IIT-JEE 2009 Linked Comprehension Type Multiple Choice Questions on Nuclear Physics

Our greatest weakness lies in giving up. The most certain way to succed is always to try just one more time.

– Thomas Alva Edison

The following three questions [(i), (ii) and (iii)] from nuclear physics were included under Linked Comprehension Type Multiple Choice Questions (single answer type) in the IIT-JEE 2009 question paper. They are simple as you will realize.

Paragraph for Questions (i), (ii) and (iii)

Scientists are working hard to develop unclear fusion reactor. Nuclei of heavy hydrogen, ₁H², known as deuteron and denoted by D, can be thought of as a candidate for fusion reactor. The D–D reaction is ₁H²+₁H² → ₂He³ + n + energy. In the core of fusion reactor, a gas of heavy hydrogen is fully ionized into deuteron nuclei and electrons. This collection of ₁H² nuclei and electrons is known as plasma. The nuclei move randomly in the reactor core and occasionally come close enough for nuclear fusion to take place. Usually, the temperatures in the reactor core are too high and no material wall can be used to confine the plasma. Special techniques are used which confine the plasma for a time t₀ before the particles fly away from the core. If n is the density (number/volume) of deuterons, the product nt₀ is called Lawson number. In one of the criteria, a reactor is termed successful if Lawson number is greater than 5 × 10¹⁴ s/cm³.

It may be helpful to use the following : Boltzmann constant k = 8.6 × 10^–5 eV/K; e²/4πε₀ = 1.44×10^–9 eVm

Question (i):

In the core of nuclear fusion reactor, the gas becomes plasma because of

(A) strong nuclear force acting between the deuterons

(B) Coulomb force acting between the deuterons

(C) Coulomb force acting between the deuteron–electron pairs

(D) the high temperature maintained inside the reactor core.

The correct option is (D) since the plasma state is achieved at high temperatures.

Question (ii):

Assume that two deuteron nuclei in the core of fusion reactor at temperature T are moving towards each other, each with kinetic energy 1.5 kT, when the separation between them is large enough to neglect Coulomb potential energy. Also neglect any interaction from other particles in the core. The minimum temperature T required for them to reach a separation of 4 × 10^–15 m is in the range

(A) 1.0×10⁹ K < T < 2×10⁹ K

(B) 2.0×10⁹ K < T < 3×10⁹ K

(C) 3.0×10⁹ K < T < 4×10⁹ K

(D) 4.0×10⁹ K < T < 5×10⁹ K

The total kinetic energy of the two deuterons is 2×1.5 kT = 3 kT. At the required separation r (equal to 4×10^–15 m), the entire kinetic energy gets converted into electrostatic potential energy:

3 kT = e²/4πε₀r

Or, 3×8.6×10^–5×T = 1.44×10^–9/(4×10^–15) since e²/4πε₀ = 1.44×10^–9 eVm

Therefore, T = 1.4×10⁹ K. [Option (A)].

Question (iii):

Results of calculations for four different designs of a fusion reactor using D–D reaction are given below. Which of these is most promising based on Lawson criterion ?

(A) deuteron density = 2.0×10¹² cm^–3, confinement time = 5.0×10^–3 s

(B) deuteron density = 8.0×10¹⁴ cm^–3, confinement time = 9.0×10^–1 s

(C) deuteron density = 4.0×10²³ cm^–3, confinement time = 1.0×10^–11 s

(D) deuteron density = 1.0×10²⁴ cm^–3, confinement time = 4.0×10^–12 s.

Lawson number is nt₀. Out of given options, Lawson number is greater than 5 × 10¹⁴ s/cm³ for option (B) since n = 8.0×10¹⁴ cm^–3 and t₀ = 9.0×10^–1 s.

Therefore the correct option is (B).

You will find some useful multiple choice questions on nuclear physics here

AIEEE 2008- An Imaginary Question on Bohr Model

The following question was included in the AIEEE 2008 question paper:

Suppose an electron is attracted towards the origin by a force k/r where ‘k’ is a constant and ‘r’ is the distance of the electron from the origin. By applying Bohr’s model to this system, the radius of the n^th orbital of the system is found to be ‘r_n’ and the kinetic energy of the electron to be ‘T_n’. Then which of the following is true?

