Multiple Choice Questions on Dimensions of Physical Quantities [Including KEAM (Engineering) 2011 question]

I believe in standardizing automobiles, not human beings

– Albert Einstein

In most of the entrance examination question papers you will find at least one question on dimensions of physical quantities. Here are a few typical multiple choice questions in this section:

(1) Which one among the following quantities is a dimensional constant?

(a) Dielectric constant of water

(b) Speed of light in free space

(c) Viscosity of water

(d) Ratio of specific heats of a diatomic gas

(e) Reynolds number

Options (a) and (c) are not constants since the dielectric constant and viscosity depend on other parameters. Options (d) and (e) are dimensionless numbers.

[Reynolds number (used in assessing the turbulence of fluids) is the ratio of inertial force to force of viscosity]

The correct option is the speed of light in free space which is a fundamental constant with dimensions LT^–1.

(2) If F denotes force and t time, then in the equation F = at^–1 + bt², the dimensions of a and b respectively are

(a) LT^–4 and LT^–1

(b) LT^–1 and LT^–4

(c) MLT^–4 and MLT^–1

(d) MLT^–1 and MLT^–4

(e) MLT^–3 and MLT^–2

The above question appeared in Kerala Engineering Entrance ((KEAM) 2011 question paper.

The dimensions of at^–1 and bt² have to be that of force which is MLT^–2. Therefore the dimensions of a must be MLT^–1 and the dimensions of b must be MLT^–4.

The correct option is (d).

The following question appeared in Karnataka CET 2004 question paper:

(3) The physical quantity having the same dimensions as Planck’s constant h is

(1) Boltzmann constant

(2) force

(3) linear momentum

(4) angular momentum

Most of you will remember that the unit of h is joule second. Therefore h has the dimensions of the product of work and time which is ML²T^–2×T = ML²T^–1.

Angular momentum is the moment of linear momentum and has dimensions L×MLT^–1 = ML²T^–1.

The correct option is (4).

The following question appeared in EAMCET 2008 Engineering Entrance Exam question paper. You will answer it in no time if you remember that the dimensions of Planck’s constant are those of angular momentum.

(4) The energy (E), angular momentum (L) and universal gravitational constant (G) are chosen as fundamental quantities. The dimensions of universal gravitational constant in the dimensional formula for Planck’s constant (h) is

(a) zero

(b) – 1

(c) 5/3

(d) 1

Since the dimensions of Planck’s constant are those of angular momentum, it follows that the dimensions of universal gravitational constant in the dimensional formula for Planck’s constant (h) is zero.

[In terms of E, L and G which are assumed as fundamental quantities in the above question, Planck’s constant (h) has zero dimension in E, one dimension in L and zero dimension in G].

EAMCET (Engineering Entrance) 2010 Questions on Capacitors

“Non-violence is the greatest force at the disposal of mankind. It is mightier than the mightiest weapon of destruction devised by the ingenuity of man”

– Mahatma Gandhi

Today we will discuss three questions which appeared in EAMCET (Engineering Entrance) 2010 question paper. Questions (1) and (2) are from electrostatics while question no.3 is from transient phenomena.

(1) Two capacitors of capacities 1 μF and C μF are connected in series and the combination is charged to a potential difference of 120 V. If the charge on the combination is 80 μC, the energy stored in the capacitor of capacity C in microjoule is

(1) 1800

(2) 1600

(3) 14400

(4) 7200

The effective capacitance of the series combination is 1×C/(1+C) μF = C/(1+C) μF.

[Remember that when capacitors C₁, C₂, C₃,….etc are connected in series, the effective capacitance C_eff is given by the reciprocal relation, 1/C_eff = 1/C₁ + 1/C₂ + 1/C₃ + …. etc.]

We retain the unit of capacitance as μF for convenience.

Since the charge carried by the combination is 80 μC, we have (from Q = CV with usual notations)

80 = [C/(1+C)]×120

[We have substituted the charge in microcoulomb itself since the capacitance is in microfarad].

