Monday, December 31, 2007

Two Questions on Specific Heat of Gases

Questions involving specific heats of gases have been discussed in some of the posts on this site. By clicking on the label ‘molar specific heat’ below this post, you will find some of the useful posts in this context. In particular, go through the post dated 1st September 2006.

Let us discuss two more questions involving specific heats of gases:

(1) An ideal diatomic gas in a container is heated so that half of the gas molecules dissociate into atoms. The molar specific heats (at constant volume) of the sample of the gas in the container before and after heating are C1 and C2. Then C1/C2 is

(a) 3/7 (b) 5/7 (c) 7/9

(d) 9/10 (e) 15/11

The important point to note here is that on dissociation, each particle (diatomic molecule) with 5 degrees of freedom produces two particles (individual atoms) with 3 degrees of freedom. Therefore you have to use the value Cv = (5/2)R for the undissociated molecule and the value Cv = (3/2)R for the atoms formed on dissociation (Cv is the molar specific heat at constant volume and R is the universal gas constant).

Assuming that there are ‘n’ moles of the diatomic gas initially, the number of moles after dissociation is (3/2)n, with (n/2)×2 = n moles of atoms and n/2 moles of molecules.

The molar specific heat (at constant volume) before dissociation,

C1 = (5/2)R, appropriate for a diatomic gas.

The molar specific heat (at constant volume) after dissociation,

C2 = (Heat supplied for increasing the temperature through 1 K) /Number of moles

= [n×(3/2)R + (n/2)×(5/2)R] / (3/2)n = (11/6)R

Therefore, C1/C2 = (5/2)R/(11/6)R = 15/11.

(2) An ideal diatomic gas is heated at constant pressure. What is the fraction of the heat energy supplied, which increases the internal energy of the gas?

(a) 2/3 (b) 3/5 (c) 5/7

(d) 7/9 (e) 1/2

This is a simple question but it has appeared in various entrance test papers.

When you heat a gas at constant pressure, the gas expands, thereby doing work. When one mole of a diatomic gas is considered, the increase in the internal energy on heating it through 1 K is equal to its molar specific heat (molar heat capacity) at constant volume, which is (5/2) R where R is the universal gas constant. The total heat energy supplied in this case for increasing the internal energy and for doing the external work is the molar specific heat at constant pressure, which is (7/2) R. The fraction required in the question is therefore [(5/2)R] /[(7/2)R] = 5/7

Monday, December 24, 2007

Merry Christmas!

Thursday, December 20, 2007

Multiple Choice Questions (MCQ) involving Inductance

(1) A long straight solenoid of cross section area10 cm2 has 10 turns per cm. A short 50 turn coaxial coil of cross section area 2 cm2 is fixed inside the solenoid at the middle. The mutual inductance of the solenoid and the coil is (in micro henry)

(a) 400 (b) 40 (c) 4π (d) π (e) ) 0.4π

Mutual inductance between an infinitely long straight solenoid and a short secondary coil placed coaxially inside at the middle is given by

M = μ0nNA where μ0 is the magnetic permeability of free space (or air), n is the number of turns per metre of the solenoid, N is the total number of turns in the secondary coil and A is the cross section area of the secondary coil. (If the secondary coil is outside the solenoid, the cross section area of the solenoid will appear in place of the area of the secondary coil).

[Note that the mutual inductance as given by the above expression is the magnetic flux (linked with the secondary coil) per unit current in the solenoid: M = NBA where B is the magnetic field (μ0n×1) produced by unit current in the solenoid]

Substituting for the known quantities in the expression for mutual inductance, we have

M = 4π×10–7×1000 ×50 ×(2×10–4) = 4π×10–6 H = μH.

(2) The secondary coil in the above question is moved with uniform velocity coaxially over a distance of 5 cm in 2 seconds when a current of 2 ampere flows through the solenoid. The emf induced in the coil during the motion is

(a) 200π μV (b) 100π μV (c) 20π μV (d) 10π μV (e) zero

No emf will be induced in the coil since the same flux is linked with the coil throughout its motion and there is no flux change.

