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

– Thomas A. Edison

Monday, November 23, 2009

EAMCET 2009 (Medical) Questions (MCQ) on Rotational Motion

The following two questions were included from rotational motion in the EAMCET 2009 (Medical) question paper. (The first question has appeared in many entrance exam question papers. It was included in the EAMCET 2009 Engineering question paper also). Here are the questions with solution:
(1) A rod of length ‘l’ is held vertically stationary with its lower end located at a position P on the horizontal plane. When the rod is released to topple about P, the velocity of the upper end of the rod with which it hits the ground is
(1) √(g/l)
(2) √(3gl)
(3) 3√(g/l)
(4) √(3g/l)
When the rod falls its gravitational potential energy mgl/2 gets converted into rotational kinetic energy of the rod. (Note that ‘m’ is the mass of the rod and initially the centre of gravity of the rod is at a height l/2 with respect to the horizontal plane).
Therefore we can write
             ½ Iω2 = mgl/2 where I is the moment of inertia of the rod about an axis passing through the end (at P) of the rod and perpendicular to the length of the rod and ‘ω’ is the angular velocity of the rod when it hits the horizontal plane.
Here I = ml2/3.
[Usually you will remember the moment of inertia of a rod about a normal axis through its middle as ml2/12. The moment of inertia about a normal axis through one end is obtained by applying the parallel axis theorem: I = ml2/12 + m(l/2)2 = ml2/3].

Substituting for I we have
             ½ (ml2/3)ω2 = mgl/2
Since ω = v/l where ‘v’ is the velocity with which the rod hits the ground, we have
             ½ (ml2/3)(v/l)2 = mgl/2
This gives v = √(3gl)

(2) A rigid uniform rod of mass M and length ‘L’ is resting on a smooth horizontal table. Two marbles each of mass ‘m’ and traveling with uniform speed ‘v’ collide with the two ends of the rod simultaneously and inelastically as shown. The marbles get stuck to the rod after the collision and continue to move with the rod. If m = M/6 and v = L ms–1, then the time taken by the rod to rotate through π/2 is
(1) 1 sec
(2) 2π sec
(3) π sec
(4) π/2 sec
Because of the collision, the rod will rotate about a normal axis through its middle with an angular velocity ω given by
             Iω = mvL/2 + mvL/2 where ‘I’ is the moment of inertia of the rod carrying the masses m and m at its ends.
[Note that we have equated the final angular momentum of the system (containing the rod and the masses) to the initial angular momentum. Before the collision the two masses have angular momentum about the central axis. These are shown on the right hand side of the above equation].
Since v = L the above equation gets modified as
             Iω = mL2
After the collision, the rod and the masses move together and the total angular momentum is given by
             Iω = [(ML2/12) + 2m(L/2)2] ω
[The first term within the square bracket above is the moment of inertia of the rod and the second term is the moment of inertia of the two masses].
From the above two equations, we have
             mL2 = [(ML2/12) + mL2/2 ] ω 
Since m = M/6 the above equation becomes
             M/6 = [(M/12) + (M/12)] ω = (M/6) ω
Therefore ω = 1 radian /sec and the time taken by the rod to rotate through π/2 radian is π/2 sec.
You will find many questions on rotational motion on this site. You can access all of them by clicking on the label ‘rotation’

You will find many useful questions with solution in this section at physicsplus and at AP Physics Resources.

Tuesday, November 17, 2009

All India Engineering/Architecture Entrance Examination 2010 (AIEEE 2010)

It’s time to apply for AIEEE 2010.
Information Bulletin containing the Application Form for applying for All India Engineering/Architecture Entrance Examination 2010 (AIEEE 2010) will be distributed from 1.12.2009 and will continue till 31.12.2009. Candidates can apply for AIEEE 2010 either on the prescribed Application Form or make application ‘online’.
Online submission of the application is possible from 16-11-2009 to 31-12-2009 at the website http://aieee.nic.in
The date of Examination is 25th April 2010.
Visit the site http://aieee.nic.in immediately for details.

You will find many old AIEEE questions (with solution) on this site. You can access all of them by typing ‘AIEEE’  in the search box at the top left of this page and clicking on the search button. Or, you may hit the enter key after feeding the search words.
You may perform a similar search for old AIEEE questions (with solution) at the site http://physicsplus.blogspot.com also.

Monday, November 02, 2009

Apply for IIT-JEE 2010

You may apply now for the Joint Entrance Examination for Admission to IITs and other Institutions (IIT-JEE 2010). The exam will be conducted on April 11th, 2010 (Sunday) as per the following schedule:
09:00 – 12:00 hrs: Paper – 1

14:00 – 17:00 hrs: Paper – 2

You can apply for IIT-JEE 2010 using either the on-line facility or the off-line facility. On-line application procedure is available from 1st November 2009 to 7th December 2009.

Off-line submission of the application using the application materials purchased from designated branches of banks (see JEE web sites of the IITs) also is possible from 16th November 2009 to 15th December 2009.

Fees required for online application are Rs. 900/- (for general category, OBC & DS students) and Rs. 450/- for female (any category), SC/ST and Physically Disabled. The fees required for offline application are Rs. 1000/- and Rs. 500 respectively.

The last date for receipt of the completed application at the IITs is 19th December, 2009 (before 17:00 hrs).

Find details at the JEE websites of the different IITs given below:
IIT Bombay: http://www.jee.iitb.ac.in

IIT Delhi: http://jee.iitd.ac.in

IIT Guwahati: http://www.iitg.ac.in/jee

IIT Kanpur: http://www.iitk.ac.in/jee

IIT Kharagpur: http://www.iitkgp.ernet.in/jee

IIT Madras: http://jee.iitm.ac.in

IIT Roorkee: http://www.iitr.ac.in/jee

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You can access all posts related to IIT-JEE (including old questions with solution) on this site by making a search for ‘IIT’ using the ‘search blog’ box provided. Similar posts can be seen at http://physicsplus.blogspot.com as well.