Apply for Joint Entrance Examination (IIT-JEE 2009) for Admission to IITs and other Institutions

The Joint Entrance Examination for Admission to IITs and other Institutions (IIT-JEE 2009) will be held on April 12, 2009 (Sunday) as per the following schedule:
Paper – 1: 09:00 hrs– 12:00 hrs

Paper – 2: 14:00 hrs– 17:00 hrs

Application materials of the Joint Entrance Examination 2009 will be issued with effect from 19^th November, 2008. On-line submission of Application will also commence on the same day.

Application materials can be purchased from designated branches of Banks and all IITs between 19.11.2008 and 24.12.2008 by paying Rs. 500/- in the case of SC/ST/PD/Female candidates, Rs.1000/- in case of male general/OBC candidates and US$ 100 in the case of candidates appearing in Dubai centre.

The last date for postal request of application materials is 16-12-2008.

The last date for receipt of the completed application at the IITs is 24^th December, 2008.

Online Submission of Application will be available between 19.11.2008 and 24.12.2008 between 8:00 hrs and 17:00 hrs. 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://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 Madras: http://jee.iitm.ac.in

IIT Roorkee: http://www.iitr.ac.in/jee

Visit the web site http://www.jee.iitb.ac.in (or any other) for more details.

Two AIPMT 2008 Questions from Thermal Physics

The following questions appeared in the All India Pre-Medical / Pre-Dental Entrance 2008 Examination:

(1) At 10º C the value of the density of a fixed mass of an ideal gas divided by its pressure is x. At 110º C this ratio is

(1) 383x/283

(2) 10x/110

(3) 283x/383

(4) x

We have PV = rT where r is the gas constant for the given mass.

Since the volume V = M/ρ where M is the mass and ρ is the density, we have

P/ρ = rT/M

Since r and M are constants for a given mass of gas, we have

(ρ₁/P₁) /(ρ₂/P₂) = T₂/T₁ where the suffix 1 is for the quantities at temperature 10º C and suffix 2 is for the quantities at temperature 110º.

It is given that (ρ₁/P₁) = x.

Therefore x/(ρ₂/P₂) = 383/283 since the temperatures are in degree kelvin.

From this (ρ₂/P₂) = 283x/383

(2) On a new scale of temperature (which is linear) and called the W scale, the freezing and boiling points of water are 39º W and 239º W respectively. What will be the temperature on the new scale, corresponding to a temperature of 239º C on the Celsius scale?

(1) 117º W

(2) 200º W

(3) 139º W

(4) 78º W

We have 0º C = 39º W and 100º C = 239º W

A temperature difference of 100º C is therefore equivalent to a temperature difference of 200º W. A temperature difference of 1º C is thus equivalent to a temperature difference of 2º W.

Therefore, a temperature 39º C is equal to [39 + (39×2)]º W = 117º W