This gives 80 C + 80 = 120 C from which C = 2 μF

When capacitors are in series, all capactors carry the same charge. Therefore, the charge on the 2 μF capacitor is 80 microcoulomb.

The energy U of the 2 μF capacitor is given by

U = Q²/2C = 80²/(2×2) = 1600 microjoule.

[Since Q is in microfarad and C is in microcoulomb, the energy is in microjoule].

(2) The potential difference between two parallel plates is 10⁴ volts. If the plates are separated by 0.5 cm, the force on an electron between the plates is

(1) 32×10^–13 N

(2) 0.32×10^–13 N

(3) 0.032×10^–13 N

(4) 3.2×10^–13 N

This is a very simple question. The electric field (E) between the plates is given by

E = V/d where V is the potential difference and d is the separation between the plates

The force F on the electron (of caharge e = 1.6×10^–19 coulomb) is given by

F = Ee = Ve/d = (10⁴×1.6×10^–19) /(0.5×10^–2) = 3.2×10^–13 N

(3) A capacitor of capacity 0.1 μF connected to a resistor of 10 MΩ is charged to a certain potential and then made to discharge through the resistor. The time in which the potential will take to fall to half its original value is (Given log₁₀ 2 = 0.3010)

(1) 2 sec

(2) 0.693 sec

(3) 0.5 sec

(4) 1 sec

If V₀ is the initial voltage across the capacitor of capacitance C, the voltage V at time t during the discharge through the resistance R is given by

V = V₀ e^–t/RC where e is the base of natural logarithms.

Therefore V/V₀ = e^–t/RC

Since V/V₀ = ½ in the question, we have

½ = e^–t/RC

Or, e^t/RC = 2

Taking logarithm, t/RC = ln 2

Therefore, t = RC ln 2 = 10×10⁶×0.1×10^–6×ln 2

If you remember that ln 2 = 0.693, you can write the answer immediately as 0.693 sec.

[You should definitely know that the relation between natural logarithm (with base e) and common logarithm (base 10) of quantity x is

ln_e x = 2.303 log₁₀ x

So ln 2 = 2.303×0.3010 = 0.693]

EAMCET 2010 Questions (MCQ) on Surface Tension

Today we will discuss two questions on surface tension which appeared in EAMCET 2010 question papers.

The following multiple choice question appeared in EAMCET 2010 (Engineering) question paper:

The excess pressure inside a spherical soap bubble of radius 1 cm is balance by a column of oil (Sp. gravity 0.8) 2 mm high. The surface tension of oil is

(1) 3.92 N/m

(2) 0.0392 N/m

(3) 0. 392 N/m

(4) 0.00392 N/m

The excess pressure ∆P insie a spherical bubble of radius r is given by

∆P = 4 T/r where T is the surface tension.

The pressure p exerted by a liquid column of height h is given by

p = hρg where ρ is the density of the liquid.

Therefore we have

4 T/r = hρg from which T = rhρg/4

Substituting for known values, T = (1×10^–2×2×10^–3×0.8×10³×10)/4 = 0.004 N/m [Option (4)].

[If you substitute g = 9.8 ms^–2 instead of 10 ms^–2 (as we did above for convenience), you will get the answer as 0.00392 N/m].

Here is the question which appeared in EAMCET 2010 (Agriculture and Medicine) question paper:

A spherical liquid drop of diameter D breaks up to n identical spherical drops. If the surface tension of the drop is ‘σ’, the change in energy in this process is

(1) πσD²(n^1/3 – 1)

(2) πσD²(n^2/3 – 1)

(3) πσD²(n – 1)

(4) πσD²(n^4/3 – 1)

The surface energy (E₁) of a drop of radius R is given by

E₁ = 4πR²σ

Or E₁ = πD²σ where D is the diameter of the drop.

