BITSAT Questions (MCQ) on Direct Current Circuits

There are two ways to live your life. One is as though nothing is a miracle; the other is as though everything is a miracle.

– Albert Einstein

We shall discuss a few interesting multiple choice questions on direct current circuits, which appeared in BITSAT question papers.

The following question appeared in BITSAT 2009 question paper:

(1) In the adjacent shown circuit, a voltmeter of internal resistance R, when connected across B and C reads (100/3) V. Neglecting the internal resistance of the battery, the value of R is

(a) 100 KΩ

(b) 75 KΩ                                                                                  

(c) 50 KΩ

(d) 25 KΩ

Since the voltage drop axross B and C is (100/3) volt, which is 1/3 of the supply voltage, the effective resistance of the parallel combination of the voltmeter resistance R and the 50 KΩ resistor must be 25 KΩ.

[2/3 of the supply voltage is dropped across the 50 KΩ resistor in the gap AB and hence the effective resistance (in the gap BC) that drops 1/3 of the supply voltage must be 25 KΩ].

Since 50 KΩ  in parallel with 50 KΩ makes 25 KΩ, the internal resistance R of the voltmeter must be 50 KΩ [Option (c)].

[If you want to make things more clear, you may write the following mathematical steps:

The current I sent by the battery is given by

            I = 100/[50 + {(50×R)/(50+R)}]

In the above equation we have written the resistances in KΩ so that we will obtain the final answer in KΩ.

Since the voltage drop across B and C is (100/3) volt, we have

            100/3 = {(50×R)/(50+R)}× I

Or, 100/3 = {(50×R)/(50+R)}×100/[50 +{(50×R)/(50+R)}]

Rearranging, (50 + R) [50 +{(50×R)/(50+R)}] = 150 R

Or, 2500 + 50 R + 50 R = 150 R

This gives R = 50 and the answer is 50 KΩ since we have written resistances in KΩ].

The following question appeared in BITSAT 2005 question paper:

(2) Two resistances are connected in two gaps of a metre bridge. The balance point is 20 cm from the zero end. A resistance of 15 Ω is connected in series with the smaller of the two. The null point shifts to 40 cm.the value of the smaller resistance in ohms is:

(a) 3

(b) 6

(c) 9

(d)12

If P is the smaller resistance (Fig.) and Q is the larger resistance, we have

            P/Q = 20/80 = ¼ ………..   (i)

After connecting 15 Ω in series with P we have

            (P + 15)/Q = 40/60 = 2/3……….(ii)

On dividing Eq.(i) by Eq.(ii) we have

P/(P + 15) = 3/8

Therefore, 8P = 3P + 45 from which P = 9 Ω

The following question also appeared in BITSAT 2005 question paper:

(3) The current in a simple series circuit is 5 A. When an additional resistance of 2 Ω is inserted, the current drops to 4 A. The original resistance of the circuit in ohms was:

(a) 1.25

(b) 8

(c) 10

(d) 20

If the emf in the circuit is V volt and the original resistance of the circuit is R ohms we have

            V/R = 5 -----------------(i)

On inserting the additional resistance of 2 Ω we have

            V/ (R+2) = 4 -----------------(ii)

On dividing Eq.(i) by Eq.(ii) we have

            (R+2)/R = 5/4 

Or, 4R + 8 = 5R from which R = 8 Ω.

The following question also appeared in BITSAT 2008 question paper:

(4) A current of 2 A flows in an electric circuit as shown in the figure. The potential difference (V_R – V_S), in volts (V_R and V_S are potentials at R and S respectively) is

(a) – 4

(b) + 2

(c) + 4

(d) – 2

Since the two branches PRQ and PSQ contain equal resistances (10 Ω), the current gets divided equally at the junction P. The same current of 1 A flows through yhe branches. Taking Q as the reference point to measure the potentials at R and S we have

            V_R = + 7 volt and

            V_S = + 3 volt

[Note that V_R is the potential drop produced across the 7 Ω resistor connected between Q and R and V_S is the potential drop produced across the 3 Ω resistor connected between Q and S].

Therefore (V_R – V_S) = + 4 volt

AIPMT (Main and Preliminary) Questions on Nuclear Physics

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 shall discuss a few questions from nuclear physics which appeared in AIPMT question papers. These questions will surely be of use to those who prepare for the National Eligibility Cum Entrance Test (NEET) 2013 for admission to MBBS and BDS Courses. Here are the questions with solution:

(1) The half life of a radioactive nucleus is 50 days. The time interval (t₂ – t₁) between the time t₂ when 2/3 of it has decayed and the time t₁ when 1/3 of it has decayed is

(1) 30 days

(2) 50 days

(3) 60 days

(4) 15 days

This question appeared in AIPMT Main 2012 question paper. You may work it out as follows:

The radioactive decay law is, N = N₀e^-λt where N₀ is the initial number of nuclei, N is the number remaining undecayed after time ‘t’ and λ is the decay constant. This  equation, modified in terms of half life can be written as  N = N₀/2ⁿ where N is number of nuclei remaining undecayed after ‘n’ half life periods.

