Two Questions (MCQ) involving Kinetic Energy and Potential Energy

The following question appeared in Kerala Engineering Entrance 2007 question paper:

A simple pendulum is released from A as shown. If m and l represent the mass of the bob and the length of the pendulum, the gain in kinetic energy at B is

(a) mgl/2

(b) mgl/√2

(c) (mgl√3)/2

(d) 2mgl/√3

(e) mgl

From position A, the bob of the pendulum has fallen through a distance l cos 30º (fig.). Therefore, the loss of potential energy is mg l cos 30º = (mgl√3)/2.

Therefore, the gain in kinetic energy by the bob (by the law of conservation of energy) is (mgl√3)/2.

Here is another question which will be useful in the present context:

The bob of a simple pendulum has mass 0.2 kg. The bob is drawn aside so that the string is horizontal and is released. When it swings through 60º its kinetic energy is 0.5 J. The kinetic energy when the sting becomes vertical will be

(a) √3 J

(b) 1/√3 J

(c) √3/2 J

(d) 2/√3 J

(e) 1 J

You can use the figure used with the previous question for working out the present question as well since the angle turned is 60º. The loss of potential energy by the bob on swinging through 60º is mg l cos30º itself. [Or, you may take it as mg l sin60º].

Since the loss of potential energy is equal to the gain of kinetic energy, we have

mg l cos30º = 0.5 J

Or, (mgl√3)/2 = 0.5 J.

The kinetic energy when the string becomes vertical will be mgl since the bob falls through a distance l and loses its entire initial potential energy mgl.

From the above equation mgl = 1/√3 J [Option (b)].

[Note that the mass of the bob given in the question just serves as a distraction].

You will find some useful questions in this section here.

IIT-JEE 2009 Multiple Choice Question (Single Answer Type) on Elastic Collision

The following question which appeared in IIT-JEE 2009 question paper is based on the principle that in an elastic collision between two particles of the same mass, the velocities get interchanged: 

Two small particles of equal masses start moving in opposite directions from a point A in a horizontal circular orbit. Their tangential velocities are v and 2v respectively as shown in the figure. Between collisions, the particles move with constant speeds. After making how many elastic collisions, other than that at A, these two particles will again reach the point A?

(a) 4

(b) 3

(c) 2

(d) 1    

Particle P₁, after traversing one third (AB) of the circular track in the anticlockwise direction, will collide with particle P₂ at position B. This collision occurs when the particle P₂  has traversed two thirds (ACB) of the circular path (since the speed of P₂ is twice that of P₁). 

         Since the collision is elastic and the particles are of equal masses, their velocities are interchanged. P₁ now travels in the clockwise direction with speed 2v and P₂ travels in the anticlockwise direction with speed v. The second collision takes place at position C. (BAC is two thirds of the circular path while BC is one third of the path). The second collision at C results in the reversal of the velocities and P₁ now travels in the anticlockwise direction with speed v where as P₂ travels in the clockwise direction with speed 2v. Therefore, the third collision will occur at A.

Since the third collision at A is not to be counted, the answer is 2 [Option (c)].

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Now suppose we modify the above question as follows:

Two small particles of equal masses start moving in opposite directions from a point A in a horizontal circular frictionless track. Their speeds are v and 3v respectively as shown in the figure. After making how many elastic collisions, other than that at A, these two particles will again reach the point A ?

(a) 4

(b) 3

(c) 2

(d) 1

[You may use the figure shown with the question above, replacing the velocity 2v with 3v]

You can easily arrive at the answer which is 3 [Option (b)].

Apply for Entrance Examination for Admission to Medical/ Agriculture/ Veterinary/ Engineering/ Architecture Degree Courses 2010 (KEAM 2010), Kerala

The Commissioner for Entrance Examinations, Govt. of Kerala, has invited applications for the Entrance Examinations for admission to the following Degree Courses in various Professional Colleges in the State for 2010-11.

(a) Medical: (i) MBBS (ii) BDS (iii) BHMS (iv) BAMS (v) BSMS

(b) Agriculture: (i) BSc. Hons. (Agriculture) (ii) BFSc. (Fisheries) (iii) BSc. Hons. (Forestry)

(c) Veterinary: BVSc. & AH

(d) Engineering: B.Tech. [including B.Tech. (Agricultural Engg.)/B.Tech. (Dairy Sc. & Tech.) courses under the Kerala Agricultural University]

(e) Architecture: B.Arch.

