Multiple Choice Questions on Thermoelectric Effect

Questions involving neutral temperature and temperature of inversion often appear in entrance examination question papers. You have to remember that the neutral temperature is a constant for a given thermo couple where as the temperature of inversion is not a constant. The temperature of inversion is dependent on the temperature of the cold junction and is always as much above the neutral temperature as the cold junction is below it.

Now, consider the following MCQ:

The temperature of the cold junction of a thermocouple is 0º C and its neutral temperature is 275º C. If the temperature of the cold is changed to 20º C, the neutral temperature and the temperature of inversion will be respectively

(a) 265º C and 550º C (b) 265º C and 530º C (c) 275º C and 530º C

(d) 275º C and 550º C (e) 275º C and 510º C

The neutral temperature will be unchanged (275º C). Since the cold junction is 255º C below the neutral temperature, the temperature of inversion has to be 255º C above the neutral temperature and will be 530º C. So, the correct option is (c).

Consider now the following simple question:

The metal which does not exhibit Thomson effect is

(a) iron (b) nickel (c) copper

(d) lead (e) bismuth

The correct option is (d). In determining thermo electric quantities, lead is often used as one member of the couple because of the absence of Thomson effect in it.

What is the unit of thermo electric power?

The term ‘thermoelectric power’ is a misnomer. Thermoelectric power is dV/dT where V is the thermo emf and T is the temperature of the hot junction. There is no ‘power’ involved in it and the unit is volt per Kelvin.

All India Pre-Medical/Pre-Dental Entrance Examination (AIPMT) 2008

Central Board of Secondary Education (CBSE), Delhi has announced the dates of All India Pre-Medical/Pre-Dental Entrance Examination (AIPMT) 2008 for admission to 15% of the total seats for Medical/Dental Courses in all Medical/Dental colleges run by the Union of India, State Governments and Municipal or other local authorities in India except in the States of ANDHRA PRADESH AND JAMMU & KASHMIR. The dates of the Examination are:

(1) Preliminary Examination : 6^th April 2008 (Sunday)

(2) Final Examination : 11^th May 2008 (Sunday)

Candidates can apply for the All India Pre-Medical/Pre-Dental Entrance Examination in the following two ways:-

(i) Online

Online submission of application may be made by accessing the Board’s website www.cbse.nic.in from 16.10.2007 (10.00 AM) to 26.11.2007 (5.00 PM). 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), Central Board of Secondary Education, Shiksha Kendra, 2, Community Centre, Preet Vihar, Delhi-110 301 by Speed Post/Registered Post only. Candidate should pay the examination fee of Rupees 400/- for General Category and Rupees 200/- for SC/ST Category through a Demand Draft in favour of the Secretary, Central Board of Secondary Education, Delhi drawn on any nationalized bank payable at Delhi. Instructions for Online submission of Application Form will be made available on the website www.cbse.nic.in.

(ii) Offline

Offline submission of Application Form may be made on the prescribed Application Form. The Information Bulletin and Application Form costing Rs.400/- for General Category candidates and Rs.200/- for SC/ST candidates inclusive of examination fee can be obtained against Cash Payment from designated branches of Canara Bank/ Regional Offices of the CBSE from 16-10-2007 to 26-11-2007. The details of Banks are given in the Admission Notice which is available on CBSE website www.cbse.nic.in.

Designated branches of Canara Bank in Kerala are:

KOTTAYAM	P.B. No.122, K.K.Road, Kottayam-686 001
QUILANDY	Fasila Buidling, Main Road, Quilandy-673 305
TRIVANDRUM	Ist Floor, Ibrahim Co. Bldg., Challai, Trivandrum-695 023
TRIVANDRUM	Plot no.2, PTP Nagar Trivandrum-695 038
TRIVANDRUM	TC No.25/1647, Devaswom Board Bldg. M.G. Road, Trivandrum-695 001
ERNAKULAM	Shenoy’s Chamber, Shanmugam Road, Ernakulam, Cochin-682 031
CALICUT	9/367-A, Cherooty Road, Calicut-673 001
TRICHUR	Trichur Main Ramaray Building, Round South, Trichur-680001
QUILON	Maheshwari Mansion, Tamarakulam, Quilon-691 001
PALGHAT	Market Road, Big Bazar, 20/68, Ist floor,Palghat-678 014.

