IIT-JEE 2007 Matrix-Match Type Question on Modern Physics

Section IV of Part I Physics question paper of IIT-JEE 2007 contained 3 Matrix-Match Type questions. Each question contained statements given in two columns. Statements A,B,C,D in Column I had to be matched with statements p,q,r,s in Column II. The answers to the questions had to be appropriately bubbled as shown at the end of this post.

Here is the question involving topics in modern physics:

Some laws / processes are given in Column I. Match these with the physical phenomena given in Column II and indicate your answer by darkening appropriate bubbles in the 4 ×4 matrix in the ORS

Column I Column II

(A) Transition between (p) Characteristic X- rays

two atomic energy levels
(B) Electron emission from (q) Photoelectric effect

a material

(C) Mosley’s law (r) Hydrogen spectrum

(D) Change of photon (s) β-decay

energy

Characteristic X-rays and hydrogen spectrum are produced by electron transition between two energy levels in an atom. So, (A) is to be matched with (p) and (r).

In photo electric effect and β-decay, electrons are emitted from a material. So, (B) is to be matched with (q) and (s).

Mosley’s law [√f α Z ] relates the frequency ‘f’ of a particular characteristic X-ray (e.g., K_α ) to the atomic number Z of the target in the X-ray tube. So, (C) is to be matched with (p).

In photoelectric effect, the energy of the incident photon is used in dislodging electrons from a photo sensitive surface. So, (D) is to be matched with (q).

The appropriate bubbles darkened are shown in figure.

Five Kerala Engineering Entrance 2007 Questions (MCQ) from Electronics

Five questions from Electronics and three questions from Communication Systems were included in the Kerala Engineering Entrance 2007 (KEAM) question paper. Here are the questions from Electronics:

(1) The real time variation of input signals A and B are as shown below. If the inputs are fed in to NAND gate, then select the output signal from the following

The NAND gate will give low output when all inputs are high. The output will be high in all other cases. The inputs A and B are high during the interval from 6 seconds to 8 seconds only. So, the output must be low during this interval only. This is indicated in option (b).

(2) In a common emitter amplifier, the current gain is 62. The collector resistance and the input resistance are 5 kΩ and 500 ohms respectively. If the input voltage is 0.01 V, the output voltage is

(a) 0.62 V (b) 6.2 V (c) 62 V (d) 620 V (e) 0.01 V

The collector current is given by I_c = βI_B with usual notations.

But, the base current I_B = V_i/Ri where V_i and R_i are the base voltage (input voltage) and input resistance respectively. Therefore, I_B = 0.01/ 500 = 2 ×10^–5 A.

The corresponding collector current is βI_B = 62×2 ×10^–5 A = 1.24×10^–3 A

The output voltage is the voltage drop across the collector resistance (output resistance) R_o.

So, output voltage = I_cR_o = 1.24×10^–3×5000 = 6.2 V.

[If you remember the simple expression for the voltage gain of a low frequency common emitter amplifier, the above result can be arrived at in a shorter time:

Voltage gain A_v = βR_o/R_i = 62×5000/500 = 620. Therefore output voltage = 620×0.01V = 6.2 V].

(3) The current gain of a transistor in common base mode is 0.995. The current gain of the same transistor in common emitter mode is

(a) 197 (b) 201 (c) 198 (d) 202 (e) 199

The expression for the current gain in common emitter mode is β = α /(1–α) where α is the current gain in common base mode.

Therefore, β = 0.995 /(1–0.995) = 199.

(4) When the forward bias voltage of a diode is changed from 0.6 V to 0.7 V, the current changes from 5 mA to 15 mA. The forward bias resistance is

(a) 0.01 Ω (b) ) 0.1 Ω (c) ) 10 Ω (d) ) 100 Ω (e) ) 0.2 Ω

Forward resistance, R = (Change in forward biasing voltage) /(Change in forward current) = (0.1 V)/ 10 mA = 0.1/(10×10^–3) Ω = 10 Ω.

(5) The energy gap between conduction band and the valence band of a material is of the order of 0.7 eV. Then it is

(a) an insulator (b) ) a conductor (c) a semiconductor

(d) an alloy (e) a superconductor

If the energy gap between the conduction band and the valence band is less than 3 eV, the substance is a semiconductor. So, the correct option is (c).

