Multiple Choice Questions (MCQ) on Elastic Collision

The figure shows three identical blocks A, B and C (each of mass m) on a smooth horizontal surface. B and C which are initially at rest, are connectd by a spring of negligible mass and force constant K. The block A is initially moving with velocity ‘v’ along the line joining B and C and hence it collides with B elastically.

Now, answer the following questions related to the system:

(1) After the collision the velocity of the block A is

(a) v/3

(b) – v

(c) – v/3

(d) v/2

(e) zero

At the moment the block A collides with the block B, you need consider these two blocks only. In one dimensional elastic collision between two bodies of the same mass, if one body is initially at rest, the moving body comes to rest and the body initially at rest moves with the velocity of the moving body. After the collision, the velocity of A is therefore zero.

[Generally, if the moving block 1 has mass m₁ and it moves with initial velocity v_1i towards the stationary block 2 of mass m₂, the final velocities of blocks 1 and 2 are given by

v_1f = v_1i(m₁ – m₂) /(m₁ + m₂) and

v_2f = 2 m₁v_1i /(m₁ + m₂)

The above relations can be easily obtained from momentum and kinetic energy conservation equations.

(2) When the compression of the spring is maximum, the velocity of block B is

(a) v

(b) – v

(c) – v/3

(d) v/2

(e) zero

On receiving the entire momentum (mv) of the block A, the block B moves towards the block C, compressing the spring. When the compression of the spring is maximum, both B and C move with the same velocity (v’). Since B and C have the same mass (m) and momentum has to be conserved, we have

mv = (m + m) v’

Therefore v’ = v/2.

(3) When the compression of the spring is maximum, the kinetic energy of the system is

(a) (½) mv²

(b) (1/4) mv²

(c) Zero

(d) mv²

(e) (1/8) mv²

The block A has come to rest after colliding with block B. When the compression of the spring is maximum, both B and C are moving with the same velocity v/2 as shown above. Therefore, the kinetic energy of the system under this condition is 2×(1/2)m(v/2)² = (1/4) mv². [Note that the spring has negligible kinetic energy since its mass is negligible].

(4) The maximum compression of the spring is

(a) √(mv²/2K)

(b) √(mv/2K)

(c) mv/2K

(d) mv²/2K

(e) √(2mv²/K)

The total energy of the system is equal to the initial kinetic energy (½ )mv² of the colliding block A. Therfore we have

(½)mv² = KE of A + KE of B + KE of C + PE of spring. The KE of A is zero after the collision. As shown above, the total KE of B and C is (1/4) mv² when the compression of the spring is maximum. Therfore, the potential energy (PE) of the spring at maximum compression is given by the equation,

(½)mv² = (1/4) mv² + (1/2)Kx² where x is the maximum compression of the spring.

This gives (1/2)Kx² = (1/4) mv² from which x = √(mv²/2K).

You will find similar questions (with solution) from different branches of physics at apphysicsresources.blogspot.com

Two Questions from Electrostatics-- Capacitor Combinations

Here are two multiple choice questions on effective capacitance of capacitor networks. [The first question was found to be some what difficult for the students (from their response)]:

(1) An unusual type of capacitor is made using four identical plates P₁, P₂, P₃ and P_4, each of area A arranged in air with the same separation d as shown in the figure. A thin wire outside the system of plates connects the plates P₂ and P₄. Wires soldered to the plates P₁ and P₃ serve as terminals T₁ and T₂ of the system. What is the effective capacitance between the terminals T₁ and T₂?

(a) (2ε₀A)/3d

(b) (3ε₀A)/2d

(c) (3ε₀A)/d

(d) (ε₀A)/d

(e) (ε₀A)/2d

The arrangement contains 3 identlcal capacitors C₁, C₂ and C₃ each of capacitance (ε₀A)/d arranged as shown in the adjoining figure. Plate P₂ is common for C₁ and C₂. Likewise, plates P₃ is common for C₂ and C₃. The capacitors C₂ and C₃ are connected in parallel (to produce an effective value 2ε₀A/d) and this parallel combination is connected in series with the capacitor C₁ of value ε₀A/d . The effective capacitance C between the terminals T₁ and T₂ is therfore given by

C = [(2ε₀A/d) ×(ε₀A/d)] / [(2ε₀A/d) +(ε₀A/d)].