Dates of Exam:

Engineering Entrance Examination (For Engineering courses except Architecture)

19.04.2010 Monday 10.00 A.M. to 12.30 P.M. Paper-I : Physics & Chemistry.

20.04.2010 Tuesday 10.00 A.M. to 12.30 P.M. Paper-II: Mathematics.

Medical Entrance Examination (For Medical, Agriculture and Veterinary Courses)

21.04.2010 Wednesday 10.00 A.M. to 12.30 P.M. Paper-I : Chemistry & Physics.

22.04.2010 Thursday 10.00 A.M. to 12.30 P.M. Paper-II: Biology.

Application form and Prospectus will be distributed from 07.12.2009 to 06.01.2010 through selected branches of Post Offices in Kerala and outside the State. The amount towards the fee of application (Rs. 700/- for general candidates and Rs.350 for SC/ST candidates) is to be remitted in cash at the Post Offices. To find the list of post offices, selected as sales centres, and details of additional fee of Rs.8500/- in the case of candidates opting Dubai as the centre of exam, visit the site http://www.cee-kerala.org/.

Online submission of application is possible in the case of candidates claiming no reservation benefit and those belonging to Non-Keralite category. See details at http://www.cee-kerala.org/.

Last Date for receipt of Application by CEE: 06-01-2010 (Wednesday) - 5 PM

Candidates seeking admission to B.Arch. course should also submit their application to the CEE. There is no state level Entrance Examination for this purpose. These candidates should write the National Aptitude Test for Architecture and should forward the NATA score and mark list of the qualifying examination to the CEE on or before 05-06-2010.

You will find complete details and information updates at http://www.cee-kerala.org/

If you want to see earlier KEAM questions discussed on this site, type in ‘Kerala’ in the search box at the top left of this page and strike the enter key or click on the search button.

Multiple Choice Questions on Diodes [Including EAMCET 2009 (Engineering) and AIEEE 2006 Questions]

Questions from electronics, especially those involving the use of diodes are generally simple at the level expected of you. Here are a few questions on diodes:

(1) Currents flowing in each of the following circuits A and B respectively are

(1) 1A, 2 A

(2) 2 A, 1 A

(3) 4 A, 2 A

(4) 2 A, 4 A

This question appeared in EAMCET 2009 (Engineering) question paper.

In circuit A both diodes are forward biased and hence the circuit reduces to two 4 Ω resistors connected across the 8 V battery. Since the parallel combined value of the two resistors is 2 Ω, the current delivered by the battery is 8 V/2 Ω = 4 A.

In circuit B one diode is forward biased and the other diode is reverse biased and hence the circuit reduces to just one 4 Ω resistor connected across the 8 V battery. The current delivered by the battery is therefore 8 V/4 Ω = 2 A. The correct option is (3).

The following questions [No. (2) and (No. (3)] were included in AIEEE 2006 question paper:

(2) In the following, which one of the diodes is reverse biased?

You should note that all potentials are with respect to the ground. Therefore the diode in circuit (1) is reverse biased.

[In circuit (2) the anode of the diode is at a higher positive potential compared to its cathode and hence it is forward biased. In circuit (3) the cathode of the diode is at a higher negative potential and hence it is forward biased. In circuit (4) the cathode of the diode is at a negative potential and hence it is forward biased].

(3) The circuit has two oppositely connected ideal diodes in parallel. What is the current flowing in the circuit?

(1) 1.33 A

(2) 1.71 A

(3) 2.00 A

(4) 2.31 A

Since the diode D₁ is reverse biased, no current will flow through D₁ and the 3 Ω resistor. The current delivered by the battery is limited by the 4 Ω and the 2 Ω resistors only and is equal to 12 V/(4+2)Ω = 2 A.

The following question is meant for checking your grasp of the behaviour of semiconductor diodes:

(4) In the circuit shown the diodes used are silicon rectifier diodes which require a forward bias of 0.7 volt for appreciable conduction. Their leakage current is negligible. The internal resistance of the battery is insignificant. The potential difference between the terminals A and B is very nearly equal to

(a) 6 V

(b) 3 V

(c) 5.3 V

(d) 0.7 V

(e) 0 V

The upper diode is reverse biased and can be ignored. The lower diode which is connected across the terminals A and B is forward biased and hence keeps the voltage across A and B at 0.7 V [Option (d)].

[You can use the voltage drop across a forward biased diode as a small reference voltage in electronic circuits just as you use the relatively larger breakdown voltages of reverse biased zener diodes].