A Question on Varying Currents

The following MCQ which appeared in KEAM 2007 (Engineering) question paper can claim to be one different from the usual type:

In a closed circuit, the current I (in ampere) at an instant of time t (in second) is given by I = 4 – 0.08t. The number of electrons flowing in 50 s through the cross section of the conductor is

(a) 1.25×10¹⁹ (b) 6.25×10²⁰ (c) 5.25×10¹⁹

(d) 2.55×10²⁰ (e) 4.25×10²⁰

To obtain the electrons flowing in 50 s, you have to calculate the total charge flowing in 50 s.

Total charge Q = ∫Idt = ∫(4–0.08t)dt = [4t–(0.08t²/2)].

Since the limits of integration are 0 and 50 seconds, Q = 4×50–(0.08×50²/2) = 100 coulomb.

Since the electronic charge is 1.6×10^–19 coulomb, the number of electrons flowing in 50 s is 100/(1.6×10^–19) = 6.25×10²⁰.

Change in Internal Energy of a Heated sphere - Two Questions

If you heat a body, its temperature rises and so its internal energy increases. You have come across the increase in internal energy of a gas on many occasions and have noted the difference between the amounts of heat to be supplied on heating the gas at constant volume and at constant pressure to undergo a given temperature rise. Here are two questions involving the work done by a solid on heating it:

(1) If a solid of volume V is heated at constant pressure so as to have a small temperature rise, the work done by the solid will be directly proportional to

(a) V^1/2 (b) V^–1/2 (c) V (d) V^–1 (e) V⁰

(2) The temperature of a copper sphere of volume ‘V’m³ and density ‘ρ’ kg m^–3 increases by a small value ∆T when it absorbs a small quantity ∆Q joule of heat at atmospheric pressure ‘P’ pascal. If the specific heat of copper is ‘C’ J kg^–1K^–1, and the cubical expansivity of copper is ‘γ’ K^–1, the increase in internal energy of the copper sphere is

(a) ∆Q–PV

(b) ∆Q

(c) ∆Q(1– γPV/ρC)

(d) ∆Q(1– γP/ρC)

(e) ∆Q(1– γP/VρC)

Let us consider the second question first since it will give us the answer for the first question also.

The entire heat energy absorbed by the sphere is not used in increasing its internal energy. A small portion is used to do work in expanding (against the atmospheric pressure). Normally we omit this small amount of energy since the expansion of a solid is small. But you have to calculate it here since you are given all data for calculating it.

The rise in temperature of the sphere ∆T = ∆Q/mC where ‘m’ is the mass of the sphere.

Increase in volume of the sphere ∆V = Vγ∆T = (m/ρ)γ(∆Q/mC) = γ∆Q/ρC. [The density ρ can be assumed to be constant as the temperature rise is small].

Therefore, work done against the atmospheric pressure = P∆V = Pγ∆Q/ρC.

The increase in internal energy of the sphere = ∆Q – (Pγ∆Q/ρC) = ∆Q(1– γP/ρC).

So, the correct option is (d).

Now the answer for question No.1 is option (e) since the above expression for increase in internal energy does not contain the volume V.

A Question on Lorentz Force

The following MCQ appeared in Kerala Engineering Entrance 2007 question paper. I have seen this question on many occasions with more or less the same options. Here is the question:

A uniform electric field and a uniform magnetic field exist in a region in the same direction. An electron is projected with a velocity pointed in the same direction. Then the electron will

(a) be deflected to the left without increase in speed

(b) be deflected to the right without increase in speed

(c) not be deflected but its speed will decrease

(d) not be deflected but its speed will increase

(e) be deflected to the right with increase in speed

This is a simple question and the most suitable option is (c) since the magnetic field (being parallel to the electron) does not exert any force, but the electric field decelerates the electron (since the charge on the electron is negative).

Now, suppose the question setter had changed the last option as “not deflected but its speed will first decrease and will then increase”. Then, this last option would be the correct option since the electron would be decelerated first and then would be accelerated in the opposite direction. [Remember speed is a scalar quantity and the modified option takes advantage of this].

Mozilla Firefox Browser

It is about a couple of months since I have been using Mozilla Firefox browser for my blog posts. I find it very convenient since it senses spelling mistakes in the posts and more importantly it saves a lot of time and effort for me when I deal with superscripts and subscripts. I should have known the possibilities of the Mozilla Firefox browser much earlier!