Two Questions on Specific Heat Capacity of a Gas

Questions similar to the following one are often found in Medical and Engineering entrance examination question papers:

One mole of an ideal mono atomic gas is mixed with two moles of an ideal diatomic gas. The ratio of specific heats of the mixture is

(a) 1.5 (b) 1.4 (c) 10/6 (d) 15/11 (e) 19/13

You should remember that the values of molar specific heats at constant volume C_v) for mono atomic and diatomic gases are respectively (3/2)R and (5/2)R where R is universal gas constant. The values of molar specific heat at constant pressure C_p) are therefore (5/2)R and (7/2)R respectively, in accordance with Meyer’s relation [C_p – C_v =R].

Therefore, C_v of the mixture = [1×(3/2)R + 2×(5/2)R]/(1+2) = (13/6)R

C_p of the mixture = C_v + R = (19/6)R.

Ratio of specific heats of the mixture, γ = C_p/C_v = 19/13.

[Generally, if n₁ moles of a gas having ratio of specific heats γ₁ is mixed with n₂ moles of a gas having ratio of specific heats γ₂, the ratio of specific heats of the mixture is given by the relation, (n₁+ n₂)/(γ– 1) = n₁/( γ₁ –1) + n₂/( γ₂ –1). You can easily arrive at this result].

If one mole of an ideal mono atomic gas is mixed with one mole of an ideal diatomic gas, the ratio of specific heats of the mixture is 1.5. As an exercise, check this.

Now consider the following MCQ:

The following sets of experimental values of C_v and C_p of a given sample of gas were reported by five groups of students. The unit used is calorie mole^–1 K^–1. Which set gives the most reliable values?

(a) C_v = 3, C_p = 4.5 (b) C_v = 2, C_p = 4 (c) C_v = 3, C_p = 4.9 (d) C_v = 2.5, C_p = 4.5 (e) C_v = 3, C_p = 4.2

Since the minimum value of C_v is (3/2)R which is the value for a mono atomic gas, when you express it in calorie mole^–1 K^–1, the minimum value is approximately 3. [R = 8.3 J mole^–1 K^–1 = 2 calorie mole^–1 K^–1, approximately]. Options (b) and (d) are therefore not acceptable. Out of the remaining three options, (c) is the most reliable since C_p – C_v = R, which should be 2 calorie mole^–1 K^–1 very nearly.

Direct current Circuits- Two IIT-JEE 2007 MCQ’s

The following MCQ’s appeared in IIT-JEE 2007 question paper:

(1) A circuit is connected as shown in the figure with the switch S open. When the switch is closed, the total amount of charge that flows from Y to X is

(a) 0 (b) 54 μC (c) 27 μC (d) 81 μC

When S is open, the P.D. across 3 μF and 6 μF capacitors are 6V and 3V respectively. [The charges on them are equal and the P.D. across them are inversely proportional to their capacities].

When S is closed, the P.D. across them become 3V and 6V respectively since the PD across the 3Ω and 6Ω resistors are 3V and 6V respectively. Positive charges have to flow from Y to X to achieve this condition.

Since the potential at point X is to be made 6V for this, the charge flowing to the 3 μF capacitor is 3μF×3V = 9μC. The charge flowing to the 6μF capacitor is 6μF×3V = 18μC. The total charge = 9 μC +18 μC =27 μC.

[You may have certain doubts regarding this solution. Once you note that the potentials of the lower potential plate of the 3 μF capacitor and the higher potential plate of the 6 μF capacitor are raised by 3 volts, by the charges flowing from Y to X, your doubts will be cleared].

(2) A resistance of 2Ω is connected across one gap of a metre bridge (the length of the wire is 100 cm) and an unknown resistance, greater than 2Ω, is connected across the other gap. When these resistances are interchanged, the balance point shifts by 20 cm. Neglecting any corrections, the unknown resistance is

(a) 3Ω (b) 4Ω (c) 5Ω (d) 6Ω

If the balancing length is measured initially on the side of the unknown resistance X, it will shift from 60 cm to 40 cm. [Remember that the balancing length is measured from the same side before and after interchanging. You might have noted that the balance point in a meter bridge shifts symmetrically with respect to the mid point of the bridge wire].

Therefore, we have X/2 = 60/40, from which X = 3Ω.