The Information Bulletin and Application Form can also be obtained by Speed Post/Registered Post by sending a written request with a Bank Draft/Demand Draft for Rs.450/- for General Category and Rs.250/- for SC/ST Category payable to the Secretary, Central Board of Secondary Education, Delhi along with a Self Addressed Envelope of size 12” x 10”. The request must reach the Deputy Secretary (AIPMT), CBSE, 2, Community Centre, Preet Vihar, Delhi-110 301 on or before 15-11-2007. The request should be super scribed as “Request for Information Bulletin and Application Form for AIPMT, 2008”.

Completed Application Form is to be dispatched by Registered Post/Speed Post only. Last Date for receipt of completed Application Forms for both Offline and Online in CBSE is 28-11-2007.

You can obtain complete information at www.cbse.nic.in

Make it a point to visit the site for information updates.

Rotational Motion of a Reel – Two Multiple Choice Questions

The following question on the acceleration of a reel carrying stitching thread is an interesting one:

A reel of mass ‘m’ in the form of a short solid cylinder carries light inextensible string wound round it. The free end of the string is tied to a hook fixed to the ceiling and the reel is allowed to roll down, unwinding the string. The linear acceleration of the reel is

(a) g (b) g/3 (c) g/2 (d) 3g/2 (e) 2g/3

If ‘a’ is the acceleration of the of the reel we have

Mg – T = ma

where ‘M’ is the mass of the reel and ‘T’ is the tension in the string.

Therefore, T = mg – a

[You can directly write the above value of T if you remember that T is equal in magnitude to the weight of the reel while moving down with acceleration ‘a’ as in a lift]

If the radius of the reel is ‘R’, the torque on the reel is TR.

Therefore, we have

TR = Iα

where ‘I’ is the moment of inertia of the reel ( I = MR²/2) and ‘α’ is its angular acceleration, which is equal to a/R.

The above equation therefore becomes

(Mg – a)R = (MR²/2) (a/R)

This gives a = 2g/3.

Now consider another MCQ:

A solid cylinder having mass M and radius R is free to rotate about its axis and has a light inextensible string wound roun it. A body of mass M/2 is attached to the free end of the string. If the system is released from rest, the mass will move down with a linear acceleration of

(a) g (b) g/3 (c) g/2 (d) 3g/2 (e) 2g/3

The tension in the string is given by

T = (M/2)(g – a).

Since TR = Iα, we have

(M/2)(g – a)R = (MR²/2) (a/R).

This equation yields a = g/2.

[It will be interesting to note that if the mass suspended were M, the acceleration would have been 2g/3, which is the acceleration of the falling reel in the previous question].

Two Questions (MCQ) on Momentum Conservation

Often you will find questions based on the law of conservation of momentum. Here is a question which may be interesting to you:

Two skaters of masses 40 kg and 50 kg respectively are 4 m apart and are facing each other, holding the ends of a light rope. They advance by pulling on the rope, thereby reducing the separation between them. The approximate distances covered by them by the time their separation has reduced to zero are respectively

(a) 2.22 m and 1.78 m (b) 2 m and 2 m (c) 1.62 m and 2.38 m

(d) 4 m and zero (e) zero and 4 m

The initial momentum of the system containing the two skaters is zero. The pulling forces they exert are internal forces (within the system) and hence the total momentum of the system must be zero throughout their motion. If v₁ and v₂ are the average velocities of the skaters respectively, we have 40 v₁ + 50 v₂ = 0.

Therefore, 40 v₁ = – 50 v₂.

Since v₁ = s₁/t and v₂ = s₂/t where s₁ and s₂ are the displacements, we have

40 s₁ = –50 s₂

Considering the magnitudes of displacements, 40 s₁ = 50 s_2.

Since s₁+ s₂ (which is the total distance covered by the two scaters) is 4 m, s₂ = 4–s₁ so that

40 s₁ = 50(4–s₁) from which s₁ = 2.22 m and s₂ = 1.78 m [Option (a)]

The following MCQ is a simple one. But be careful not to be distracted to arrive at a wrong answer!