IIT-JEE 2007 Linked Comprehension type Questions (MCQ) on Properties of Matter

Among the IIT-JEE 2007 physics questions there were 4 Linked Comprehension Type questions, each paper carrying two of them. Each Linked Comprehension Type question contained a paragraph based on which three multiple choice questions were asked. Here is the question on properties of matter:

Paragraph for the multiple choice questions:

A fixed thermally conducting cylinder has a radius R and height L₀. The cylinder is open at the bottom and has a small hole at its top. A piston of mass M is held at a distance L from the top surface, as shown in the figure. The atmospheric pressure is P₀.

Questions:

(1) The piston is now pulled out slowly and held at a distance 2L from the top. The pressure in the cylinder between its top and the piston will then be

(A) P₀ (B) P₀/2 (C) P₀/2 + Mg/πR² (D) P₀/2 –Mg/πR²

The correct option is (A) since the air will enter the cylinder through the hole at the top and will keep the pressure at P_०

(2) While the piston is at a distance 2L from the top, the hole at the top is sealed. The piston is then released, to a position where it can stay in equilibrium. In this condition, the distance of the piston from the top is

(A)[2P₀πR²/(πR²P₀ +Mg)](2L)

(B) [(P₀πR²–Mg)/πR²P₀](2L)

(C)[(P₀πR² +Mg)/πR²P₀](2L)

(D) [P₀πR²/( πR²P₀ –Mg)](2L)

If P is the pressure inside (the region between the piston and the top of the cylinder), the weight of the piston is balanced by the thrust (P₀ – P)πR² arising due to the pressure difference between the top and the bottom of the cylinder. Therefore,

(P₀ –P)πR² = Mg from which P = P₀ –Mg/πR² = (πR²P₀ –Mg)/πR²

Now, substitute this value of P in the relation, P₀×2L = P×L’, where L’ is the final distance of the piston from the top. [Note that this equation follows from Boyle’s law].

Thus, P₀×2L = [(πR²P₀ –Mg)/ πR²] × L' from which

L' = [P₀πR²/( πR²P₀ –Mg)](2L).

(3) The piston is taken completely out of the cylinder. The hole at the top is sealed. A water tank is brought below the cylinder and put in a position so that the water surface in the tank is at the same level as the top of the cylinder as shown in the figure. The density of water is ρ. In equilibrium, the height H of the water column in the cylinder satisfies

(A) ρg(L₀–H)² +P₀(L₀–H)+ L₀P₀ = 0

(B) ρg(L₀–H)² –P₀(L₀–H)– L₀P₀ = 0

(C) ρg(L₀–H)² +P₀(L₀–H)– L₀P₀ = 0

(D) ρg(L₀–H)² –P₀(L₀–H)+ L₀P₀ = 0

The correct option is obtained by applying Boyle’s law. The initial pressure of air column of length L₀ is the atmospheric pressure P₀. The final pressure of the air column of length (L₀–H) is [(P₀+(L₀–H)ρg] since the pressure of air inside the cylinder exceeds the atmospheric pressure by the hydrostatic pressure exerted by water column of height (L₀–H).

Applying Boyle's law, P₀L₀ = [(P₀ +(L₀–H)ρg](L₀–H).

Rearranging this you will get the equation in option (c).

MCQ on Drift Velocity of Electrons

Most of you may be aware of the fact that the drift speed of electrons in a current carrying conductor is very small compared to the speed with which a current flows in a circuit. The drift speed is typically of the order of millimeter per second while the speed with which a current flows in a circuit is nearly equal to the speed of light. Understand that the electron need not move from one end of the conductor to the other end for a current to flow. What happens is that an electric field is established in each region of a conductor (at nearly the speed of light) and the electrons in the region drift under its influence.

Now, consider the following MCQ:

A current of 4.4 A is flowing in a copper wire of radius 1 mm. Density of copper is 9×10³ kg m^–3 and its atomic mass is 63.5 u. If every atom of copper contributes one conduction electron then the drift velocity of electrons is nearly

(a) 0.1 mm s^–1 (b) 0.5 mm s^–1 (c) 1 mm s^–1 (d) 1.5 mm s^–1 (e) 5 mm s^–1

The current in the wire is given by

I = nAve where ‘n’ is the number of conduction electrons per unit volume, ‘A’ is the area of cross section of the wire, ‘v’ is the drift velocity of the electrons and ‘e’ is the electronic charge.

