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%.

Two AIEEE 2005 Questions on Direct Current Circuits

(1) In the circuit, the galvanometer G shows zero deflection. If the batteries A and B have negligible internal resistance, the value of the resistor R will be

(a) 1000Ω (b) 500Ω (c) 100Ω (d) 200Ω

As there is no current through the galvanometer, the voltage drop across R must be 2V ( so as to balance the emf 2V of the battery B). The voltage drop (produced by the current sent by battery A) across 500Ω is therefore 10V. The same current flows through R and hence R must be 100Ω.

(2) An energy source will supply a constant current in to the load if its internal resistance is

(a) non zero but less than the resistance of the load

(b) zero

(c) very large compared to the load resistance

(d) equal to the resistance of the load

The correct option is (c) since the current will not change with the load resistance if the total resistance of the circuit is almost equal to the source resistance. Ideally, a constant current generator should have infinite resistance.

[You may note also that an ideal constant voltage generator should have zero resistance].

Two IIT-JEE 2007 Questions on X-rays

The following multiple choice question appeared in IIT-JEE 2007 question paper:

Electrons with de Broglie wave length λ fall on the target in an X-ray tube. The cut off wave length of the emitted X-rays is

(a) λ₀ = 2mcλ²/h (b) λ₀ = 2h/mc (c) λ₀ = 2m²c² λ³/h² (d) λ₀ = λ

X-rays of cut off (minimum) wave length for a given target voltage are obtained when the entire kinetic energy of the incident electron is converted into X-ray photon energy. The kinetic energy of the electron is p²/2m where ‘p’ is its momentum and ‘m’ is its mass.

But p = h/λ where ‘h’ is Planck’s constant [de Broglie relation] so that we have

h²/2mλ² = hc/λ₀

[The term on the right hand side is the energy of the X-ray photon of cut off wave length λ₀].

From the above, λ₀ = 2mcλ²/h

Here is an assertion-reason type multiple choice question which appeared in IIT-JEE 2007 question paper:

STATEMENT-1

If the accelerating potential in an X-ray tube is increased, the wave lengths of the characteristic X-rays do not change

because

STATEMENT-2

When an electron beam strikes the target in an x-ray tube, part of the kinetic energy is converted in to X-ray energy.

(a) Statement-1 is True, Statement-2 is True; Statement-2 is a correct

explanation for Statement-1

(b) Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct

explanation for Statement-1

(c) Statement-1 is True, Statement-2 is False

(d) Statement-1 is False, Statement-2 is True

Wave lengths of characteristic X-rays are characteristic of the target material (since they depend only on the energy levels of the atoms in the target material). Therefore, their wave lengths do not change when the accelerating voltage is increased. Statement-1 is therefore true.

Statement-2 is true. But, it is not a correct explanation for statement-1. Characteristic X-rays originate because of electron transitions between discrete energy levels of the atoms in the target. Statement-2 is only a general statement regarding the process of X-ray production.

Two Multiple Choice Questions on Moment of Inertia

(1) Three identical thin rods each having mass 2 kg and length 1 m are joined to form an equilateral triangle. The moment of inertia of this system about an axis perpendicular to the plane of the triangle and passing through one corner of the triangle is (in kg m²)
(a) 1 (b) 2 (c) 3 (d) 4 (e) 5

The moment of inertia of the rods AB and CB about an axis perpendicular to the plane of the triangle and passing through the corner B is ML²/3 + ML²/3 = 2 ML²/3 where M is the mass of each rod and L is the length. [Note that the moment of inertia of a thin rod about a normal axis through its centre of mass is ML²/12 and hence its moment of inertia about a normal axis through one end, according to the theorem of parallel axes, is (ML²/12) + M(L/2)² = ML²/3].
The moment of inertia of the rod AC about the axis through B is
ML²/12 + M[(√3/2)L]² = ML²/12 + 3 ML²/4 = 5ML²/6, in accordance with the theorem of parallel axes. [The distance of AC from B is L sin60° = (√3/2)L].
The moment of inertia of the entire triangle about the axis through B is the sum of the moments of inertia of the three rods and is equal to 2 ML²/3 +5ML²/6 = 9ML²/6 = 3ML²/2. Substituting for M (=2kg) and L (=1m), we obtain the answer as 3 kgm².

(2) The moment of inertia of a body does not depend upon its
(a) mass (b) axis of rotation (c) shape (d) angular velocity (e) size
This is a very simple question and you should not have any doubt in picking out option (d) as the correct answer.

