## IIT JAM PHYSICS 2011Previous Year Question Paper with Solution.

1. The line integral , where , along the semi-circular path as shown in the figure below is:

(a) –2

(b) 0

(c) 2

(d) 4

Ans. (b)

Sol.

2. Six simple harmonic oscillation each of same frequency and equal amplitude are superposed. The phase difference between any two consecutive oscillations i.e. is constant, where is the phase of the nth oscillation. If the resultant amplitude of the superposition is zero, what is the phase difference ?

(a)

(b)

(c)

(d)

Ans. (b)

Sol.

3. A particle of mass 'm' is moving in a potential

where w0 and 'a' are positive constants. The angular frequency of small oscillations for the simple harmonic motion of the particle about a stable minimum of the potential V(x) is:

(a)

(b)

(c)

(d)

Ans. (b)

Sol.

4. Intensity of three different light beams after passing through an analyzer is found to vary as shown in the following graphs. Identify the option giving the correct states of polarization of the incident beams from the graphs.

(a) Graph 1: Linear polarization, Graph 2: Circular polarization, Graph 3: Elliptic polarization

(b) Graph 1: Circular polarization, Graph 2: Linear polarization, Graph 3: Elliptic polarization

(c) Graph 1: Unpolarized, Graph 2: Circular polarization, Graph 3: Linear polarization

(d) Graph 1: Unpolarized, Graph 2: Elliptic polarization, Graph 3: Circular polarization.

Ans. (b)

Sol. Figure(a) is circularly or unpolarized light as there is no change in intensity. But figure (b) is linear polarized light as there is change in intensity from minimum to maximum. However, in figure (c) there is change in intensity around 0.5 which means this light is always eliptically polarized light.

5. Which of the following circuit does not satisfy the Boolean expression .

(a)

(b)

(c)

(d)

Ans. (d)

Sol.

6. Light described by the equation E = (90V / m) [sin (6.28 × 1015 s–1)t + sin (12.56 × 1015 s–1)t] is incident on a metal surface. The work function of the metal is 2.0 eV. Maximum kinetic energy of the photoelectrons will be

(a) 2.14 eV

(b) 4.28 eV

(c) 6.28 eV

(d) 12.56 eV

Ans. (c)

Sol. Given light consists of two frequencies, and

Photon of lower frequency will give how K.E. and photon of higher frequency will give high K.E.

7. A gas of molecular mass 'm' is at temperature T. If the gas obeys Maxwell-Boltzmann velocity distribution, the average speed of molecules is given by

(a)

(b)

(c)

(d)

Ans. (d)

Sol. According to Maxwell-Boltzmann velocity distribution. The number of particles having velocity in the range

8. Consider free expansion of one mole of an ideal gas in an adiabatic container from volume V1 to V2. The entropy change of the gas, calculated by considering a reversible process between the original state (V1, T) to the final state (V2, T) where T is the temperature of the system, is denoted . The corresponding change in the entropy of the surrounding is . Which of the following combinations is correct?

(a)

(b)

(c)

(d)

Ans. (c)

Sol.

There will not be any entropy change in the surrounding

9. Equipotential surfaces corresponding to a particular charge distrubution are given by 4x2 + (y – 2)2 + z2 = V1, where the values of V1 are constants. The electric field at the origin is

(a)

(b)

(c)

(d)

Ans. (d)

Sol.

10. The wave function of a quantum mechanical particle is given by where and are eigenfunctions with corresponding energy eigenvalues –1 eV and –2eV, respectively. The energy of the particle in the state is:

(a)

(b)

(c)

(d)

Ans. (a)

Sol. Given wave function for a particle

Energy eigenvalue of eigenfunction and that of

Then energy of the particle in the state will be

11. A rain drop falling vertically under gravity gathers moisture from the atmosphere at a rate given by , where 'm' is the instantaneous mass 't' is time and 'k' is a constant. The equation of motion of the rain drop is:

If the drop starts falling at t = 0, with zero initial velocity and initial mass m0 (given: m0 = 2 gm, k = 12 gm/s3 and g = 1000 cm/s2), the velocity (b) of the drop after one second is:

(a) 250 cm/s

(b) 500 cm/s

(c) 750 cm/s

(d) 1000 cm/s

Ans. (b)

Sol.

