AIEEE 2008 and AIPMT 2008 Questions on Digital Circuits

Questions from digital circuits at your level are simple and you should never omit them. The following question appeared in All India Engineering/Architecture Entrance Examination 2008 question paper:

In the circuit below, A and B represent two inputs and C represents the output.

The circuit represents

(1) AND gate

(2) NAND gate

(3) OR gate

(4) NOR gate

If the inputs A and B are low, the diodes nwill not conduct and the potential at the output point C will be low (ground potential). If either A or B is high (at high voltage level), the corresponding diode will conduct and will place the output point C at the high voltage level. If both A and B are high, again the output will be high. The circuit therefore implements OR operation [Option (3)].

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In the above question we have taken, as usual, the high voltage level as the logic 1 level and the low voltage level as the logic 0 level. This system is the positive logic system.

If the low voltage is taken as logic 1 level and the high voltage is taken as logic 0 level, the systerm is called negative logic system. Unless specified otherwise, all systems are positive logic systems.

In the negative logic system, the above circuit will become AND gate.

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The following question appeared in All India Pre-Medical/Pre-Dental Entrance Examination (AIPMT) 2008 question paper:

The circuit

is equivalent to

(1) NAND gate

(2) NOR gate

(3) OR gate

(4) AND gate

The second gate is a NAND gate. Since its inputs are shorted, it functions as a NOT gate. So the circuit is a NOR gate followed by Two NOT gates. NOR followed by NOT is OR and OR followed by NOT is once again NOR. So the correct option is (2).

AIPMT 2008 Questions on Newton’s Laws of Motion

The following questions which appeared in All India Pre-Medical/Pre-Dental Entrance Examination (AIPMT) 2008 are worth noting:

(i) A shell of mass 200 g is ejected from a gun of mass 4 kg by an explosion that generates 1.05 kJ of energy. The initial velocity of the shell is

(1) 80 ms^–1

(2) 40 ms^–1

(3) 120 ms^–1

(4) 100 ms^–1

This question is meant for testing your grasp of the laws of conservation of momentum and energy. If v and V represent the initial velocity of the shell of mass m and the recoil velocity of the gun of mass M respectively, we have (from the law of conservation of momentum)

mv + MV = 0 from which V = – (m/M)v = – (0.2/4)v = – v/20 ms^–1

[The negative sign shows that the recoil velocity of the gun is opposite to the velocity of the shell].

The energy of the explosion is used to impart kinetic energy to the shell and the gun so that we have

½ (mv² + MV²) = 1.05×10³

Or, ½ [0.2 v² + (4v²/400 )] = 1.05×10³

This gives v² = 10⁴ so that v = 100 ms^–1.

(ii) Sand is being dropped on a conveyor belt at the rate of M kg/s. The force necessary to keep the belt moving with a constant velocity of v ms^–1 will be

(1) 2Mv newton

(2) Mv/2 newton

(3) zero

(4) Mv newton

Since there is a time rate of change of momentum equal to Mv per second, the force required is Mv Newton.

IIT – JEE 2008 Question on Errors in Measurement

Today we will discuss a few questions on errors in measurement. Students are generally found to commit mistake in working out questions involving the calculation of error. The following question appeared in IIT – JEE 2008 question paper:

Students I, II and III perform an experiment for measuring the acceleration due to gravity (g) using a simple pendulum. They use different lengths of the pendulum and/or record time for different number of oscillations. The observations are shown in the table.

Least count for length = 0.1 cm.

Least count for time = 0.1 s.

Student .Length of pend(cm).No.of osc (n).Total time (s)..Time period (s)

....I.............64.0............ .........8...................128.0 ...............16.0....

....II ...........64.0......................4.....................64.0................16.0....

....III..........20.0......................4.....................36.0..................9.0.....

If E_I, E_II and E_III are the percentage errors in g, i.e., (∆g/g)×100 for students I, II and III respectively,

(a) E_I = 0

(b) E_I is minimum

(c) E_I = E_II

(d) E_II is maximum

The period of oscillation (T) of a simple pendulum of length ℓ is given by

T = 2π√(ℓ/g)

Therefore, g = 4π² ℓ/T² so that the fractional error in g is given by

∆g/g = (∆ℓ/ℓ) + 2(∆T/T)

[The above expression is obtained by taking logarithm of both sides and then differentiating. Note that the sign of the second term on the RHS is changed from negative to positive since we have to consider the maximum possible error].

