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