Multiple Choice Questions on Dimensions of Physical Quantities

The world is a dangerous place, not because of those who 

do evil, but because of those who look on and do nothing. 

–Albert Einstein 

Today we will discuss a few questions on units and dimensions:

(1) Van der Waals equation of state for real gases is [P + (a/V²)](V – b) = RT where P is the pressure, V is the volume of one mole of gas, R is the universal gas constant, T is the temperature and a and b are constants. The dimensional formula of a is

(a) [M⁰L⁶T⁰]

(b) [ML^–1T^–2]

(c) [ML⁵T^–2]

(d) [ML^–2T^–2]

(e) [ML³T^–2]

This is a question popular among question setters and has appeared many times in various entrance test papers. Since a/V² appears as added to P, the dimensions of a/V² must be the same as those of P. Therefore, ‘a’ has the dimensions of PV².

Note that pressure P is force per unit ares and hence the dimensional formula for P is [ML^–1T^–2]. The dimensional formula for V² is [L⁶]. Therefore, the dimensional formula for ‘a’ is [ML⁵T^–2].

(2) The dimensional formula for Stefan’s constant is

(a) [MT^–3K^–4]

(b) [ML²T^–2K^–4]

(c) [ML²T^–2]

(d) [MT^–2L⁰]

(e) [MT⁴L⁰]

This question appeared in Kerala Engineering Entrance (KEAM) 2009 question paper. You will find similar questions in other entrance test papers also.

We have E = σT⁴ where E is the energy radiated per second from unit area, σ is Stefan’s constant and T is the temperature of the black body.

Therefore, σ = E/T⁴

Note that energy has dimensional formula [ML²T^–2]. Sine E is the energy radiated per second from unit area, the dimensional formula for E is [MT^–3].

Therefore, the dimensional formula for Stefan’s constant σ is [MT^–3K^–4].

(3) Pick out the correct dimensions in length of the quantity μ₀ε₀ from the following:

(a) 2

(b) 1

(c) – 1

(d) – 2

(e) zero

You know that the speed ‘c’ of electromagnetic waves in free space is given by

c = 1/√(μ₀ε₀)

Therefore μ₀ε₀ = 1/c² and the dimensional formula for μ₀ε₀ is [L^–2T²]. The quantity μ₀ε₀ therefore has – 2 dimensions in length [Option (d)].

(4) Which of the following is dimensionless?

(a) Stress

(b) Gas constant

(c) Frequency of sound

(d) Efficiency of heat engine

(e) Thermal conductivity

The correct option is (d) since the efficiency is the ratio of the output power to the input power.

Applications invited for Kerala Entrance Examinations for Admission to Medical/ Agriculture/ Veterinary/ Engineering/ Architecture Degree Courses 2010 (KEAM 2010)

The Commissioner for Entrance Examinations, Govt. of Kerala, has invited applications for the Entrance Examinations for admission to the following Degree Courses in various Professional Colleges in Kerala for 2010-2011.

(a) Medical (i) MBBS (ii) BDS (iii) BHMS (iv) BAMS (v) BSMS

(b) Agriculture (i) BSc. Hons. (Agriculture) (ii) BFSc. (Fisheries) (iii) BSc. Hons. (Forestry)

(c) Veterinary BVSc. & AH

(d) Engineering B.Tech. [including B.Tech. (Agricultural Engg.)/B.Tech. (Dairy Sc. & Tech.) courses under the Kerala Agricultural University]

(e) Architecture B.Arch.

[Candidates aspiring for B.Arch. course should also submit their application to the CEE. Even though there is no state level Entrance Examination for this purpose, these candidates should write the National Aptitude Test for Architecture and should forward the NATA score and mark list of the qualifying examination to the CEE on or before 05.06.2010].

Dates of Exam:

Engineering Entrance Examination (For Engineering courses except Architecture)

19.04.2010 Monday 10.00 A.M. to 12.30 P.M. Paper-I : Physics & Chemistry.

20.04.2010 Tuesday 10.00 A.M. to 12.30 P.M. Paper-II: Mathematics.

Medical Entrance Examination (For Medical, Agriculture and Veterinary Courses)

21.04.2010 Wednesday 10.00 A.M. to 12.30 P.M. Paper-I : Chemistry & Physics.

22-04.2010 Thursday 10.00 A.M. to 12.30 P.M. Paper-II: Biology.

Sale of Application Form through selected post offices: From  07-12-2009 to 06-01-2010

To see the list of selected post offices, visit http://www.cee-kerala.org/
Last Date for Receipt of Application by CEE: 06-01-2010 (Wednesday), before 5 P.M.

Facility for online submission of application in the case of non-reservation category and Non-Keralite category is available.

Find details at http://www.cee-kerala.org/

In addition to the multiple choice questions (of the type expected to appear in KEAM 2010) on this site, you will find many similar useful questions with solution at the site  http://physicsplus.blogspot.com.  If you want to see earlier KEAM questions only, type in ‘Kerala’ in the search box at the top left of this page and click on the search button or press the enter key.

