## IIT JAM PHYSICS 2023

Previous Year Question Paper with Solution.

1. For a cubic unit cell, the dashed arrow in which of the following figures represents the direction [220]?

(a)

(b)

(c)

(d)

Ans. (c)

Sol. From the concept of Miller indices, the plane is in x-y direction.

2. Which of the following fields has non-zero curl?

(a)

(b)

(c)

(d)

Ans. (d)

Sol.

3. Which of the following statements about the viscosity of a dilute ideal gas is correct?

(a) It is independent of pressure at fixed temperature

(b) It increases with increasing pressure at fixed temperature

(c) It is independent of temperature

(d) It decreases with increasing temperature

Ans. (a)

Sol. Viscosity is defined as . Thus, n is independent of pressure at a fix temperature.

4. The plot of the function f(x) = ||x| – 1| is

(a)

(b)

(c)

(d)

Ans. (b)

Sol. At x = 1, |x| = 1

f(x) = |1 – 1| = 0

and at x = 0, |x| = 0

f(x) = |0 – 1| = 1

Also at x = –1, |x| = 1

f(x) = |1 – 1| = 0

We see that only option (b) is satisfying the condition.

5. A system has N spins, where each spin is capable of existing in 4 possible states. The difference in entropy of disordered states (where all possible spin configurations are equally probable) and ordered states is

(a) 2(N – 1)k_{B} ln2

(b) (N – 1)k_{B} ln2

(c) 4k_{B} ln2

(d) Nk_{B} ln2

Ans. (a)

Sol. 4 possible states and disorder w = 4^{N}

Entropy S_{1} = k_{B} ln4 = 2Nk_{B} ln2

For disorder w = 4, S_{2} = k_{B} ln4

Difference in entropy, S_{1} – S_{2} = Nk_{B} ln4 – k_{B} ln4

S_{1} – S_{2} = (N – 1)k_{B} ln4

= 2(N – 1)k_{B} ln2

6. Temperature (T) dependence of the total specific heat (C_{v}) for a two dimensional metallic solid at low temperatures is

(a)

(b)

(c)

(d)

Ans. (a)

Sol. Electronic part of heat capacity is C_{e} = AT

Photon heat capacity C_{Ph} = BT^{d}

where, d Dimension

For 2 dimension, d = 2

then, C_{Ph} = BT^{2}

Total heat capacity C_{V} = AT + BT^{2}

= A + BT

Hence, option (a) will correct.

7. For the following circuit, choose the correct waveform corresponding to the output signal (V_{out}). Given V_{in} = 5 V, forward bias voltage of the diodes (D and Z) = 0.7 V and reverse Zener voltage = 3 V.

(a)

(b)

(c)

(d)

Ans. (a)

Sol. Applying up the voltage of 3 Volt, both diode and zener-diode is open (conducting). Above 3 Volt, input the zener will not conduct so the voltage will be fixed again below 3 volt input, the output feature will be similar as input.

In the negative cycle of input, the diode is reverse and zener is forward bias. In the negative cycle the out will be varies as input. However, above 0.7 volt input, the output will be constant.

Therefore, above 0.7 input, the output will be constant.

8. If the ground state energy of a particle in an infinite potential well of width L_{1} is equal to the energy of the second excited state in another infinite potential well of width L_{2}, then the ratio is equal to

(a) 1

(b) 1/3

(c)

(d) 1/9

Ans. (b)

Sol. The energy of 1-D infinite square box is E =

In ground state, n = 1

and in second excited state n = 3

Acc. to given condition,

9. In the given circuit, with an ideal op-amp for what value of the output of the amplifier V_{out} = V_{2} – V_{1}?

(a) 1

(b) 1/2

(c) 2

(d) 3/2

Ans. (a)

Sol. For given configuration,

V_{out} = V_{1} – V_{2} and v_{+} =

Output from +ve terminal if –ve terminal made ground is :

output from –ve terminal if +ve terminal made ground is :

V_{01} = = –V_{1}

Now, output voltage is :

V_{out} =

Put V_{out} = V_{2} – V_{1}

R_{1} + R_{2} = 2R_{2}

R_{2} = R_{1}

10. A projectile of mass m is moving in the vertical x-y plane with the origin on the ground and y-axis pointing vertically up. Taking the gravitational potential energy to be zero on the ground, the total energy of the particle written in planar polar coordinates (r, ) is (here g is the acceleration due to gravity)

(a)

(b)

(c)

(d)

Ans. (c)

Sol.