Here ∆ℓ = 0.1 cm and ∆T = 0.1 s

The percentage error is 100 times the fractional error so that

E_I = ∆g/g = [(0.1/64) + 2(0.1/128)]×100 = 5/16 %,

E_II = ∆g/g = [(0.1/64) + 2(0.1/64)]×100 = 15/32 % and

E_III = ∆g/g = [(0.1/20) + 2(0.1/36)]×100 = 19/18 %

Thus E_I is minimum so that the correct option is (b).

Now, consider the following MCQ:

A firm manufacturing electronic watches claims that the maximum error in the time indicated by their watches is 5 seconds in six months. What will be maximum possible difference between the times indicated by two watches in one year?

(a) 5 s

(b) 10 s

(c) 20 s

(d) 40 s

(e) zero

This is a very simple question. But you must remember that both positive and negative errors are possible. The maximum possible difference between the times indicated by the watches will be obtained when one watch loses time (slow) and the other watch gains time (fast).

The maximum time lost in one year is 10 s. The maximum time gained in one year also is 10 s. Therefore, the maximum possible difference between the times indicated by the clocks in one year will be 20 s.

Here is another question:

A 200 Ω carbon film resistor with 1% tolerance is connected in series with a 100 Ω carbon film resistor with 5% tolerance. The effective resistance of the combination is (in Ω)

(a) 300 ± 7 Ω

(b) 300 ± 6 Ω

(c) 300 ± 3 Ω

(d) 300 + 18 Ω

(e) 300 +7 Ω

The effective resistance (R) is given by

R = R₁ + R₂ where R₁ = (200 ± 2) Ω and R₂ = (100 ± 5) Ω

Thus R =(200 ± 2) Ω + (100 ± 5) Ω = 300 ± 7 Ω

Multiple Choice Questions involving Lens Maker’s Equation

While attempting to solve problems in optics related to spherical mirrors and lenses, many among you will have some confusion about the application of the inevitable sign convention. Get your confusion cleared by going through the post on this site here. You can find all posts in optics on this site by clicking on the label ‘OPTICS’ below this post. Your doubts regarding the application of the sign convention can be cleared completely only by working out questions involving the sign convention.

We will now discuss a few questions involving lens maker’s equation:

(1) Crown glass of refractive index 1.5 is used to make a plano concave lens of focal length 40 cm in air. What should be the radius of curvature of the curved face?

(a) 20 cm

(b) 40 cm

(c) 60 cm

(d) 80 cm

(e) 120 cm

This is a situation where you have to use the lens maker’s equation,

1/f = (n₂/n₁ – 1)(1/R₁ – 1/R₂) where ‘f’ is the focal length of the lens, R₁ and R₂ are its radii of curvature, n₂ is its refractive index and n₁ is the refractive index of the medium in which the lens is placed.

Since the lens is plano concave and hence diverging, its focal length is negative according to the Cartesian sign convention. The above equation on substituting the known values (n₁ = 1, n₂ = 1.5 and R₂ = ∞) becomes

–1/40 = (1.5 – 1)( 1/R₁ – 0)

This gives R₁ = –20 cm [Option (a)].

You obtain the answer as negative in accordance with the sign convention, indicating that the curved face is concave towards the incident ray. If you had taken the second face as curved, you would have put R₁ = ∞ and the answer would be R₂ = 20 cm. This too conforms to the Cartesian sign convention, indicating that the second face is convex towards the incident ray and hence the radius of curvature is positive.

In the present problem, there will be no need of worrying too much about the sign convention if you remember that the focal length (in air) and the radius of curvature will have the same value in the case of biconvex and biconcave lenses of equal radii of curvature if the refractive index of the lens is 1.5. In the case of plano convex and plano concave lenses made of material of refractive index 1.5, the radius of curvature of the curved face will be half the value of the focal length.