EAMCET 2009 (Medical) Questions (MCQ) on Rotational Motion

The following two questions were included from rotational motion in the EAMCET 2009 (Medical) question paper. (The first question has appeared in many entrance exam question papers. It was included in the EAMCET 2009 Engineering question paper also). Here are the questions with solution:

(1) A rod of length ‘l’ is held vertically stationary with its lower end located at a position P on the horizontal plane. When the rod is released to topple about P, the velocity of the upper end of the rod with which it hits the ground is

(1) √(g/l)

(2) √(3gl)

(3) 3√(g/l)

(4) √(3g/l)

When the rod falls its gravitational potential energy mgl/2 gets converted into rotational kinetic energy of the rod. (Note that ‘m’ is the mass of the rod and initially the centre of gravity of the rod is at a height l/2 with respect to the horizontal plane).

Therefore we can write

             ½ Iω² = mgl/2 where I is the moment of inertia of the rod about an axis passing through the end (at P) of the rod and perpendicular to the length of the rod and ‘ω’ is the angular velocity of the rod when it hits the horizontal plane.

Here I = ml²/3.

[Usually you will remember the moment of inertia of a rod about a normal axis through its middle as ml²/12. The moment of inertia about a normal axis through one end is obtained by applying the parallel axis theorem: I = ml²/12 + m(l/2)² = ml²/3].

Substituting for I we have

             ½ (ml²/3)ω² = mgl/2

Since ω = v/l where ‘v’ is the velocity with which the rod hits the ground, we have

             ½ (ml²/3)(v/l)² = mgl/2

This gives v = √(3gl)

(2) A rigid uniform rod of mass M and length ‘L’ is resting on a smooth horizontal table. Two marbles each of mass ‘m’ and traveling with uniform speed ‘v’ collide with the two ends of the rod simultaneously and inelastically as shown. The marbles get stuck to the rod after the collision and continue to move with the rod. If m = M/6 and v = L ms^–1, then the time taken by the rod to rotate through π/2 is

(1) 1 sec

(2) 2π sec

(3) π sec

(4) π/2 sec

Because of the collision, the rod will rotate about a normal axis through its middle with an angular velocity ω given by

             Iω = mvL/2 + mvL/2 where ‘I’ is the moment of inertia of the rod carrying the masses m and m at its ends.

[Note that we have equated the final angular momentum of the system (containing the rod and the masses) to the initial angular momentum. Before the collision the two masses have angular momentum about the central axis. These are shown on the right hand side of the above equation].

Since v = L the above equation gets modified as

             Iω = mL²

After the collision, the rod and the masses move together and the total angular momentum is given by

             Iω = [(ML²/12) + 2m(L/2)²] ω

[The first term within the square bracket above is the moment of inertia of the rod and the second term is the moment of inertia of the two masses].

From the above two equations, we have

             mL² = [(ML²/12) + mL²/2 ] ω 

Since m = M/6 the above equation becomes

             M/6 = [(M/12) + (M/12)] ω = (M/6) ω

Therefore ω = 1 radian /sec and the time taken by the rod to rotate through π/2 radian is π/2 sec.

You will find many questions on rotational motion on this site. You can access all of them by clicking on the label ‘rotation’

You will find many useful questions with solution in this section at physicsplus and at AP Physics Resources.

All India Engineering/Architecture Entrance Examination 2010 (AIEEE 2010)

It’s time to apply for AIEEE 2010.

Information Bulletin containing the Application Form for applying for All India Engineering/Architecture Entrance Examination 2010 (AIEEE 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

The date of Examination is 25^th April 2010.

Visit the site http://aieee.nic.in immediately for details.

You will find many old AIEEE questions (with solution) on this site. You can access all of them by typing ‘AIEEE’  in the search box at the top left of this page and clicking on the search button. Or, you may hit the enter key after feeding the search words.

You may perform a similar search for old AIEEE questions (with solution) at the site http://physicsplus.blogspot.com also.

Apply for IIT-JEE 2010

You may apply now for the Joint Entrance Examination for Admission to IITs and other Institutions (IIT-JEE 2010). The exam will be conducted 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 for IIT-JEE 2010 using either the on-line facility or the off-line facility. 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 (see JEE web sites of the IITs) also is possible from 16^th November 2009 to 15^th December 2009.

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

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

Find details at the JEE websites of the different IITs 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

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You can access all posts related to IIT-JEE (including old questions with solution) on this site by making a search for ‘IIT’ using the ‘search blog’ box provided. Similar posts can be seen at http://physicsplus.blogspot.com as well.

Apply for 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 (AIPMT 2010) as per the following schedule for admission to 15% of the merit seats for the Medical/Dental Courses.

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

For the Preliminary Examination there will be one paper containing 200 objective type questions (four options with one correct answer) from Physics, Chemistry and Biology (Botany & Zoology).

The final examination will consist of one paper containing 120 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.

