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

– Thomas A. Edison

Tuesday, June 19, 2007

KEAM 2007(Medical) Question on Electric Dipole

The following MCQ which appeared in Kerala Medical Entrance 2007 question paper is straight forward, requiring your knowledge of the expression for torque on a dipole:

An electric dipole consists of two opposite charges each 0.05 μC separated by 30 mm. The dipole is placed in an uniform external electric field of 106 NC–1. The maximum torque exerted by the field on the dipole is

(a) 6×10–3 Nm (b) 3×10–3 Nm (c) 15×10–3 Nm

(d) 1.5×10–3 Nm (e) 9×10–3 Nm

The torque on a dipole of moment p in an electric field E is the vector product p×E. The magnitude of the torque (τ) is pEsinθ where θ is the angle between dipole moment vector and the electric field vector. The maximum value of the torque is pE (when θ = 90º).

Note that p = q×2a where ‘q’ is the charge and 2a is the separation between the charges.

Therefore, maximum torque = (0.05×10–6×30×10–3) ×106 = 1.5×10–3 Nm.

Wednesday, June 13, 2007

Electrostatic Potential- An IIT-JEE 2007 MCQ

The following MCQ which appeared in IIT-JEE 2007 question paper is meant for checking whether you have the correct idea about electric potential and potential difference:

Positive and negative point charges of equal magnitude are kept at (0,0, α/2) and(0,0,α/2), respectively. The work done by the electric field when another positive point charge is moved from (α,0,0) to (0,α,0) is

(a) positive

(b) negative

(c) zero

(d) depends on the path connecting the initial and final positions

You should have a three dimensional mental view of the Cartesian coordinate system to work out this problem in no time. Even without drawing a figure, some of you will be able to work this out. But you may draw a quick figure to help you.

In the figure shown, the positions of the equal positive and negative charges and the movement of the third positive charge from P to Q are indicated. The path shown is straight, but it can be any curve, since the work done is independent of the path in an electrostatic field and is dependent on the initial and final positions only.

Since P and Q are equidistant from the positive and negative charges, the resultant potentials at P as well as Q due to these charges is zero. Therefore, no work need be done for moving the third charge from P to Q [Option (c)].

Friday, June 08, 2007

Kerala Medical Entrance 2007 Questions on Simple harmonic Motion

The following two questions were included from simple harmonic motion, in KEAM (Medical) 2007 question paper:

(1) A simple pendulum has a time period T in vacuum. Its time period when it is completely immersed in a liquid of density one-eighth of the density of material of the bob is

(a) √(7/8) T (b) √(5/8) T (c) √(3/8) T (d) √(8/7) T (e) √(8/5) T

The case of a pendulum immersed in a liquid was briefly discussed in the post dated 16th August 2006.

The mass of the bob is vρ where ‘v’ is the volume of the bob and ‘ρ’ is the density of the material of the bob. The apparent weight of the bob is v(ρ-σ)g where ‘σ’ is the density of the liquid so that the net value of the downward acceleration of the bob is v(ρ-σ)g /vρ = (1- σ/ρ)g = (1- 1/8)g = (7/8)g.

The period of oscillation of the pendulum suspended in vacuum (or, as usual, in air), is given by

T = 2π√(L/g) where 'L’ is the length and ‘g’ is the acceleration due to gravity.

When the pendulum bob is immersed in the liquid, the net acceleration is (7/8)g and the period becomes

Tliquid = 2π√(8L/7g) = √(8/7) T.

You may see the earlier post on simple pendulum containing other interesting questions, by clicking on the label ‘simple pendulum’ below this post or by making use of the search facility at the top of this page.

(2) A body of mass 20 g connected to a spring of constant K executes SHM with a frequency (5/π) Hz. The value of spring constant is

(a) 4 Nm–1 (b) 3 Nm–1 (c) 2 Nm–1 (d) 5 Nm–1 (e) 2.5 Nm–1

This is a standard direct question. The frequency of oscillation of a spring-mass system is given by

N = (1/2π)√(K/m) where K is the spring constant and ‘m’ is the mass.

Therefore, 5/π = (1/2π)√(K/0.02), from which

K = 2 Nm–1

Monday, June 04, 2007

Online Physics Text Books from NCERT

One advice I used to give my students when I had occasion to teach the pre-degree students was to have a copy of NCERT Physics Text Book. Even now the best Physics text books available for 11th and 12th classes are those published by NCERT. The authors have taken pains to explain the fundamentals in detail and it is this aspect that makes the books different from nany other books in the market.

NCERT Text Books are now available online. For downloading the books you will find a link “e-Books from NCERT” on the right side of this page or you will find it here.

Saturday, June 02, 2007

KEAM 2007 (Medical) Question on Young’s Double Slit

Here is a question on Young’s double slit which appeared in Kerala Medical Entrance 2007 question paper:

In a two slit experiment, with monochromatic light, fringes are obtained on a screen placed at some distance from the slits. If the screen is moved by 5×10–2 m towards the slits, the change in fringe width is 10–3 m. Then the wave length of light used is (given that distance between the slits is 0.03 mm)

(a) 4000 Ǻ (b) 4500 Ǻ (c) 5000 Ǻ (d) 5500 Ǻ (e) 6000 Ǻ

It is enough to write two equations for the fringe width and solve for the wave length λ. If β is the initial fringe width, we have

β = λD/d where ‘D’ is the initial distance of the screen from the double slit and ‘d’ is the distance between the slits.

When D is decreased, β is decreased. Therefore, when the screen is moved by 5×10–2 m towards the slits, we have

β –10–3 = λ(D – 5×10–2 )/d

Subtracting this equation from the first one, we obtain

10–3 = λ×(5×10–2)/d

Since d = 0.03 mm = 0.03×10–6 m,

λ = (0.03×10–6)/(5×10–2) m = 6000×10–10 m = 6000 Ǻ. [Option (e)].

Friday, June 01, 2007

Mozilla Firefox Browser

It is my belief that the spelling mistakes in my posts are currently reduced thanks to the Mozilla Firefox Browser. This browser senses spelling mistakes and the user can easily spot out the incorrect words. Now I need not resort to time consuming HTML editing for subscripts and superscripts. These are supported by the browser when pasted in to the browser window. I could have saved a lot of time if I had used the Mozilla Firefox Browser earlier!