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

– Thomas A. Edison

Thursday, September 27, 2007

Three AIEEE 2003 Questions on Waves

The following MCQ which appeared in AIEEE 2003 question paper is worth noting:

A metal wire of linear mass density 9.8 g/m is stretched with a tension of 10 kg wt between two rigid supports 1m apart. The wire passes at its middle point between the poles of a permanent magnet, and it vibrates in resonance when carrying an alternating current of frequency ‘n’. The frequency ‘n’ of the alternating current is

(a) 25 Hz (b) 50 Hz (c) 100 Hz (d) 200 Hz

The wire vibrates because of the magnetic force on it. When the alternating current completes one cycle, the wire completes one oscillation and hence the frequency of oscillation of the wir is the same as the frequency of the alternating current (n). Therefore we have

n = (1/2L)√(T/m) where L is the length of the segment of the wire betweenetween the supports, T is the tension and ‘m’ is the linear density (mass per unit length) of the wire. Substituting for L, T and m we have

n = (½)√[(10×9.8)/(9.8×10–3 )] = 50 Hz.

The following MCQ is a conventional type:

The displacement ‘y’ of a wave traveling in the X-direction is given by

y =10–4 sin(600t – 2x + π/3) metre,

where x is expressed in metre and t in seconds. The speed of the wave motion in ms–1 is

(a) 200 (b) 300 (c) 600 (d) 1200

It will be useful to remember that the velocity of the wave, v = Coefficient of t /Coefficient of x. So, the correct option is (b).

Now consider the following MCQ on beats, which is of the type popular among question setters:

A tuning fork of known frequency 256 Hz makes 5 beats per second with the vibrating string of a piano. The beat frequency decreases to 2 beats per second when the tension in the piano string is slightly increased. The frequency of the piano string before increasing the tension was

(a) 256 + 5 Hz (b) 256 + 2 Hz (c) 256 2 Hz (d) 256 5 Hz

Since the beat frequency before increasing the tension in the piano wire was 5 Hz, the frequency of vibration of the piano wire was (256 ± 5) Hz. When the tension in the piano wire is increased, its frequency increases. If the original frequency of the piano wire was (256 + 5) Hz, the beat frequency would have increased on increasing the tension in the piano wire. Therefore, the original frequency of the piano wire was (256 – 5) Hz .

You will find more multiple choice questions (with solution) on waves at physicsplus: Multiple Choice Questions on Waves

Wednesday, September 19, 2007

KEAM (Engineering) 2007 Questions on Elasticity

You will often find questions involving elastic potential energy in entrance examinations for admission to professional and other degree courses. Here is a question which appeared in Kerala Government Engineering Entrance 2007 test paper:

A wire of natural length L, Young’s modulus Y and area of cross section A is extended by x. Then the energy stored in the wire is given by

(a) (YA/2L)x2 (b) (YA/3L)x2 (c) (YL/2A)x2 (d) (YA/2L2)x2 (e) (A/2YL)x2

The elastic potential energy per unit volume is ½ stress×strain. Since the Young’s modulus Y = stress/strain, we can modify the above expression as (½)Y×strain2. The elastic potential energy stored in the entire wire is (½)Y×strain2 × volume of the wire = (½)Y×strain2 × AL = (½)Y×(x/L)2 × AL = (YA/2L)x2.

Here is another question which appeared in Kerala Government Engineering Entrane 2007 test paper:

The length of a rubber cord is l1 metre when the tension is 4N and l2 metre when the tension is 6N. The length when the tension is 9N is

(a) (2.5 l2 1.5 l1) m (b) (6 l2 1.5 l1) m (c) (3 l1 2 l2) m

(d) (3.5 l2 2.5 l1) m (e) (2.5 l2 + 1.5 l1) m

You can use Hooke’s law to solve this question. Since the increase in length is directly proportional to the force (tension) applied, in accordance with Hooke’s law, we have

l2 – l1 = K(6 – 4) where K is the constant of proportionality.

[Note that the increase in length from l1 to l2 is produced by the increase in tension from 4N to 6 N].

If the length of the rubber cord is l3 when the tension is 9 N, we have

l3 – l1 = K (9 – 4)

Dividing the first equation by the second, we obtain

(l2 – l1)/(l3 – l1) = 2/5, from which l3 = (2.5 l2 1.5 l1) metre.

You will find more questions on elasticity on clicking on the label ‘elasticity’ below this post or on the on the side of this page.