Multiple Choice Questions on Centre of Mass

A metallic sphere of mass 1kg and radius 5 cm is welded to the end B of a thin rod AB of length 50 cm and mass 0.5 kg as shown. This rod with the sphere will balance horizontally on a knife edge placed at a distance ‘x’ from the end A of the rod if ‘x’ is equal to

(a) 35cm (b) 40 cm (c) 45 cm (d) 48 cm (e) 52 cm

This system can be imagined to be made of two point masses of 0.5 kg and 1 kg located at the middle (centre of mass) of the original rod and the centre (centre of mass) of the sphere respectively. The distance between these point masses is 30 cm (half the length of rod + radius of the sphere).

The centre of mass of the system is at a distance ‘d’ from the 0.5 kg mass of the rod such that 0.5×d = (30–d)×1.

This gives d = 20 cm. The distance ‘x’ of this centre of mass from the end A of the rod is given by x = 25+d = 25+20 = 45 cm.

Therefore, the knife edge is to be placed at 45 cm from the end A [Option (c)].

Consider now the following MCQ which appeared in Kerala Medical Entrance 2005 question paper:

A cricket bat is cut at the location of its centre of mass as shown. Then,

(a) the two pieces will have the same mass

(b) the bottom piece will have larger mass

(c) the handle piece will have larger mass

(d) mass of handle piece is double the mass of bottom piece

(e) cannot say

There is more material near the bottom end of the bat and you can compare it to the rod with the sphere we considered in the above question. The centre of mass of the bat is nearer the bottom end of the bat.

The bat will balance horizontally on a knife edge placed at the centre of mass. If the centre of mass of the handle piece and the bottom piece are at distances d₁ and d₂ respectively from the centre of mass of the entire bat, we have M₁d₁ = M₂d₂ where M₁ and M₂ are the masses of the handle piece and the bottom piece respectively. The distance d₁ is greater than d₂ since the handle piece has thinner regions compared to the bottom piece.Therefore, M₂ > M₁ [Option (b)].

You will find more multiple choice questions on centre of mass at physicsplus: Questions on Centre of Mass

Two Questions on Internal Energy of a Gas

The molar heat capacity (molar specific heat) of a gas at constant volume is given by

C_v = (n/2)R where ‘n’ is the number of degrees of freedom of the gas molecule and ‘R’ is the universal gas constant.

The molar heat capacity at constant pressure is given by

C_p = C_v + R = (n/2)R + R = [(n+2)/2]R, in accordance with Meyer’s relation.

The number of degrees of freedom of a mono atomic gas molecule is 3. Therefore, in the case of a mono atomic gas, C_v = (3/2)R and C_p = (5/2)R.

The number of degrees of freedom of diatomic gas molecules is 5 and that of triatomic and polyatomic gas molecules is 6 if we do not consider the possible vibrational modes. Therefore, in the case of a diatomic gas, C_v = (5/2)R and C_p = (7/2)R where as in the case of triatomic and polyatomic gases, C_v = (6/2)R = 3R and C_p = 4R.

Now, consider the following MCQ:

When an ideal diatomic gas without vibrational modes is heated at constant pressure, the fraction of heat energy supplied which increases the internal energy of the gas is

(a) 5/7 (b) 3/5 (c) 2/5 (d) 2/3 (e) 1/3

In the case of a diatomic gas, the molar specific heats at constant volume and constant pressure are respectively (5/2)R and (7/2)R. This means that the increase in internal energy of one mole of the gas when we supply (7/2)R joule of heat at constant pressure is (5/2)R joule. The remaining R joule of heat is spent on doing work during the expansion of the gas.

The fraction of the total energy supplied which increases the internal energy of the gas is therefore (5/2)R/(7/2)R = 5/7 [Option(a)].

[Note that this fraction in the case of a mono atomic gas will be 3/5].

Let us consider another question:

Two moles of an ideal mono atomic gas is heated from 20°C to 120°C. The increase in internal energy of the gas is nearly (Universal gas constant = 8.3 J mol^–1 K^–1).

(a) 1660 J (b) 2490 J (c) 3320 (d) 4150 J (e) 4980 J

Some of you may be inclined to wonder which molar specific heat you have to use here (C_p or C_v?). Don’t have any doubt. The increase in internal energy depends on the temperature rise and not on the conditions of constant volume or constant pressure. When you heat a gas at constant volume, the entire heat supplied is used to increase the internal energy. When you heat the gas at constant pressure, a part of the heat supplied is used for doing external work since the gas has to expand to keep the pressure constant. In both cases, the increase in the internal energy of the gas is the same if the temperature rise is the same. You have to use the molar specific heat at constant volume, which is equal to (3/2)R in the case of a mono atomic gas, for calculating the increase in internal energy.

So, increase in internal energy = No. of moles×C_v×Temperature rise = 2× (3/2)R(120–20) = 2490 J, on substituting the value 8.3 for the universal gas constant R.