IIT-JEE 2010 – Multiple Choice Questions (Straight Objective Single Correct Answer Type) on Direct Current Circuits

Today we will discuss three multiple choice questions which appeared in IIT-JEE 2010 question paper:

(1) Consider a thin square sheet of side L and thickness t, made of a material of resistivity ρ. The resistance between two opposite faces, shown by the shaded areas in the figure is

(A) directly proportional to L

(B) directly proportional to t

(C) independent of L

(D) independent of t

We have R = ρL/A where A is the cross section area which is equal to L t.

Therefore, R = ρL/Lt = ρ/t, which is independent of L [Option (C)].

(2) Incandescent bulbs are designed by keeping in mind that the resistance of their filament increases with increase in temperature. If at room temperature, 100 W, 60 W and 40 W bulbs have filament resistances R₁₀₀, R₆₀ and R₄₀ respectively, the relation between these resistances is :

(A) 1/R₁₀₀ = 1/R₄₀ + 1/R₆₀

(B) R₁₀₀ = R₄₀ + R₆₀

(C) R₁₀₀ > R₆₀ > R₄₀

(D) 1/R₁₀₀ > 1/R₆₀ > 1/R₄₀

The power P dissipated in a resistance R is given by

P = V²/R where V is the voltage applied across the resistance.

This shows that the resistance of the filament is smaller if the wattage of the incandescent lamp is greater. In other words, the reciprocal of the filament resistance is greater if the wattage of the incandescent lamp is greater. This is true at all temperatures including room temperature. Therefore we have

1/R₁₀₀ > 1/R₆₀ > 1/R₄₀

(3) To verify Ohm's law, a student is provided with a test resistor R_T, a high resistance R₁, a small resistance R₂, two identical galvanometers G₁ and G₂, and a variable voltage ource V. The correct circuit to carry out the experiment is

To verify Ohm’s law, the current through R_T is to be measured by connecting an ammeter in series with it and the voltage across R_T is to be measured by connecting a voltmeter in parallel with it. One galvanometer is to be converted into an ammeter by connecting the small resistance R₂ across it. The other galvanometer is to be converted into a voltmeter by connecting the high resistance R₁ in series with it. So the correct circuit to carry out the experiment is the one shown in option (C).

You will find some multiple choice questions (with solution) on direct current circuits here.

Alternating Current Circuits – Two Questions (MCQ) on Resonance

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"We have to do the best we can. This is our sacred human responsibility."

–Albert Einstein

The phenomenon of resonance in alternating current circuits is an interesting one with many practical applications and questions involving resonance are often seen in entrance question papers. You may try searching for ‘resonance’ on this blog (making use of the search box provided on this page) to access all previous posts related to resonance.

Today we will discuss two multiple choice questions meant for high lighting the behaviour of AC circuits at resonance:

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(1) An alternating current source of negligible impedance has an emf 6 volt. It is connected in series with a series LCR circuit consisting of a 72 Ω resistor and an inductor-capacitor combination. The inductor offers a reactance of 120 Ω at the operating frequency and the power dissipated in the LCR circuit is 0.5 W. What is the reactance offered by the capacitor?

(a) 60 Ω

(b) 72 Ω

(c) 120 Ω

(d) 192 Ω

(e) Zero

It is always a good idea to check whether the circuit is at resonance (Question setters have a craze to set questions involving resonance).

Since the power is dissipated in the resistor alone, we have

V²/72 = 0.5 where V is the voltage across the resistor.

[Remember the power equation, P = V²/R]

We obtain V² = 36 from which V = 6 volt.

The circuit is indeed at resonance since the entire supply voltage appears across the resistance.

At resonance the inductive reactance is equal to the capacitive reactance and hence the correct option is 120 Ω.

(2) In a series LCR circuit R = 200 Ω and the voltage and the frequency of the main supply are 220 V and 50 Hz respectively. On taking out the capacitance from the circuit the current lags behind the voltage by 30º. On taking out the inductor from the circuit the current leads the voltage by 30º. The power dissipated in the LCR circuit is

(1) 242 W

(2) 305 W

(3) 210 W

(4) 0 W

This question appeared in AIEEE 2010 question paper.

The phase lag of current produced by the inductance is equal in value to the phase lead of current produced by the capacitor. This means that in the LCR circuit connected to the 220 V, 50 Hz supply, there is no phase difference between the supply voltage and the current and the circuit behaves as a pure resistor. Indeed the circuit is at resonance and the entire supply voltage appears across the resistor.

Therefore, the power P dissipated in the LCR circuit is given by

P = V²/R = 220²/200 = 242 W.