(1) T_n α 1/n², r_n α n²

(2) T_n independent of n, r_n α n

(3) T_n α 1/n, r_n α n

(4) T_n α 1/n, r_n α n²

The force k/r supplies the centripetal force for the circular motion of the electron so that we have

k/r = mv²/r where ‘m’ is the mass and ‘v’ is the speed of the electron.

Therefore, mv² = k which is independent of the quantum number ‘n’. The kinetic energy T_n of the electron is ½ mv² which is therefore independent of the quantum number ‘n’.

Also, v = √(k/m).

The angular momentum of the electron in the n^th orbit of radius r_n is mvr_n and in the Bohr model mvr_n = nh/2π where ‘h’ is Planck’s constant. Substituting for ‘v’ we have

m×√(k/m) ×r_n = nh/2π

This gives r_n α n. So the correct option is (2).

In the above question the attractive force on the electron was imagined to be inversely proportional to the distance just for the sake of testing your problem solving skill. In a real hydrogen atom the force is certainly inversely proportional to the square of the distance. You will find questions on real Bohr model on this site by clicking on the label ‘Bohr model ‘ below this post.

All India Pre-Medical/Pre-Dental Entrance Examination (Preliminary) 2009 (AIPMT 2009) Questions on Nuclear Physics

Everything should be made as simple as possible, but not simpler

– Albert Einstein

The following questions on nuclear physics appeared in the All India Pre-Medical/Pre-Dental Entrance Examination (Preliminary) 2009 question paper:

1. In the nuclear decay given below:

^AX_Z → ^AY_Z+1→ ^{A– 4} B* _{Z– 1}→ ^{A– 4} B _{Z– 1},

the particles emitted in the sequence are

(1) γ, β, α

(2) β, γ, α

(3) α, β, γ

(4) β, α, γ

The nucleus ^AX_Z emits a β-particle and its atomic number increases by 1 to transform to the nucleus ^AY_Z+1. The nucleus ^AY_Z+1 emits an α-particle so that its mass number reduces by 4 and atomic number reduces by 2 to become the unstable nucleus ^{A– 4}B* _{Z– 1}. It then emits a γ photon which does not produce any change in mass number and atomic number. The correct option is (4).

2. The number of β-particles emitted by a radioactive substance is equal to twice the number of α -particles emitted by it. The resulting daughter is an

(1) isomer of parent

(2) isotone of parent

(3) isotope of parent

(4) isobar of parent

When an α -particles emitted the mass number reduces by 4 and atomic number reduces by 2. When two β-particles are emitted the atomic number increases by 2. Therefore, when the number of β-particles emitted by a radioactive substance is equal to twice the number of α -particles emitted by it there is no change in the atomic number, but there is a reduction in mass number.

The resulting daughter is therefore an isotope of the parent nucleus [Option (3)].

3. In a Rutherford scattering experiment when a projectile of charge z₁ and mass M₁ approaches a target nucleus of charge z₂ and mass M₂,the distance of closest approach is r₀. The energy of the projectile is

(1) directly proportional to z₁z₂

(2) inversely proportional to z₁

(3) directly proportional to mass M₁

(4) directly proportional to M₁M₂

At the distance of closest approach the entire kinetic energy (E) of the projectile gets converted into electrostatic potential energy of the system. Therefore we have

E = (1/4πε₀) (z₁z₂/r₀)

Therefore, the energy of the projectile is directly proportional to z₁z₂ [Option (1)].

Two Questions (MCQ) on Doppler Effect in Sound

You will find the formulae to be remembered in connection with Doppler effect and some useful multiple choice questions (with solution) on Doppler effect at this location on this site.

[If you want all questions on Doppler effect on this site, you may click on the label ‘Doppler effect’ below this post].

Today I give you two more multiple choice questions on Doppler effect:

(1) A whistle producing sound waves of frequencies 9500 Hz and above is approaching a stationary person with speed v ms^–1. The velocity of sound in air is 300 ms^–1. If the person can hear frequencies up to 10000 Hz, the maximum value of v up to which he can hear the whistle is

(1) 30 ms^–1

(2) 15√2 ms^–1

(3) 15/√2 ms^–1

(4) 15 ms^–1

The apparent frequency (n’) in terms of all possible variables is given by

n’ = n(v+w–v_L)/(v+w–v_S) where ‘n’ is the real frequency of sound, ‘v’ is the velocity of sound, ‘w’ is the velocity of wind, ‘v_L’ is the velocity of listener and ‘v_S’ is the velocity of the source of sound. Note that in the above expression all velocities are in the same direction and the source is behind the listener and is therefore approaching the listener.

