Two Kerala Engineering Entrance 2007 Questions from Nuclear Physics

The following MCQ (on radioactivity) which appeared in Kerala Engineering Entrance 2007 question paper is simple, but it differs slightly from the conventional type:

Radium has half life of 5 years. The probability of decay of a radium nucleus in 10 years is

(a) 50% (b) 75% (c) 100% (d) 60% (e) 25%

Since the half life is 5 years, the amount getting decayed in 10 years will be 75%.

[This can be found very easily: After 5 years half the initial amount will be decayed; after another 5 years, half of the remaining will be decayed. If the half life and the time period given are not so simply related, you will have to calculate the number of nuclei undecayed (N) using the equation, N = N₀/2ⁿ where N₀ is the initial number and ‘n’ is the number of half lives in the given time. The percentage decayed in the given time is then calculated].

The probability for decay of any given nucleus in 10 years is therefore 75%.

The following question involving the relative abundance of isotopes is a popular one and it has found place in Kerala Engineering Entrance 2007 question paper:

The natural boron of atomic weight 10.81 is found to have two isotopes B¹⁰ and B¹¹. The ratio of abundance of isotopes in natural boron should be

(a) 11:10(b) 81:19 (c) 10:11 (d) 15:16 (e) 19:81

This is a simple arithmetical problem. If there are n₁ boron atoms of atomic weight 10 and n₂ boron atom of atomic weight 11 in a sample of natural boron, the mean atomic weight (which is given as 10.81) is related to n₁ and n₂ as

(10n₁ + 11n₂)/ (n₁ + n₂) = 10.81.

Rearranging, 0.81n₁ = 0.19n₂, from which n₁/n₂ = 19/81.

AP Physics Exams.

Many among the regular visitors of this site might have already noted that the posts here are useful for preparing for the AP (Advanced Placement) Physics Exams. Multiple choice questions discussed here have five choices as in the AP Physics Exams and are designed and expected to be worked out without using calculators and physical and mathematical tables. Even though multiple choice questions are discussed here, you will find the posts useful for answering the free-response section of the AP Physics Exams also, since the questions are answered with all necessary theoretical details.

You will find more questions (with solution) covering the content of AP Physics at physicsplus

Two Questions (MCQ) on X-rays

(1) The atomic numbers of the target materials in two X-ray tubes are in the ratio 1:3. The X-ray photon energies of the K_α lines of these materials are in the ratio

(a) 1:3 (b) 3:1 (c) 1:√3 (d) 1:9 (e) 1:1

According to Mosley’s law, the frequency ‘ν’ of a particular characteristic X-ray (such as K_α) is directly proportional to the square of the atomic number of the target. Since the photon energy is hν, it follows that the energy is directly proportional to the square of the atomic number. Therefore, the energies are in the ratio 1:9.

(2) The minimum wave length of X-rays produced by an X-ray tube operating at an anode voltage of 24.8 kV is very nearly

(a) 1.8 Ǻ (b) 1.5 Ǻ (c) 1 Ǻ (d) 0.8 Ǻ (e) 0.5 Ǻ

The minimum wave length X-ray photon will have the entire energy of the impinging electron, which is 24.8 keV.

Since the product of the photon energy in eV and the wave length in Angstrom is 12400, the minimum wave length is 12400 ÷ 24800 = 05 Ǻ.

[You may use the equation hc/λ = 24800 eV= 24800×1.6×10^–19 joule to calculate λ in metre after substituting for Planck’s constant ‘h’ and the speed of light ‘c’, but it will be time consuming].

Two Kerala Medical Entrance 2007 Questions on Work and Energy

The following two questions appeared in Kerala Medical Entrance 2007 question paper which contained relatively simple questions:

(1) When a bullet is fired at a target, its velocity decreases by half after penetrating 30 cm in to it. The additional thickness it will penetrate before coming to rest is

(a) 30 cm (b) 40 cm (c) 10 cm (d) 50 cm (e) 20 cm

Questions similar to this one often find place in admission test question papers. This can be worked out using the work-energy principle or by using equations of one-dimensional motion. Let us use the work-energy principle:

Since the decrease in kinetic energy is equal to the work done against the retarding force in the target, we can write

½ m[v² – (v/2)²] = F×0.3, where ‘m’ is the mass of the bullet, ‘v’ is its initial velocity and F is the retarding force, which has to be assumed to be constant.

If ‘s’ is the additional distance penetrated (for the velocity to decrease to zero from the value v/2),

½ m(v/2)² = Fs

Dividing the first equation by the second, 3 = 0.3/s, from which s = 0.1 m = 10 cm.

[ If you use the equation of uniformly retarded linear motion, v² = u² –2as where ‘u’ and ‘v’ are the initial and final velocities respectively and ‘a’ is the retardation, we have the two equations,

(u/2)² = u² – 2a×0.3 and

0 = (u/2)² – 2as

The first equation is 3u²/4 = 2a×0.3.

The second equation is u²/4 = 2as.

Dividing the first one by the second, 3 = 0.3/s from which s = 01 m = 10 cm].

(2) A body constrained to move in the Y-direction is subjected to a force F = 2i + 15j + 6k newton. The work done by this force in moving the body through a distance of 10 m along the Y-axis is

(a) 100 J (b) ) 150 J (c) ) 120 J (d) ) 200 J (e) 50 J

The work (W) done by a force F in producing a displacement S is given by the scalar product of the vectors F and S.

Therefore, W = F.S = (2i + 15j + 6k) . 10j = 150 J

[The displacement vector is 10j since it is along the Y-axis].

You may work this out also by arguing that the component of the force along the direction of displacement of 10 m is 15 newton so that the work done is 15×10 = 150 J.

A Question (MCQ) Involving Lorentz Force

Here is an interesting question which requires some imagination on your part:

Parallel Electric and magnetic fields act at a certain of space. If a positive point charge is projected into this region with a velocity ‘v’, making a small angle with respect to the fields, the path of the point charge within the fields will be

(a) a straight line

(b) a helix of constant pitch and radius

(c) a helix of increasing pitch and radius

(d) a helix of constant pitch and increasing radius

(e) a helix of increasing pitch and constant radius

Since the charge is projected at a small angle, it will move along a helical path. This is because of the fact that the velocity component perpendicular to the magnetic field forces the particle to move along a circle and the velocity component parallel to the magnetic field makes it move forward in the direction of the magnetic field. In the absence of the electric field, the particle will move along a helix of constant pitch and radius.

Since there is an electric field parallel to the magnetic field, the forward component of the velocity goes on increasing while the perpendicular component remains unchanged in magnitude. The particle will therefore move along a helix of increasing pitch and constant radius under the combined action of the electric and magnetic Lorentz forces.