If 1/3 of the radioactive nucleus decays (and therefore 2/3 of it remains undecayed) in x half life periods, we can write

2N₀/3 = N₀/2^x

Therefore, 2^x = 3/2 so that x = (log 3 – log 2)/log 2 = [{(log 3)/(log 2)} – 1]

The half life of the radioactive nucleus is given as 50 days.

Therefore t₁ = 50[{(log 3)/(log 2)} – 1] days

If 2/3 of the radioactive nucleus decays (and therefore 1/3 of it remains undecayed) in y half life periods, we can write

N₀/3 = N₀/2^y

Therefore, 2^y = 3 so that y = (log 3)/(log 2)

Therefore t₂ = 50(log 3)/(log 2) days

Therefore t₂ – t₁ = 50 days

[You may use the decay law N = N₀e^-λt as such to work out the above problem as follows:

At time t₁ we have

            2N₀/3 = N₀e^-λt1……………(i)

At time t₂ we have

            N₀/3 = N₀e^-λt2……………..(ii)

From the above equations we have

            2 = e ^λ(t2-t1)

Therefore λ(t₂ – t₁) = ℓn 2

Or, (t₂ – t₁) = ℓn 2/λ

But ℓn2/λ is the half life which is 50 days in the present case].     

(2) A radioactive nucleus of mass M emits a photon of frequency ν and the nucleus recoils. The recoil energy will be

(1) hν

(2) Mc² – hν

(3) h²ν²/2Mc²

(4) Zero

This question appeared in AIPMT Preliminary 2011 question paper.

The magnitude of the recoil momentum p of the nucleus is the same as that of the photon and is therefore equal to hν/c where c is the speed of light in free space. The kinetic energy of the nucleus is p²/2M = h²ν²/2Mc²

(3) A nucleus  _nX^m emits one α–particle and two β–particles. The resulting nucleus is

(1) _n–2Y^m–4

(2) _n–4Z^m–6

(3) _nZ^m–6

(4) _nX^m–4

This question also appeared in AIPMT Preliminary 2011 question paper.

When an α–particle is emitted the mass number decreases by 4 and the atomic number decreases by 2. When two β–particles are emitted the atomic number increases by 2 but the mass number is unaffected. The resultant nucleus is X itself since the atomic number is unchanged. But it has mass number (m–4). The correct option is (4).

(4) The decay constant of a radio isotope is λ. If A₁ and A₂ are its activities at times t₁ and t₂ respectively the number of nuclei which have decayed during the time (t₂ – t₁) is

(1) A₁t₁ – A₂t₂

(2) A₁ – A₂

(3) (A₁ – A₂)/λ

(4) λ(A₁ – A₂)

This question appeared in AIPMT Main 2010 question paper.

We have N = N₀e^-λt where N₀ is the initial number of nuclei, N is the number remaining undecayed after time ‘t’ and λ is the decay constant.

The activity A at time t is dN/dt = – λ N₀e^-λt = – λ N.

The negative sign just inucates that the activity decreases with time.

Ignoring the negative sign, the activities A₁ and A₂ at times t₁ and t₂ are given by

            A₁ = λN₁ and

            A₂ = λN₂ where N₁ and N₂ are the number of nuclei at times t₁ and t₂.

Therefore N₁ = A₁/λ and N₂ = A₂/λ 

The number of nuclei which have decayed during the time (t₂ – t₁) is

            N₁ – N₂ = (A₁– A₂)/λ, as given in option (3).

(5) If the nuclear radius of ²⁷Al is 3.6 Fermi, the approximate nuclear radius of ⁶⁴Cu in Fermi is

(1) 2.4

(2) 1.2

(3) 4.8

(4) 3.6

This question also appeared in AIPMT Preliminary 2012 question paper.

We have nuclear radius R = R₀(A)^1/3 where R₀ is a constant an A is the mass number.

If R₁ and R₂  are the nuclear radii of Al and Cu we have

            R₁/R₂ = (27/64)^1/3 = 3/4

Therefore R₂ = 4R₁/3 =  (4×3.6)/3 = 4.8 Fermi

Questions (MCQ) Involving Gravitation

"Men often become what they believe themselves to be. If I believe I cannot do something, it makes me incapable of doing it. But when I believe I can, then I acquire the ability to do it even if I didn't have it in the beginning."