The total energy can be written as follows E = T + V, T =
(x^{2} + y^{2})

In polar coordinate

Thus,

Potential energy V = mgy = mgr sin

Now, E = T + V =

E = T + V =

11. A small bar magnet is dropped through different hollow copper tubes with same length and inner diameter but with different outer diameter. The variation in the time (t) taken for the magnet to reach the bottom of the tube depends on its wall thickness (d) as

(a)

(b)

(c)

(d)

Ans. (c)

Sol. All conducting materials, including copper, create their own magnetic field when a current is passed through them, as the eddy currents are created. As gravity pulls the magnet downwards through the pipe, the magnetic field created by the eddy currents resists the magnetic field produced by the magnet, slowing it down.

Hence, correct option is (c)

12. Two digital inputs A and B are given to the following circuit. For A = 1, B = 0, the values of X and Y are:

(a) X = 0, Y = 0

(b) X = 1, Y = 0

(c) X = 0, Y = 1

(d) X = 1, Y = 1

Ans. (b)

Sol. Rule of binary algebra

A + B + AB = A + B(1 + A) = A + B

and (A + B)AB = AB + AB = AB

By applying this, we will get

X = A + B and Y = AB

When A = 1, B = 0, X = 1, Y = 0

13. The Jacobian matrix for transforming from (x, y) to another orthogonal coordinates system (u, v) as shown in the figure is

(a)

(b)

(c)

(d)

Ans. (c)

Sol.

From eq. (1) and (2), we can conclude that

(By comparing coefficients of 'u' and 'v')

14. A rotating disc is held in front of a plane mirror in two different orientations which are (i) angular momentum parallel to the mirror and (ii) angular momentum perpendicular to the mirror. Which of the following schematic figures correctly describes the angular momentum (solid arrow) and its mirror image (shown by dashed arrows) in the two orientations?

(a)

(b)

(c)

(d)

Ans. (b)

Sol. From the concept of mirror image and also we have to apply velocity direction of image w.r.t mirror

U_{o} – U_{M} = v_{i} – v_{o}

15. Inverse of the matrix is

(a)

(b)

(c)

(d)

Ans. (b)

Sol. We know A^{–1} =

Given, A =

After calculation, we get

|A| = 1 and adj A =

A^{–1} =

16. Suppose the divergence of magnetic field is nonzero and is given as = µ_{0}, where µ_{0} is the permeability of vacuum and is the magnetic charge density. If the corresponding magnetic current density is , then the curl of the electric field is

(a)

(b)

(c)

(d)

Ans. (d)

Sol. As we know, = 0

Given

We know, and

Also, we know,

On taking gradient, we have

Thus,

17. For a thermodynamic system, the coefficient of volume expansion and compressibility , where V, T, and P are respectively the volume, temperature, and pressure. Considering that is a perfect differential, we get

(a)

(b)

(c)

(d)

Ans. (c)

Sol.

18. A linearly polarized light of wavelength 590 nm is incident normally on the surface of a 20 µm thick quartz film. The plane of polarization makes an angle 30° with the optic axis. Refractive indices of ordinary and extraordinary waves differ by 0.0091, resulting in a phase difference of f between them after transmission. The value of f (rounded off to two decimal places) and the state of polarization of the transmitted light is

(a) 0.62 and linear

(b) 0.62 and elliptical

(c) –0.38 and elliptical

(d) 0.5 and circular

Ans. (b)

Sol. From the given information we can say that the amplitude of component of O-Ray E_{y} = E_{0} sin (30) = and the amplitude of component of E-Ray E_{X} = E_{0} cos (30) = .

So, the both amplitudes are different. Hence, it will be elliptical if the phase differences between emerging rays are not equal to 0, , 2, ....