From this equation, v = I/nAe.

Since the atomic mass of copper is given as 63.5 g, the number of copper atoms in 0.0635 kg is 6.02 ×10²³. The density of copper being 9×10³ kg m^–3, unit volume of copper contains (9×10³/0.0635) ×6.02×10²³ = 8.5×10²⁸ atoms.

Therefore, number of conduction electrons per unit volume, n = 8.5×10²⁸ (since each atom contributes one conduction electron).

Therefore, drift velocity, v = I/nAe = 4.4/[8.5×10²⁸×π(1×10^–3)²×1.6×10^–19] = 0.1×10^–3 ms^–1(nearly).

The following MCQ appeared in Kerala Engineering Entrance 2007 question paper:

When a current I flows through a wire, the drift velocity of the electrons is ‘v’. When current 2I flows through another wire of the same material having double the length and double the area of cross section, the drift velocity of the electrons will be

(a) v/8 (b) v/4 (c) v/2 (d) v (e) 2v

The drift velocity is given by v = I/nAe.

Since the wires are of the same material, the number density (n) of conduction electrons is the same. The electronic charge (e) is a constant. The current and area of cross section are doubled. Therefore, the drift velocity is unchanged.

Two Questions (MCQ) from Thermal Physics

The following simple questions are for checking whether you have understood certain fundamental principles in Thermal Physics:

(1) Three bodies A, B, and C with thermal capacities in the ratio 1:2:3 are at temperatures T₁, T₂, and T₃ respectively. When A and B are kept in contact, the common temperature is T. When A, B and C are kept in contact, the common temperature is T itself. Then T is equal to

(a) (T₁+ T₂ +T₃)/3 (b) (T₁ – T₂ +T₃)/3 (c) 2(T₁ + T₂)/3 (d) T₂ (e) T₃

Simple questions often make even fairly intelligent students commit mistake and this is one such question.

The crucial point you must note is that the common temperature of A and B is unchanged when C also is kept in contact with them. So, the common temperature of A and B must be the same as the temperature of C which is T₃.

(2) Equal masses of three liquids A, B,and C with specific heats c₁, c₂ and c₃ at temperatures t₁, t₂ and t₃ (all in degree Celsius) respectively are thoroughly mixed. The resulting temperature is

(a) (c₁t₁ + c₂t₂ + c₃t₃)/3

(b) (c₁t₁ + c₂t₂ + c₃t₃)/(c₁ + c₂ + c₃)

(c) 3(c₁t₁ + c₂t₂ + c₃t₃)/(c₁ + c₂ + c₃)

(d) 3(t₁ + t₂ + t₃)/(c₁ + c₂ + c₃)

(e) (t₁ + t₂ + t₃)/ (c₁ + c₂ + c₃)

The initial heat content is (mc₁t₁ + mc₂t₂ + mc₃t₃) where ‘m’ is the mass of each liquid.

The final heat content is (mc₁ + mc₂ + mc₃)t where ‘t’ is the common temperature after mixing. [We have taken the reference heat energy level at zero degree Celsius].

Equating these two, we obtain t = (c₁t₁ + c₂t₂ + c₃t₃)/(c₁ + c₂ + c₃)

Kerala Govt Engineering Entrance 2007-Questions on Alternating Currents

Kerala Government Engineering Entrance 2007 question paper contained four questions from alternating currents out of 72 questions in the physics part. Here are the questions:

(1) A square coil of side 25 cm having 1000 turns is rotated with uniform speed in a magnetic field about an axis perpendicular to the direction of the field. At an instant t the e.m.f. induced in the coil is e = 200 sin100πt. The magnetic induction is

(a) 0.50 T (b) ) 0.02 T (c) ) 10^–3 T (d) ) 0.1 T (e) ) 0.01 T

If you remember the production of alternating voltage when a plane coil rotates in a magnetic field, you will be able to answer this question in no time. The induced alternating e.m.f. is

e = naBω sin (ωt) and the maximum emf induced is naBω where ‘n’ is the number of turns, ‘a’ is the area (of the coil), B is the magnetic induction and ‘ω’ is the angular speed. So, we have

1000×(25×10^–2)²×B×100π = 200. [The value of ‘ω’ is obtained from the form of the equation for the emf].