– Mahatma Gandhi

Today we shall discuss a few multiple choice questions involving gravitation. The first question pertains to a simple binary star system.

In a simple binary star system two stars orbit around their centre of mass under their mutual gravitational attractive force. The two stars constituting the binary star system need not be of the same mass. Note that Kepler’s laws are applicable to binary stars also. Click to see this.

Star systems with more than two stars (multiple star systems) are also present. About half of the stars in our galaxy, the milky way, are part of binary star systems or multiple star systems. Here is the first question:

(1) Consider a simple binary star system (Fig.) in which the member stars are identical, orbiting in a circle of radius r. If each star has mass m, the orbital speed of each star is

(a) √(Gm/r)

(b) √(Gm/2r)

(c) √(Gm/4r)

(d) √(2Gm/r)

(e) √(4Gm/r)

The gravitational force between the stars supplies the centripetal force required for the circular motion. Therefore we have

             Gmm/(2r)² = mv²/r where G is the gravitational constant and v is the orbital speed.

The above equation gives

            v = √(Gm/4r)

The following question is related to a simple multiple star system: 

(2) Consider a multiple star system containing four stars of the same mass m arranged symmetrically and orbiting around their centre of mass C (Fig.) in the clockwise direction in a circle of radius r. What is the orbital speed of each star?

(a) 4[(Gm/r)(1+2√2)]^1/2

(b) 2[(Gm/2r)(1+2√2)]^1/2

(c) [(Gm/r)(1+2√2)]^1/2

(d) (½)[(Gm/r)(1 + 2√2)]^1/2

(e) (¼)[(Gm/r)(1+2√2)]^1/2

The gravitational force F₁ on a star due to the diametrically opposite star is given by

            F₁ = Gmm/(2r)² = Gm²/4r²

[Note that the distance between the diametrically opposite stars is 2r]

Evidently F₁ is directed towards the centre of mass C of the star system.

The other two stars exert gravitational forces F₂ and F₃ as shown in the figure. These forces are directed at 45º with respect to F₁ and they have the same magnitude given by

                        F₂= F₃ = Gmm/(r√2)² = Gm²/2r²

[Note that the distance between adjacent stars is (√2)r]

            F₂ and F₃ have equal components (Gm²/2r²)cos 45º directed towards the centre of mass C of the star system. The net radial force acting on a star is therefore equal to F₁ + F₂ cos 45º + F₃ cos 45º = Gm²/4r² + (Gm²/2r²)cos 45º + (Gm²/2r²)cos 45º.

The radial force thus works out to (Gm²/r²)[(1/4) + (1/√2)] = (Gm²/r²)[(1 + 2√2)/4]

            Since the above radial force supplies the centripetal force required for circular motion, we have

            mv²/r =  (Gm²/r²)[(1 + 2√2)/4]  

This gives the orbital speed, v = (½)[(Gm/r)(1 + 2√2)]^1/2, as given in option (d).

(3) A particle of mass 4m is kept fixed at point P (Fig.) in the xy-plane. Another free particle of mass m at the origin is found to be unperturbed when a third fixed particle of mass 6m also is present in the xy-plane. If the x-coordinate and y-coordinate of the particle of mass 4m are –√2 and +√2 respectively, what are the x-coordinate and y-coordinate respectively of the particle of mass 6m?

(a) +√3 and –√3

(b) –√3 and +√3

(c) –√6 and +√6

(d) +√2 and –√2

(e) +3 and –3

The free particle of mass m at the origin is unperturbed since the gravitational forces on it due to the fixed particles are equal in magnitude and opposite in direction. This can happen only if the three particles are on the same straight line, with the free particle of mass m in between the fixed particles.

Equating the magnitudes of the gravitational forces, we have

            G(4m)m/2² = G(6m)m/r² where G is the constant of gravitation and r is the distance of the particle of mass 6m from the origin.

[Note that the distance of the particle of mass 6m from thr origin is {(√2)² + (√2)²}^1/2 = 2

The above equation gives

            Gm² = 6Gm²/r² from which r = √6

Since the line joining the particles evidently bisects the angle XOY’ the x-coordinate and y-coordinate of the particle of mass 6m must be of equal value d such that

            √(d² + d²) = √6

Therefore d =√3

Since the particle of mass 6m is located in the 4^th quadrant, its x-coordinate and y-coordinate are +√3 and –√3 respectively [Option (a)].