(3) In question No.1, suppose the solenoid carries a current of 4 A. If this current is switched off in 100 ms, the emf induced in the secondary coil will be

(a) 0.2π μV (b) 0.4π μV (c) π mV (d) 0.08π mV (e) 0.16π mV

We have ε = –M(dI/dt) = = 4π×10–6 [(4 – 0)/(100×10–3)] = 160π×10–6 volt = 0.16π mV.

4. A battery of emf V volt is connected in series with a coil of inductance L and resistance R at the instant t = 0. The current in the circuit when t = 2τ where τ is the time constant of the circuit is (base of natural logarithm = e)

(a) (V/R)[1 (1/e2)]

(b) (V/R (1/e2)

(c) (V/R)[1 (1/e)]

(d) (V/R)(1/e)]

(e) 2e(V/R)

The expression for the growth of current (with time) in an LR circuit is

I = I0[1– e–Rt/L]

where I0 is the final steady (maximum possible) current which is equal to V/R.

The time constant of the LR circuit is L/R. After 2 time constants, the current will be

I = I0[1– e–R× 2(L/R)/L] = (V/R) [1– e– 2] = (V/R)[1– (1/e2)]

You will find more questions (with solution) at AP Physics Resources: AP Physics C - Multiple Choice Questions on Inductance

Friday, December 14, 2007

Two Questions on Force of Buoyancy

How does a balloon filled with helium rise in air? The weight of the balloon with helium is less than the force of buoyancy on the balloon and hence the balloon rises. Here is a question high lighting this principle:

A balloon of mass 6 kg is to be filled with helium so as to lift an instrument weighing 24 kg. The minimum volume of helium to be filled in the balloon is nearly (Density of air = 1.29 kg m–3, density of helium = 0.18 kg m–3 approximately)

(a) ) 22 m3 (b) 27 m3 (c) 30 m3 (d) ) 133 m3 (e) 166 m3

The balloon will start rising when the force of buoyancy on the balloon just exceeds the weight of the balloon with helium and the instrument. The force of buoyancy is equal to the weight of displaced air. Therefore, the minimum volume V of helium to be filled is given by

(6 + 0.18 V + 24)g = 1.29 Vg.

This gives V = 27 m3 approximately.

[The weight of air displaced by the instrument and the empty balloon is negligible and therefore not considered here].

Now consider the following MCQ:

A block of cork floats on the surface of water (in a container) with 30 cm outside water. If the system is placed on the moon where there is no atmosphere and the value of acceleration due to gravity is approximately one sixths that on the earth, the portion outside water will be

(a) 5 cm

(b) slightly less than 5 cm

(c) slightly less than 30 cm

(d) slightly greater than 30 cm

(e) slightly greater than 5 cm

If the acceleration due to gravity has a non-zero value, the exposed portion will remain unchanged since the weight of the floating body and the weight of displaced fluid are directly proportional to the value of the acceleration due to gravity. So, the exposed portion (or, the immersed portion) will remain the same whether you take the system to the moon or mars or inside a coal mine, when you consider the effect of ‘g’ only. But, you have to consider the effect of the atmosphere also. Since there is no atmosphere on the moon, there is no force of buoyancy due to the atmosphere and the net force of buoyancy (which is due to water in the container and due to the air above) is slightly reduced. Therefore, the block of cork gets immersed a little more and the exposed portion is slightly less than 30 cm.

[Note that if ‘g’ is zero, there is no weight and therefore no question of floatation].