This gives B = 0.01 T

(2) When a d.c. voltage of 200 V is applied to a a coil of self inductance 2√3/π H, a current of 1 A flows through it. But by replacing d.c. source with a.c. source of 200 V, the current in the coil is reduced to 0.5 A. Then the frequency of a.c. supply is

(a) 100 Hz (b) 75 Hz (c) 60 Hz (d) 30 Hz (e) 50 Hz

In an LR circuit the direct current is limited by resistance only. On applying a direct voltage of 200V, the current flowing is 1 A which means that the resistance of the coil is 200 Ω.

When the a.c. source is connected, the current is given by I = V/Z where the impedance Z is given by Z = √(R²+L² ω²)

Therefore 0.5 = 200/√[200²+(2√3/π)² ω²] = 200/√[200²+(2√3/π)² (2πf)²], from which f = 50 Hz.

(3) In a LR circuit, the value of L is (0.4/π) henry and the value of R is 30 Ω. If in the circuit, an alternating emf of 200 V at 50 cycles per second is connected, the impedance of the circuit and the current will be

(a) 11.4 Ω, 17.5 A (b) 30.7 Ω, 6.5 A (c) 40.4 Ω, 5 A

(d) 50 Ω, 4 A (e) 35 Ω, 6.5 A

This question is quite simple and straight forward.

Impedance, Z = √(R²+L² ω²) = √[(30²+(0.4/π)² (100π)²)] since ω = 2πf =2π×50 = 100π.

This gives Z = 50 Ω.

Current I = V/Z = 200/50 = 4 A.

(4) A transformer has an efficiency of 80%. It is connected to a power input of 5 kW at 200 V. If the secondary voltage is 250 V, the primary and secondary currents are respectively

(a) 25 A, 20 A (b) 20 A, 16 A (c) 25 A, 16 A (d) 40 A, 25 A (e) 40 A, 16 A

This question also is quite simple and can be worked out within a minute.

Primary current I_P = (5000 watts)/(200 volts) = 25 A.

Secondary current = (5000×0.8 watts)/(250 volts) = 16 A.

[Note that the output power from the secondary side is efficiency times the input power].

Two Questions Involving Friction in Rotational Motion

A coin is placed on a horizontal turn table, at a distance‘r’ from the axis of rotation of the turn table. Starting from rest, the turn table rotates with a constant angular acceleration ‘α’. If the coefficient of friction between the coin and the turn table is ‘μ’, the time after which the coin will begin to slip is

(a) √(μ/α) (b) √(μg/α²) (c) √(μg/αr) (d) √(μgα²/r) (e) √(μg/α²r)

The coin will start slipping when the centrifugal force begins to exceed the frictional force. In the limiting case therefore, mrω² = μmg where ‘m’ is the mass of the coin and ‘ω’ is the angular velocity of the turn table when the coin begins to slip.

But, ω = ω₀ + αt = αt since the initial angular velocity (ω₀) is zero.

[Note that the above equation is similar to the equation, v = v₀ + at in the case of linear motion].

Substituting this value of ω in the condition for slipping, we have

mr(αt)² = μmg from which t = √(μg/α²r).

Now, let us modify the above question as follows:

In a region of space where gravitational force is negligible, a coin is kept in contact with a rod AB as shown, at a distance ‘r’ from the end A. If the rod starts from rest and rotates with a constant angular acceleration ‘α’ about an axis XX’ perpendicular to the rod and passing through the end A, the time after which the coin will begin to slip is

(a) √(μα) (b) √(μα/r) (c) √(μ/αr) (d) √(μ/α) (e) negligibly small

This problem is very much different since gravity is negligible. In the previous problem, the normal force that produced friction was the result of the weight of the coin where as here it is produced because of the inertial force produced by the acceleration of the rod. When the rod rotates with an angular acceleration α, the coin presses against the rod with an inertial force equal to ma where ‘m’ is the mass and ‘a’ is the tangential acceleration of the mass.

As in the previous problem, the condition for slipping to start is obtained by equating the centrifugal force to the frictional force:

mrω² = μma

Therefore, ω = √(μa/r).

But, ω = ω₀ + αt = αt (since the initial angular velocity ω₀ is zero); a = αr.

Substituting these we obtain αt = √(μα) so that t =√(μ/α) .