At t = 0; m = m0

We have m0 = 2, k = 12, g = 1000 and t = 1

Putting the numerical values in equation (2)

12. Given two (n × n) matrices and such that is hermitian and is skew (anti)-hermitian. Which one of the following combinations of and is necessarily a Hermitian matrix?

(a)

(b)

(c)

(d)

Ans. (c)

Sol.

13. An X-ray diffraction (XRD) experiment is carried out on a crystalline solid having FCC structure at room temperature. The solid undergoes a phase transformation on cooling to –20ºC and shows orthorhombic structure with small decrease in its unit cell lengths as compared to the FCC unit cell lengths. As a result, the (311) line of the XRD pattern corresponding to the FCC system

(a) will split into a doublet

(b) will split into a triplet

(c) will remain unchanged

(d) will split into two separate doublets

Ans. (b)

Sol.

(3, 1, 1), (1, 3, 1) (1, 1, 3). Hence the peak will be splitted into 3 peak. (d) will not same for all cases.

14. A closed Gaussian surface consisting of a hemisphere and a circular disc of radius R, is placed in a uniform electric field, , as shown in the figure. The circular disc makes an angle = 30º with the vertical. The flux of the electric field vector coming out of the curved surface of the hemisphere is:

(a)

(b)

(c)

(d)

Ans. (b)

Sol. There is no charge enclosed inside hemisphere.

15. In experiment, the resistance of a rectangular slab of a semiconductor is measured as a function of temperature. The semiconductor shows a resistance of 300W at 200K and at 250 K. Its energy band gap is [Given: ln(15) = 2.708, ln(10) = 2.303]

(a) 0.138 eV

(b) 0.431 eV

(c) 0.690 eV

(d) 0.862 eV

Ans. (d)

Sol. Near room temperature resistance of semi conductor varies as

16.

(a) Calculate the line integral from point P O along the path P Q R O as shown in the figure.

(b) Using Stoke's theorem appropriately. Calculate for the same path P Q R O.

Sol.

17. An infinitely long solid cylindrical conductor of radius r1, carries a uniform volume current density . It is uniformly surrounded by another coaxial cylinder of a linear magnetic medium with permeability µ, up to radius r1 as shown in the figure.

(a) Determine the magnetic field in the region, r < r1 and magnetic induction in the regions, r1 < r < r2 and r > r2, where r is the radial distance from the axis of the cylinder.

(b) Sketch the variation of with r in all the three regions.

Sol. (a) According to Ampere circuital law.

Let us consider a cylinderical surface of radius r < r1 then current enclosed by the surface is

(b) Sketch of H in all the region,

18. (a) Consider heat conduction in a medium. Let T(x, y, z, t) denote the temperature of the medium at time 't' and position (x, y, z). Consider a volume V enclosed by a surface S inside the medium. The decrease in heat energy per unit volume per unit time is a an outward flux of heat per unit area of the surface per unit is , where a and b are material dependent constants. If there is no generation or loss of heat, show that T satisfies the equation.

(b) Now consider a thin annular shaped material enclosed between two concentratric circles of radii r0 and 2r0 as shown in the figure. The temperature is 2T0 at r = r0 and T0 at r = 2r0. Assuming steady state condition, Find T as a function of radial distance r from the centre O, for r0 < r < 2r0.

Sol.

Using boundary condition (T = 2T0, r = r0; T = T0, r = 2r0) we get,

19. An ideal gas reversible engine operates in a closed cycle. The P-V diagram is shown below.

(a) Find out the efficiency of the reversible engine assuming both specific heats, CP and CV as constants.

(b) Identify the thermodynamic processes and draw the corresponding T-S diagram schematically.

Sol. Let temperature at A, B and C be TA, TB and TC respectively.

(b) In adiabatic process Tds = 0 i.e. S is constant. Consider isobaric process where

Equation indicate that for a reversible isobaric process the curve (in Ts diagram) has a slope.