[If the balancing length is measured on the side of the known resistance, it will change from 40 cm to 60 cm. In this case, 2/X = 40/60, from which X =3Ω].

Three Karnataka CET 2005 Questions on Electrostatics

The following three questions which appeared in Karnataka CET 2005 are aimed at testing your understanding of basic principles in electrostatics.

(1) The electric flux for Gaussian surface A that encloses the charged particles in free space is ( given that q₁ = –14 nC, q₂ = 78.85 nC, q₃ = – 56 nC)

(a) 10³ Nm²C^–1 (b) 10³ CN^–1m^–2

(c) 6.32×10³ Nm²C^–1 (d) 6.32×10³ CN^–1m^–2

The Gaussian surface B is just for distracting you. Remember, Gaussian surface is an imagined surface and it has no action on the charge configuration. When you consider the Gaussian surface A, the net charge enclosed by the surface is –14 nC + 78.85 nC + (– 56 nC) = 8.85 nC.

The electric flux over the closed surface A, according to Gauss theorem, is q/ε₀ where q is the net charge enclosed by the surface.

Therefore, electric flux = 8.85×10^–9/8.85×10^–12 = 10³ Nm²C^–1 [Option (a)].

[You should note that the electric flux is the product of electric field and area so that its unit is (N/C)×m² = Nm²C^–1].

(2) The work done in carrying a charge ‘q’ once round a circle of radius ‘r’ with a charge ‘Q’ at the centre is

(a) qQ/4πε₀r (b) qQ/4πε₀²r² (c) qQ/4πε₀r² (d) none of these

The work done is zero since there is no potential difference between the initial and final positions of the charge q. The correct option therefore is (d). Note that this is the case for any closed path of any shape.

(3) An air filled parallel plate condenser has a capacity of 2 pF. The separation of the plates is doubled and the inter space between the plates is filled with wax. If the capacity is increased to 6 pF, the dielectric constant of wax is

(a) 2 (b) 3 (c) 4 (d) 6

The capacitance of air cored capacitor is ε₀A/d where A is the area of the plates and ‘d’ is the separation between the plates. When the separation is doubled, the capacitance is halved and becomes 1 pF.

The capacitance when the inter space is filled with a dielectric of dielectric constant ‘K’, the capacitance is Kε₀A/d so that it is increased to K times the value with air as the dielectric. Since the increment is from 1 to 6, K = 6.

Two Questions (MCQ) on Angular Momentum

The following two questions are similar in that both require the calculation of angular momentum in central field motion under inverse square law forces.

(1) An artificial satellite of mass ‘m’ is orbiting the earth of mass ‘M’ in a circular orbit of radius ‘r’. If ‘G’ is the gravitational constant, the orbital angular momentum of the satellite is

(a) [GMm²r]^1/2 (b) [GMmr]^1/2 (c) [GMm/r]^1/2 (d) [GMm²/r²]^1/2 (e) [GMm²r³]^1/2

The orbital angular momentum of a satellite is mvr where ‘v’ is the orbital speed. [Angular momentum = Iω = mr²ω = mr²v/r = mvr where ‘I’ is the moment of inertia and ‘ω’ is the angular velocity of the satellite].

The centripetal force required for the circular motion of the satellite is supplied by the gravitational pull so that we have

mv²/r = GMm/r²

From this, m²v²r² = GMm²r so that angular momentum mvr = [GMm²r]^1/2

(2) In a hydrogen atom in its ground state, the electron of mass ‘m’ is moving round the proton in a circular orbit of radius ‘r’. The orbital angular momentum of the electron is (with usual meaning for symbols)

(a) [m²e²r/4πε₀]^1/2 (b) [m²er/4πε₀]^1/2 (c) [me²r²/4πε₀]^1/2

(d) [me²r/4πε₀]^1/2 (e) [me²r³/4πε₀]^1/2

The steps for finding the orbital angular momentum of the electron are similar to those in question No.1, with the difference that the centripetal force is supplied in this case by the electrostatic attractive force between the proton and the electron.