(2) A biconvex lens has the same radii of curvature for its faces. If its focal length in air is equal to the radii of curvature of its faces, its focal power when immersed in a liquid of refractive index 1.5 will be (in dioptre)

(a) 0.667
(b) 1.5
(c) 0.5
(d) zero
(e) data insufficient.
You may substitute the same values for the focal length in air and the radii of curvature of the faces in the lens maker’s equation and satisfy yourself that the refractive index of the lens is 1.5.[Since it is a biconvex lens, f is positive, R₁ is positive and R₂ is negative. Also, put the numerical values of R₁ and R₂ equal to f].Since the refractive index of the liquid is the same as that of the lens, there is no convergence or divergence for the rays of light and the focal power will be zero.
(3) The radius of curvature of the convex face of a plano convex lens is 15 cm and the refractive index of the material is 1.4. Then the power of the lens in dioptre is
(a) 1.6
(b) 1.566
(c) 2.6
(d) 2.66
(e) 1.4
The above question appeared in Kerala Engineering Entrance 2008 question paper.

You have to use the lens maker’s equation,

1/f = (n₂/n₁ – 1)(1/R₁ – 1/R₂) where ‘f’ is the focal length of the lens, R₁ and R₂ are its radii of curvature, n₂ is its refractive index and n₁ is the refractive index of the medium in which the lens is placed.

Note that you have to substitute the focal length in metre in this equation which gives you the focal power 1/f itself. Substituting the given values,

1/f = (1.4 – 1)(1/0.15 – 0) = 2.66 nearly.

You will find more useful questions at apphysicsresources.

I have no special talents. I am only passionately curious. 

                                                                    – Albert Einstein 

Electronics: Multiple Choice Questions on Zener Diodes

In electronics, occasionally you will get questions meant for high lighting the principle of operation and use of Zener diodes. Questions in this section are usually simple. Consider the following question which appeared in KEAM (Medical) 2008 question paper:

In the given circuit, the current through the resistor 2 kΩ is

(a) 2 mA

(b) 4 mA

(c) 6 mA

(d) 1 mA

(e) 10 mA

This is a very simple question. The voltage across the reverse biased Zener diode is its breakdown voltage, which is indicated in the figure as 12 V. The load resistor 2 kΩ is connected across the Zener diode and hence the current through it is 12 V/2 kΩ = 6 mA.

Now consider the following question pertaining to the points you have to remember in using a Zener diode as a simple shunt voltage regulator (of the type in the above question):

In the simple shunt voltage regulator circuit shown in the adjoining figure, the unregulated supply voltage varies between 8 V and 10 V. If the maximum reverse current that the Zener diode can safely handle is 100 mA and the Zener diode should draw a minimum current of 5mA (for reliable operation of the voltage regulator), what is the expected maximum current drawn by the load resistor R_L?

(a) 95 mA

(b) 105 mA

(c) 200 mA

(d) 10 mA

(e) 110 mA

Since the maximum current allowed through the Zener diode is 100 mA, the maximum current drawn by the Zener diode and the load resistor together is 100 mA itself. If the load is disconnected, the entire current will flow through the Zener diode and this is to be limited to 100 mA.

As the Zener diode is expected to draw at least 5 mA (which happens when the load draws the maximum current), the maximum expected load current is (100 mA – 5 mA) = 95 mA.

[Note that in the shunt voltage regulator, the Zener diode and the load together draws the same current irrespective of the input voltage variation and the load current variation (within the design limits)].

In the above question, what is the value of the series resistance R?

(a) 10 Ω

(b) 20 Ω

(c) 40 Ω

(d) 60 Ω

(e) 95 Ω

You have to consider the maximum unregulated input voltage to determine the value of R. Here it is 10 V. Since the regulated voltage (break down voltage of the Zener diode) is 6 V, the voltage drop across R must be 10 V– 6 V = 4 V.

Since the current through R in all situations (within design limits) is 100 mA, the value of R is given by

R = 4 V/ 100 mA = 4 V/ 0.1 A = 40 Ω.

IIT-JEE 2008 Multiple Choice Questions on Electrostatics

The following two questions from Electrostatics were included under Straight Objective Type (Multiple choice, single answer type) in the IIT-JEE 2008 question paper:

(1) Consider a system of three charges q/3, q/3 and –2q/3 placed at points A, B and C respectively as shown in the figure. Take O to be the centre of the circle of radius R and angle CAB = 60º

(a) The electric field at point O is q/8πε₀R²

(b) The potential energy of the system is zero

(c) The magnitude of the force between the charges at C and B is q²/54πε₀R²

(d) The potential at point O is q/12πε₀R

The electric field at O is due to the negative charge at C only since the equal positive charges situated at A and B will produce equal and opposite fields at O (and they will mutually cancel). The field at O is therefore negative and the option (a) is obviously wrong. Option (b) also is obviously wrong.