AP (Advanced Placement) Physics Examinations

Many among the visitors of this site might have already noted that the posts here are useful for preparing for the AP (Advanced Placement) Physics Exams. Even though you find multiple choice questions here, you will definitely find the posts useful for answering the free-response section of the exams since the questions are answered with all necessary theoretical details. In many cases the discussions touch even minute details. This is done for helping students who are just above average. The needs of those who appear for the AP Physics Exams will be considered while discussing questions here. You may make use of the facility for comments for communications in this regard.

Two Multiple Choice Questions on Elastic Collision

You should remember that momentum and kinetic energy are conserved in elastic collisions where as momentum alone is conserved in inelastic collisions. Let us consider two questions (MCQ) involving elastic collision in one dimension:

(1) A system consisting of two identical blocks A and B, each of mass ‘m’, connected by a light spring of force constant ‘k’ is resting on a smooth horizontal surface. A third identical block C of mass ‘m’ moving with a velocity ‘v₀’along the direction of the line joining A and B collides elastically with A and compresses the spring. The maximum compression of the spring is

(a) √(mv₀/k) (b) √(mv₀/2k) (c) √(m/v₀k) (d) v₀√(m/k) (e) v₀√(m/2k)

At the instant of collision, you need consider the two identical colliding masses C and A only. As the collision is elastic, the entire momentum and kinetic energy of C are transferred to A and C comes to rest. The block A then moves towards B, compressing the spring. When the compression is maximum, both A and B move with the same velocity.

Since the momentum is always conserved,

mv₀ = (m + m)v where ‘v’ is the common velocity with which the masses A and B move. Therefore, v = v₀/2.

The entire kinetic energy of C is transferred to A, which then compresses the spring and pushes B so that we have

½ mv₀² = ½ (m + m)(v₀/2)² + ½ kx², where ‘x’ is the maximum compression.

This gives x = v√(m/2k).

(2) An α-particle of mass ‘m’ suffers a one-dimensional elastic collision with a nucleus of unknown mass at rest. After the collision the α-particle is scattered directly backwards, losing 75% of its kinetic energy. Then, the mass of the nucleus is

(a) m (b) 2m (c) 3m (d) (3/2)m (e) 5m

This MCQ appeared in the Kerala Engineering Entrance 2003 question paper.

The total momentum of the system is equal to the initial momentum p₁ of the α-particle. Equating the total initial and final momenta,

p₁ = P – p₂ where P and p₂ are respectively the final momenta of the nucleus and the α-particle. (The negative sign for p₂ is because of the backward motion of the α-particle after the collision).

But the magnitude of p₂ is p₁/2 since the final kinetic energy of the α-particle is ¼ of its initial kinetic energy. [Remember that K.E. = p²/2m and hence the momentum p is directly proportional to the square root of the kinetic energy].

Therefore, P = p₁ + (p₁/2) = (3/2) p₁.

Since the collision is elastic, kinetic energy also is conserved.

Therefore, K.E. lost by the α-particle = K.E gained by the nucleus so that

P²/2M = (3/4) p₁² /2m where ‘M’ is the mass of the nucleus.

Substituting for P, [(3/2)p₁]²/ 2M = [(3/4)p₁²]/2m

This gives M = 3m.

IIT-JEE 2007 Matrix Match Type Question from Units and Dimensions

In the post dated 30^th April 2007, a Matrix Match Type question based on topics in Modern Physics, which appeared in the IIT-JEE 2007 question paper, was discussed. Let us discuss now the question (which appeared in the same examination) based on units and dimensions. Here is the question:

Some physical quantities are given in column I and some possible SI units in which these quantities may be expressed are given in Column II. Match the physicsl quantities in column I with the units in Column II and indicate your answer by darkening appropriate bubbles in the 4×4 matrix given in the ORS.

column I column II

(A) GM_eM_s (p) (volt) (coulomb) (metre)

G – universal gravitational constant,

M_e – mass of the earth,

M_s– mass of the sun

(B) 3RT / M (q) (kilogram)(metre)³(second)^–2

R – universal gas constant,

T – absolute temperature,

M – molar mass

(C) F²/q²B² (r) (metre)²(second)^–2

F – force,

q – charge,

B – magnetic field

(D) GM_e/R_e (s) (farad) (volt)²(kg)^–1

G – universal gravitational constant,

M_e – mass of the earth,

R_e – radius of the earth

(A) has to be matched with (p) and (q). This can be done in various ways. But it will be convenient to do it by noting that the quantity GM_eM_s is force×distance² since the gravitational force between the earth and the sun is GM_eM_s/r², with usual notations.

