If we did all things we are capable of, we would literally astound ourselves.

– Thomas A. Edison

Monday, May 28, 2007

A Question (MCQ) on Rolling

Here is a question which checks your understanding of basic things in angular motion and simple harmonic motion:

A solid cylinder of mass M and radius R is resting on a horizontal platform which is parallel to the XZ plane. The cylinder can rotate freely about its own axis, which is along the X-direction. The platform is given a linear simple harmonic motion of angular frequency ‘ω’ and amplitude ‘A’ in the Z-direction. If there is no slipping between the cylinder and the platform, the maximum torque acting on the cylinder is

(a) 2MR22 (b) MR22 (c) 2MRAω2

(d) MRAω2 (e) MRAω2/2

You may be remembering the expression for maximum acceleration of a simple harmonic motion: amax = ω2A

[ If you don’t remember the above expression, you may use the simplest form of simple harmonic motion of amplitude ‘A’ and angular frequency ‘ω’ and differentiate it twice:

z = A sin ωt (We write the displacement as ‘z’ since it is in the Z-direction).

a = d2z/dt2 = –ω2A sin ωt

Therefore, the maximum acceleration is ω2A].

The maximum angular acceleration (αmax) of the cylinder is given by

αmax = amax/R = ω2A/R.

The maximum torque (τmax) on the cylinder is given by

τmax = αmaxI, where ‘I’ is the moment of inertia of the cylinder about its own axis (which is equal to MR2/2).

Therefore, maximum torque τmax = (ω2A/R) (MR2/2) = MRAω2/2

You will find an interesting MCQ on rolling at physicsplus: MCQ on Rolling Bodies

Friday, May 25, 2007

KEAM 2007: Kerala Engineering/Medical Entrance Examination 2007 Results Announced

The results of Kerala Engineering/Medical Entrance Examinations conducted during the last week of April 2007 by the Commissioner for Entrance Examinations, Govt. of Kerala are announced. The results are available at www.keralaresults.nic.in and at www.results.kerala.nic.in

Wednesday, May 23, 2007

IIT-JEE 2007 Assertion-Reason Type MCQ on Friction

The following Assertion-Reason Type MCQ which appeared in IIT-JEE 2007 question paper is meant for checking your understanding of the conservative nature of gravitational force and the non-conservative nature of the frictional force:


A block of mass m starts moving on a rough horizontal surface with a velocity v. It stops due to friction between the block and the surface after moving through a certain distance. The surface is now tilted to an angle of 30º with the horizontal and the same block is made to go up on the surface with the same initial velocity v. The decrease in the mechanical energy in the second situation is smaller than that in the first situation



The coefficient of friction between the block and the surface decreases with the increase in the angle of inclination.

(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

In both situations the initial kinetic energy of the block is reduced to zero. In the first situation, the initial K.E. is entirely used up in doing work against friction and is irrecoverably lost since friction is a non-conservative force. In the second situation, part of the initial K.E. is used in doing work against gravitational force and is stored as gravitational potential energy in the block since gravitational force is a conservative force. The decrease in mechanical energy in the second situation is therefore smaller than that in the first situation.

Statement-1 is therefore true.

Statement-2 is false since the coefficient of friction is independent of the inclination and is dependent only on the nature of the two surfaces in contact.

The correct option therefore is (C).

Sunday, May 20, 2007

MCQ on Wien’s Law

The following question on Wien’s law is a typical one and is of the type often found in Medical and Engineering Entrance test papers:

A black body emits radiations of maximum intensity for the wave length of 7000Ǻ, when the temperature of the body is 3927ºC. If the temperature of the body is increased by 1000ºC, the maximum intensity would be observed at nearly

(a) 3650Ǻ (b) 4650Ǻ (c) 5650Ǻ

(d) 6000Ǻ (e) 6650Ǻ

The wave length λm corresponding to the maximum intensity emission from a black body at a temperature T is given by Wien’s law,

λmT = constant [= 0.29 cm K, if you express λm in cm].

In other words, λm1T1 = λm2T2

[Remember, you have to substitute the absolute (Kelvin) temperatures in this equation].

Therefore, 7000×4200 = λm2×5200, from which λm2 =5650Ǻ, nearly.

[For question setters: When a question is set using Wien’s law, it should be remembered that the product λmT satisfies the condition, λmT = 0.29 cmK, if λm is expressed in cm. I write this because I have seen questions in which this point is ignored]

Monday, May 14, 2007

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.

Sunday, May 13, 2007

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

Sunday, May 06, 2007

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 Kα 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 Kα) 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].

Friday, May 04, 2007

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.

Tuesday, May 01, 2007

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.