NEET 2013 Questions (MCQ) on Rotational Motion

God used beautiful mathematics in creating the world.

– Dirac, Paul  Adrien Maurice

The following two questions on rotational motion, which were included in the NEET 2013 question paper, are worth noting:

(1) A small object of uniform density rolls up a curved surface with an initial velocity “v”. It reaches up to a maximum height of 3v²/4g with respect to the initial position. The object is

(1) solid sphere

(2) hollow sphere

(3) disc

(4) ring

Initially the object has rotational and translational kinetic energy but zero gravitational potential energy. At the maximum height of 3v²/4g it has zero kinetic energy since the entire kinetic energy is converted into gravitational potential energy. Thus we have

            ½ I ω² + ½ mv² + 0 = mg(3v²/4g) where ‘I’ is the moment of inertia of the object about the axis of rotation, ‘ω’ is the angular velocity, ‘m’ is the mass and ‘v’ is the linear velocity of the object.

Simplifying, ½ I ω² = ¼  mv²

Since ω =v/R the above equation becomes

            Iv²/R² = ½ mv²

Therefore I = mR²/2

This means that the object is a disc [Option (3)].

(2) A rod PQ of M  and length L is hinged at end P. The rod is held horizontally by a massless string tied to point Q as shown in the figure. When the string is cut, the initial acceleration of the rod is

(1) g/L

(2) 2g/L

(3) 2g/3L

(4) 3g/2L

When the string is cut, the rod rotates about the end P and the torque responsible for the rotation is MgL/2.

[The weight Mg of the rod acts through the centre of gravity located at the middle of the rod. The distance of the line of action of the weight from the axis of rotation (the lever arm for the torque) is L/2]

Since torque is equal to Iα where I is the moment of inertia and α is the angular acceleration we have

            Iα = MgL/2 ……………..(i)

The moment of inertia of the rod about the axis of rotation through its end is ML²/3 as given by the parallel axes theorem.

[Moment of inertia of a uniform rod about a central axis perpendicular to its length is ML²/12. Moment of inertia about a parallel axis through the middle is ML²/12 + M(L/2)² = ML²/3].

Substituting for I in equation (i),  we have

            (ML²/3)α = MgL/2

Therefore α = 3g/2L

JEE Main 2013 Questions on Geometric Optics

Science distinguishes a Man of Honor from one of those Athletic Brutes whom undeservedly we call Heroes.

– John Dryden (English poet, critic, dramatist)

The following questions on geometrical optics appeared in JEE Main 2013 question paper:

(1) Diameter of a plano - convex lens is 6 cm and thickness at the centre is 3mm. If speed of light in material of lens is 2 × 10⁸ m/s, the focal length of the lens is :

(1) 15 cm

(2) 20 cm

(3) 30 cm

(4) 10 cm

In the adjoining figure the plano-convex lens is represented by ABCDA. The thickness of the lens is BD (which is equal to 0.3 cm) and the radius of the curved surface of the lens is R.

We have

            R² = (R – 0.3)² + 3²

This gives R ≈ 15 cm.

The refractive index n of the material of the lens is the ratio of the speed of light in free space to the speed in the material of the lens:

             n = (3×10⁸) /(2×10⁸) = 1.5

The focal length f of the lens is given by the lens maker’s equation.

            1/f = (n – 1)(1/R  – 0)

[Since the second surface is plane, the reciprocal of its radius of curvature is zero]

Therefore 1/f = (1.5 – 1)(1/15) from which f = 30 cm

(2) The graph between angle of deviation (δ) and angle of incidence (i) for a triangular prism is represented by :

This is a very simple question. Most of you might have drawn the i - δ  curve by experimentally determining the values of δ for various values of i during your lab session. On increasing the angle of incidence i from a small value, the angle of deviation δ  decreases, attains a minimum value and then increases. The non-linear increase and decrease are in accordance with the relation,

            δ = i + e – A where e is the angle of emergence and A is the angle of the prism.

The correct option is (3). 

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