You will find similar questions (with solution) at AP Physics Resources: Fluid Mechanics- Questions on Force of Buoyancy

Thursday, December 06, 2007

Two Questions on Oscillations

Today we will discuss two multiple choice questions on simple harmonic motion:
(1) A simple pendulum of period 2 s has a small bob of mass 50 g. The amplitude of oscillation of the bob is 10 cm and it is at a height of 45 cm from the ground in its mean position. While oscillating, the string breaks just when the bob is in its mean position. The horizontal distance R `from the mean position where the bob will strike the ground is nearly

(a) 35.2 cm (b) 23 cm (c) 15.3 cm

(d)12.4 cm (e) 9.4 cm

The angular frequency (ω) of oscillation of the pendulum is given by

ω = 2π/T = 2π/2 = π rad/s.

The bob has maximum velocity (vmax) in the mean position and is given by

vmax = ωA where A is the amplitude.

Therefore, vmax = π × 0.1 = 0.1 π ms–1

On getting detached from the string, the bob moves like a projectile shot horizontally from a height of 0.45 m with a velocity of 0.1 π ms–1. Its time of flight (t) is obtained from the vertical displacement of 0.45 m:

0.45 = 0 + ½ gt2. (Note that the initial vertical velocity is zero and the vertical acceleration is g, which we may take as 10 ms–2). This gives t2 = 0.09 so that t = 0.3 s.

The horizontal distance covered by the bob during this time is 0.1 π × 0.3 = 0.094 m = 9.4 cm.

[Note that the mass of the bob does not come into the picture and it just serves as a distraction].

Now, consider the following MCQ :

A large horizontal surface moves up and down simple harmonically with an amplitude of 1 cm. If a mass of 3 kg (which is placed on the surface) is to remain continually in contact with it, the maximum frequency of the SHM will be

(a) 5 Hz (b) 2 Hz (c) 8 Hz (d) 10Hz (e) 15 Hz

The mass will remain in contact with the surface if the maximum acceleration produced in the simple harmonic motion does not exceed the acceleration due to gravity (g).

Therefore, we have

ω2A = g where ω is the angular frequency of the SHM.

From this ω = √(g/A) = √(10/ 0.01) = 10×√10.

The maximum frequency (linear) of oscillations is therefore given by

n = ω/2π = (10×√10)/2π = 5 Hz.

[Note that the mass of the body placed on the surface does not come into the picture and it just serves as a distraction].

Saturday, December 01, 2007

All India Engineering/Architecture Entrance Examination 2008 (AIEEE 2008)

Application Form and the Information Bulletin in respect of the All India Engineering/Architecture Entrance Examination 2008 (AIEEE 2008) to be conducted by Central Board of Secondary Education (CBSE) on 27th April 2008 (Sunday) are being distributed from 30.11.2007 (Friday) and will continue till 5.1.2008 (Saturday) through selected Branches of Syndicate Bank, Regional Offices of CBSE and designated institutions.

Important dates in this regard are given below:


Date of Examination



Sale of AIEEE Information Bulletin containing Application Form

30.11.2007 to 05.01.2008


Online submission of application on website

30.11.2007 to 05.01.2008


Last date for


Receipt of request for Information Bulletin and Application Form by Post atAIEEE Unit,CBSE,PS1-2,Institutional Area,IP Extension,Patparganj,Delhi-110092



Sale of Information Bulletin at designated branches of Syndicate Bank, Regional Offices of the CBSE and designated institutions



Online submission of applications



Receipt of complete applications “by post” including Registration Forms with Bank Draft at AIEEE Unit, CBSE, PS1-2,Institutional Area,IP Extension,Patparganj,Delhi-110092



Date of dispatch of Admit Card

10.03.2008 to 31.03.2008


Issue/dispatch of duplicate admit card(or request only with fee of Rs. 50/- + postal charges of Rs. 30/- extra for out station candidate.

11.04.2008 to 27.04.2008 (By Hand)

11.04.2008 to 21.04.2008 (By Post)


Dates of Examination

PAPER – 1 27.04.2008 (0930-1230 hrs)
PAPER – 2 27.04.2008 (1400-1700 hrs)

Paper-1 is for B.E./B.Tech and Paper-2 is for B.Arch/B.Planning.

By visiting the site complete information in this regard can be obtained. Visit the site without any delay.