Similarly to isochoric process the slope will be

Since CP > CV at a particular temperature slope of isobaric process is less than isochoric process.

20. (a) A solid having a simple cubic structure at room temperature with lattice a parameter a and one valence electron per atom, is assumed to show free electron behaviour. Calculate the magnitude of the Fermi wave vector and the corresponding equivalent temperature.

(b) Find the ratio of the magnitude of the Fermi wave vector to the radius of the largest sphere that can be inscribed within the first Brillouin zone of the solid.

Sol. (a) For a simple cubic structure number of electrons per unit volume.

Where 'a' is lattice parameter.

(b) First Brillouin zone of simple cubic solid is also simple cubit with sides.

Sphere of largest radius that can be fitted Into the cubic of side is .

So, ratio of fermi wave vector to radius of largest sphere .

21. For the given circuit using an operational amplifier, the input is a sinusoidal signal of amplitude Vm = 1 mV (peak - to - peak).

(a) What is the lower cut-off frequency at which the gain is down by 3 dB as compared to the gain at midband? If the bandwidth of the amplifier is 1 MHz for unity gain, what will be the bandwidth of the given circuit?

(b) What is the output voltage (V0) at 15 kHz?

Sol. (a) For AC signal

Band width (B.W.) of given circuit = gain × B.W. = 1 MHz; 10 × B.W. = 1 MHz; B.W. = 100 KHz

The output voltage due to D.C. is 7.5 V. So, net output voltage is D.C. + a.c.

22. A particle of mass 'm' and angular momentum l is moving under the action of a central force f(r) along a circular path of radius 'a' as shown in the figure. The force centre 'O' lies on the orbit.

(a) Given the orbit equation in a central field motion.

determine the form of the force in terms of l, m, a and r.

(b) Calculate the total energy of the particle assuming that the potential energy V(r) 0 as r .

Sol. (a) From figure,

Therefore, using given equation we get,

23. A particle of mass 'm' moves in a potential given by

(a) Write down the general solutions for wave functions in regions I and II, if the energy of the particle E < V0. Using appropriate boundary conditions, find the equation that relates E to V0, m and L.

(b) Now, set V0 = 0 and assume that a beam of particles in incident on the infinite step potential (from x > 0) with energy E(> 0). Using the general solution for the wave function, calculate the reflection coefficient.

Sol.

Boundary condition,

From equation (i) and (ii), we get

(b) V(x) = V´0 x < 0

= 0 x > 0

Therefore, solution and Schrodinger equation are

Now the reflection coefficient.

Boundary condition:

From equation (i) and (ii), we get

3

Applying boundary conditions,

24. A diffraction grating having N slits, each of width 'b' and period d, is illuminated normally by a monochromatic plane wave of wavelength

(a) Obtain an expression for the highest diffraction order that can be observed. What is the phase difference between waves from first and Nth in the highest diffraction.

(b) If alternate slits are covered with a retarder that retards the wave by obtain an expression for the intensity distribution of the Fraunhofer diffraction pattern?

Sol. (a) Condition for maxima in case of grating (b + d) sin

The highest diffraction order can be observed when , (b + d) = max.

The electric field due to 1st slit reaching P,

Due to Nth slit,

The phase difference between 1st and Nth,

For highest diffraction order

(b) If alternate slits are covered with retarder which retards wave by the resultant dielectric field at P is:

25. Unpolarized light is incident on an air-dielectric interface. The interface is the x-y plane and the plane of incidence is y-z plane. The electric field of the reflected light is given by , where 'k' is the propagation constant constant in air and is the angular frequency of the light. Assume magnetic permeability µ = µ0.

(a) Determine the dielectric constant of the second medium.

(b) Determine the direction of the Poynting vector in the dielectric medium.

Sol. (a) Reflected light has electric field in direction. i.e., it is plane polarised. Therefore angle of incidence is equal to Brewster angle.

Direction of propagation of reflected beam is given by wave vector,

Therefore, angle of reflection is given by

We know that, tan = n (refractive index)

(b) For angle of reflection (r´) we use Snell's law.

Therefore, direction of poynting vector is given by