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.