Two spheres of the same material have radii R and 2R. They are released in free space with initial separation between their centres equal to 15R. If the only force between the spheres is the gravitational pull between them, the distance covered by the smaller sphere before collision is approximately

(a) 6.67R (b) 10.67R (c) 13.33R (d) ) 2.33R (e) 4.33R

As in the previous question, the forces acting are internal forces and hence the total momentum of the system remains unchanged. The initial momentum of the system containing the two spheres is zero and hence the total momentum of the system must be zero throughout their motion.

Proceeding as before, we have m₁s₁ = m₂s₂ where m₁ and m₂ are the masses of the spheres and s₁ and s₂ are the distance covered by them before collision. But you have to be careful to note that when the spheres collide, the distance between their centres is (R+ 2R) = 3R so that the total distance covered by the spheres is 15R – 3R = 12R.

Therefore, s₁+s₂ = 12R so that m₁s₁ = m₂(12R–s₁).

Since the spheres are of the same material, their masses are directly proportional to their volumes which are proportional to the cubes of their radii. Therefore we have

R³s₁ = (2R)³(12R–s₁) from which s₁ = 10.67 R.

Kerala Engineering Entrance 2007 Questions on Heating Effect of Electric Current

Two questions on heating effect of electric current appeared in KEAM (Engineering) 2007 question paper:

(1) The resistance of a wire at room temperature 30º C is found to be 10 Ω. Now to increase the resistance by 10%, the temperature of the wire must be [Temperature coefficient of resistance of the material of the wire is 0.002 per º C]

(a) 36º C (b) ) 83º C (c) ) 63º C (d) ) 33º C (e) ) 66º C

The resistance (R_t) at tº C can be expressed in terms of the resistance (R₀) at 0º C and the temperature coefficient (α) of resistance as

R_t = R₀(1 + αt).

Therefore, we have

R₃₀ = 10 = R₀(1 + 0.002×30) and

R_t = 11 = R₀(1 + 0.002 t)

[The resistance R_t at the unknown temperature ‘t’ is greater by 10% and is therefore equal to (10 + 1) Ω = 11 Ω]

From the above equations we have

10/11 = (1 + 0.06)/ (1 + 0.002t) from which t = 83º C.

(2) If R₁ and R₂ be the resistances of the filaments of 200 W and 100 W electric bulbs operating at 220 V, then R₁/R₂ is

(a) 1 (b) 2 (c) 0.5 (d) 4 (e) 0.25

Since the power is V²/R where V is the operating voltage and R is the resistance, we have

220²/R₁ = 200 and

220²/R₂ = 100.

Dividing the second equation by the first, we obtain

R₁/R₂ = 0.5

Kerala Engineering Entrance 2007 Questions on Rotational Motion

The following questions appeared in KEAM (Engineering) 2007 question paper:

(1) A sphere of mass ‘m’ and radius ‘r’ rolls on a horizontal plane without slipping with speed ‘u’. Now if it rolls up vertically, the maximum height it would attain would be

(a) 3u²/4g (b) 5u²/2g (c) 7u²/10g (d) u²/2g (e) 11u²/9g

The initial kinetic energy (E) of the sphere while rolling on the horizontal surface is given by

E = ½ mu² + ½ Iω² where I is the moment of inertia of the sphere about its diameter[(which is equal to (2/5)mr²] and ‘ω’ is the angular velocity (which is u/r).

[Note that the first term is the translational kinetic energy and the second term is the rotational kinetic energy].

Therefore, E = ½ mu² + ½ ×(2/5)mr²u²/r² = 7mu²/10.

Since this energy is converted into gravitational potential energy (mgh) to attain the maximum height h, we have

7mu²/10 = mgh, from which h = 7u²/10g.

(2) If the earth were to contract such that its radius becomes one quarter, without change in its mass, the duration of the full day would be

(a) 3 hours (b) 1.5 hours (c) 6 hours (d) 4 hours (e) 2 hours

If the radius of the earth becomes ‘n’ times he present value, without change in the mass, the duration of the day will become 24n² hours so that the answer to the above question is 24×(1/4)² = 1.5 hours.

