## Monday, November 26, 2007

### Two All India Institute of Medical Sciences (AIIMS) Questions on Optics

The following questions which appeared in All India Institute of Medical Sciences (AIIMS) 2005 entrance question paper for admitting students to the MBBS Degree course are simple as usual. They are meant for checking your knowledge and understanding of fundamentals.

(1) The apparent depth of water in a cylindrical water tank of diameter 2R cm is reducing at the rate of x cm/minute when water is being drained out at a constant rate. The amount of water drained in c.c. per minute is (n1= refractive index of air, n2 = refractive index of water)

(a) x π R2 n1/n2 (b) x π R2 n2/n1 (c) 2 π R n1/n2 (d) π R2x

Since the refractive index is the ratio of real depth to the apparent depth, we have

Real depth = Apparent depth × refractive index.

Therefore, the rate at which the real depth is decreasing = xn2/n1 cm per minute.

The amount of water drained in c.c. per minute is therefore equal to x π R2 n2/n1, given in option (b).

(2) A telescope has an objective lens of focal length 200 cm and an eye piece with focal length 2 cm. If the telescope is used to see a 50 m tall building at a distance of 2 km, what is the length of the image of the building formed by the objective lens?

(a) 5 cm (b) 10 cm (c) 1 cm (d) 2cm

At the first glance this question may seem to be one involving the magnification produced by a telescope; but, this is quite simple since you are asked to consider the objective only.

The objective will produce the image of the building at the focus (which is at 2 m from the lens) and hence from the expression for magnification (M) we have

M = Distance of image/ Distance of object = Height of image/ Height of object

so that 2/ 2000 = x/50 where ‘x’ is the height of image in metre.

Therefore, x = 2×50/2000 = 0.05 m = 5 cm.

## Thursday, November 15, 2007

### AP Physics Exam Resources- Two Questions (MCQ) on Moment of Inertia

Multiple choice questions discussed on this site will be useful for entrance examinations for admission to various degree courses including professional courses. They will be suitable for those preparing for AP Physics Examination, as can be judged by working out the following two questions:

(1) Three circular discs of radii R, R and 2R are cut from a metallic sheet of uniform thickness and the smaller discs are placed symmetrically on the larger disc as shown in the figure. If the mass of a smaller disc is M, the moment of inertia of the system about an axis at right angles to the plane of the discs and passing through the centre of the larger disc is (a) 5MR2 (b) 7MR2 (c) 9MR2

(d) 11MR2 (e) 12MR2

The mass of the larger disc is 4M (since its radius is twice that of the smaller disc) and its moment of inertia is (4M)×(2R)2/2 = 8MR2.

The moment of inertia of each smaller disc about the axis passing through the centre of the larger disc (as given by the parallel axis theorem) is MR2/2 + MR2 = 3MR2/2.

Note that the moment of inertia is a scalar quantity. Therefore, the total moment of inertia of the system of three discs is 8MR2 + 2×3MR2/2 = 11MR2.

(2) A small body of regular shape made of iron rolls up with an initial velocity ‘v’ along an inclined plane. It reaches a maximum height of 7v2/10g where ‘g’ is the acceleration due to gravity. The body is a

(a) ring (b) disc (c) solid sphere

(d) hollow sphere (e) cylindrical rod

The initial kinetic energy of the body is ½ Mv2 + ½ I ω2 where M is its mass and I is its moment of inertia about its axis (of rolling). The first term is its translational kinetic energy and the second term is its rotational kinetic energy.

Since the entire kinetic energy is used in gaining gravitational potential energy, we have

½ Mv2 + ½ I ω2 = Mgh where ‘h’ is the maximum height reached.

Therefore, ½ Mv2 + ½ I v2/R2 = Mg×7v2/10g, from which

I = (2/5)MR2.

The body is therefore a solid sphere.

## Sunday, November 11, 2007

### Joint Entrance (2008) Examination for Admission to IITs and other Institutions—(IIT-JEE 2008)

Application form and Prospectus of the Joint Entrance Examination 2008 (JEE-2008) for admission to the seven IITs at Bombay, Delhi, Guwahati, Kanpur, Kharagpur, Madras and Roorkee as well as to Institute of Technology, Banaras Hindu University, Varanasi and the Indian School of Mines University, Dhanbad, will be issued with effect from November 23, 2007 (Friday). On-line submission of Application will also commence on the same day.

Application form along with the Information Brochure can be purchased from any one of the selected branches of Canara Bank/State Bank of India/ Union Bank/ Punjab National Bank (visit site: http://www.iitkgp.ernet.in/jee/advt.html for the list of branches) or from any of the IITs between 23.11.2007 (Friday) and 4.1.2008 (Friday). by paying Rs. 500/- in case of SC/ST/Female candidates and Rs. 1000/- in case of all other candidates by way of cash. SC/ST/Female candidates will get the materials in a WHITE coloured envelope while the other candidates will get the materials in a BLUE coloured envelope.