The optical path difference

(n_{0} – n_{E})t = 0.0091 × 20 × 10^{–6}

Phase diff. × 0.0091 × 20 × 10^{–6
}

= 0.62

Obviously, the phase difference is not integer multiple of thus it will be elliptical with = 0.62

19. The phase velocity v_{p} of transverse waves on a one-dimensional crystal of atomic separation d is related to the wavevector k as

The group velocity of these waves is

(a)

(b) C cos (kd/2)

(c)

(d)

Ans. (b)

Sol. We know that v_{g} = , where = v_{p}.k

Given, v_{p} =

Thus,

and

= C cos (kd/2)**
**

20. In a dielectric medium of relative permittivity 5, the amplitudes of the displacement current and conduction current are equal for an applied sinusoidal voltage of frequency f = 1 MHz. The value of conductivity (in of the medium at this frequency is

(a) 2.78 × 10^{–4
}

(b) 2.44 × 10^{–4
}

(c) 2.78 × 10^{–3
}

(d) 2.44 × 10^{–3}

Ans. (a)

Sol. Frequency, f = 1 MHz = 1 × 10^{6} Hz

Permittivity in free space,

= 8.85 × 10^{–12} Nm^{2}c^{–2}

Permittivity in a medium,

Angular freq.

Conductivity of medium,

= 2 × 3.14 × 10^{6} × 5 × 8.85 × 10^{–12}

= 2.78 × 10^{–4}

21. For a given vector the surface integral over the surface S of a hemisphere of radius R with the centre of the base at the origin is

(a)

(b)

(c)

(d)

Ans. (a)

Sol.

22. In the circuit shown, assuming the current gain = 100 and V_{BE} = 0.7 V, what will be the collector voltage V_{C} in V?

Given : V_{CC} = 15 V, R_{1} = 100 , R_{2} = 50 , R_{C} = 4.7 and R_{E} = 3.3

**
**

(a) 8.9

(b) 5.1

(c) 4.3

(d) 3.2

Ans. (a)

Sol. Given, = 100

V_{BE} = 0.7 V

V_{CC} = 15V

R_{1} = 100

R_{2} = 50

R_{C} = 4.7

R_{E} = 3.3

So,

V_{BE} = 5V

Now, applying KVL,

V_{E} = V_{BE} – V_{B}

V_{E} = 5 – 0.7

V_{E} = 4.3V

For the circuit, I_{C}

So,

Again applying KVL,

V_{C} = V_{CC} – I_{C}R_{C}

V_{C} = 15 – (1.3 × 4.7)

V_{C} = 8.9V

Option (a) is correct.

23. A uniform stick of length *l* and mass m pivoted at its top end is oscillating with an angular frequency . Assuming small oscillations, the ratio /, where is the angular frequency of a simple pendulum of the same length, will be

(a)

(b)

(c)

(d)

Ans. (b)

Sol. We know, Angular velocity of stick

where *l*' is the distance from the centre of gravity to point of suspension (end point here) and I is moment of inertia about point of suspension.

From the given information we can say

24. An oil film in air of thickness 255 nm is illuminated by white light at normal incidence. As a consequence of interference, which colour will be predominantly visible in the reflected light

Given the refractive index of oil = 1.47

(a) Red (~ 650 nm)

(b) Blue (~ 450 nm)

(c) Green (~ 500 nm)

(d) Yellow (~ 560 nm)

Ans. (c)

Sol. Condition for constructive interference in thin film

2µt cos (r) =

For normal incident r = 0°

We have to compare here with path difference

25. Water from a tank is flowing down through a hole at its bottom with velocity 5 ms^{–1}. If this water falls on a flat surface kept below the hole at a distance of 0.1 m and spreads horizontally, the pressure (in kNm^{–2}) exerted on the flat surface is closest to

Given: acceleration due to gravity = 9.8 m s^{–2} and density of water = 1000 kg m^{–3}

(a) 13.5

(b) 27.0

(c) 17.6

(d) 6.8

Ans. (b)

Sol. Mass flow rate = Density × Velocity × Area

Momentum flow rate = Force = Density × Velocity^{2} × Area

Pressure = Density × Velocity^{2}

Now, Velocity^{2} = 5^{2} + 2gh = 5^{2} + 2 × 10 × 0.1 = 27

Pressure = Density × Velocity^{2}

= 1000 × 27

26. At the planar interface of two dielectrics, which of the following statements related to the electric field , electric displacement and polarization is true?