Solution to Multiple Choice Questions on Centre of Mass

Two multiple choice questions on centre of mass were given to you for practice yesterday. Here is the solution along with the questions:

(1) A boy weighing 40 kg is standing on a wooden log of mass 500 kg floating on still water in a lake. The distance of the boy from the shore is 12 m. The viscous force exerted by water on the wooden log may be neglected. If the boy walks slowly along the wooden log through 2 m towards the shore, the centre of mass of the system (wooden log and the boy) will move with respect to the shore through a distance

(a) 2 m (b) 1.25 m (c) 0.25 m (d) 0.16 m (e) zero

You should have worked this out within seconds. The centre of mass will be unaffected with respect to external fixed points if there are no external forces acting on the system. So, the correct option is (e).

(2) A T-shaped object with dimensions shown in the figure, is lying on a smooth floor. A force F is applied at the point P parallel to AB, such that the object has only translational motion without rotation. Find the location of P with respect to C

(a) L (b) 4L/3 (c) 3L/2 (d) 2L/3

The point P must be the centre of mass of the T-shaped object since the force F does not produce any rotational motion of the object. So, we have to find the distance of the centre of mass from the point C.

The horizontal part of the T-shaped object has length L. If the mass of the horizontal portion is ‘m’, the mass of the vertical portion of the T- shaped object is 2m since its length is 2L. For finding the centre of mass of the T shaped object, it is enough to consider two point masses m and 2m located respectively at the mid points of the horizontal and vertical portions of the T.

Therefore, the T-shaped object reduces to two point masses m and 2m at distances 2L and L respectively from the point C. The distance ‘r’ of the centre of mass of the system from the point C is given by

r = (m₁r₁ + m₂r₂)/(m₁ + m₂) = (m×2L + 2m×L)/(m + 2m) = 4L/3

[ Note that we have used the equation, r = (m₁r₁ + m₂r₂)/(m₁ + m₂) for the position vector r of the centre of mass in terms of the position vectors r₁ and r₂ of the point masses m₁ and m₂. We could use the simple equation involving the distances from C since the points are collinear].

Two Multiple Choice Questions on Centre of Mass (For practice)

Here are two questions on centre of mass. These are meant for checking whether you have a clear idea of the concept of centre of mass:

(1) A boy weighing 40 kg is standing on a wooden log of mass 500 kg floating on still water in a lake. The distance of the boy from the shore is 12 m. The viscous force exerted by water on the wooden log may be neglected. If the boy walks slowly along the wooden log through 2 m towards the shore, the centre of mass of the system (wooden log and the boy) will move with respect to the shore through a distance

(a) 2 m (b) 1.25 m (c) 0.25 m (d) 0.16 m (e) zero

(2) A T-shaped object with dimensions shown in the figure, is lying on a smooth floor. A force F is applied at the point P parallel to AB, such that the object has only translational motion without rotation. Find the location of P with respect to C

(a) L
(b) 4L/3
(c) 3L/2
(d) 2L/3

This MCQ appeared in AIEEE 2005 question paper.

Try to find the answer to the above questions. If you have clear understanding of the centre of mass, you will be able to find the answers in a couple of minutes. I’ll be back with the solution shortly.

Multiple Choice Questions on Work and Energy

Here is a simple question which is meant for gauging your understanding of the work-energy principle. This MCQ appeared in AIEEE 2005 question paper:

The block of mass M moving on the frictionless horizontal surface collides with the spring of spring constant K and compresses it by length L. The maximum momentum of the block after collision is

(a) ML²/K (b) zero (c) KL²/2M (d) √(MK).L

If the maximum momentum of the block after the collision is ‘p’ the maximum kinetic energy is p²/2M. This must be equal to the maximum potential energy of the spring so that we have

p²/2M = ½ KL², from which p = L√(MK)., given in option (d).

Now consider the following MCQ which appeared in Kerala Medical Entrance 2006 question paper:

The work done by a force F = –6x³ î newton, in displacing a particle from x = 4m to x = –2m is

(a) 360 J (b) 240 J (c) – 240 J (d) – 360 J (e) 408 J

This is a case of variable force (in the X-direction), the point of application of which is moved in the X-direction. The work done is therefore given by

W = ∫F.dx = ₄ ∫^–2 (–6x³)dx = –6[x⁴/4] with x between limits 4 and –2.

Therefore, W = –(6/4)(16 – 256) = 360 joule, given in option (a).