Therefore, C = (2ε₀A)/3d

(2) Four 2 μF capacitors and a 3 μF capacitor are connected as shown in the figure. The effective capacitance between the points A and B is
(a) 3 μF

(b) 2 μF

(c) 1.5 μF

(d) 1 μF

(e) 0.5 μF

This is a very simple question once you identify the circuit to be a balanced Wheatstone’s bridge. The 3 μF capacitor is connected between equipotential points and you can ignore it. Without the diagonal branch, there are four 2 μF capacitors only. The capacitors in the upper pair are in series and produce a value of 1 μF. The capacitors in the lower pair also produce a value of 1 μF. Since they are in parallel the effective capacitance between the points A and B is 2 μF.

Two Karnataka CET Multiple Choice Questions on Direct Current Circuits

Questions on direct current circuits have been discussed on this site on various occasions. You can find all those posts by clicking on the label ‘direct current circuit’ below this post. The search box at the top of this page can also be used to access topics of your choice.

The following multiple choice questions appeared in Karnataka CET 2004 question paper:

(1) In the circuit shown, the internal resistance of the cell is negligible. The steady state current in the 2 Ω resistor is

(a) 0.6 A

(b) 1.2 A

(c) 0.9 A

(d) 1.5 A

In steady state, the capacitor is fully charged and no current flows into its branch. So, you can ignore the branch containing the capacitor and the 4 Ω resistor. The circuit thus reduces to the 6 V cell in series with the 2.8 Ω resistor and the parallel combined value of 2 Ω and 3 Ω [ which is 2×3/(2+3) = (6/5) Ω = 1.2 Ω].

The current sent by the cell is 6 V/ (2.8 + 1.2) Ω = (6/4) A = 1.5 A.

This current gets divided between the 2Ω and 3Ω resistor paths. The current (I) through the 2Ω resistor is given by

I = (Main current × Resistance of the other branch)/ Total resistance

[Remember the above relation for the branch current when current gets divided between two branches].

Therefore, we have

I = (1.5×3)/(2+3) = 0.9 A.

(2) An unknown resistance R₁ is connected in series with a resistance of 10 Ω. This combination is connected to one gap of a metre bridge while a resistance R₂ is connected in the other gap. The balance point is at 50 cm. Now, when the 10 Ω resistor is removed and R₁ alone is connected, the balance point shifts to 40 cm. The value of R₁ is (in ohms)

(a) 20

(b) 10

(c) 60

(d) 40

The condition for the balance of a metre bridge (which is basically a Wheatstone bridge) is

P/Q = L/(100 – L) where P and Q are the resistances in the two gaps and the balancing length L cm is measured from the side of P.

In the above problem we have

(R₁ + 10) /R₂ = 1 since the balance point is at the middle of the wire when one gap contains the series combination of R₁ and 10 Ω and the other gap contains R₂.

With R₁ alone in the gap (on removing the 10 Ω resistance), the balance condition is

R₁/R₂ = 40/60

From the above two equations (on dividing one by the other), we have

(R₁ + 10) /R₁ = 60/40, from which R₁ = 20 Ω.

Two EAMCET Questions from Nuclear Physics

The following MCQ which appeared in EAMCET 2005 question paper highlights the law of conservation of momentum as applied to nuclear processes:

A nucleus of mass 218 amu in free state decays to emit an α-particle. Kinetic energy of the α-particle emitted is 6.7 MeV. The recoil energy (in MeV) of the daughter nucleus is

(a) 1

(b) 0.5

(c) 0.25

(d) 0.125

The kinetic energy (K) is given by

K = p²/2m where p is the momentum and m is the mass. The recoil momentum of the daughter nucleus is equal and opposite to the momentum of the α-particle.

We have p²/(2×4 amu) = 6.7 MeV, since the mass of the α-particle is 4 amu.

[You need not convert amu into kilogram and energy into joule and waste your time. You may imagine that p has proper unit to get the energy in MeV]

In the case of the daughter nucleus (of mass 218 – 4 = 214 amu), we have

p²/(2×214 amu) = K, whetre K is the recoil energy of the daughter (in MeV).

From the above two equations we obtain K = 0.125 MeV.

The following MCQ appeared in EAMCET 2000 question paper:

In a nuclear reactor using U²³⁵ as fuel the power output is 4.8 mega watts. The number of fissions per second is (energy released per fission of U²³⁵ = 200 MeV).

(a) 1.5×10¹⁷

(b) 3×10¹⁹

(c) 1.5×10²⁵

(d) 3×10²⁵

This is a very simple question. The energy produced per second is 4.8×10⁶ joule. Since 1 electron volt is 1.6×10^–19 joule, 200 MeV = 200×10⁶×1.6×10^–19 joule.

Therefore, the number of fissions per second = (4.8×10⁶)/(200×10⁶×1.6×10^–19) = 1.5×10^17.

You can find all related posts on this site by clicking on the label ‘nuclear physics’ below this post.