We have mv²/r = (1/4πε₀)e²/r² from which m²v²r² = (1/4πε₀) ×me²r, so that orbital angular momentum, mvr = [me²r/4πε₀]^1/2

Two Multiple Choice Questions on Digital Circuits

Most of you will like the section on digital circuits, especially because at the class 12 level you have to study very simple circuits. Consider the following MCQ:

For the circuit shown, logic level 1 is +5 volts and logic level 0 is 0 volt. This circuit is

(a) an EXOR gate (b) an AND gate (c) an INVERTER

(d) an OR gate (e) a NOR gate

If both inputs A and B are zero, the diodes will not conduct and the output point will be at ground potential so that the output Y = 0.

If at least one input is at logic 1 level (+5 volts), the diode connected to that input will conduct. The diode connected to the output also will conduct making the output high (+5V). Thus the output Y=1. The same thing happens if both inputs are high.

So, the circuit is an OR gate.

Now, consider the following question:

The digital circuit shown in the figure implements

(a) OR operation (b) EXOR operation

(c) NAND operation (d) NOR operation (e) AND operation

The first gate is a NAND gate. The second gate also is a NAND gate whose inputs are shorted. But when the inputs of a NAND gate are shorted, it becomes an inverter (NOT gate). So, the circuit is a NAND followed by an inverter which is altogether an AND gate [Option (e)].

An IIT-JEE 2007 Question on Centre of Mass

Here is an assertion-reason type MCQ (involving centre of mass motion) which appeared in IIT-JEE 2007 question paper:

STATEMENT-1

If there is no external torque on a body about its centre of mass, then the velocity of the centre of mass remains constant

because

STATEMENT-2

The linear momentum of an isolated system remains constant.

(a) Statement-1 is True, Statement-2 is True; Statement-2 is a correct

explanation for Statement-1

(b) Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct

explanation for Statement-1

(c) Statement-1 is True, Statement-2 is False

(d) Statement-1 is False, Statement-2 is True

This is a very simple question meant for checking your understanding of basic principles. But if you are not careful, you are liable to pick out a wrong answer!

Generally, if there are external torques or forces or both, the velocity of the centre of mass will change. So, the condition of no external torque alone is not sufficient to ensure the constancy of the velocity of the centre of mass. Statement-1 is therefore false.

[You should also note that an external torque about the centre of mass will not change the velocity of the centre of mass].

Statement-2 is evidently true.

The correct option therefore is (d).

IIT-JEE 2007 Questions on Rotational Motion

The following straight objective type multiple choice question appeared in IIT-JEE 2007 question paper:

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

(a) ring (b) solid sphere (c) hollow sphere (d) disc

The body has translational and rotational kinetic energies and these are completely converted in to gravitational potential energy at the maximum height so that we can write

½ Mv² + ½ Iω² = Mgh Where M is the mass, I is the moment of inertia and ω is the angular velocity of the body and h is the maximum height reached. Since ω = v/R where R is the radius of the rolling body, the above equation can be rewritten as

½ Mv² + ½ I(v²/R²) = Mg×(3v²/4g)

From this, I = MR²/2, which is the value for a disc [Option (d)].

The following three questions which appeared in IIT-JEE 2007 question paper are Linked Comprehension Type multiple choice questions:

Two discs A and B are mounted coaxially on a vertical axle. The discs have moments of inertia I and 2I respectively about the common axis. Disc A is imparted an initial angular velocity 2ω using the entire potential enargy of a spring compressed by a distance x₁. Disc B is imparted an angular velocity ω by a spring having the same spring constant and compressed by a distance x₂. Both the discs rotate in the clockwise direction.

Question(1):

The ratio x₁/x₂ is

(a) 2 (b) ½ (c) √2 (d) 1/√2

Equating the potential energies of the springs to the kinetic energies of the discs, we have

½ k x₁² = ½ I×4ω² and

½ k x₂² = ½ ×2I×ω² for the two cases. Here ‘k’ is the spring constant.

From these equations, x₁/x₂ = √2

Question 2:

When disc B is brought in contact with disc A, they acquire a common angular velocity in time t. The average frictional torque on one disc by the other during this period is

(a) 2Iω/3t (b) 9Iω/2t (c) 9Iω/4t (d) 3Iω/2t

Since the angular momentum is conserved, we have

I×2ω + 2I×ω = (I+2I)×ω’ where ω’ is the common angular velocity of the discs. [We have added the angular momenta since they are in the same direction].

From this, ω’ = (4/3)ω

Disc A will have an angular retardation of magnitude ‘α₁’ during the time ‘t’ where as disc B will have an angular acceleration of different magnitude ‘α₂’ during the time ‘t’.