Application Form and Prospectus by Post from IITs:

The request for Application Form and Prospectus by Post from any of the IITs will also be accepted from November 23, 2007. Application Form and Prospectus can be obtained by post from any of the IITs by sending a request along with two self- addressed slips and a Demand Draft for Rs.500/- (in case of SC/ST/Female applicants) and for RS.1000/- in case of other applicants, payable to the “CHAIRMAN, JEE” of the respective IIT, at the corresponding city.[For example, those applying to IIT, Madras, should take the DD in favour of “Chairman, JEE, IIT Madras” and payable at Chennai]. Such requests will be accepted from 23.11.2007(Friday) to 21.12.2007 (Friday).

Submission of filled up Application Forms: Duly completed Application Form, refolded only along the original fold should be inserted in the envelope supplied, along with the attested copy of the 10th Class Pass, or Equivalent Examination Certificate, and the Acknowledgement Card. These items should not be stapled or pasted with the Application form. Irrespective of the Bank/Institute from where the Application has been obtained, they should be re-submitted along with the contents by Registered Post/Speed Post only to the IIT located in the Zone where the centre of the examination chosen by the applicant is located. They may also submitted in person at the JEE office of the IIT concerned.

The last date for receipt of the completed application at the IITs is 17:00 hours on January 4, 2008 (Friday).

Online Submission of Application: This facility will be available between 23.11.2007 (Friday) and 5 pm on 28.12.2007 (Friday) through the JEE websites of the different IITs. The JEE websites of the different IITs are given below:
IIT Bombay: http://www.jee.iitb.ac.in
IIT Delhi: http://www.jee.iitd.ac.in
IIT Guwahati: http://www.iitg.ac.in/jee
IIT Kanpur: http://www.iitk.ac.in/jee
IIT Kharagpur: http://www.iitkgp.ernet.in/jee
IIT Roorkee: http://www.iitr.ac.in/jee

Schedule of the Examination: The examination will be held on April 13, 2008 (Sunday) as per the following schedule:
09:00 – 12:00 hrs Paper - 1
14:00 – 17:00 hrs Paper – 2

## Thursday, November 08, 2007

### Multiple Choice Questions on Elasticity

Here are three questions (MCQ) on elasticity which appeared in Kerala Engineering and Medical Entrance 2004 Examination question papers:

(1) Wires A and B are made from the same material. A has twice the diameter and three times the length of B. If the elastic limits are not reached when each is stretched by the same tension, the ratio of energy stored in A to that in B is

(a) 2:3 (b) 3:4 (c) 3:2 (d) 6:1 (e) 12:1

The work (W) done in increasing the length of a wire or rod by ‘l’ by applying a force ‘F’ is given by

W = ½ Fl

[ Here is the proof for the above: The work dW done for increasing the length by dl is

F×dl. The total work done for increasing the length by ‘l’ is ∫F×dl where the limits of

integration are zero and ‘l’. Since the Young’s modulus, Y = (F/A)(L/l) where A is the

area of cross section and L is the length of the wire, we have F = YAl/L. The total work

done is therefore 0l (Yal/L)dl = ½ (Yal2/L) = ½ (YAl/L)×l = ½ Fl ]

The energy stored (which is equal to the work done) in a wire is therefore directly proportional to the increase in length. The ratio of energy stored is therefore W1/W2 = l1/l2 where l1 and l2 are the increases in length of the wires. But l1 = FL1/A1Y and l2 = FL2/A2Y so that

W1/W2 = l1/l2 = (L1/L2)×(A2/A1) = (3/1)×(1/4) = 3/4 [Option (b)].

(2) A wire of cross section 4 mm2 is stretched by 0.1 mm. How far will a wire of the

same material and length but of area 8 mm2 stretch under the action of the same

force?

(a) 0.05 mm (b) 0.01 mm (c) 0.15 mm

(d) 0.2 mm (e) 0.25 mm

This question as well as the previous one appeared in Kerala Medical Entrance 2004 question paper.

Since the increase in length is inversely proportional to the area of cross section of the wire, the correct option is 0.05 mm.

(3) Compressibility of water is 5×10–10 m2/N. The change in volume of 100 ml of water subjected to 15×10–6 Pa pressure will be

(a) no change (b) increase by 0.75 ml (c) increase by 1.5 ml

(d) decrease by 1.5 ml (e) decrease by 0.75 ml

This question appeared in Kerala Engineering Entrance 2004 question paper. You must definitely remember the expression for bulk modulus ‘B’ as

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

Compressibility is the reciprocal of bulk modulus. Therefore we have

1/(5×10–10) = (15×10–10 × 100×10–6)/dV, from which

dV = 0.75×10–6 m3 = 0.75 ml.

Since an increase in pressure produces a decrease in volume, the correct option is (e).