100 and the corresponding wavenumber K with K_{0} = 5 as given in the table below
What would be the ratio v_{g}/v_{p} of the group velocities v_{g} and the phase velocity v_{p} ?
Given ω_{0} = 100
_{0} = 5
Phase velocity
dω = 110.25  90.25
dω = 20
dK = 6.25  4.25 = 2
Water freezes at 0°C at atmospheric pressure (1.01 x 10^{5} Pa). The density of water and ice at this temperature and pressure are 1000 kg/m^{3} and 934 kg/ m^{3} respectively. The latent heat of fusion is 3.34 x 10^{5} J/kg. The pressure difference required for depressing the melting temperature of ice by 10°C is ______ (G Pa).
According to clausius clapeyron equation
dP = 1.73 x 10^{8} Pa
dP = 0.173 x 10^{9} Pa
dP = 0.173 GPa
If the K_{α}radiation of Mo (Z = 42) has a wavelength of 0.71 Å , then evaluate the wavelength of the corresponding radiation of Cu (Z = 29).
From Moseley’s law for K_{α} line, we have
1 docs34 tests

1 docs34 tests