Considering disc A, we have ω’ = 2ω – α₁t from which α₁ = (2ω – ω’)/t = 2ω/3t since ω’ = (4/3)ω.

The average frictional torque exerted on A by B = Iα₁ = 2Iω/3t

[An equal and opposite torque will be exerted on B by A. Check by finding α₂ and hence 2Iα₂].

Question 3:

The loss of kinetic energy during the above process is

(a) Iω²/2 (b) Iω²/3 (c) Iω²/4 (d) Iω²/6

Loss of kinetic energy = Initial kinetic energy – Final kinetic energy

= ½ I×(2ω)² + ½ ×2I×ω² – ½ ×3I×(4ω/3)² = (Iω²)/3

Sound – Two Questions on Beats

Questions similar to the following one have appeared in various entrance tests:

A set of 31 tuning forks are so arranged that each gives 5 beats per second with the previous one. If the frequency of the last tuning fork is double that of the first, the frequency of the third tuning fork is

(a) 250 Hz (b) 220 Hz (c) 180 Hz (d) 160 Hz (e) 150 Hz

Don’t be scared by the relatively large number of forks. This is a very simple question.

If n₁, n₃ and n₃₁ are the frequencies of the 1^st, 3^rd and 31^st forks, we have

n₃₁ = 2n₁ since the frequency of the last fork is double that of the first.

Further, n₃₁ = n₁ + 30×5 since there are 30 increments in frequency (each of 5 Hz) from the first fork to the 31^st fork.

Thus we have 2n₁ = n₁ + 150 from which n₁ = 150 Hz.

The frequency of the 3^rd fork, n₃ = n₁ + 2×5 = 150 + 10 = 160 Hz.

Now, consider the following MCQ:

When a tuning fork is excited, molecules of air vibrate in accordance with the equation x = A cos(512πt). When this tuning fork and another identical tuning fork loaded with a little wax are excited together, 4 beats are heard. The frequency of the second fork loaded with wax is

(a) 516 Hz (b) 508 Hz (c) 384 Hz (d) 260 Hz (e) 252 Hz

When the first tuning fork is excited, the vibrations of the air molecules are simple harmonic with angular frequency ω = 512π as is evident from the form of the equation, x = A cos(512πt).

The linear frequency of vibration of the first fork is n = ω/2π = 512π/2π = 256 Hz.

The frequency of the 2^nd tuning fork before loading with wax was therefore 256 Hz. After loading with wax, its frequency is lowered. Since the beat frequency is 4 Hz, its frequency (after loading) is 256 – 4 = 252 Hz.

Bohr Model of Hydrogen Atom – An IIT-JEE 2007 question

The following MCQ appeared in IIT-JEE 2007 question paper:

The largest wave length in the ultraviolet region of the hydrogen spectrum is 122 nm. The smallest wave length in the infra red region of the hydrogen spectrum (to the nearest integer) is

(a) 802 nm (b) 823 nm (c) 1882 nm (d) 1648 nm

The wave number 1/λ, which is the number of waves per meter length in the case of hydrogen spectrum is given by Rydberg’s relation,

1/λ = R(1/n₁² – 1/n₂²) where R is Rydberg’s constant and n₁ and n₂ are integers.

Ultraviolet radiations are obtained in the Lyman series of hydrogen spectrum when electron transitions take place from higher orbits (of quantum number n>1)to the innermost orbit (of quantum number n=1). So, for the Lyman series, n₁=1 and n₂ = 2,3,4,…etc. The largest wave length in the Lyman series is obtained when the transition is from 2^nd orbit (n₂=2) to the first orbit (n₁=1).

The smallest wave length in the infra red region is obtained when electron transition occurs from the outermost orbit (n₂ = ∞) to the third orbit (n₁ = 3) and this spectral line is the shortest wave length line in the Paschen series ( for which n₁ = 3 and n₂ = 4,5,6….etc.).

For the largest wave length ultraviolet line we have (expressing the wave length in nanometer),

1/122 = R(1/1² – 1/2²) = 3R/4

For the smallest wave length(λ') infrared line we have

1/λ' = R(1/3² – 1/∞) = R/9

Dividing the first equation by the second, λ'/122 = 3×9/4, from which λ' = 823 nm.