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. Visit the web site www.aipmt.nic.in for details

Online Submission:

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-110301 by Speed Post/Registered Post in such a way that it should reach the Board on or before 04.12.2009 which is the last date stipulated.

For online submission, the 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.

Detailed 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.

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

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Some old AIPMT questions with solution can be seen on this site. You can access them by searching for ‘AIPMT’, making use of the search box provided on this site. Old AIPMT questions with solution can be found at http://physicsplus.blogspot.com also.

EAMCET (Medical) 2009 Questions on Work and Energy

The following questions which appeared in the EAMCET (Medical) 2009 question paper are worth noting:

(1) A block of mass ‘m’ is connected to one end of a spring of spring constant ‘k’. The other end of the spring is fixed to a rigid support. If the mass is released slowly so that the total energy of the system is then constituted by only the potential energy, then ‘d’ is the maximum extension of the spring. Instead, if the mass is released suddenly from the same initial position, the maximum extension of the spring now is (g = acceleration due to gravity)

(1) mg/k

(2) 2d

(3) mg/3k

(4) 4d

The mass m is suspended by means of the spring. Since the spring is extended through a distance d, we have

mg = kd so that k = mg/d

When the mass is suddenly released, suppose the spring extends through an additional distance x. The total extension then is d+x.

The spring mass system momentarily comes to rest in the condition of maximum extension and then tries to return to the initial extension of d, executing simple harmonic oscillations. In the condition of maximum extension (equal to d+x) the gravitational potential energy mg(d+x) of the mass is converted into elastic potential energy of the spring so that we have

mg(d+x) = (½) k(d+x)²

Or, mg(d+x) = ½ (mg/d)(d+x)² since k = mg/d

This gives 2 = (d+x)/d from which x = d

The total extension d+x is therefore equal to 2d [Option (2)]

(2) A particle is projected up from a point at an angle θ, with the horizontal direction. At any time ‘t’, if ‘p’ is its linear momentum, ‘y’ is the vertical displacement and ‘x’ is the horizontal displacement, the graph among the following, which does not represent the variation of kinetic energy of the projectile is

(1) Graph (A)

(2) Graph (B)

(3) Graph (C)

(4) Graph (D)

The kinetic energy of a projectile has to decrease with the increase in its vertical displacement since its gravitational potential energy increases at the cost of its kinetic energy. Therefore graph (A) is incorrect.

[Graphs (B) and (C) are correct since the kinetic energy decreases with the increase in the horizontal displacement x, becomes a minimum at half the horizontal range (corresponding to the maximum height) and then increases. Graph (D) also is correct since the kinetic energy k is given by

k = p²/2m where p is the linear momentum and m is the mass of the particle.

Therefore, k is directly proportional to p², yielding a straight line graph].

Kerala Engineering Entrance 2009 Multiple Choice Questions on Work and Energy

In the KEAM (Engineering ) 2009 question paper three questions were included from the section ‘work, energy and power’. Here are those questions with solution:

(1) A particle is acted upon by a force F which varies with position x as shown in figure. If the particle at x = 0 has kinetic energy of 25 J, then the kinetic energy of the particle at x = 16 m is

(a) 45 J

(b) 30 J

(c) 70 J

(d) 135 J

(e) 20 J

The work done by a variable force acting along the direction of displacement can be found by drawing a force-displacement graph. The area under this graph gives the work done. Since the force has positive and negative values, the area also has positive (+50 units) and negative (– 30 units) values. The net area is +20 units and hence the gain in kinnetic energy during the movement from position x = 0 to the position x = 16 m is 20 J.

The particle has kinetic energy of 25 J at position x = 0. Therefore the kinetic energy at position x = 16 m is 25 + 20 = 45 J.

(2) Two springs P and Q of force constants k_P and k_Q [k_Q = k_P/2] are stretched by applying force of equal magnitude. If the energy stored in in Q is E, then the energy stored in P is

(a) E

(b) 2 E

(c) E/8

(d) E/4

(e) E/2

The potential energy (U) of a spring stretched (by a force F) through distance x is given by

U = ½ kx² where k is the spring constant (force constant) given by k = F/x

This can be rewritten as

U = ½ (F²/k) since x = F/k

The nergy stored in P and Q are respectively given by U_P = ½ (F²/k_P) and U_Q = ½ (F²/k_Q).

Therefore, U_P/U_Q = k_Q/k_P = ½ as given in the question.

This gives U_P = U_Q/2 = E/2.

(3) A rod of mass m and length l is made to stand at an angle of 60º with the vertical. Potential energy of this rod in this position is

(a) mgl

(b) mgl/2

(c) mgl/3

(d) mgl/4

(e) mgl/√2

When the rod is kept inclined at an angle of 60º with the vertical, its centre of gravity is raised (from the ground level) by a height h = (l/2)cos 60º = l/4.

The gravitational potential energy of the rod in this position is mgh = mgl/4.

You will find many useful multiple choice questions on work, energy and power at physicsplus and at AP Physics Resources