(a) Normal component of both and are continuous

(b) Normal component of both and are discontinuous

(c) Normal component of is continuous and that of is discontinuous

(d) Normal component of both and are continuous

Ans. (c)

Sol.

and Boundary condition

27. Consider a system of large number of particles that can be in three energy states with energies 0 meV, 1 meV, and 2 meV. At temperature T = 300 K, the mean energy of the system (in meV) is closest to

Given: Boltzmann constant k_{B} = 0.086 meV K^{–1}

(a) 0.12

(b) 0.97

(c) 1.32

(d) 1.82

Ans. (b)

Sol.

28. For the Maxwell-Boltzmann speed distribution, the ratio of the root-mean-square speed (v_{rms}) and the most probable speed (v_{max}) is

Given: Maxwell-Boltzmann speed distribution function for a collection of particles of mass m is

where, v is the speed and k_{B}T is the thermal energy.

(a)

(b)

(c)

(d)

Ans. (a)

Sol.

29. In an extrinsic p-type semiconductor, which of the following schematic diagram depicts the variation of the Fermi energy level (E_{F}) with temperature (T)?

(a)

(b)

(c)

(d)

Ans. (a)

Sol. The Fermi level of P-type semiconductor can be written as follows :

E_{FP} = E_{F} – k_{B}T ln

As we know that Fermi level at normal temperature is close to the valence band. However if we increase the temp. beyond the range; intrinsic carrier Conc^{n} n_{i} will increase. As it becomes n_{i} = N_{A} then E_{FP} = E_{F}. Thus, correct representation for Fermi level for P-type is shown in figure (a).

30. A container is occupied by a fixed number of non-interacting particles. If they are obeying Fermi-Dirac, Bose-Einstein, and Maxwell-Boltzmann statistics, the pressure in the container is P_{FD}, P_{BE} and P_{MB}, respectively. Then

(a) P_{FD} > P_{MB} > P_{BE
}

(b) P_{FD} > P_{MB} = P_{BE
}

(c) P_{FD} > P_{BE} > P_{MB
}

(d) P_{FD} = P_{MB} = P_{BE}

Ans. (a)

Sol. In Fermi gas, PV = NKT

In Bose gas, PV = NKT

In MB gas, PV = NKT. Thus, P_{BE} < P_{MB} < P_{FD}.

31. The spectral energy density u_{T}() vs wavelength () curve of a black body shows a peak at = _{max}. If the temperature of the black body is doubled, then

(a) the maximum of u_{T}() shifts to _{max}/2

(b) the maximum of u_{T}() shifts to 2_{max
}

(c) the area under the curve becomes 16 times the original area

(d) the area under the curve becomes 8 times the original area

Ans. (a, c)

Sol.

32. A periodic function f(x) = x^{2} for – < x < is expanded in a Fourier series. Which of the following statement(s) is/are correct?

(a) Coefficients of all the sine terms are zero

(b) The first term in the series is

(c) The second term in the series is –4cos x

(d) Coefficients of all the cosine terms are zero

Ans. (a, b, c)

Sol. f(x) = x^{2} for – < x <

= –4

= –4

Since, the function is even. Thus, sine term will be zero.

Also, the function is even. Thus, sine term will be non-zero.

Hence, option (a, b, c) are correct.

33. The state of a harmonic oscillator is given as , where and are the normalized wave functions of ground, first excited, and second excited states, respectively. Which of the following statement(s) is/are true?

(a) A measurement of the energy of the system yields E = with non-zero probability

(b) A measurement of the energy of the system yields E = with non-zero probability

(c) Expectation value of the energy of the system

(d) Expectation value of the energy of the system

Ans. (a, c)

Sol. Given,

Total probability,

For Harmonic oscillator,

and

So, probability of a specific state is

Now,

Hence, option (a), (c) are correct

34. A rod of mass M, length L and non-uniform mass per unit length , is held horizontally by a pivot, as shown in the figure, and is free to move in the plane of the figure. For this rod, which of the following statements are true?