Two Questions (MCQ) on Bohr Atom Model

You can find the earlier questions (with solution) on Bohr atom model posted on this site by clicking on the label ‘Bohr model’ or ‘hydrogen atom’ below this post. You can get them also by using the search option at the top of this page. Today I give you two questions which are meant for gauging the depth of your understanding of Bohr’s theory.

(1) When the electron in a hydrogen atom of mass M undergoes transition from an orbit of higher quantum number n₂ to an orbit of lower quantum number n₁, the recoil velocity acquired by the atom is (Rydberg’s constant = R, Planck’s constant = h)

(a) (R/hM) (1/n₁² – 1/n₂²)

(b) (hR/M) (n₂ – n₁)

(c) 1/hRM (1/n₁² – 1/n₂²)

(d) h/RM) (1/n₁² – 1/n₂²)

(e) (hR/M) (1/n₁² – 1/n₂²)

The wave number of the photon emitted because of the electron transition is

ν' = 1/λ = R(1/n₁² – 1/n₂²) where λ is the wave length of the photon and R is Rydberg’s constant.

The momentum of the photon is p = h/λ = hR(1/n₁² – 1/n₂²) where h is Planck’s constant.

When the photon is emitted with this momentum, the atom recoils (like a gun firing a bullet) with an equal and opposite momentum. Therefore, the recoil velocity of the atom is given by

v = p/M = (hR/M)(1/n₁² – 1/n₂²).

(2) If the radius of the innermost electron orbit in a hydrogen atom is R₁, the de Broglie wave length of the electron in the second excited state is

(a) πR₁

(b) 3πR₁

(c) 4πR₁

(d) 6πR₁

(e) 9πR₁

The wave length of the electron in the n^th orbit is given by

λ = 2πR_n/n where R_n is the radius of the n^th orbit.4

[This follows because the angular momentum of the electron in the n^th orbit is

mvR_n = nh/2π.

Therefore, de Broglie wave length, λ = h/mv = 2πR_n/n ]

The second excited state has quantum number n = 3 (Third orbit). The radius of the 3^rd orbit in terms of the radius R₁ of the first orbit is given by (remembering R_n = n² R₁)

R₃ = 9R₁

Therefore, λ = 2πR_n/n = 2π×9R₁/3 = 6πR₁

[It will be convenient to remember that the de Broglie wave length of the electron in the n^th orbit is n times the the wave length in the innermost orbit].

You will find some useful posts on Atomic Physics and Quantum effects at apphysicsresources

The Atomic Bomb – A Surprise Question

When I thought of posting a few questions on nuclear fission, a funny incident came to my mind. In the Physics Department we used to conduct ‘surprise colloquium’ for the benefit of our post graduate students. Without giving any prior information, three or four teachers would go to the class with bits of paper on which different topics were written. The paper bits (contained in a small box) would be shuffled and the students asked one by one to pick a bit at random. After drawing a paper bit the student was required to think of the topic written on the paper bit for a couple of minutes and then start talking on the topic. The maximum time allowed for the talk was five minutes. This was followed by questions by students and teachers and answer by the ‘victim’, for few more minutes.

During one such ‘surprise colloquium’ one of the students picked out a paper bit carrying the topic nuclear fission. As usual, he thought of the topic for a while and started presenting the details such as fissionable material, critical mass, chain reaction, Einstein’s mass-energy relation, energy released in fission, moderators and control rods in nuclear reactors and the uncontrolled chain reaction in the atomic bomb. After the torrent of words for well over five minutes, it was question time when one of the listeners (a close friend of the speaker) raised a question:

“We know that the critical mass is only of the order of kilograms. The earth contains tons and tons of fissionable U²³⁵. Yet the earth does not explode. Why?”

The speaker thought for a while and to the surprise of all of us retorted: “Do you know?”

The entire class burst into laughter and the students and the teachers enjoyed the day’s ‘surprise colloquium’ very much…

Most of you know the answer to the above question. You require the fissionable material in a concentrated form, with mass greater than the critical mass, so that the neutrons produced by the fission are able to produce further fissions in neighbouring nuclei. Then only a sustained chain reaction is possible.

Two Questions (MCQ) Involving Electric Power

The following questions are simple and you are expected to solve them in a couple of minutes:

(1) A current I ampere gets divided and flows into a parallel network of resistors as shown in the adjoining figure. If the power dissipated in the 1 Ω resistor is P watt, what is the total power (in watt) dissipated in the branch containing the two 3 Ω resistors?

(a) P

(b) 6P

(c) 3P

(d) P/2

(e) 3P/2

If the current through the 1 Ω resistor is I₁, the current through the branch containing the two 3 Ω resistors is I₁/2.

