Kerala Engineering Entrance 2007 Questions on Heating Effect of Electric Current

Two questions on heating effect of electric current appeared in KEAM (Engineering) 2007 question paper:

(1) The resistance of a wire at room temperature 30º C is found to be 10 Ω. Now to increase the resistance by 10%, the temperature of the wire must be [Temperature coefficient of resistance of the material of the wire is 0.002 per º C]

(a) 36º C (b) ) 83º C (c) ) 63º C (d) ) 33º C (e) ) 66º C

The resistance (R_t) at tº C can be expressed in terms of the resistance (R₀) at 0º C and the temperature coefficient (α) of resistance as

R_t = R₀(1 + αt).

Therefore, we have

R₃₀ = 10 = R₀(1 + 0.002×30) and

R_t = 11 = R₀(1 + 0.002 t)

[The resistance R_t at the unknown temperature ‘t’ is greater by 10% and is therefore equal to (10 + 1) Ω = 11 Ω]

From the above equations we have

10/11 = (1 + 0.06)/ (1 + 0.002t) from which t = 83º C.

(2) If R₁ and R₂ be the resistances of the filaments of 200 W and 100 W electric bulbs operating at 220 V, then R₁/R₂ is

(a) 1 (b) 2 (c) 0.5 (d) 4 (e) 0.25

Since the power is V²/R where V is the operating voltage and R is the resistance, we have

220²/R₁ = 200 and

220²/R₂ = 100.

Dividing the second equation by the first, we obtain

R₁/R₂ = 0.5

Kerala Engineering Entrance 2007 Questions on Rotational Motion

The following questions appeared in KEAM (Engineering) 2007 question paper:

(1) A sphere of mass ‘m’ and radius ‘r’ rolls on a horizontal plane without slipping with speed ‘u’. Now if it rolls up vertically, the maximum height it would attain would be

(a) 3u²/4g (b) 5u²/2g (c) 7u²/10g (d) u²/2g (e) 11u²/9g

The initial kinetic energy (E) of the sphere while rolling on the horizontal surface is given by

E = ½ mu² + ½ Iω² where I is the moment of inertia of the sphere about its diameter[(which is equal to (2/5)mr²] and ‘ω’ is the angular velocity (which is u/r).

[Note that the first term is the translational kinetic energy and the second term is the rotational kinetic energy].

Therefore, E = ½ mu² + ½ ×(2/5)mr²u²/r² = 7mu²/10.

Since this energy is converted into gravitational potential energy (mgh) to attain the maximum height h, we have

7mu²/10 = mgh, from which h = 7u²/10g.

(2) If the earth were to contract such that its radius becomes one quarter, without change in its mass, the duration of the full day would be

(a) 3 hours (b) 1.5 hours (c) 6 hours (d) 4 hours (e) 2 hours

If the radius of the earth becomes ‘n’ times he present value, without change in the mass, the duration of the day will become 24n² hours so that the answer to the above question is 24×(1/4)² = 1.5 hours.

Solution to Multiple Choice Questions on Centre of Mass

Two multiple choice questions on centre of mass were given to you for practice yesterday. Here is the solution along with the questions:

(1) A boy weighing 40 kg is standing on a wooden log of mass 500 kg floating on still water in a lake. The distance of the boy from the shore is 12 m. The viscous force exerted by water on the wooden log may be neglected. If the boy walks slowly along the wooden log through 2 m towards the shore, the centre of mass of the system (wooden log and the boy) will move with respect to the shore through a distance

(a) 2 m (b) 1.25 m (c) 0.25 m (d) 0.16 m (e) zero

You should have worked this out within seconds. The centre of mass will be unaffected with respect to external fixed points if there are no external forces acting on the system. So, the correct option is (e).

(2) A T-shaped object with dimensions shown in the figure, is lying on a smooth floor. A force F is applied at the point P parallel to AB, such that the object has only translational motion without rotation. Find the location of P with respect to C

(a) L (b) 4L/3 (c) 3L/2 (d) 2L/3

The point P must be the centre of mass of the T-shaped object since the force F does not produce any rotational motion of the object. So, we have to find the distance of the centre of mass from the point C.