The following MCQ appeared in Kerala Engineering entrance 2006 question paper:

A running man has the same kinetic energy as that of a boy of half the mass. The man speeds up by 2 ms^–1 and the boy changes his speed by ‘x’ ms^–1 so that the kinetic energies of the boy and the man are again equal. Then ‘x’ in ms^–1 is

(a) – 2√2 (b) + 2√2 (c) √2 (d) 2 (e) 1/√2

If the mass of the man is ‘m’, the mass of the boy is m/2. If v₁ and v₂ are the initial velocities of the man and boy respectively, we have

½ mv₁² = ½ (m/2)v₂²

Therefore, v₂ = v₁√2.

On changing the speeds, we have

½ m(v₁+2)² = ½ (m/2)(v₂+x)²

On substituting for v₂ (=v₁√2), the above equation simplifies to

(v₁+2)² = ½ (v₁√2+x)² from which x = 2√2 ms^–1, given in option (b).

Multiple Choice Questions on Direct Current Circuits

The following MCQ which appeared in All India Pre-Medical/Dental Entrance 2005 Examination (AIPMT) is one requiring the application of Kirchoff’s laws:

Two batteries, one of emf 18 volt and internal resistance 2 Ω and the other of emf 12 volt and internal resistance 1 Ω, are connected as shown. The voltmeter V will record a reading of

(a) 30 volt (b) 18 volt (c) 15 volt (d) 14 volt

The net emf in the closed path containing the two batteries is 18volt – 12 volt = 6 volt

Since this must be equal to the net voltage drop across the internal resistances of the batteries, we have

I(2 + 1) = 6 volt where ‘I’ is the current through the batteries. [Note that the voltmeter will draw negligible current (ideally, zero current) and hence the same current flows through the batteries].

Therefore, I = 2 ampere.

The voltage across the voltmeter is the same as the terminal voltages of the batteries. If you consider the 18 volt battery, its terminal voltage is 18 – (2×2) = 14 volt.

[If you consider the 12 volt battery, its terminal voltage is 12 + (1×2) = 14 volt, which is the same as the above value]. So, the correct option is (d).

Here is another question which you can answer in no time if you had worked out a similar one earlier:

Five identical cells each of emf 1.5 V and internal resistance ‘r’ send the same current through an external resistance of 1 Ω whether the cells are connected in series or in parallel. The internal resistance ‘r’ of each cell is

(a) 0.2Ω (b) 0.5 Ω (c) 1 Ω (d) 1.5 Ω (e) 3 Ω

The condition depicted in the above question occurs when the internal resistance of each cell is equal to the external resistance. The answer therefore is 1 Ω.

The proof is simple: Let there be ‘n’ cells. If the emf of each cell is V and the external resistance is R, the current through R on connecting the cells in series is nV/(nr + R). When the cells are in parallel, the current through R is V/[(r/n) + R].

Since the currents are equal, nV/(nr + R) = V/[(r/n) + R].

Therefore, r + nR = nR + R, from which r = R.

Now, let us consider the following MCQ which appeared in All India Pre-Medical/Dental Entrance 2004 Examination question paper:

Five equal resistances each of resistance R are connected as shown in the figure. A battery of V volts is connected between A and B. The current flowing in AFCEB will be

(a) 3V/R (b) V/R (c) V/2R (d) 2V/R

Once you realise that C and D are equipotential points with respect to point A, the resistance R connected directly between C and D can be ignored (since it will not carry any current). The current through the path AFCEB is therefore V/2R.

Science is a wonderful thing if one does not have to earn
one's living at it

–Albert Einstein 

Two Questions (MCQ) on Capacitors

Here are two multiple choice questions which will be interesting to you:

(1) Six identical capacitors C₁, C₂, C₃, C₄, C₅ and C₆ each having capacitance C are connected as shown. The equivalent capacitance between the points A and B is

(a) C (b) 2C (c) 2.5C (d) 3C (e) 3.5C

This may appear to be a difficult question for some of you. In questions of this type try to identify equipotential points. Circuit elements connected between equipotential points can be ignored. In most cases the circuit will then become simple.

In the present case, the junction between C₁ and C₆ is at the same potential as that at the junction between C₂ and C₅. [This follows since the capacitors are identical. Generally, the points will be equipotential points if the balance condition for Wheatstone bridge is satisfied: C₁/C₆ = C₂/C₅]. The capacitor C₃ connected between the equipotential points can therefore be ignored.

