## Saturday, November 21, 2009

### Questions (MCQ) on Nuclear Physics

The following three questions from nuclear physics are simple but are useful in your preparation for entrance tests:
(1) Two radioactive samples S1 and S2 have half lives 3 hours and 7 hours respectively. If they have the same activity at a certain instant t, what is the ratio of the number of atoms of S1 to the number of atoms of S2 at the instant t?
(a) 9 : 49
(b) 49 : 9
(c) 3 : 7
(d) 7 : 3
(e) 1 : 1
If The number of atoms present at the instant t is N, we have
N = N0eλt where N0 is the initial number, e is the base of natural logarithms and λ is the decay constant.
Therefore, activity, dN/dt = λ N0eλt = λN
If N1 and N2 are the number of atoms of S1 and S2 respectively when the activities are the same, we have
λ1N1 = λ2N2 from which N1/N2 = λ2/λ1
But the decay constant λ is related to the half life T as T = 0.693/λ.
Therefore, N1/N2 = λ2/λ1 = T1/T2 = 3/7 [Option (c)].
(2) A nucleus ZXA has mass M kg. If Mp and Mn denote the mass (in kg) of proton and neutron respectively, the binding energy in joule is
(a) [ZMp + (A – Z)MnM]c2
(b) [ZMp + ZMnM]c2
(c) M – ZMp – (A – Z)Mn
(d) [M– ZMp – (A – Z)Mn]c2
(e) [AMn M]c2
Total mass of the Z protons is ZMp. Since the total number of nucleons is A, the total number of neutrons is (A – Z) and the total mass of the neutrons is (A – Z)Mn.
The mass defect M is the difference between the total mass of the nucleons (protons and neutrons together) and the mass of the nucleus: M =[ZMp + (A – Z)MnM].
Therefore, binding energy.= Mc2 where ‘c’ is the speed of light in free space.
Thus binding energy = [ZMp + (A – Z)MnM]c2
(3) If the aluminium nucleus 13Al27 has nuclear radius of about 3.6 fm, then the tellurium nucleus 52Te125 will have radius approximately equal to
(a) 3.6 fm
(b) 16.7 fm
(c) 8.9 fm
(d) 6.0 fm.
(e) 4.6
The nuclear radius R is given by
R = R0A1/3 where R0 is a constant (equal to 1.2×10–15 m, nearly) and A is the mass number of the nucleus.
If Rl and R2 are the radii of the given Al and Te nuclei respectively, we have
Rl = R0 (27)1/3 = 3R0 and
R2 = R0 (125)1/3 = 5R0
Dividing, Rl/R2 = 3/5
Therefore, R2 = 5R1/3 = (5×3.6)/3 fm = 6 fm.
By clicking on the label ‘nuclear physics’ below this post, you can access all posts related to nuclear physics on this site.
You can find useful posts in this section here.

## Tuesday, November 17, 2009

### Apply for All India Engineering/Architecture Entrance Examination 2010 (AIEEE 2010)

Application Form and the Information Bulletin in respect of the All India Engineering/Architecture Entrance Examination 2010 (AIEEE 2010) to be conducted on 25-4-2010 will be distributed from 1.12.2009 and will continue till 31.12.2009. Candidates can apply for AIEEE 2010 either on the prescribed Application Form or make application ‘Online’.
Online submission of the application is possible from 16-11-2009 to 31-12-2009 at the website http://aieee.nic.in

You may visit the site http://aieee.nic.in for details and information updates.

You will find many old AIEEE questions (with solution) on this blog. You can access all of them by typing ‘AIEEE’ in the search box at the top left of this page and then hitting the enter key (or clicking the search button).