Since the wind velocity is zero and the listener is stationary in the first case, the above expression reduces to

n’ = nv/(v–v_S)

[The apparent frequency is therefore greater than the real frequency].

As the person can hear frequencies up to 10000 Hz only, the maximum value of v_S upto which he can hear the whistle is given by

10000 = 9500×300/(300 – v_S)

Therefore, 300 – v_S = 285 so that v_S = 15 ms^–1

[The above question appeared in AIEEE 2006 question paper]

(2) A person moves away with constant velocity ‘v₀’ from a stationary train which blows its whistle. The ratio of the real frequency of the whistle to the apparent frequency as measured by the person is 1.25. If the person is stationary and the whistle is moving away from the person with the same velocity ‘v₀’, the ratio of the real frequency of the whistle to the apparent frequency as measured by the person will be

(a) 1.2

(b) 1.25

(c) 1.4

(d) 1.45

(e) 1.5

The apparent frequency (n₁) as measured by the person moving away from the whistle with velocity v_L is given by

n₁ = n(v–v_L)/v where ‘n’ is the real frequency of the whistle and ‘v’ is the velocity of sound.

Therefore, n/n₁ = v/(v–v_L) = v/(v–v₀).

Since n/n₁ = 1.25,we have v/(v–v₀) =1.25 so that 1– (v₀/v) = 0.8 from which v₀ = 0.2v.

If the person is stationary and the whistle is moving away from the person with the same velocity ‘v₀’, the apparent frequency (n₂) as measured by the person is given by

n₂ = nv/(v +v_S) where v_S = v₀

Therefore, n/n₂ = (v+v₀)/v.

Substituting for v₀ (= 0.2v) we obtain n/n₂ = (v+0.2v)/v = 1.2

Geometric Optics- Multiple Choice Questions involving Refraction at Plane Surfaces

I am neither especially clever nor especially gifted. I am only very, very curious.

– Albert Einstein

The following question at the first glance may appear to be a difficult one to many of you; but, you will realise how easy it is when you apply basic points you studied in geometric optics:

A glass jar has the plane inner surface PQ of its bottom silvered and contains water (of refractive index n = 4/3) column of height t = 6 cm. A small light emitting diode (LED) is arranged at O at a height d = 8 cm from the water surface AB (Fig.). The silvered bottom of the jar acts as a plane mirror. At what distance from the free surface (AB) of water will this plane mirror form the image of the light emitting diode?

(a) 11 cm

(b) 14 cm

(c) 17 cm

(d) 18 cm

(e) 20 cm

When you look into the plane mirror (silvered surface) PQ from the position O of the LED, the plane mirror will appear to be located at P₁Q₁ (fig.) at a distance t/n from the free surface of water (because of normal refraction at the water surface). The distance of the LED from this refracted image P₁Q₁ of the plane mirror is therefore equal to (d + t/n) as shown in the adjoining figure.

The image of the LED must be formed at O₁ which is at the same distance (d + t/n) from the effective plane mirror P₁Q₁.

As is clear from the adjoining figure, the distance of the image O₁ from the free surface (AB) of water is (d + t/n) + t/n which is equal to (d + 2t/n) = 8 + 2×6/(4/3) = 17 cm.

The following question appeared in EAMCET (Engineering) 2003 question paper:

One of the refracting surfaces of a prism of refractive index √2 is silvered. The angle of the prism is equal to the critical angle of a medium of refractive index 2. A ray of light incident on the unsilvered surface passes through the prism and retraces its path after reflection at the silvered face. Then the angle of incidence on the unsilvered surface is

(a) 0º

(b) 30º

(c) 45º

(d) 60º

The angle A of the prism (as mentioned in the question) is given by n = 1/sin A where n = 2.

[Remember n = 1/sin C where n is the refractive index and C is the critical angle].