Let us consider another similar question which appeared in Kerala Medical Entrance 2001 question paper:

Given that the longest wave length in Lyman series is 1240 Ǻ, the highest frequency emitted in Balmer series is

(a) 8×10¹⁴ Hz (b) 8×10¹² Hz (c) 8×10¹⁰ Hz (d) 8×10³ Hz (e) 8×10² Hz

In the Rydberg’s relation, 1/λ = R(1/n₁² – 1/n₂²), n₁=1 and n₂ = 2,3,4,…etc., for the Lyman series. For the Balmer series (which is in the visible region), n₁=2 and n₂ = 3,4,5….etc.

The longest wave length in Lyman series is obtained for n₂ = 2 and the highest frequency (shortest wavelength) in Balmer series is obtained for n₂ = ∞.

Rydberg’s relation for the above two cases are (expressing wave lengths in Angstrom),

1/1240 = R(1/1² – 1/2²) = 3R/4 and

1/ λ' = R(1/2² – 1/∞) = R/4

Dividing the first equation by the second,

λ'/1240 = 3, from which λ' = 3720 Ǻ.

The frequency (ν) of this line is given by

ν = c/λ' = (3×10⁸) /(3720×10^–10) = 8×10¹⁴ Hz.

IIT-JEE 2007- Three Questions from Optics

The following three questions appeared in IIT-JEE (Part I) 2007 question paper:

(1) In an experiment to determine the focal length (f) of a concave mirror by the u-v method,a student places the object pin A on the principal axis at a distance x from the pole P. The student looks at the pin and its inverted image from a distance, keeping his/her eye in line with PA. When the student shifts his/her eye towards left, the image appears to the right of the object pin. Then,

(a) x less than f (b) f less than x less than 2f (c) x = 2f (d) x greater than 2f

This question is a simple one and is intended to check your understanding of techniques used in experimental physics in addition to your theoretical knowledge. Since the image appears to the right of the object when the student shifts his eye towards the left, the image is nearer to the student and hence the image distance ‘v’ is greater than the object distance ‘u’. This is possible only if the object is placed between f and 2f. So the correct option is (b).

[The image is real since it is inverted as mentioned in the question. The problem can be worked out even if this fact is not mentioned in this problem].

(2) A ray of light traveling in water is incident on its surface open to air. The angle of incidence is θ, which is less than the critical angle. Then there will be

(a) only a reflected ray

(b) only a refracted ray and no reflected ray

(c) a reflected ray and a refracted ray and the angle between them would be less

than 180° – 2θ

(d) A reflected ray and a refracted ray and the angle between them would be

greater than 180° –2θ

If the angle of incidence is not equal to or greater than the critical angle, there will be partial transmission and partial reflection. As is evident from the figure, the angle between the reflected ray and the refracted ray (angle SQR) is less than 180° – 2θ. The correct option is (c).

(3) STATEMENT-1

The formula connecting u,v and f for a spherical mirror is valid only for mirrors
whose sizes are very small compared to their radii of curvature

because

STATEMENT-2

Laws of reflection are strictly valid for plane surfaces, but not for large spherical surfaces.

(a) Statement-1 is True, Statement-2 is True; Statement-2 is a correct

explanation for Statement-1

(b) Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct

explanation for Statement-1

(c) Statement-1 is True, Statement-2 is False

(d) Statement-1 is False, Statement-2 is True

This is an assertion-reason type MCQ. You might have noted that the formula connecting u, v and f was derived considering rays close to the principal axis so that only a small portion of the mirror surrounding the pole is involved. Statement-1 is therefore true. Statement-2 is false since the laws of reflection are applied to a ray which is incident at a given point. The size of the mirror and its curvature are not involved here.

You can find all the posts in Optics on this site by clicking on the label 'optics' below this post.

MCQ involving Magnetic Force on Current carrying Conductors

Questions on magnetic force on current carrying conductors are interesting and often simple. Occasionally you may find questions which may baffle you. See the following question:

A copper wire is bent in the form of a sine wave of wave length ‘λ’ andpeak to peak value ‘a’ as shown in figure. A magnetic field of flux density ‘B’ tesla acts perpendicular to the plane of the figure in the entire region. If the wire carries a steady current ‘I’ ampere, the magnetic force on the wire is

(a) I√(a²+λ²) B (b) IaB (c) I(a+λ)B (d) I(λ– a)B (e) IλB

This is a very simple question and the correct option is (e).