When a drop of radius R breaks into n identical droplets, the radius r of each droplet is given by (on equating the volumes)

(4/3) πR³ = n×(4/3) πr³

Therefore, r = R/n^1/3

The total surface energy (E₂) of all the n droplets is given by

E₂ = n×4πr²σ = n×4π (R/n^1/3) ²σ = 4πR²σn^1/3

Or, E₂ = πD²σn^1/3

The change in energy is E₂ – E₁ = πσD²(n^1/3 – 1)

Two EAMCET (Medical) 2009 Questions on Photons and Electrons

Today we will discuss two multiple choice questions from modern physics. These questions are from the section ‘electrons and photons’ and they were included in the EAMCET (Medical) 2009 question paper:

(1) In Millikan’s oil drop experiment, a charged oil drop of mass 3.2×10^–14 kg is held stationary between two parallel plates 6 mm apart, by applying a potential difference of 1200 V between them. How many electrons does the oil drop carry?

(1) 7

(2) 8

(3) 9

(4) 10

The electric field ‘E’ between the plates is V/d where V is the potential difference and d is the separation between the plates.

Therefore E = 1200/(6×10^–3) = 2×10⁵ volt per metre.

The electric force on the drop is Eq where ‘q’ is the charge on the drop. Since the drop is stationary, the electric force balances the weight mg of the drop.

Therefore Eq = mg so that

q = mg/E = (3.2×10^–14×10)/( 2×10⁵) = 1.6×10^–18 coulomb.

Since the electronic charge is 1.6×10^–19 coulomb, the number of excess electrons carried by the drop is (1.6×10^–18) /(1.6×10^–19) = 10.

(2) Electrons accelerated by a potential of ‘V’ volts strike a target material to produce ‘continuous X-rays’. Ratio between the de Broglie wave length of the electrons striking the target and the shortest wave length of the ‘continuous X-rays’ emitted is

(1) h/√(2Vem)

(2) (1/c) √(2m/Ve)

(3) (1/c) √(Ve/2m)

(4) he/√(Ve/2m)

The de Broglie wave length λ of the electron is given by

λ = h/p where h is Planck’s constant and p is the momentum of the electron.

The shortest wave length λ_min of the ‘continuous X-rays’ produced by the striking electrons is given by

hc/λ_min = Ve where c is the speed of light in free space and e is the electronic charge.

[Note that we have equated the entire energy of the electron to the energy of the X-ray photon for obtaining minimum wave length (or, the maximum frequency ν_max):

eV = hν_max = hc/λ_min].

Therefore, λ_min = hc/Ve.

The ratio of wave lengths required in the problem is

λ /λ_min = (h/p)/(hc/Ve ) = Ve/pc………..(i)

The kinetic energy of the electron is eV and the momentum p of the electron is related to its kinetic energy as

eV = p²/2m where m is the mass of the electron.

Therefore, p = √(2mVe)

Substituting this in Eq (i), we obtain

λ /λ_min = (1/c) √(Ve/2m)

Multiple Choice Questions on Diodes [Including EAMCET 2009 (Engineering) and AIEEE 2006 Questions]

Questions from electronics, especially those involving the use of diodes are generally simple at the level expected of you. Here are a few questions on diodes:

(1) Currents flowing in each of the following circuits A and B respectively are

(1) 1A, 2 A

(2) 2 A, 1 A

(3) 4 A, 2 A

(4) 2 A, 4 A

This question appeared in EAMCET 2009 (Engineering) question paper.

In circuit A both diodes are forward biased and hence the circuit reduces to two 4 Ω resistors connected across the 8 V battery. Since the parallel combined value of the two resistors is 2 Ω, the current delivered by the battery is 8 V/2 Ω = 4 A.

In circuit B one diode is forward biased and the other diode is reverse biased and hence the circuit reduces to just one 4 Ω resistor connected across the 8 V battery. The current delivered by the battery is therefore 8 V/4 Ω = 2 A. The correct option is (3).

The following questions [No. (2) and (No. (3)] were included in AIEEE 2006 question paper:

(2) In the following, which one of the diodes is reverse biased?

You should note that all potentials are with respect to the ground. Therefore the diode in circuit (1) is reverse biased.

[In circuit (2) the anode of the diode is at a higher positive potential compared to its cathode and hence it is forward biased. In circuit (3) the cathode of the diode is at a higher negative potential and hence it is forward biased. In circuit (4) the cathode of the diode is at a negative potential and hence it is forward biased].

