Two Questions (MCQ) on Elasticity

Have you ever thought why a tree does not grow beyond a limiting height? It is the breaking stress of the tree that limits its height. Here is a question that high lights this point:

A tree has a breaking stress of 4.5×10⁵ Nm^–2. If the density of the tree is 900 kg m^–3, the maximum height up to which it can grow is (assuming that the tree is cylindrical in shape)

(a) 100 m (b) 75 m (c) 50 m (d) 30 m (e) 25 m

The weight of the tree is AHρg where A is the area of cross section, H is the height, ρ is the density and ‘g’ is the acceleration due to gravity.

Maximum stress will be developed at the base of the tree and will be equal to AHρg/A = Hρg.

Since the breaking stress is 4.5×10⁵ Nm^–2, we can write

Hρg = 4.5×10⁵ Nm^–2 so that H = 4.5×10⁵ Nm^–2/ρg = 4.5×10⁵ Nm^–2/9000 nearly.

This works out to 50 m.

Now, consider the following MCQ involving compressibility:

A solid sphere made of copper having compressibility K is placed in an air chamber and the pressure of air is reduced by P. The fractional change in the radius of the sphere will be

(a) PK (b) 3PK (c) P/3K (d) K/3P (e) PK/3

Compressibility (K) is the reciprocal of bulk modulus (B) and we have

B = –P/(dV/V) where dV is the change in volume V due to a pressure change P. The negative sign indicates that a decrease in pressure will produce a increase in volume.

In terms of compressibility, the above equation can be written as

B = 1/K = –P/(dV/V) so that the fractional change in volume is given by

dV/V = PK.

But V = (4/3) πR³ where R is the radius of the sphere so that dV/V = 3 dR/R. The fractional change in radius is therefore given by dR/R = (dV/V)/3 = PK/3.

You may find all posts on elasticity on this site by clicking on the label ‘elasticity’ below this post or on the side of this page.

CBSE announces dates of All India Pre-Medical/Pre-Dental Entrance Examination (AIPMT) 2008

Central Board of Secondary Education (CBSE), Delhi has announced the dates of All India Pre-Medical/Pre-Dental Entrance Examination (AIPMT) 2008 for admission to 15% of the total seats for Medical/Dental Courses in all Medical/Dental colleges run by the Union of India, State Governments and Municipal or other local authorities in India except in the States of Andhra pradesh and Jammu&Kashmir. The dates of the Examination are:

(1) Preliminary Examination : 6^th April 2008 (Sunday)

(2) Final Examination : 11^th May 2008 (Sunday)

Candidates can apply for the All India Pre-Medical/Pre-Dental Entrance Examination in the following two ways:-

(i) Online submission

Online submission of application may be made by accessing the website www.cbse.nic.in from 16.10.2007 (10.00 AM) to 26.11.2007 (5.00 PM). Candidates are required to take a print out of the Online Application after successful submission of data. The print out of the computer generated application, complete in all respect as applicable for Offline submission should be sent to the Deputy Secretary (AIPMT), Central Board of Secondary Education, Shiksha Kendra, 2, Community Centre, Preet Vihar, Delhi-110 301 by Speed Post/Registered Post only. Candidate should pay the examination fee of Rupees 400/- for General Category and Rupees 200/- for SC/ST Category through a Demand Draft in favour of the Secretary, Central Board of Secondary Education, Delhi drawn on any nationalized bank payable at Delhi. Instructions for Online submission of Application Form will be made available on the website www.cbse.nic.in.

(ii) Offline submission

Offline submission of Application Form may be made on the prescribed Application Form. The Information Bulletin and Application Form costing Rs.400/- for General Category candidates and Rs.200/- for SC/ST candidates inclusive of examination fee can be obtained against Cash Payment from designated branches of Canara Bank/ Regional Offices of the CBSE from 16-10-2007 to 26-11-2007. The details of Banks are given in the Admission Notice which is available on CBSE website www.cbse.nic.in.

Designated branches of Canara Bank in Kerala are:

KOTTAYAM	P.B. No.122, K.K.Road, Kottayam-686 001
QUILANDY	Fasila Buidling, Main Road, Quilandy-673 305
TRIVANDRUM	Ist Floor, Ibrahim Co. Bldg., Challai, Trivandrum-695 023
TRIVANDRUM	Plot no.2, PTP Nagar Trivandrum-695 038
TRIVANDRUM	TC No.25/1647, Devaswom Board Bldg. M.G. Road, Trivandrum-695 001
ERNAKULAM	Shenoy’s Chamber, Shanmugam Road, Ernakulam, Cochin-682 031
CALICUT	9/367-A, Cherooty Road, Calicut-673 001
TRICHUR	Trichur Main Ramaray Building, Round South, Trichur-680001
QUILON	Maheshwari Mansion, Tamarakulam, Quilon-691 001
PALGHAT	Market Road, Big Bazar, 20/68, Ist floor,Palghat-678 014.

The Information Bulletin and Application Form can also be obtained by Speed Post/Registered Post by sending a written request with a Bank Draft/Demand Draft for Rs.450/- for General Category and Rs.250/- for SC/ST Category payable to the Secretary, Central Board of Secondary Education, Delhi along with a Self Addressed Envelope of size 12” x 10”. The request must reach the Deputy Secretary (AIPMT), CBSE, 2, Community Centre, Preet Vihar, Delhi-110 301 on or before 15-11-2007. The request should be super scribed as “Request for Information Bulletin and Application Form for AIPMT, 2008”.

Completed Application Form is to be dispatched by Registered Post/Speed Post only. Last Date for receipt of completed Application Forms for both Offline and Online in CBSE is 28-11-2007.

You can obtain complete information at www.cbse.nic.in

Visit the site regularly for information updates.

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