Thursday, October 27, 2011

IIT-JEE 2011 and IIT-JEE 2010 Questions on Doppler Effect

"I am a friend of Plato, I am a friend of Aristotle, but truth is my greater friend."
– Sir Isaac Newton

Questions on Doppler Effect were discussed on this site earlier. You can access them by clicking on the label ‘Doppler effect’ below this post or by trying a search for ‘Doppler effect’ using the search box provided on this page. Questions on Doppler effect appear frequently in Entrance examination question papers. Today we will discuss two questions in this section Here is the first question which appeared in IIT-JEE 2011 question paper as a single answer type multiple choice question:

(1) A police car with a siren of frequency 8 kHz is moving with uniform velocity 36 km/hr towards a tall building which reflects the sound waves. The speed of sound in air is 320 m/s. The frequency of the siren heard by the car driver is

(A) 8.50 kHz

(B) 8.25 kHz

(C) 7.75 kHz

(D) 7.50 kH

The general expression for the apparent frequency (n’) produced due to Doppler effect is

n’ = n(v+w–vL)/(v+w–vS) where ‘v’ is the velocity of sound, ‘w’ is the wind velocity ‘vL’ is the velocity of the listener, ‘vS’ is the velocity of the source of sound and ‘n’ is the actual frequency of the sound emitted by the source. Note that all the velocities in this expression are in the same direction and the source is behind the listener. In other words, the listener is moving away from the source and the source is moving towards the listener.

[It will be helpful to remember that if the source moves towards the listener or the listener moves towards the source, the apparent frequency increases. If they move away, the apparent frequency decreases].

In the present case the source of sound which produces the Doppler effect is the reflected image of the siren. (The car driver will not detect any change in the frequency of the direct sound from the siren since he is moving along with the siren). The reflected image (source) of the siren moves towards the car driver with a speed of 36 km/hr and the car driver (listener) moves towards the reflected image (source) with the same speed.

Thus in the expression for apparent frequency we have ro substitute

n = 8 kHz, vL = 36 km/hr = 36×(5/18) ms–1 = –10 ms–1, w = 0, vS = + 10 ms–1.

Therefore, n’ = 8×(320 + 10)/(320 – 10) = 8×(33/31) kHz = 8.5 kHz.

The following question appeared in IIT-JEE 2010 question paper as an integer type question (in which the answer is a single-digit integer, ranging from 0 to 9):

(2) A stationary source is emitting sound at a fixed frequency f0, which is reflected by two cars approaching the source. The difference between the frequencies of sound reflected from the cars is 1.2% of f0. What is the difference in the speeds of the cars (in km per hour) to the nearest integer? The cars are moving at constant speeds much smaller than the speed of sound which is 330 ms–1.

This is similar to question no.1 above. The difference between the frquencies of the sound reflected from the cars is given by

n1 n2’ = [f0(v+v1)/(v–v1)] – [f0(v+v2)/(v–v2)] where v1 and v2 are the speeds of the two cars.

Therefore (n1 n2’)/f0 = [(v+v1)/(v–v1)] – [(v+v2)/(v–v2)]

The quantity on the left hand side of the above equation is 1.2/100 as given in the question.

Therefore, 0.012 = [(v+v1) (v–v2) (v+v2) (v–v1)] / [(v–v1)(v–v2)]

Or, 0.012 = (2 vv1 2 vv2)/ [(v–v1)(v–v2)] = 2v(v1 –v2)/v2 since (v–v1) (v–v2) ≈ v

[The cars are moving at speeds much smaller than the speed of sound].

Therefore we have

0.012 = 2(v1 –v2)/v from which (v1 –v2) = 0.006×330 ms–1

The difference in the speeds of the cars in km per hour is 0.006×330×(18/5) = 7.128

The answer, which is the nearest integer, is 7.