(a) Moment of inertia of the rod about an axis passing through the pivot is ML^{2}

(b) Moment of inertia of the rod about an axis passing through the pivot is ML^{2}

(c) Torque on the rod about the pivot is MgL

(d) If the rod is released, the point at a distance from the pivot will fall with acceleration g

Ans. (a, c)

Sol.

dM = dx

Then,

Hence, option (a, c) are correct.

35. Which of the following schematic plots correctly represent(s) a first order phase transition occurring at temperature T = T_{c}? Here g, s, v are specific Gibbs free energy, entropy and volume, respectively.

(a)

(b)

(c)

(d)

Ans. (b, c)

Sol. We know that Gibbs free energy can be written as dG = VdP – SdT

In first order, first derivative of Gibbs free energy will be discontinuous.

Option (b, c) are correct

36. A particle (p_{1}) of mass m moving with speed v collides with a stationary identical particle (p_{2}). The particles bounce off each other elastically with p_{1} getting deflected by an angle = 30° from its original direction. Then, which of the following statement(s) is/are true after the collision?

(a) Speed of p_{1} is

(b) Kinetic energy of p_{2} is 25% of the total energy

(c) Angle between the directions of motion of the two particles is 90°

(d) The kinetic energy of the centre of mass of p_{1} and p_{2} decreases

Ans. (a, b, c)

The angle between the direction of motion of two particles are 90°

The conservation of linear momentum along x(chosen) is :

mv + 0 = mv_{1} cos(30°) + mv_{2} cos(60°)

The conservation of linear momentum along y (chosen) is

0 + 0 = –mv_{1} sin 30° + mv_{2} sin 60°

Now,

Now, from eq. (2),

Hence, option (a, b, c) are correct.

37. A wave travelling along the x-axis with y representing its displacement is described by (v is the speed of the wave)

(a)

(b)

(c)

(d)

Ans. (a, b, d)

Sol. Differential form of wave equation is :

Its solution will be

y = A sin (wt – kx) ...(2)

and y = A sin (wt + kx) ...(3)

From eq. (1), we have

From eq. (2),

From eq. (3),

Hence, option (a, b, d) are correct.

38. An objective lens with half angular aperture is illuminated with light of wavelength . The refractive index of the medium between the sample and the objective is n. The lateral resolving power of the optical system can be increased by

(a) decreasing both and

(b) decreasing and increasing

(c) increasing both and n

(d) decreasing and increasing n

Ans. (b, c, d)

Sol. R.P. =

Thus, resolving power decreasing and increasing .

It is increasing both and n.

It is decreasing and increasing n.

So, option (b, c, d) are correct.

39. Which of the following statement(s) is/are true for a LC circuit with L = 25 mH and C = 4 µF?

(a) Resonance frequency is close to 503 Hz

(b) The impedance at 1 kHz is 15

(c) At a frequency of 200 Hz, the voltage lags the current in the circuit

(d) At a frequency of 700 Hz, the voltage lags the current in the circuit

Ans. (a, c)

Sol. c = 4 µF

Resonance frequency(w) =

Thus, option (a) is correct.

Since, f_{R} = 503 Hz

So, f > f_{R} i.e., current lags the voltage and f < f_{R} i.e., current leads the voltage.

Hence, option (c) is correct because at 200 Hz, voltage lags the current. But option (d) is not correct because at 700 Hz, current lags the voltage.

Impedance, Z = and

Z = J(117)

Hence, option (a, c) are correct.

40. For a particle moving in a general central force field, which of the following statement(s) is/are true?

(a) The angular momentum is a constant of motion

(b) Kepler's second law is valid

(c) The motion is confined to a plane

(d) Kepler's third law is valid

Ans. (a, b, c)

Sol. (a) The angular momentum is a constant of motion : This statement is always true for central force problem.

(b) Kepler's second law is valid : This statement is always true because = constant.

(c) Motion is confined to a plane : This statement is always true.

(d) Kepler's third law is valid : This statement is not always true. In third law, we have T^{2} = a^{3} which is true for potential like V(r) = .

Hence, option (a, b, c) are correct.