The horizontal part of the T-shaped object has length L. If the mass of the horizontal portion is ‘m’, the mass of the vertical portion of the T- shaped object is 2m since its length is 2L. For finding the centre of mass of the T shaped object, it is enough to consider two point masses m and 2m located respectively at the mid points of the horizontal and vertical portions of the T.

Therefore, the T-shaped object reduces to two point masses m and 2m at distances 2L and L respectively from the point C. The distance ‘r’ of the centre of mass of the system from the point C is given by

r = (m₁r₁ + m₂r₂)/(m₁ + m₂) = (m×2L + 2m×L)/(m + 2m) = 4L/3

[ Note that we have used the equation, r = (m₁r₁ + m₂r₂)/(m₁ + m₂) for the position vector r of the centre of mass in terms of the position vectors r₁ and r₂ of the point masses m₁ and m₂. We could use the simple equation involving the distances from C since the points are collinear].

Two Multiple Choice Questions on Centre of Mass (For practice)

Here are two questions on centre of mass. These are meant for checking whether you have a clear idea of the concept of centre of mass:

(1) A boy weighing 40 kg is standing on a wooden log of mass 500 kg floating on still water in a lake. The distance of the boy from the shore is 12 m. The viscous force exerted by water on the wooden log may be neglected. If the boy walks slowly along the wooden log through 2 m towards the shore, the centre of mass of the system (wooden log and the boy) will move with respect to the shore through a distance

(a) 2 m (b) 1.25 m (c) 0.25 m (d) 0.16 m (e) zero

(2) A T-shaped object with dimensions shown in the figure, is lying on a smooth floor. A force F is applied at the point P parallel to AB, such that the object has only translational motion without rotation. Find the location of P with respect to C

(a) L
(b) 4L/3
(c) 3L/2
(d) 2L/3

This MCQ appeared in AIEEE 2005 question paper.

Try to find the answer to the above questions. If you have clear understanding of the centre of mass, you will be able to find the answers in a couple of minutes. I’ll be back with the solution shortly.

Multiple Choice Questions on Work and Energy

Here is a simple question which is meant for gauging your understanding of the work-energy principle. This MCQ appeared in AIEEE 2005 question paper:

The block of mass M moving on the frictionless horizontal surface collides with the spring of spring constant K and compresses it by length L. The maximum momentum of the block after collision is

(a) ML²/K (b) zero (c) KL²/2M (d) √(MK).L

If the maximum momentum of the block after the collision is ‘p’ the maximum kinetic energy is p²/2M. This must be equal to the maximum potential energy of the spring so that we have

p²/2M = ½ KL², from which p = L√(MK)., given in option (d).

Now consider the following MCQ which appeared in Kerala Medical Entrance 2006 question paper:

The work done by a force F = –6x³ î newton, in displacing a particle from x = 4m to x = –2m is

(a) 360 J (b) 240 J (c) – 240 J (d) – 360 J (e) 408 J

This is a case of variable force (in the X-direction), the point of application of which is moved in the X-direction. The work done is therefore given by

W = ∫F.dx = ₄ ∫^–2 (–6x³)dx = –6[x⁴/4] with x between limits 4 and –2.

Therefore, W = –(6/4)(16 – 256) = 360 joule, given in option (a).

The following MCQ appeared in Kerala Engineering entrance 2006 question paper:

A running man has the same kinetic energy as that of a boy of half the mass. The man speeds up by 2 ms^–1 and the boy changes his speed by ‘x’ ms^–1 so that the kinetic energies of the boy and the man are again equal. Then ‘x’ in ms^–1 is

(a) – 2√2 (b) + 2√2 (c) √2 (d) 2 (e) 1/√2

If the mass of the man is ‘m’, the mass of the boy is m/2. If v₁ and v₂ are the initial velocities of the man and boy respectively, we have

½ mv₁² = ½ (m/2)v₂²

Therefore, v₂ = v₁√2.