Sunday, October 16, 2011

Joint Entrance Examination 2012 (IIT-JEE 2012) for Admission to IITs

Notification in respect of the Joint Entrance Examination 2012 (IIT-JEE 2012) for Admission to IITs has been issued. This Joint Entrance Examination is for admission to the under graduate programmes in the IITs at Bhubaneswar, Bombay, Delhi, Gandhinagar, Guwahati, Hyderabad, Indore, Kanpur, Kharagpur, Madras, Mandi, Patna, Rajasthan, Roorkee and Ropar as well as at IT-BHU Varanasi and ISM Dhanbad.

Important Dates for IIT JEE 2012:

Online application process: October 31 to December 10, 2011
Sale of off-line application forms: November 11 to December 5, 2011.

Last date for receiving application forms at IITs: Thursday, December 15, 2011

Date of Exam of IIT JEE 2012:

Sunday, April 8, 2012 Paper 1: 09:00 to 12:00 hrs;

Paper 2 : 14:00 to 17:00 hrs

All information and information updates regarding online application and offline application and other important details about IIT JEE 2012 can be obtained from the following websites:

IIT Bombay: http://www.jee.iitb.ac.in
IIT Guwahati: http://www.iitg.ernet.in/jee/
IIT Delhi: http://jee.iitd.ac.in
IIT Kharagpur: http://www.iitkgp.ernet.in/jee
IIT Kanpur: http://www.jee.iitk.ac.in
IIT Roorkee: http://jee.iitr.ernet.in/

You will find many earlier IIT-JEE questions with solution on this blog. Click on the label ‘IIT’ below this post or try a search for ‘IIT’ using the search box provided on this page.

Sunday, October 02, 2011

Multiple Choice Questions on Unusual Combinations of Capacitors

"It is unwise to be too sure of one's own wisdom. It is healthy to be reminded that the strongest might weaken and the wisest might err."

– Mahatma Gandhi

Today we will consider a few questions on unconventional combinations of capacitors which occasionally find place in entrance examination question papers conducted for admitting students to medical, engineering and other degree courses.

(1) Four parallel metal plates are arranged in air as shown, with the same separation ‘d’ between neighbouring plates. If the area (of one surface) of each plates is A, the capacitance between the terminals T1 and T2 is

(a) 4ε0A/d

(b) ε0A/d

(c) 2ε0A/d

(d) 3ε0A/d

(e) ε0A/3d

Let us number the plates starting from the top as 1, 2, 3, and 4 respectively. The lower surface of plate No.1 and the upper surface of plate No.2 with air in between makes a capacitor of capacitance ε0A/d.

The lower surface of plate No.2 and the upper surface of plate No.3 with air in between makes another capacitor of capacitance ε0A/d.

The lower surface of plate No.3 and the upper surface of plate No.4 with air in between makes a third capacitor of capacitance ε0A/d.

This arrangement therefore makes three capacitors in parallel. The capacitance value of each capacitor is ε0A/d so that the parallel combined value which appears between the terminals T1 and T2 is 3ε0A/d [Option (d)].

(2) If the connection of metal plates in the above question is modified as shown in the adjoining figure, the capacitance between terminals T1 and T2 is

(a) ε0A/2d

(b) 3ε0A/2d

(c)0A/d

(d) 0A/d

(e) 0A/3d

In this case the capacitor at the centre (having capacitance ε0A/d) appears directly across the terminals T1 and T2. The upper and lower capacitors (each of value ε0A/d) are connected in series and this series combination having effective capacitance ε0A/2d also appears between the terminals T1 and T2. Therefore, the effective capacitance between T1 and T2 is ε0A/d + ε0A/2d = 3ε0A/2d [Option (b)].

(3) If a fifth plate is added to the combination shown in question No.1, as shown in the figure, what is the effective capacitance between T1 and T2?

(a) ε0A/4d

(b) ε0A/5d

(c) 0A/d

(d) 0A/5d

(e) 0A/d

Once you answer question No.1, this question becomes quite simple. You can easily see that there are four identical capacitors (each having capacitance ε0A/d) connected in parallel. Therefore, the effective capacitance between T1 and T2 is 4ε0A/d [Option (e)].

You can find a few more questions (with solution) in this section here.