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

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.

Two Kerala Engineering Entrance 2007 Questions from Electrostatics

When you calculate the electrostatic potential energy of a pair of charged particles, it is enough to consider the energy of one particle in the electric field of the other. If you have three charged particles (as for example A, B & C), you will have three pairs (AB, AC & BC) and the total energy will be the algebraic sum of the energies due to these three pairs. But, if you have four charged particles (A, B, C & D), you will have six pairs (AB, AC, AD, BC, CD & BD).

Consider the following MCQ which appeared in KEAM (Engineering) 2007 question paper:

The electrostatic potential energy between proton and electron separated by a distance 1 Ǻ is

(a) 13.6 eV (b) 27.2 eV (c) 14.4 eV (d) 1.44 eV (e) 28.8 eV

This is a simple question since we have just one pair of charges. The electrostatic potential energy (EPE) of the system is (1/4πε₀)(q₁q₂/r) where q₁ and q₂ are the charges of proton and electron.

Therefore, EPE = (1/4πε₀)(1.6×10^–19)²/10^–10 joule

= 9×10⁹×1.6×10^–19/10^–10 electron volt

= 14.4 eV.

[When you find the EPE of more than two charged particles, you will have to be careful to consider the sign of the charges since the total EPE is the algebraic sum of the EPE’s of all the pairs of charges].

The following simple MCQ also is from the KEAM (Engineering) 2007 question paper:

The work done in moving an alpha particle between two points having potential difference 25 volt is

(a) 8×10^–18 J (b) 8×10^–19 J (c) 8×10^–20 J (d) 8×10^–10 J (e) 4×10^–18 J

This question is intended to test your basic understanding of the work done in moving a charge in an electric field. An electric field will exist only if there is a potential gradient and the wok done in moving a charge is the product of the charge and the potential difference. The alpha particle has two protons in it, each carrying positive charge of 1.6×10^–19 coulomb.

Therefore work done = (2×1.6×10^–19)×25 joule = 8×10^–18 J

Imagination is more important than knowledge.

– Albert Einstein   

Two Questions (MCQ) on Electric Potential

(1) An infinite number of point charges each equal to +Q coulomb are arranged at random around a point P such that the distances of the charges from the point P are 1m, 2m, 4m, 8m, 16m,…….etc... The electric potential at P is

(a) zero (b) infinite (c) negligibly small

(d) Q/2πε₀ (e) Q/4πε₀

Note that the electric potential is a scalar quantity. Therefore, the direction of the location of the charge does not matter and the potentials simply add up. The resultant potential (V) at P is given by

V = (Q/4πε₀) × [(1/1) + (1/2) + (1/4) + (1/8) + ………]

The infinite series within the square bracket yields a value equal to 2 so that V = Q/2πε_0.

(2) A thin spherical conducting shell of radius R has a charge +Q. Another point charge –q is placed at the centre of the shell. The electrostatic potential at a point P distant R/2 from the centre of the shell is

(a) (Q/4πε₀R) – (2q/4πε₀R) (b) (q/4πε₀R) – (2Q/4πε₀R)

(c) – (2q/4πε₀R) (d) (Q/4πε₀R) (e) zero
The electrostatic potential at any point within the shell due to the charge Q on the shell is constant and is equal to Q/4πε₀R.

The potential at distance R/2 due to the charge –q placed at the centre of the shell is – q/4πε₀(R/2) = – 2q/4πε₀R.

Therefore, the net potential (V) at the point P distant R/2 from the centre of the shell is given by

V = (Q/4πε₀R) – (2q/4πε₀R), given in option (a).

[Note that positive charges will produce positive potential where as negative charges will produce negative potential].

You can find more posts on electrostatics by clicking on the label electrostatics below this post or on the left side of this page.

You will find similar multiple choice questions with solution at physicsplus also.

Two KEAM (Engineering) 2007 Multiple Choice Questions on Gravitation

The following questions which appeared in KEAM 2007 question paper are typical and of the type repeatedly asked in various entrance examinations:

(1) The change in potential energy when a body of mass ‘m’ is raised to a height nR from earth’s surface is (R = radius of the earth)

(a) mgR(n/n–1) (b) mgR (c) mgR(n/n+1)

(d) mgR(n²/n² +1) (e) mgR/n

This question appears often in entrance question papers. There may be slight change in the wording (as for example, “what is the work done in lifting a body of mass m, from the earth’s surface, through a height nR?”).

Gravitational potential energy of a mass ‘m’ at a height ‘h’ is given by

U = – GMm/(R+h) where M is the mass of the earth and G is the Gravitational constant.

Since h = nR, the change in potential energy is

– GMm/(R+nR) – (–GMm/R) = (GMm/R)[1 – 1/(1+n)] = (GMm/R)[n/(1+n)]

Since g = GM/R², the change in potential energy becomes mgRn/(n+1), given in option (c).

(2) The escape velocity of a body on the surface of the earth is 11.2 km/s. If the mass of the earth is doubled and its radius is halved, the escape velocity becomes

(a) 5.6 km/s (b) 11.2 km/s (c) 22.4 km/s

(d) 44.8 km/s (e) 67.2 km/s

The escape velocity (v_e) of a body on the surface of the earth is given by

v_e = √(2GM/R)

Therefore we have √(2GM/R) = 11.2 km/s

If the mass (M) of the earth is doubled and its radius (R) is halved, the escape velocity becomes √(2×4GM/R) = 2×√(2GM/R) = 22.4 km/s.