(3) The circuit has two oppositely connected ideal diodes in parallel. What is the current flowing in the circuit?

(1) 1.33 A

(2) 1.71 A

(3) 2.00 A

(4) 2.31 A

Since the diode D₁ is reverse biased, no current will flow through D₁ and the 3 Ω resistor. The current delivered by the battery is limited by the 4 Ω and the 2 Ω resistors only and is equal to 12 V/(4+2)Ω = 2 A.

The following question is meant for checking your grasp of the behaviour of semiconductor diodes:

(4) In the circuit shown the diodes used are silicon rectifier diodes which require a forward bias of 0.7 volt for appreciable conduction. Their leakage current is negligible. The internal resistance of the battery is insignificant. The potential difference between the terminals A and B is very nearly equal to

(a) 6 V

(b) 3 V

(c) 5.3 V

(d) 0.7 V

(e) 0 V

The upper diode is reverse biased and can be ignored. The lower diode which is connected across the terminals A and B is forward biased and hence keeps the voltage across A and B at 0.7 V [Option (d)].

[You can use the voltage drop across a forward biased diode as a small reference voltage in electronic circuits just as you use the relatively larger breakdown voltages of reverse biased zener diodes].

MCQs on Magnetism including EAMCET 2009 (Medical) Question

Some multiple choice questions on magnetism have already been posted on this site. You can access them by clicking on the label ‘magnetism’ below this post. Today we will discuss a few more multiple choice questions on magnetism.

(1) The period of oscillation of a magnetic needle in a magnetic field is T. If an identical bar magnetic needle is tied at right angles to it to form a cross (fig), the period of oscillation in the same magnetic field will be

(a) 2^1/4T

(b) 2^1/2T

(c) 2T

(d) T√3

(e) T/2

The period of oscillation (T) of the single magnetic needle is given by

T = 2π√(I/mB) where ‘I’ is the moment of inertia of the magnetic needle about the axis of rotation, ‘m’ is the magnetic dipole moment of the needle and ‘B’ is flux density of the magnetic field.

When two magnetic needles are tied together to form a cross, the moment of inertia becomes 2I and the magnitude of the magnetic dipole moment becomes √(m² + m²) = m√2.

[Note that magnetic dipole moment is a vector quantity. Two identical vectors (each of magnitude m) at right angles will yield a resultant magnitude m√2].

The resultant magnetic moment will be directed along the bisector of the angle between the axes of the individual magnets since the magnets are identical. In the absence of a deflecting torque, the resultant dipole moment vector will align along the applied magnetic field B. On deflecting from this position, the system will oscillate with period T₁ given by

T₁ = 2π√(2I/mB√2) = 2π√(I√2/mB) = 2^1/4T

(2) Three identical magnetic needles each L metre long and of dipole moment m ampere metre are joined as shown without affecting their magnetisation. At points B and C unlike poles are in contact. The dipole moment of this system is

(a) m

(b) 2m

(c) 3m

(d) 3m/2

(e) 5m/2

The distance (AD) between the ends of the compound magnet is 2L. Since the pole strength is m/L, the dipole moment of the compound magnet is (m/L)2L = 2m

(3) A magnet of length L and moment M is cut into two halves (A and B) perpendicular to its axis. One piece A is bent into a semicircle of radiur R and is joined to the other piece at the poles as shown in the figure below:

Assuming that the magnet is in the form of a thin wire initially, the moment of the resulting magnet is given by

(1) M/2π

(2) M/π

(3) M(2 + π)/2π

(4) Mπ/(2 + π)

The above question appeared in EAMCET 2009 (Medicine) question paper.

The distance between the poles of the resulting magnet is (L/2) + 2R

Since the semicircular portion of radius R is made of the magnetised wire of length L/2, we have L/2 = πR so that R = L/2π and 2R = L/π

Therefore, length of the resulting magnet (L/2) + (L/π)

The pole strength (p) of the magnet is given by

p = M/L

Therefore, the dipole moment of the resulting magnet = Pole strength×Length = (M/L)[ (L/2) + (L/π)] = M(2 + π)/2π