All India Engineering/Architecture Entrance Examination 2010 (AIEEE 2010) Questions Involving Electromagnetic Induction

The following interesting multiple choice questions involving electromagnetic induction were included in the AIEEE 2010 question paper:

(1) A rectangular loop has a sliding connector PQ of length l and resistance R Ω and it is moving with a speed v as shown. The set-up is placed in a uniform magnetic field going into the plane of the paper. The three currents I₁, I₂ and I are

(1) I₁ = I₂ = Blv/6R, I = Blv/3R

(2) I₁ = – I₂ = Blv/R, I = 2Blv/R

(3) I₁ = I₂ = Blv/3R, I = 2Blv/3R

(4) I₁ = I₂ = I = Blv/R

The emf induced in the sliding connector PQ is ε = Blv, which is shown as a battery of internal resistance R in the adjoining figure.

The two external resistances R and R appear in parallel with this battery. The effective external resistance is therefore R/2 and the battery has to drive the current I through a total resistance R + (R/2), which is 3R/2

Therefore, I = ε/(3R/2) = 2Blv/3R.

The current I gets divided equally between the two identical branches and hence I₁ = I₂ = Blv/3R.

(2) In the circuit shown below, the key K is closed at t = 0. The current through the battery is

(1) V(R₁ + R₂)/R₁R₂ at t = 0 and V/R₂ at t = ∞

(2) VR₁R₂/√(R₁²+ R₂²) at t = 0 and V/R₂ at t = ∞

(3) V/R₂ at t = 0 and V(R₁ + R₂)/R₁R₂ at t = ∞

(4) V/R₂ at t = 0 and VR₁R₂/√(R₁²+ R₂²) at t = ∞

The current through the inductance branch is zero at the instant the circuit is switched on. [Remember the exponential growth of current in an LR circuit]

Therefore, at the instant t = 0 the current is limited by R₂ alone and is equal to V/R₂.

At t = ∞ the current has become steady, there is no induced voltage to oppose the flow of current and the inductor functions as a piece of conductor. The current is now limited by R₁ and R₂ which are in parallel with the battery. The current in this case is V/[R₁R₂/(R₁ + R₂)] = V(R₁ + R₂)/R₁R₂, as given in option (3).

You will find similar useful multiple choice question (with solution) here.

AIPMT 2010 Multiple Choice Questions (MCQ) on Electronics

Questions on electronics at the level expected from you in most of the entrance examinations are generally simple. (You won't find questions in this section in IIT-JEE and AP Physics Exam question papers). Here are the five questions on electronics (including digital circuits) which appeared in the AIPMT 2010 question paper:

(1) A common emitter amplifier has a voltage gain of 50, an input impedance of 100 Ω and an output impedance of 200 Ω. The power gain of the amplifier is

(1) 50

(2) 500

(3) 1000

(4) 1250

In the case of common emitter amplifier, the voltage gain A_v is given by

A_v = – β_ac(R_o/R_i) where β_ac is the a.c. current gain, R_o is the output resistance and R_i is the intput resistance. (Instead of R_o and R_i you may use Z_o and Z_i, the respective impedances). The negative sign just indicates the 180º phase shift of the output signal with respect to the input signal.

Therefore we have 50 = β_ac×200/100 from which β_ac = 25.

Power gain = current gain×voltage gain = 25×50 = 1250.

(2) Which one of the following bonds produces a solid that reflects light in the visible region and whose electrical conductivity decreases with temperature and has high melting point?

(1) Covalent bonding

(2) Metallic bonding

(3) van der Waal's bonding

(4) Ionic bonding

This is a very simple question and the answer is metallic bonding.

(3) The device that can act as a complete electronic circuit is

(1) Zener diode

(2) Junction diode

(3) Integrated circuit

(4) Junction transistor

This too is a very simple question and the answer is integrated circuit.

(4) Which one of the following statement is FALSE ?

(1) The resistance of intrinsic semiconductor decreases with increase of temperature

(2) Pure Si doped with trivalent impurities gives a p-type semiconductor

(3) Majority carriers in a n-type semiconductor are holes

(4) Minority carries in a p-type semiconductor are electrons

The FALSE statement is given in option (3).

(5) To get an output Y = 1 from the circuit shown below, the input must be

---A B C---

(1) 1 0 0

(2) 0 1 0

(3) 0 0 1

(4) 1 0 1

The correct option is (4) since the circuit implements the operation Y = (A+B).C

You will find the same question and some other questions on digital circuits with solution here: Karnataka CET Questions (MCQ) on Digital Circuits

IIT-JEE 2010 Questions on Gravitation

I would never die for my belief because I might be wrong.

– Bertrand Russell

Two questions on gravitation were included in the IIT-JEE 2010 question paper. They are given below with solution. The first question is a ‘single correct choice type’ (single answer type multiple choice) question where as the second question is an ‘integer type’ question which has a single digit integer ranging from 0 to 9 as answer.

(1) A thin uniform annular disc (see figure) of mass M has outer radius 4R and inner radius 3R. The work required to take a unit mass from point P on its axis to infinity is

(A) (2GM /7R) (4√2 – 5)

(B) – (2GM /7R) (4√2 – 5)

(C) GM /4R

(D) (2GM /5R) (√2 – 1)

The work required to take a unit mass from infinity to point P is the gravitational potential at P and it will be negative. Therefore, the work required to take a unit mass from point P to infinity will be numerically equal to the gravitational potential at P but it will be positive.

