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.

An IIT-JEE 2007 Question on Resonance Column

The following unusually simple MCQ appeared in IIT-JEE 2007 question paper:

In the experiment to determine the speed of sound using resonance column,

(a) prongs of the tuning fork are kept in a vertical plane

(b) prongs of the tuning fork are kept in a horizontal plane

(c) in one of the two resonances observed, the length of the resonating air column is close to the wave length of sound in air

(d) in one of the two resonances observed, the length of the resonating air column is close to half of the wave length of sound in air

As maximum possible displacement is to be imparted to the molecules of air in the resonance tube (for making the judgment of resonance condition easier), the prongs of the fork are to be kept such that they vibrate along the vertical direction (upwards and downwards). So, the correct option is (a).

You will find more multiple choice questions with solution at

physicsplus

Two EAMCET 2004 Questions on Magnetism

Even though isolated magnetic monopoles are not available, you will find questions involving isolated north poles and south poles. The practical method of obtaining a pole of the required polarity at a point is to use a long bar magnet so that the effect of the other pole at the point can be neglected. Consider the following MCQ which appeared in EAMCET 2004 question paper

Two magnetic isolated north poles each of strength ‘m’ ampere-metre are placed one at each of two vertices of an equilateral triangle of side ‘a’. the resultant magnetic induction at the third vertex is

(a) μ₀m/4π a² (b) μ₀√2 m/4π a² (c) μ₀√3 m/4π a² (d) μ₀m/4π a

The magnetic inductions (flux density) at the third vertex due to the two isolated poles have the same magnitude B = μ₀m/4π a², but their directions are inclined at 60º. The magnitude of the resultant flux density at the third vertex is

[B² + B² + 2B² cos 60º]^1/2 = √3 B = μ₀√3 m/4π a²

[If the triangle is right angled and isosceles with short sides of length ‘a’ and the poles are placed at the ends of the long side, the flux density at the third vertex will be μ₀√2m/4πa²]

The following MCQ also appeared in EAMCET 2004 question paper:

A bar magnet used in a vibration magnetometer is heated so, as to reduce the magnetic moment by 36%. The time period of magnet (neglecting the changes in dimensions of magnet)

(a) increases by 25 % (b) decreases by 36 %

(c) increases by 16 % (d) decreases by 16 %

The period (T) of angular oscillations of the magnet in a magnetic field of flux density ‘B’ is given by

T = 2π√(I/mB) where ‘I’ is the moment of inertia of the magnet about the suspension fibre and ‘m’ is the magnetic dipole moment.

When the dipole moment is reduced by 36%, ‘m’ is to be replaced by 0.64m in the above equation. The period of oscillation then becomes T/(0.8) = 1.25T.

The period thus increases by 25%.