[Note that the P.D across the three branches is the same and hence the current through the 6 Ω branch must be half the current through the 3 Ω branch].

The power dissipated in the 1 Ω resistor is

I₁² ×1= P

Therefore, the power dissipated in the branch containing the two 3 Ω resistors is

(I₁/2) ²×6 = 6P/4 = 3P/2 watt.

(2) A battery of internal resistance R ohm and output V volt is connected across a variable resistor. The heat generated in the variable resistor is maximum when the current in it is

(a) V/2R

(b) V/4R

(c) 4V/R

(d) V/R

(e) None of the above

You may be remembering the maximum power transfer theorem which states that a direct current source will transfer maximum power to a load when the resistance of the load is equal to the internal resistance of the source.

In the above problem, since the internal resistance of the battery is R, the power dissipated in the variable resistor is maximum when its resistance also is R. The total resistance in the circuit in this condition is R+R = 2R and the current is V/2R.

Electrostatics- Multiple Choice Questions Involving Capcitors

The following questions will help you to understand the calculation of common potential difference and the energy loss on connecting capacitors.

(1) A 2 µF capacitor is charged by a 10 V battery. It is then disconnected from the battery and connected to an uncharged 3 µF capacitor (fig.). The total electrostatic energy stored by the system (of the two capacitors) is (in microjoule)

(a) 40

(b) 60

(c) 80

(d) 100

(e) 200

The charge will be strictly conserved and hence the total charge on the two capacitors will be the same as the initial charge on the 2 μF capacitor. Therefore, we have (from Q = C₁ V₁ = C₂ V₂),

2×10^–6×10 = (2+3)×10^–6×V , where V is the common potential difference between the plates of the capacitors on connecting together. Therefore,

V = 4 volts.

The electrostatic energy stored by the system is ½ (C₁+C₂)V² = (½)×5×10^–6×4² = 40 μJ

(2) In the above question, what is the difference between the initial and final electrostatic energies of the system?

(a) 40

(b) 60

(c) 80

(d) 100

(e) zero

The initial electrostatic energy in the system of the two capacitors is

(½) C₁V₁² = (½)×2×10^–6×10² = 100 μJ

The final energy, as w have seen above, is 40 μJ. Therefore the difference between the initial and final electrostatic energies of the system is 100 – 40 = 60 μJ

Kerala Engineering, Architecture & Medical Entrance Examinations 2008

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 the State for 2008-09.

(a) Medical: (i) MBBS (ii) BDS (iii) B.Pharm. (iv) B.Sc. (Nursing) (v) B.Sc. (MLT)

(vi) BAMS (vii) B.H.M.S (viii) B.S.M.S (Siddha) (ix) B.Sc. Nursing (Ayurveda)

(x) B.Pharm. (Ayurveda) and (xi) B.P.T (Physiotherapy).

(b) Agriculture: (i) B.Sc. Hons. (Agriculture) (ii) B.F.Sc. (Fisheries) (iii) B.Sc. Hons. (Forestry)

(c) Veterinary: B.V.Sc. & 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.

The last date for the receipt of completed applications by the Commissioner for Entrance Examinations is 29-2-2008 Friday (before 5 PM).

Click here for complete details.

Two Kerala Engineering Entrance 2007 Questions on Linear Motion

The following questions appeared in Kerala Engineering Entrance 2007 question paper:

(1) Two balls are dropped to the ground from different heights. One ball is dropped two seconds after the other but they both strike the ground at the same time. If the first ball takes 5 s to reach the ground, then the difference in initial heights is (g = 10 ms^–2)

(a) 20 m

(b) 80 m

(c) 170 m

(d) 40 m

(e) 160 m

The initial height of the first ball is the distance travelled by it in 5 seconds and is given by

x₁ = 0×5 + (½)×10×5², using the relation, x = v₀t + (½)at²

Therefore, x₁ = 125 m.

The time of travel of the second ball is 3 seconds and hence its initial height is given by

x₂ = 0×3 + (½)×10×3² = 45 m.

The difference in initial heights is x₁– x₂ =125 – 45 = 80 m.

(2) A ball is thrown vertically upwards with a velocity of 25 ms^–1 from the top of a tower of height 30 m. How long will it travel before it hits the ground?

(a) 6 s

(b) 5 s

(c) 4 s

(d) 12 s

(e) 10 s

Let us take all downward vectors positive. The displacement is 30 m and is positive. The acceleration due to gravity, g (equal to 10 ms^–2) also is positive. But, the velocity of projection is negative (being upwards).