The magnitude of the force (F) between the charges at C and B is given by

F = (1/4πε₀) [(2q/3)(q/3)/(2Rsin60º)²].

Thus F = q²/54πε₀R² as given in option (c).

(2) A parallel plate capacitor with plates of unit area and separation d is filled with a liquid of dielectric constant K = 2. The level of liquid is d/3 initially. Suppose the liquid level decreases at a constant speed V. The time constant as a function of time t is

(a) 6ε₀R/(5d + 3Vt)

(b) (15d + 9Vt) ε₀R/(2d² – 3dVt – 9V²t²)

(c) 6ε₀R/(5d – 3Vt)

(d) (15d – 9Vt) ε₀R/(2d² + 3dVt – 9V²t²)

When a dielectric slab of thickness t₁ and dielectric constant K is introduced in between the plates of a parallel plate air capacitor of plate area A and plate separation d, the effective capacitance changes from ε₀A/d to ε₀A/[d–t₁ + (t₁/K)]

The effective capacitance C of a parallel plate capacitor with parallel dielectric slabs of thickness t₁, t₂, t₃ etc. of dielectric constants K₁, K₂, K₃ etc. respectively is given by the series combined value of capacitors with these dielectrics. Therefore,

1/C = 1/(K₁ε₀A/t₁) + 1/(K₂ε₀A/t₂) + 1/(K₃ε₀A/t₃) + etc.

With a single slab of thickness t₁ and dielectric constant K introduced between the plates, we have

1/C = 1/(Kε₀A/t₁) + 1/[ε₀A/(d – t₁)] from which C = ε₀A/[d–t₁ + (t₁/K)]

In the present problem, the thickness of the layer of liquid (which serves as the dielectric slab) at the instant t is (d/3) – V t and the thickness of the air space is (2d/3) + V t.

The effective capacitance is therefore given by

C = ε₀A/[{(2d/3) + Vt} + {(d/3) – Vt}/k)]

Substituting for A = 1 and K = 2, we obtain C = 6ε₀/(5d + 3Vt).

The time constant of the circuit = RC = 6ε₀R/(5d + 3Vt).

Reality is merely an illusion, albeit a very persistent one. 

–Albert Einstein 

All India Pre-Medical/Pre-Dental Entrance Examination (AIPMT) 2008 Questions on Rotational Motion

The following questions appeared in All India Pre-Medical/Pre-Dental Entrance Examination (AIPMT) 2008 question paper:

(1) A thin rod of length L and mass M is bent at its mid point into two halves so that the angle between them is 90º. The moment of inertia of the bent rod about an axis passing through the bending point and perpendicular to the plane defined by the two halves of the rod is

(a) ML²/12

(b) ML²/6

(c) √2 ML²/24

(d) ML²/24

You should note that moment of inertia is a scalar quantity so that the total moment of inertia of the bent rod is the sum of the moments of inertia of the two halves about the common axis. [The axis must be the same when you add the moments of inertia, for obvious reasons].

The moment of inertia of each half is (M/2)(L/2)²/3 so that the moment of inertia of the entire bent rod is 2×[(M/2)(L/2)²]/3 = ML²/12

[You will certainly remember that the moment of inertia of a thin uniform rod of mass m and length ℓ about a perpendicular axis through its mid point is mℓ²/12. The moment of inertia about a perpendicular axis through the end, as give by the parallel axis theorem is (mℓ²/12) + m(ℓ/2)² = mℓ²/3]

(2) The ratio of the radii of gyration of a circular disc to that of a circular ring, each of the same mass and radius, about their respective axes is

(a) 1 : √2

(b) √2 : 1

(c) √2 : √3

(d) √3 : √2

Since the moments of inertia of circular disc and circular ring about their axes are respectively MR²/2 and MR², their radii of gyration about their axes are R/√2 and R respectively.

[Remember that I = Mk² where I is the moment of inertia and k is the radius of gyration].

Therefore, the ratio of the radii of gyration = (R/√2)/ R = 1/√2.

