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 = N₀e^–λt where N₀ is the initial number, e is the base of natural logarithms and λ is the decay constant.

Therefore, activity, dN/dt = – λ N₀e^–λt = 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 _ZX^A has mass M kg. If M_p and M_n denote the mass (in kg) of proton and neutron respectively, the binding energy in joule is

(a) [ZM_p + (A – Z)M_n – M]c²

(b) [ZM_p + ZM_n – M]c²

(c) M – ZM_p – (A – Z)M_n

(d) [M– ZM_p – (A – Z)M_n]c²

(e) [AM_n – M]c²

Total mass of the Z protons is ZM_p. 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)M_n.

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 =[ZM_p + (A – Z)M_n – M].

Therefore, binding energy.= ∆Mc² where ‘c’ is the speed of light in free space.

Thus binding energy = [ZM_p + (A – Z)M_n – M]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 = R₀A^1/3 where R₀ is a constant (equal to 1.2×10^–15 m, nearly) and A is the mass number of the nucleus.

If R_l and R₂ are the radii of the given Al and Te nuclei respectively, we have

R_l = R₀ (27)^1/3 = 3R₀ and

R₂ = R₀ (125)^1/3 = 5R₀

Dividing, R_l/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.

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).

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.

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) 2^1/4T

(b) 2^1/2T

(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) = 2^1/4T

(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

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 11^th, 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 1^st November 2009 to 7^th December 2009.

Off-line submission of the application using the application materials purchased from designated branches of banks also is possible from 16^th November 2009 to 15^th 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

IIT Madras: http://jee.iitm.ac.in

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

The online application fees are Rs. 900/- (for general category, OBC & DS students) and Rs. 450/- [for female (any category), SC/ST and Physically Disabled]. The respective offline application fees are Rs. 1000/- and Rs. 500

The last date for receipt of the completed application at the IITs is 19^th December, 2009 (Saturday), before 17:00 hrs.

Visit one of the web sites given above (such as http://jee.iitm.ac.in) for more details.

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Some old IIT-JEE questions with solution can be seen on this site. You can access all posts related to IIT-JEE by searching for ‘IIT’, making use of the search box provided.

All India Pre-Medical / Pre-Dental Entrance Examination -2010 (AIPMT 2010)

Central Board of Secondary Education (CBSE), Delhi has invited applications in the prescribed form for All India Pre-Medical / Pre-Dental Entrance Examination -2010 as per the following schedule for admission to 15% of the merit positions for the Medical/Dental Courses of India

1. Preliminary Examination – 3^rd April, 2010 (Saturday) 10 AM to 1 PM

2. Final Examination – 16^th May, 2010 (Sunday) 10 AM to 1 PM

The Preliminary Examination will consist of one paper containing 200 objective type questions (four options with one correct answer) from Physics, Chemistry and Biology.

The Final Examination is only for those who qualify in the Preliminary Examination. It will consist of one paper containing 120 objective type questions (four options with one correct answer) from Physics, Chemistry and Biology.

Candidate can apply for the All India Pre-Medical/Pre-Dental Entrance Examination either offline or online as explained below:

Offline (On prescribed application form)

Offline submission of Application Form may be made using the prescribed application form. The Information Bulletin and Application Form costing Rs.800 for General & OBC Category Candidates and Rs.450/- for SC/ST Category Candidates can be obtained against cash payment from any of the specified branches of Canara Bank/ Regional Offices of the CBSE. Find details at www.aipmt.nic.in.

Online

Online submission of application may be made by accessing the Board’s website www.aipmt.nic.in. Candidates are required to take a print of the Online Application after successful submission of data. The print out of the computer generated application, complete in all respect as applicable for Offline submission should be sent to the Deputy Secretary (AIPMT Unit), CBSE, Shiksha Kendra, 2, Community Centre, Preet Vihar, Delhi-110 301 by Speed Post/Registered Post in such a way that it should reach the Board on or before the last date. Fee of Rs.800/- for General and OBC Category Candidates and Rs.450/- for SC/ST category candidates may be remitted in the following ways :

1. By credit card, or

2. Through Demand Draft in favour of the Secretary, CBSE, Delhi drawn on any Nationalized Bank payable at Delhi.

Instructions for Online submission of Application Form are available on the website www.aipmt.nic.in.

The last date of receipt of Application Form for both offline and online is 04.12.2009.

In case the application is submitted online, printout of the computer generated form complete in all respects as applicable for offline submission must reach The Deputy Secretary (AIPMT Unit), CBSE, Shiksha Kendra, 2, Community Centre, Preet Vihar, Delhi-110301 on or before the last date. A grace period of 15 days will be allowed from the last date of submission of the application to the candidates belonging to remote areas viz. Mizoram, Assam, Meghalaya, Arunachal Pradesh, Manipur, Nagaland, Tripura, Sikkim, Lahaul and Spiti Districts and Pangi sub-division of Chamba District of Himachal Pradesh, Andaman & Nicobar Islands and Lakshadweep.