## Tuesday, November 10, 2009

### IIT-JEE 2010 – Candidates to get Performance Score Cards

Leading news papers have flashed a welcome news item which will be of great interest to candidates appearing for IIT-JEE 2010. Here is the gist of the news item:

The JEE Board will issue performance cards specifying the marks and the ranks secured by candidates who will be appearing for IIT-JEE 2010. The performance cards can be considered as certificates by many other institutions wanting to give admission to JEE candidates. A decision for issuing such performance cards has been taken by the Joint Admission Board (JAB) for IITs in its meeting on August 23. The performance score cards will be issued two weeks after the results are declared. The marks of the students will be published on the IIT-JEE websites just a week after the results are declared. The next JAB meeting in April 2010 will give the final approval to the scheme.

## Thursday, November 05, 2009

### MCQs on Magnetism including EAMCET 2009 (Medical) Question

Some multiple choice questions on magnetism have already been posted on this site. You can access them by clicking on the label ‘magnetism’ below this post. Today we will discuss a few more multiple choice questions on magnetism. (1) The period of oscillation of a magnetic needle in a magnetic field is T. If an identical bar magnetic needle is tied at right angles to it to form a cross (fig), the period of oscillation in the same magnetic field will be

(a) 21/4T

(b) 21/2T

(c) 2T

(d) T√3

(e) T/2

The period of oscillation (T) of the single magnetic needle is given by

T = 2π√(I/mB) where ‘I’ is the moment of inertia of the magnetic needle about the axis of rotation, ‘m’ is the magnetic dipole moment of the needle and ‘B’ is flux density of the magnetic field.

When two magnetic needles are tied together to form a cross, the moment of inertia becomes 2I and the magnitude of the magnetic dipole moment becomes √(m2 + m2) = m√2.

[Note that magnetic dipole moment is a vector quantity. Two identical vectors (each of magnitude m) at right angles will yield a resultant magnitude m√2].

The resultant magnetic moment will be directed along the bisector of the angle between the axes of the individual magnets since the magnets are identical. In the absence of a deflecting torque, the resultant dipole moment vector will align along the applied magnetic field B. On deflecting from this position, the system will oscillate with period T1 given by

T1 = 2π√(2I/mB√2) = 2π√(I√2/mB) = 21/4T (2) Three identical magnetic needles each L metre long and of dipole moment m ampere metre are joined as shown without affecting their magnetisation. At points B and C unlike poles are in contact. The dipole moment of this system is

(a) m

(b) 2m

(c) 3m

(d) 3m/2

(e) 5m/2

The distance (AD) between the ends of the compound magnet is 2L. Since the pole strength is m/L, the dipole moment of the compound magnet is (m/L)2L = 2m

(3) A magnet of length L and moment M is cut into two halves (A and B) perpendicular to its axis. One piece A is bent into a semicircle of radiur R and is joined to the other piece at the poles as shown in the figure below: Assuming that the magnet is in the form of a thin wire initially, the moment of the resulting magnet is given by

(1) M/2π

(2) M/π

(3) M(2 + π)/2π

(4) Mπ/(2 + π)

The above question appeared in EAMCET 2009 (Medicine) question paper.

The distance between the poles of the resulting magnet is (L/2) + 2R

Since the semicircular portion of radius R is made of the magnetised wire of length L/2, we have L/2 = πR so that R = L/2π and 2R = L/π

Therefore, length of the resulting magnet (L/2) + (L/π)

The pole strength (p) of the magnet is given by

p = M/L

Therefore, the dipole moment of the resulting magnet = Pole strength×Length = (M/L)[ (L/2) + (L/π)] = M(2 + π)/2π

## Monday, November 02, 2009

### Apply for Joint Entrance Examination for Admission to IITs and other Institutions- (IIT-JEE 2010)

The Joint Entrance Examination for Admission to IITs and other Institutions (IIT-JEE 2010) will be held on April 11th, 2010 (Sunday) as per the following schedule:
09:00 – 12:00 hrs: Paper – 1

14:00 – 17:00 hrs: Paper – 2

You can apply either on-line or off-line for IIT-JEE 2010. On-line application procedure is available from 1st November 2009 to 7th December 2009.

Off-line submission of the application using the application materials purchased from designated branches of banks also is possible from 16th November 2009 to 15th December 2009. Designated branches of banks can be found by visiting the web sites of the IIT’s.

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