But, force×distance² = energy×distance. The quantity (volt) (coulomb) (metre) also is energy×distance (since ‘Vq’ is energy). The quantity (kilogram)(metre)³(second)^–2 also is energy×distance (remembering F = ma and energy = Fs).

(B) has to be matched with (r) and (s). This can be proved by noting that 3RT / M is the square of the r.m.s. velocity of molecules. Its unit is therefore (metre)²(second)^–2.

The quantity (farad) (volt)²(kg)^–1 also has the dimensions of velocity² since farad metre² has the dimensions of energy (remember E = ½ CV²) and energy per kg has dimensions of velocity² (E = p²/2m so that E/m = p²/2m² = m²v²/2m² which has dimensions of velocity²).

(C) is to be matched with (r) and (s) since F²/q²B² has dimensions of velocity². (Remember the expression F = qvB for magnetic force on a moving charge).

(D) is to be matched with (r) and (s) since GM_e/R_e is gR_e which has dimensions of [(metre/sec²) ×metre] and therefore those of velocity².

It will be difficult for you to remember the dimensions of all the physical quantities, especially those you come across in branches of physics other than mechanics. But you will remember the important expressions you derive in all the branches. The Matrix Match Type question we discussed is aimed at checking your knowledge of those expressions and your understanding of the significance of dimensions.

KEAM 2007- Engineering Entrance Examination Questions from Optics

The following questions on Optics were asked in the Kerala Government Engineering Entrance test of 2007:

(1) The position of final image formed by the given lens combination from the third lens will be at a distance of

(a) 15 cm (b) infinity (c) 45 cm (d) 30 cm (e) 35 cm

Applying the law of distances [1/f = 1/v – 1/u] to the first lens,

1/10 = 1/v₁ –1/(–30), noting that the object distance ‘u’ is negative in accordance with the Cartesian sign convention. From this, the distance of the image produced by the first lens is v₁ = 15 cm.

The image produced by the first lens is therefore on the right side of the concave lens and the object distance for it is 15 – 5 = 10 cm. This is positive according to the sign convention. The image distance for the second lens (v₂) is given by

1/(–10) = 1/v₂ – 1/10, from which 1/v₂ = 0 so that v₂ = infinity.

This means that the rays emerging from the second lens are parallel to the principal axis. Therefore, the image produced by the third lens is at its focus, at a distance of 30 cm from it.

(2) A slit of width ‘a’ is illuminated by red light of wave length 6500 Ǻ. If the first minimum falls at θ = 30º, the value of ‘a’ is

(a) 6.5×10^–4 mm (b) 1.3 micron (c) 3250 Ǻ (d) 2.6×10^–4 mm (e) 1.3×10^–4 mm

In the single slit diffraction pattern, the first minimum is obtained at an angle θ given by

sin θ = λ/a where λ is the wave length of light used and ‘a’ is the width of the slit.

[Usually, the angle θ is small so that sinθ is approximated to θ and the above relation is written as θ = λ/a].

Therefore, sin 30º = (6500×10^–10)/a, from which a = 1.3×10^–6 m = 1.3 micron

[Even if you use the approximated relation, θ = λ/a, you will arrive at option (b) since 30º = 30 × π/180 radian and hence 30 × π/180 = (6500×10^–10)/a, from which a = 1.2414 micron, which is nearest to the value given in option (b)].

(3) Two beams of light of intensity I₁ and I₂ interfere to give an interference pattern. If the ratio of maximum intensity to minimum intensity is 25/ 9, then, I₁/ I₂ is

(a) 5/3 (b) 4 (c) 81/ 625 (d) 16 (e) ½

If a₁ and a₂ are the amplitudes of the two interfering waves,

I_max/ I_min = (a₁ + a₂)² / (a₁ – a₂)², since the intensity is directly proportional to the square of the amplitude.

Therefore, (a₁ + a₂) / (a₁ – a₂) = √(25/9) = 5/3

From this, a₁/a₂ = 4 so that I₁/I₂ = a₁²/a₂² = 16.

(4) Magnification at least distance of distinct vision of a simple microscope having its focal length 5 cm is

(a) 2 (b) 4 (c) 5 (d) 6 (e) 7

The answer to this very simple question ( which is asked to boost your morale!) is 6 since m = 1 + D/f where D is the least distance of distinct vision (which is 25 cm).