Visit the web site www.aipmt.nic.in for all details and information updates.

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You will find some old AIPMT questions with solution on this site. You can access these questions by searching for ‘AIPMT’, making use of the search box provided on the site.

EAMCET (Engineering) 2009 Questions on Photoelectric Effect

Questions on photoelectric effect are usually interesting. One or two questions in this section often find place in entrance examination question papers. I give below two questions on photoelectric effect which were included in the EAMCET (Engineering) 2009 question paper:

(1) The work function of a certain metal is 3.31×10^–19 J. Then the maximum kinetic energy of photo electrons emitted by incident photons of wave length 5000 Ǻ is (given h = 6.62×10^–34 Js, c = 3×10⁸ ms^–1, e = 1.6×10^–19 coulomb)

(1) 2.48 eV

(2) 0.41 eV

(3) 2.07 eV

(4) 0.82 eV

It will be useful (as I have mentioned on earlier occasions) to remember that a photon of wave length 1000 Ǻ has energy of 12.4 eV. In other words, the product of the wavelength in Angstrom (Ǻ) and the energy in electron volt (eV) of any photon is 12400 units. Therefore, the energy of 5000 Ǻ photon is 12400/5000 = 2.48 eV.

[You can calculate the energy (E) of the photon in joule using the equation, E = hc/λ and then convert it into electron volt by dividing it by 1.6×10^–19 as 1 eV = 1.6×10^–19 joule. But you will have to spend too much of your valuable time for this].

The work function of the metal is 3.31×10^–19 J = (3.31×10^–19)/(1.6×10^–19) eV = 2.07 eV nearly.

The maximum kinetic energy of photoelectrons emitted = Energy of incident photon – Work function = 2.48 eV – 2.07 eV = 0.41 eV.

(2) A photon of energy ‘E’ejects a photo electron from a metal surface whose work function is W₀. If this electron enters into a magnetic field of induction ‘B’ in a direction perpendicular to the field and describes a circular path of radius ‘r’, then the radius ‘r’ is given by (in the usual notation)

(1) √[2m(E – W₀)/eB]

(2) √[2m(E – W₀) eB]

(3) (1/mB)√[2e(E – W₀)]

(4) (1/eB)√[2m(E – W₀)]

The kinetic energy (k) of the photo electron is given by

k = E – W₀

The momentum p of the photo electron is related to its kinetic energy as

k = p²/2m,

Therefore, momentum p =√(2mk) = √[2m(E – W₀)]

The radius ‘r’ of the circular path followed by the electron in the magnetic field is given by

r = mv/qB

[This is obtained by equating the magnetic force to the centripetal force:

qvB = mv²/r]

Since mv = p, we have r = p/qB = p/eB = (1/eB)√[2m(E – W₀)].

You can access all posts related to photoelectric effect on this blog by clicking on the label ‘photoelectric effect’ below this post.

You will find more questions (with solution) at AP Physics Resources.

AIPMT 2009 Multiple Choice Questions on Work, Energy & Power

How strange is the lot of we mortals! Each of us is here for a brief sojourn; for what purpose we know not, though sometimes sense it. But we know from daily life that we exist for other people first of all for whose smiles and well-being our own happiness depends.

–Albert Einstein

Three questions from the section ‘work, energy & power’ were included in the AIPMT 2009 question paper. Here are those questions with solution:

(1) A block of mass M is attached to the lower end of a vertical spring. The spring is hung from the ceiling and has force constant value k. The mass is released from rest with the spring initially unstretched. The maximum extension produced in the length of the spring will be:

(1) 2 Mg/k

(2) 4 Mg/k

(3) Mg/2k

(4) Mg/k

If x is the maximum extension produced, we have

Mgx = ½ kx², on equating the decrease in the gravitational potential energy of the mass M to the increase in the elastic potential energy of the spring.

Therefore, x = 2Mg/k

(2) A body of mass 1 kg is thrown upwards with a velocity 20 m/s. It momentarily comes to rest after attaining a height of 18 m. How much energy is lost due to air friction? (g = 10 ms^–2)

(1) 30 J

(2) 40 J

(3) 10 J

(4) 20 J

The initial kinetic energy of the 1 kg mass is given by

½ mv² = ½ ×1×20² = 200 J.

The gravitational potential energy of the 1 kg mass at its maximum height is given by

Mgh = 1×10×18 = 180 J.

The energy lost due to air friction is therefore equal to 200 – 180 = 20 J.

(3) An engine pumps water continuously through a hose. Water leaves the hose with a velocity v and m is the mass per unit length of the water jet. What is the rate at which kinetic energy is imparted to water?

(1) mv²

(2) ½ mv²

(3) ½ m²v²

(4) ½ mv³

Mas of water flowing out per second through the hose is mv. Therefore, kinetic energy imparted per second to the water is ½ ×(mv)×v² = ½ mv³ [Option (3)].

You will find useful posts in this section here.