Mozilla Firefox Browser

It is my belief that the spelling mistakes in my posts are currently reduced thanks to the Mozilla Firefox Browser. This browser senses spelling mistakes and the user can easily spot out the incorrect words. Now I need not resort to time consuming HTML editing for subscripts and superscripts. These are supported by the browser when pasted in to the browser window. I could have saved a lot of time if I had used the Mozilla Firefox Browser earlier!

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) 2MR²Aω² (b) MR²Aω² (c) 2MRAω²

(d) MRAω² (e) MRAω²/2

You may be remembering the expression for maximum acceleration of a simple harmonic motion: a_max = 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 (α_max) of the cylinder is given by

α_max = a_max/R = ω²A/R.

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

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

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

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

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

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:

STATEMENT-1

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

because

STATEMENT-2

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).

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

λ_mT = constant [= 0.29 cm K, if you express λ_m in cm].

In other words, λ_m1T₁ = λ_m2T₂

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

Therefore, 7000×4200 = λ_m2×5200, from which λ_m2 =5650Ǻ, nearly.

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