Now we have three parallel connections across the points A and B:

(i) Series combination of C₁ and C₆ giving a capacitance C/2.

(ii) Series combination of C₂ and C₅ giving a capacitance C/2.

(iii) Capacitor C₄ of value C.

The parallel combined value of the abve three connections is (C/2) + (C/2) + C = 2C.

(2) In a circuit, capacitors C₁ and C₂ (in series) are connected between two points A and B (Fig). If the potentials at A and B are V₁ and V₂ respectively, what is the potential at the junction point P between C₁ and C₂?

(a) C₁(V₁+V₂)/(C₁+C₂)

(b) (V₁ + V₂)/2

(c) V₁ – V₂

(d) (C₁V₂ + C₂V₁)/ (C₁ + C₂)

(e) (C₁V₁ + C₂V₂)/ (C₁ + C₂)

Since the capacitors are in series, they carry the same charge. If the potential at the junction point P is ‘V’, the charges on the capacitors C₁ and C₂ are respectively C₁(V – V₁) and C₂(V₂ – V).

Therefore, we have

C₁(V – V₁) = C₂(V₂ – V), from which

V = (C₁V₁ + C₂V₂)/ (C₁ + C₂).

You can find more posts on electrostatics by clicking on the label ELECTROSTATICS below this post or on the left side of this page.

Three KEAM 2007 Multiple Choice Questions from Heat and Thermodynamics

Four questions from Heat and Thermodynamics appeared in the Kerala Engineering Entrance 2007 question paper. One question was quite simple, meant for testing your knowledge of the temperature at which water has minimum volume, which most of you know, is 4º C. Here are the remaining three questions:

(1) A closed gas cylinder is divided into two parts by a piston held tight. The pressure and volume of gas in two parts respectively are (P, 5V) and (10 P, V). If now the piston is left free and the system undergoes isothermal process, then the volume of the gas in two parts respectively are

(a) 2 V, 4 V (b) 3 V, 3 V (c) 5 V, V (d) 4 V, 2 V (e) 2.5 V, 3.5 V

Since the temperature is constant (isothermal process), we can apply Boyle’s law. The piston will move towards the lower pressure side until the pressures on the two sides are the same. If this common pressure is P’, we have

P×5V + 10P×V = P’×6V, since the total volume is 6 V.

From this, P’ = (15/6) P.

Now, applying Boyle’s law to the gas on the two sides separarely, we have

P×5V = (15/6) P×V₁ and

10P×V = (15/6) P×V₂

where V₁ and V₂ are the volumes of the gas in the two parts respectively.

From the above equations, we obtain V₁ = 2V and V₂ = 4V [Option (a)].

(2) A Carnot engine with sink’s temperature at 17º C has 50% efficiency. By how much should its source temperature be changed to increase its efficiency to 60%?

(a) 225 K (b) 128º C (c) 580 K (d) 145 K (e) 145º C

We have efficiency η = (T₁ – T₂)/T₁ where T₁ and T₂ are the temperatures of the source and sink respectively.

Here, η = 0.5 and T₂ = 17º C = 290 K. On substituting these values, we obtain

1– (290/T₁) = 0.5, from which T₁ = 580 K.

When the efficiency is 60%, we have

1– (290/T₁^') = 0.6, from which T₁^' = 725 K.

Therefore, the change in source temperature = 725 – 580 = 145 K.

(3) Two moles of oxygen is mixed with eight moles of helium. The effective specific heat of the mixture at constant volume is

(a) 1.3 R (b) 1.4 R (c) 1.7 R (d) 1.9 R (e) 1.2 R

Oxygen is diatomic and hence its molar specific heat at constant volume is (5/2) R (corresponding to 5 degrees of freedom).

Helium is mono-atomic and hence its molar specific heat at constant volume is (3/2) R (corresponding to 3 degrees of freedom).

Therefore, the total quantity of heat required to raise the temperature of two moles of oxygen and eight moles of helium through 1 K is

2×(5/2)R + 8×(3/2)R = 17 R.

Since the number of moles in the mixture is 10, the effective molar specific heat of the mixture is (17/10)R = 1.7 R.