The annular disc can be considered to be made of a large number of concentric rings of radii ranging from 3R to 4R. Consider one such ring of radius r and small thickness dr. Its mass dm is given by

dm = [M/π(16R² – 9R²)] 2πr dr = 2Mrdr/7R²

If the rim of the ring is at distance x from the point P, the gravitational potential (dV) at P due to the ring is given by

dV = – Gdm/x = – (2GM/7R²)(rdr/x)

But x = (16R² + r²)^1/2 and so we have dx= (½ )×(16R² + r²)^–1/2 ×2rdr

Or, dx = rdr/(16R² + r²)^1/2 = rdr/x

Therefore dV = – (2GM/7R²)dx

The gravitational potential at P due to the entire annular disc is given by

V = ∫ dV = – ∫(2GM/7R²)dx = – (2GM/7R²)[x]

The integration is between the appropriate limits of x, which are (16R² + 9R²)^1/2 and (16R² + 16R²)^1/2, on substituting the values r = 3R and r = 4R respectively in the expression, x = (16R² + r²)^1/2. Therefore we have

V = – (2GM/7R²)[(16R² + 16R²)^1/2 – (16R² + 9R²)^1/2]

= – (2GM/7R²)[R√(32) – R√(25)]

Thus V = – (2GM/7R) (4√2 – 5)

The negative sign is to be dispensed with to obtain the work required to take a unit mass from point P to infinity and the correct choice is (2GM/7R) (4√2 – 5).

(2) Gravitational acceleration on the surface of a planet is (√6/11)g, where g is the gravitational acceleration on the surface of the earth. The average mass density of the planet is 2/3 times that of the earth. If the escape speed on the surface of the earth is taken to be 11 kms^–1, the escape speed on the surface of the planet in kms^–1 will be :

[This is is an ‘integer type’ question. You have to work out this question and the single digit integer answer is to be shown in the Objective Response Sheet (ORS) by darkening the appropriate bubble].

We have v_esc = √(2gR) where g is the acceleration due to gravity and R is the radius of the planet (or earth or star).

Therefore v_esc α √(gR).

If g_p and g_e are the accelerations due to gravity on the surface of the planet and the earth, R_p and R_e are their radii, and g_p and g_e the escape velocities respectively, we have

v_p/v_e = √( g_pR_p)/ √( g_eR_e) = √[( g_p/g_e)(R_p/R_e)] ………(i)

Since the gravitational acceleration on the surface is given by g = GM/R² where G is the gravitational constant and M is the mass of the planet (or earth or star), we can write

g = G[(4/3)πR³ρ] /R² where ρ is the average mass density of the planet (or earth or star).

Therefore g α Rρ and g_p/g_e = (R_p/R_e)(ρ_p/ρ_e).

From this R_p/R_e = (g_p/g_e)/(ρ_p/ρ_e).

But g_p/g_e = √6/11 and ρ_p/ρ_e = 2/3 as given in the question. Therefore we have

R_p/R_e = (√6/11)×3/2

Substituting these in Eq(i),

v_p/v_e = √[(√6/11) (√6/11)×3/2] = (√6/11) ×√(3/2) = 3/11

Since v_e = 11 kms^–1, the escape speed on the surface of the planet in kms^–1 will be

v_p = 11×(3/11) = 3 kms^–1

The answer integer is 3.

AIPMT 2010 Questions on Friction

The following multiple choice questions involving friction were included in the AIPMT 2010 question paper:

(1) A block of mass m is in contact with the cart C as shown in the figure. The coefficient of static friction between the block and the cart is μ. The acceleration α of the cart that will prevent the block from falling statisfies

(1) α < g/μ

(2) α > mg/μ

(3) α > g/μm

(4) α ≥ g/μ

When the cart moves forwards with acceleration α, the block of mass m presses against the vertical surface of the cart with a horizontal force of magnitude mα. Therefore, the normal force acting on the block is mα and the maximum frictional force trying to prevent the block from falling is μmα. This must be greater than or equal to the weight mg of the block.

Therefore we have

μmα ≥ mg from which α ≥ g/μ

(2) A gramophone record is revolving with an angular velocity ω. A coin is placed at a distance r from the centre of the record. The static coefficient of friction is μ. The coin will revolve with the record if

(1) r ≥ μg/ω²

(2) r = μgω²

(3) r < ω²/μg

(4) r ≤ μg/ω²

Let the mass of the coin be m. The coin will revolve with the record if the frictional force which supplies the centripetal force (required for the circular motion of the coin) is sufficient. Since the maximum value of the frictional force is μmg, we have

μmg ≥ mrω²

Therefore r ≤ μg/ω².

You will find more questions (with solution) on friction on this site. One of the posts on friction is here

Similar useful posts in this section can be seen here