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

Two Kerala Engineering/Medical Entrance 2007 Questions from Electrostatics

The following MCQ appeared in Kerala Engineering Entrance 2007 question paper:

The plates of a parallel plate capacitor with air as medium are separated by a distance of 8 mm. A medium of dielectric constant 2 and thickness 4 mm having the same area is introduced between the plates. For the capacitance to remain the same, the distance between the plates is

(a) 8 mm (b) 6 mm (c) 4 mm (d) 12 mm (e) 10 mm

If a dielectric slab of thickness ’t’ is introduced between the plates, the electric fields in the air space and in the dielectric space become respectively q/ε₀A and q/Kε₀A where ‘q’ is the charge on each plate (+q on one, –q on the other) and A is the area of the plate. The P.D. between the plates is V = (q/ε₀A)(d–t)+(q/Kε₀A)t = (q/ε₀A)[d–t+(t/K)]. The capacitance of the system on introducing the dielectric is C = q/V = ε₀A/[d–t+(t/K)] = ε₀A/[d–(t – t/K)].

Since the capacitance of the capacitor with air filling the entire space between the plates is ε₀A/d, the effect of introducing the dielectric slab of thickness ‘t’ is to reduce the thickness of air by t– (t/K). In order to restore the capacitance to the original value, the separation between the plates is to be increased by t– (t/K).

Therefore, the separation between the plates is to be increased by t–(t/K) = 4 – (4/2) = 2 mm.

The distance between the plates should be 8 + 2 = 10 mm.

The following question appeared in Kerala Medical Entrance 2007 question paper:

The capacitance of a parallel plate capacitor with air as medium is 3 μF. With the introduction of a dielectric medium between the plates, the capacitance becomes 15 μF. The permittivity of the medium is

(a) 5 (b) 15 (c) 0.44×10^–10C²N^–1m^–2

(d) 8.854×10^–10C²N^–1m^–2 (e) 5C²N^–1m^–2

The capacitance of a parallel plate capacitor with air as the medium between the plates is ε₀A/d and the capacitance with a medium of dielectric constant K instead of air is Kε₀A/d where ε₀, A and d are the relative permittivity of free space (or air very nearly), the area of each plate and the separation between the plates respectively. The capacitance therefore increases to K times the original value on replacing air with the dielectric medium.

Since the increase here is from 3 μF to 15 μF, the value of K is 5.

The permittivity of the medium is ε₀K = 8.85×10^–12×5 = 0.44×10^–10C²N^–1m^–2.

[The value of ε₀ was not given in the question paper. Some of you may not remember the value 8.854×10^–12C²N^–1m^–2, but all of you are expected to remember the value of 1/4πε₀, which is 9×10⁹, very nearly. This will be enough to get the value of ε₀].

Two Questions Involving Kinetic Theory of Gases

Questions based on specific heat and energy of gases often find place in all Medical and Engineering Entrance Examination question papers. See the following two typical questions:

(1) In an isobaric process, the increase in temperature of 0.4 mole of oxygen is 200K. The work done by the gas is (Universal gas constant, R = 1.99 calorie-mol ^–1K^–1)

(a) 88 cal (b) 124.6 cal (c) 79.6 cal (d) 159.2 cal (e) 318.4 cal

Some of you may get confused on seeing the unit of work in calories. Further, R also is given in terms of calorie. Work and energy can be expressed in calorie as well as joule.

Since the pressure is constant (isobaric process), the work done by one mole of gas on getting heated through 1K is the difference (C_p – C_v) between its molar specific heats. But C_p – C_v = R, so that the work done by 0.4 mole of the gas on getting heated through 200 K is 0.4×200×R = 0.4×200×1.99 = 159.2 calorie.

(2) At what temperature will the translational kinetic energy of an ideal gas molecule be half of the value at 100º C?

(a) 323 K (b) – 179.75º C (c) –186.5º C (d) –86.5º C (e) 50º C

This is a very simple question. The translational kinetic energy of any ideal gas molecule is (3/2) kT where ‘k’ is Boltzman constant and T is the absolute (Kelvin) temperature. So, the translational kinetic energy is directly proportional to the absolute temperature. [The total kinetic energy also is directly proportional to the absolute temperature since total kinetic energy = (n/2)kT where ‘n’ is the number of degrees of freedom of the gas molecule].

Now, 100º C = 373 K. So, the energy becomes half at 186.5 K = (186.5 – 273)º C = – 86.5º C.