(3) A roller coaster is designed such that the riders experience “weightlessness” as they go round the top of a hill whose radius of curvature is 20 m. The speed of the car at the top of the hill is between

(a) 15 m/s and 16 m/s

(b) 16 m/s and 17 m/s

(c) 13 m/s and 14 m/s

(d) 14 m/s and 15 m/s

When the rider feels weightless, the weight of the rider is equal in magnitude to the centrifugal force.

Therefore, we have mg = mv²/r where m is the mass of the rider, v is the velocity and r is the radius of the path. From this

v = √(rg) = √(20×10) = √200 = 14.14 m/s.

The correct option therefore is (4).

You can find all posts related to rotational motion on this site by clicking on the label ‘rotational motion’ below this post.

You will find a useful post on the equations to be remembered in rotational motion and circular motion at aphysicsresources

IIT-JEE 2008: Linked Comprehension Type Multiple Choice Questions on Bohr Model of Hydrogen-like Atoms

The IIT-JEE Physics question paper is no more a nightmare to students aspiring for a seat in one of the IIT’s, consequent on the changes effected from 2007. If you have good grasp of fundamentals, you should definitely take up the challenge for brightening your future. Students appearing for other entrance examinations such as Physics GRE, AP Physics, AIEEE, KEAM and the like also will find the IIT-JEE question paper very useful for their preparation.

Here are three multiple choice questions on Bohr atom model which appeared under Linked Comprehension Type in IIT-JEE 2008 question paper:

Paragraph for Question Nos. 1 to 3

In a mixture of H – He⁺ gas (He⁺ is singly ionized He atom) H atoms and He⁺ ions are excited to their respective first excited states. Subsequently, H atoms transfer their total excitation energy to He⁺ ions (by collisions). Assume that the Bohr model of atom is exactly valid.

(1) The quantum number n of the state finally populated in He⁺ ions is

(A) 2

(B) 3

(C) 4

(D) 5

The singly ionised helium atom is hydrogen like since it contains a single electron. In a hydrogen like atom the energy (E) of the electron in the n^th orbit is given by

E = – 13.6 Z²/n² electron volt where Z is the atomic number.

For hydrogen atom the energies are – 13.6 eV, – 3.4 eV, – 1.51 eV and – 0.85 eV for values of n equal to 1, 2, 3 and 4 respectively

For the helium ion the energies are four fold, equal to – 54.4 eV, – 13.6 eV, – 6.04 eV and – 3.4 eV for values of n equal to 1, 2, 3 and 4 respectively

The energy difference between the electrons in first excited states of hydrogen atom and helium ion is (– 3.4 eV) – (– 13.6 eV) = 10.2 eV.

On collision, the helium ion absorbs this energy and occupies the 3^rd excited state (n = 4)

of energy – 13.6 eV + 10.2 eV = –3.4 eV.

Therefore the correct option is (C).

(2) The wave length of light emitted in the visible region by He⁺ ions after collision with H atoms is

(A) 6.5×10^–7 m

(B) 5.6×10^–7 m

(C) 4.8×10^–7 m

(D) 4.0×10^–7 m

The visible photons have wave lengths in the range 4000 Ǻ to 7000 Ǻ and their energies lie in the range 3.1 eV to 1.771 eV.

[It will be useful to remember the range of energies in eV. You will definitely remember the wave length range of visible photons which you can convert into energy range using the equation, λ E = 12400  where λ is the wave length in Angstrom and E is the energy in electron volt].

The energy difference between the He⁺ ion levels for n = 3 and n = 4 correspond to visible photon energy: – 3.4 eV– (– 6.04 eV) =  2.64 eV

The wave length of the light emitted in the visible region is therefore 124000/2.64 = 4800 Ǻ, nearly. The answer is thus 4.8×10^–7 m [Option (C)].

(3) The ratio of the kinetic energy of the n = 2 electron for the H atom to that of the He⁺ ion is

(A) ¼

(B) ½

(C) 1

(D) 2

The kinetic energy of the electron in the n^th orbit is 13.6 Z²/n².

[Note that the kinetic energy and the total energy of the electron are equal in amount. But the kinetic energy is positive where as the total energy is negative].

The required ratio is Z_H²/Z_He² = ¼.

All posts on Bohr model of hydrogen atom on this site can be accessed by clicking on the label ‘Bohr model’ below this post.

You can find a few posts on hydrogen atom at apphysicsresources.blogspot.com also. One such post is here.