This is a very simple question and the correct option is (e).

Imagine the wire to be made of a large number of horizontal and vertical elements as shown. There are as many vertical elements carrying current upwards as there are those carrying currents downwards. The magnetic forces on them will be leftwards and rightwards and they will get canceled. But the magnetic forces on the horizontal elements will be in the same direction ( either upwards or downwards, depending in the directions of the magnetic field and the current) and they will get added to produce a net force IλB.

The following MCQ appeared in IIT–JEE Screening 2003 question paper:

A conducting loop carrying a current ‘I’ is placed in a uniform magnetic field pointing in to the plane of the paper as shown. The loop will have a tendency to

(a) contract

(b) expand

(c) move towards positive X-axis

(d) move towards negative X-axis

The magnetic force everywhere on the loop is radially outwards as given by Fleming’s left hand rule. So, the loop has a tendency to expand [Option (b)].

In the above question, suppose that in place of the option ‘expand’, you had the option, ‘move towards the positive Z-direction’. In that case also, the correct option would be (b) because the loop will act as a magnetic dipole whose south pole is the nearer face. The loop will therefore move towards the reader.

MCQ on Viscosity – Poiseuille’s Flow

Poiseuille’s formula for the volume ‘V’ of a liquid of density ‘ρ’ flowing in time ‘t’ through a capillary tube of length ‘L’ and radius ‘r’ under a pressure difference ‘P’ between the ends of the tube is

V = (πPr⁴t)/ (8Lη) where ‘η’ is the coefficient of viscosity of the liquid. [You should note that this formula holds good only if the flow is slow and steady (stream-lined)].

The rate of flow (volume flowing per second) ‘Q’ is given by

Q = (πPr⁴)/ (8Lη)

Here P = hρg where ‘h’ is the height of the liquid column which produces the pressure difference so that

Q = (π hρg r⁴)/ (8Lη)

You will often find questions based on Poiseuille’s formula in entrance tests for admitting students to various courses. Consider the following question:

A capillary tube of length ‘L’ and radius ‘r’ is joined to another capillary tube of length L/4 and radius r/2 A liquid flows through this series combination. If the pressure difference between the ends of the first tube is P, that between the ends of the second tube is

(a) P (b) 4P (c) 8P (d) 16P (e) P/4

Since the tubes are in series, the rates of flow through the tubes are equal so that from Poiseuille’s equation

πPr⁴/8Lη = πP'(r/2)⁴/[8(L/2)η], where P’ is the pressure difference between the ends of the second tube.

From this P' = 16P/2 = 8P.

Now, consider the following MCQ:

Under a constant pressure head, the volume of a liquid flowing per second through a capillary tube of radius 1 mm and length 16 cm is 4 cm³. If another tube of radius 0.5 mm and length 8 cm is connected in series with it and the same pressure head is applied across the combination, the volume of liquid flowing per second will be ( in cm³)

(a) 5/3 (b) 5/6 (c) 4/3 (d) 4/5 (e) 4/9

When tubes are connected in series, the net rate of flow (Q_net) under a given pressure head is given by the reciprocal relation,

1/Q_net = 1/Q₁ + 1/Q₂ + 1/Q₃ +…..etc. where Q₁, Q₂, Q₃.....etc. are the individual rates of flow when the tubes are connected separately to the same pressure heaed.

In the present case, since there are two tubes only, 1/Q_net = 1/Q₁ + 1/Q₂ so that

Q_net = Q₁Q₂/(Q₁+Q₂). Here Q₁ = 4 cm³.

Since the rate of flow through a tube is given by Q = (πPr⁴)/ (8Lη), we have Q α r⁴/L.

Since the radius of the second tube is half that of the first, the rate of flow is reduced to (½)⁴ = 1/16 of that through the first tube on account of this. Since the length of the second tube is half that of the first, the rate of flow is increased to twice that through the first tube on this account. Therefore, rate of flow through the second tube is given by

Q₂ = Q₁×(2/16) = Q₁/8 = 4/8 cm³ = 0.5 cm³.

The net rate of flow is therefore given by Q_net = (4×0.5)/(4+0.5) = 4/9 cm³ [Option (e)].