Therefore, sin A = ½ so that A = 30º

Since the ray retraces its path after reflection at the silvered face, it is incident normally at the silvered face (at the point N in the figure). With reference to the figure, angle QNA in the triangle QNA is 90º.

Since the angle A is 30º it follows that angle AQN = 60º so that the angle of refraction (r) at Q is 30º.

The angle of incidence (i) at the unsilvered face is given by

n = sin i/sin r from which sin i = n sin r = √2 sin 30º.

This gives sin i =1/√2 so that i = 45º.

You may search for ‘optics’ using the ‘search blog’ facility at the top left of this page to find all related posts on this site.

A useful post on the equations to be remembered in Geometric Optics can be found here.

Kerala Engineering Entrance 2008 and other Multiple Choice Questions (MCQ) on Work and Energy

Setting an example is not the main means of influencing others; it is the only means.

– Albert Einstein

Let us discuss a few multiple choice questions on work and energy.

(1) A small sphere of mass 20 g is projected vertically up with a velocity of 10 ms^–1. If air resistance is negligible, what is the total work done by gravity during the upward trip of the sphere?

(a) 10 J

(b) –10 J

(c) – 1 J

(d) 1 J

(e) 9.8 J

When the sphere rises up its kinetic energy goes on decreasing because of the work done by the gravitational force against the motion of the sphere. (The gravitational potential energy of the sphere goes on increasing by an equal amount). The work done by gravity is evidently negative. When the sphere reaches the maximum height the entire kinetic energy gets converted into gravitational potential energy. The total work done by gravity during the the upward trip of the sphere is numerically equal to the initial kinetic energy (½ mv²) of the sphere but its sign is negative.

Therefore, the answer is – ½ mv² = – ½ ×0.020×10² = –1 J.

(2) A world class athlete covers a distance of 100 m in 10 s. His kinetic energy can be estimated to be in the range

(a) 2 × 10⁵ J to 5 × 10⁵ J

(b) 2 × 10⁴ J to 5 × 10⁴ J

(c) 2 × 10³ J to 5 × 10³ J

(d) 200 J to 500 J

(e) 20 J to 50 J

This question has appeared in various entrance tests with slight differences in the wording and in the options. Last year it appeared in the All India Engineering Entrance Examination.

In the 100 m dash the velocity v is almost constant so that we have

v = 100 m/10 s = 10 ms^–1

We may take the mass of the athlete to be 70 kg to 80 kg. Let us use 80 kg.

His kinetic energy will be ½ mv² = ½ ×80×10² = 4000 J so that the correct option is (c).

Questions (3) and (4) appeared in Kerala Engineering Entrance Examination 2008 question paper.

(3) Two bodies A and B have masses 20 kg and 5 kg respectively. Each one is acted upon by a force of 4 kg wt. If they acquire the same kinetic energy in times t_A and t_B, then the ratio t_A/ t_B is

(a) ½

(b) 2

(c) 2/5

(d) 5/6

(e) 1/5

The concept of impulse will be very useful here. The impulse received by A and B in the times t_A and t_B are respectively F t_A and F t_B where F is the force acting on them (4 kg wt. here). But impulse is the change in momentum so that the momenta acquired by A and B are F t_A and F t_B.

Since the kinetic energy is p²/2m where p is the momentum, we have

(F t_A)²/2m_A = (F t_B)²/2m_B

Therefore, t_A/ t_B = √( m_A/ m_B) = √(20/5) = 2.

(4) A particle acted upon by constant forces 4i + j – 3k and 3i + j – k is displaced from the point i + 2j + 3k to the point 5i + 4j + k. The total work done by the forces in SI units is

(a) 20

(b) 40

(c) 50

(d) 30

(e) 35

The resutant force (F) on the particle is the sum of the forces given by

F = (4i + j – 3k) + (3i + j – k) = 7i + 2j – 4k.

The displacement (s) of the particle is given by

s = (5i + 4j + k) – (i + 2j + 3k) = (4i + 2j – 2k)

The work done (W) is given by

W = F.s = (7i + 2j – 4k) . (4i + 2j – 2k) = 28 + 4 + 8 = 40.

You will find similar useful multiple choice questions (with solution) at AP Physics Resources.