41. The lattice constant (in Å) of copper, which has FCC structure, is _________ (rounded off to two decimal places).

Given: density of copper is 8.91 g cm^{–3} and its atomic mass is 63.55 g mol^{–1}; Avogadro's number = 6.023 × 10^{23} mol^{–1}

Sol. = 8.94 g/cm^{3}

M_{Cu} = 63.5 g/mol

N_{A} = 6.023 × 10^{23}

We know,

For FCC,

= 3.6Å

42. Two silicon diodes are connected to a battery and two resistors as shown in the figure. The current through the battery is __________ A (rounded off to two decimal places).

Given: The forward voltage drop across each diode = 0.7 V

Sol..

By applying KVL,

–5 + 0.7 + 10I = 0

I = = 0.43A

43. The absolute error in the value of sin if approximated up to two terms in the Taylor's series for = 60° is ___________ (rounded off to three decimal places).

Sol.

In Taylor series,

Error, = 0.866 – 0.855 = 0.011

44. A single pendulum hanging vertically in an elevator has a time period T_{0} when the elevator is stationary. If the elevator moves upward with an acceleration of a = 0.2g, the time period of oscillations is T_{1}. Here g is the acceleration due to gravity. The ratio is ____________ (rounded off to two decimal places).

Sol.

45. A spacecraft has speed v_{s} = fc with respect to the earth, where c is the speed of light in vacuum. An observer in the spacecraft measures the time of one complete rotation of the earth to be 48 hours. The value of f is __________ (rounded off to two decimal places).

Sol. From the concept of time dilation,

46. The sum of the x-components of unit vectors and for a particle moving with angular speed 2 rad s^{–1} at angle = 215° is__________ (rounded off to two decimal places)

Sol. Given that = 215

and = 2rad/sec

At = 215 and = 2 rad/sec

47. Consider a spring mass system with mass 0.5 kg and spring constant k = 2 N m^{–1} in a viscous medium with drag coefficient b = 3 kg s^{–1}. The additional mass required so that the motion becomes critically damped is _________ kg (rounded off to three decimal places).

Sol. b^{2} = 4mk

Excess mass, m_{excess} = 1.12 – 0.5 = 0.62 Kg

48. Unit vector normal to the equipotential surface of V(x, y, z) = 4x^{2} + y^{2} + z at (1,2,1) is given by . The value of |b| is _____________ (rounded off to two decimal places).

Sol. V(x, y, z) = 4x^{2} + y^{2} + z

Normal unit vector,

b = 0.43

49. A rectangular pulse of width 0.5 cm is travelling to the right on a taut string (shown by full line in the figure) that has mass per unit length µ_{1}. The string is attached to another taut string (shown by dashed line) of mass per unit length µ_{2}. If the tension in both the strings is the same, and the transmitted pulse has width 0.7 cm, the ratio µ_{1}/µ_{2} is ___________(rounded off to two decimal places).

Sol. Frequency of pulse will not change while on moving. Given that rectangular pulse of width 0.5 cm, we can see its wavelength.

f_{1} = f_{2
}

_{
}

Given, = w_{1} = 0.5 cm

and = w_{2} = 0.7 cm

Velocity, v =

where, T Tension and µ =

then, (T_{1} = T_{2})

50. An particle with energy of 3 MeV is moving towards a nucleus of ^{50}Sn. Its minimum distance of approach to the nucleus is f × 10^{–14} m. The value of f is ________ (rounded off to one decimal place).

Sol. From the conservation of energy, the distance of closest approach can be written as :

= 4.8 × 10^{–14} m

51. In a X-Ray tube operating at 20 kV, the ratio of the de-Broglie wavelength of the incident electrons to the shortest wavelength of the generated X-rays is ________ (rounded off to two decimal places).

Given: e/m ratio for an electron = 1.76 × 10^{11} C kg^{–1} and the speed of light in vacuum is 3 × 10^{8} m s^{–1}

Sol. The De-Broglie wavelength of incident electron can be written as :

The shortest wavelength of X-ray is :

The given ratio

52. A point source emitting photons of 2 eV energy and 1 W of power is kept at a distance of 1m from a small piece of a photoelectric material of area 10^{–4} m^{2}. If the efficiency of generation of photoelectrons is 10%, then the number of photoelectrons generated are f × 10^{12} per second. The value of f is __________ (rounded off to two decimal places).

Given: 1eV = 1.6 × 10^{–19} J

Sol.