From the equation, x = v₀t + (½)at², we have

30 = – 25 t + (½)×10×t²

Or, t² – 5 t – 6 = 0, from which t = [5±√(25+24)]/ 2 = + 6 s or –1 s.

Since negative time is impossible, the answer is 6 s.

[You could have taken upward quantities as positive if you wanted to, but you would get the same answer].

Communication Systems: MCQ on Amplitude modulation

Here are some questions (on amplitude modulation) of the type you are most likely to encounter in your entrance exam:

(1) A carrier wave of peak value 16 V is used to transmit a note of frequency 1000 Hz. What should be the peak value of the modulating signal to have a modulation index of 60% using amplitude modulation?

(a) 12 V

(b) 11.2 V

(c) 10.4 V

(d) 9.6 V

(e) 7.2 V

This question demands from you merely a basic knowledge of the modulation process and you are expected to calculate the modulation index (μ) using the equation,

μ = A_m/A_c where A_m and A_c are the amplitudes of the modulating signal and the carrier respectively.

Therefore, 0.6 = A_m/16, from which A_m = 9.6 V. (The frequency of the modulating signal is just a distraction in this question).

[In a practical modulator, the modulating signal amplitude may not be μ times the carrier amplitude. For instance, if the carrier is applied on the base side and the modulating signal is applied on the collector side of a transistor, the carrier amplitude (at the base) will be much less than the modulating signal amplitude, even though the modulation index is less than 100%. Of course, the carrier will appear in the amplified form at the collector. The modulation index (μ) in all situations will be correctly obtained if you consider it as the ratio of the amplitude of variation of the envelope of the amplitude modulated carrier to the amplitude of the unmodulated carrier. But problems with the above type of statement are often seen].

(2) The maximum amplitude of an amplitude modulated wave is 12 V and the minimum amplitude is 4V. The modulation percentage is

(a) 20

(b) 30

(c) 50

(d) 60

(e) 80

The modulation percentage is the modulation index expressed as percentage.

The maximum amplitude of an amplitude modulated (AM) wave is the sum of the amplitude of the unmodulated carrier and the amplitude of variation of the envelope of the AM wave.

The minimum amplitude of an AM wave is the difference between the amplitude (A_c) of the unmodulated carrier and the amplitude (A_m) of variation of the envelope of the AM wave.

Therefore, we have (A_c + A_m) = 12 V

(A_c – A_m) = 4 V

From these equations, A_c = 8 V and A_m = 4 V so that

μ = A_m/A_c = 4/8 = 0.5 = 50%

(3) An unmodulated carrier has a peak value of V volt. When it is amplitude modulated with a sine wave, the maximum peak to peak value (of the modulated carrier) is 3 V volt. The modulation index is

(a) 0.33

(b) 0.5

(c) 0.6

(d) 0.66

(e) 0.8

The maximum peak value (maximum amplitude) of the modulated carrier is 3V/2 volt.

Therefore, (A_c + A_m) = 3V/2

Since the amplitude of the unmodulated carrier (A_c) is V volt, the amplitude (A_m) of the variation of the envelope of the modulated carrier is V/2. Therefore, The modulation index is given by

μ = A_m/A_c = (V/2)/V = 0.5

(4) An amplitude modulated wave has modulation index 0.6. If the power carried by the carrier component in the modulated wave is P_c , what is the power carried by the upper and lower side bands (together)?

(a) 0.6 P_c

(b) 0.54 P_c

(c) 0.36 P_c

(d) 0.46 P_c

(e) 0.18 P_c

The total power (P) carried by an amplitude modulated wave is given by

P = P_c[1+ (μ²/2) where P_c is the carrier power and μ is the modulation index.

Therefore, the power carried by the two side bands together is

P_c ×μ²/2 = P_c×(0.6)²/2 = 0.18 P_c.

[This shows how much power is wasted in the carrier component which does not carry any information. Further, we need only one side band to extract the modulating signal. This points to the benefit of the suppressed carrier single side band transmission].

(5) An AM radio station using double side band transmitted carrier system employs a carrier of frequency 1.2 MHz. If the modulating signal frequency is 2 kHz, the frequencies present in the modulated carrier are

(a) 1.2 MHz and 1.4 MHz

(b) 1 MHz, 1.2 MHz and 1.4 MHz

(c) 1.198 MHz, 1.2 MHz and 1.202 MHz

(d) 1.198 MHz and 1.202 MHz

(e) 1.2 MHz and 1.202 MHz

The double side band transmitted carrier system is the one employed by ordinary broadcast stations. The amplitude modulated wave will contain the carrier frequency (f_c) as well as the upper and lower side frequencies f_c + f_m and f_c – f_m. Therefore, the correct option is (c).