The energy falling on material per unit time = × Area of material = × 10^{–4}

No. of photo electronics generated =

= × 0.1 = 2.48 × 10^{12}

Hence, f = 2.48

53. Consider the -decay ^{90}Th^{232}
^{88}Ra^{228}. In an experiment with one gram of ^{90}Th^{232}, the average count rate (integrated over the entire volume) measured by the -detector is 3000 counts s^{–1}. If the half life of ^{90}Th^{232} is given as 4.4 × 10^{17} s, then the efficiency of the -detector is ___________ (rounded off to two decimal places).

Given: Avogadro's number = 6.023 × 10^{23} mol^{–1}

Sol. The activity or decay rate is defined as : A = N_{0}

The value of N_{0} in one gram of Th^{232} is

and

Thus,

A = 4088

Now, Efficiency =

54. In the Thomson model of hydrogen atom, the nuclear charge is distributed uniformly over a sphere of radius R. The average potential energy of an electron confined within this atom can be taken as V = . Taking the uncertainty in position to be the radius of the atom, the minimum value of R for which an electron will be confined within the atom is estimated to be f × 10^{–11} m. The value of f is ___________ (rounded off to one decimal place).

Given: The uncertainty product of momentum and position is . = 1 × 10^{–34} J s^{–1}, e = 1.6 × 10^{–19}C, and = 9 × 10^{9} Nm^{2}C^{–2}

Sol. 2.2

55. The sum of the eigenvalues and of matrix B = I + A + A^{2}, where A = is____________ (rounded off to two decimal places).

Sol. B = I + A + A^{2
}

^{ }A =

|A – I| = 0

(2 – ) (0.5 – ) + 0.5 = 0

1 – 2 – 0.5 + ^{2} + 0.5 = 0

^{2} – 2.5 + 1.5 = 0

^{2} – – 1.5 + 1.5 = 0

( – 1) – 1.5( – 1) = 0

= 1, 1.5

A^{2} =

=

I_{r} = 2

A^{2} – 2.5A + 1.5I = 0

A^{2} + A – A + I – I – 2.5A + 1.5I = 0

A^{2} + A + I = B = 3.5A – 0.5I

Thus, sum of eigen values of B

= (3.5) (2.5) – (0.5) (2.5) = 7.75

56. A container of volume V has helium gas in it with N number of He atoms. The mean free path of these atoms is . Another container has argon gas with the same number of Ar atoms in volume 2V with their mean free path being . Taking the radius of Ar atoms to be 1.5 times the radius of He atoms, the ratio / is__________ (rounded off to two decimal places).

Sol. We know that, mean free path can be written as :

where, n =

We can say,

Thus,

57. Three frames F_{0}, F_{1} and F_{2} are in relative motion. The frame F_{0} is at rest, F_{1} is moving with velocity with respect to F_{0} and F_{2} is moving with velocity with respect to F_{1}. A particle is moving with velocity with respect to F_{2}. If v_{1} = v_{2} = v_{3} = c/2, where c is the speed of light, the speed of the particle with respect to F_{0} is fc. The value of f is ____________ (rounded off to two decimal places).

Sol. Speed of F_{2} w.r.t. F_{0} is

Now, speed of particle w.r.t. F_{0} is :

So, f = 0.93 c

58. A fission device explodes into two pieces of rest masses m and 0.5m with no loss of energy into any other form. These masses move apart respectively with speeds and , with respect to the stationary frame. If the rest mass of the device is fm then f is________ (rounded off to two decimal places).

Sol. From conservation of energy,

59. A conducting wire AB of length m has resistance of .6 It is connected to a voltage source of 0.5 V with negligible resistance as shown in the figure. The corresponding electric and magnetic fields give Poynting vectors all around the wire. Surface integral is calculated over a virtual sphere of diameter 0.2 m with its centre on the wire, as shown. The value of the integral is ______W (rounded off to three decimal places).

Sol.

60. A metallic sphere of radius R is held at electrostatic potential V. It is enclosed in a concentric thin metallic shell of radius 2R at potential 2V. If the potential at the distance R from the centre of the sphere is fV, then the value of f is _______ (rounded off to two decimal places).

Sol.

Now, 2V = A_{0}(2R)^{0} +

f = 1.66