On changing the speeds, we have

½ m(v₁+2)² = ½ (m/2)(v₂+x)²

On substituting for v₂ (=v₁√2), the above equation simplifies to

(v₁+2)² = ½ (v₁√2+x)² from which x = 2√2 ms^–1, given in option (b).

Multiple Choice Questions on Direct Current Circuits

The following MCQ which appeared in All India Pre-Medical/Dental Entrance 2005 Examination (AIPMT) is one requiring the application of Kirchoff’s laws:

Two batteries, one of emf 18 volt and internal resistance 2 Ω and the other of emf 12 volt and internal resistance 1 Ω, are connected as shown. The voltmeter V will record a reading of

(a) 30 volt (b) 18 volt (c) 15 volt (d) 14 volt

The net emf in the closed path containing the two batteries is 18volt – 12 volt = 6 volt

Since this must be equal to the net voltage drop across the internal resistances of the batteries, we have

I(2 + 1) = 6 volt where ‘I’ is the current through the batteries. [Note that the voltmeter will draw negligible current (ideally, zero current) and hence the same current flows through the batteries].

Therefore, I = 2 ampere.

The voltage across the voltmeter is the same as the terminal voltages of the batteries. If you consider the 18 volt battery, its terminal voltage is 18 – (2×2) = 14 volt.

[If you consider the 12 volt battery, its terminal voltage is 12 + (1×2) = 14 volt, which is the same as the above value]. So, the correct option is (d).

Here is another question which you can answer in no time if you had worked out a similar one earlier:

Five identical cells each of emf 1.5 V and internal resistance ‘r’ send the same current through an external resistance of 1 Ω whether the cells are connected in series or in parallel. The internal resistance ‘r’ of each cell is

(a) 0.2Ω (b) 0.5 Ω (c) 1 Ω (d) 1.5 Ω (e) 3 Ω

The condition depicted in the above question occurs when the internal resistance of each cell is equal to the external resistance. The answer therefore is 1 Ω.

The proof is simple: Let there be ‘n’ cells. If the emf of each cell is V and the external resistance is R, the current through R on connecting the cells in series is nV/(nr + R). When the cells are in parallel, the current through R is V/[(r/n) + R].

Since the currents are equal, nV/(nr + R) = V/[(r/n) + R].

Therefore, r + nR = nR + R, from which r = R.

Now, let us consider the following MCQ which appeared in All India Pre-Medical/Dental Entrance 2004 Examination question paper:

Five equal resistances each of resistance R are connected as shown in the figure. A battery of V volts is connected between A and B. The current flowing in AFCEB will be

(a) 3V/R (b) V/R (c) V/2R (d) 2V/R

Once you realise that C and D are equipotential points with respect to point A, the resistance R connected directly between C and D can be ignored (since it will not carry any current). The current through the path AFCEB is therefore V/2R.

Science is a wonderful thing if one does not have to earn
one's living at it

–Albert Einstein 

Two Questions (MCQ) on Capacitors

Here are two multiple choice questions which will be interesting to you:

(1) Six identical capacitors C₁, C₂, C₃, C₄, C₅ and C₆ each having capacitance C are connected as shown. The equivalent capacitance between the points A and B is

(a) C (b) 2C (c) 2.5C (d) 3C (e) 3.5C

This may appear to be a difficult question for some of you. In questions of this type try to identify equipotential points. Circuit elements connected between equipotential points can be ignored. In most cases the circuit will then become simple.

In the present case, the junction between C₁ and C₆ is at the same potential as that at the junction between C₂ and C₅. [This follows since the capacitors are identical. Generally, the points will be equipotential points if the balance condition for Wheatstone bridge is satisfied: C₁/C₆ = C₂/C₅]. The capacitor C₃ connected between the equipotential points can therefore be ignored.

Now we have three parallel connections across the points A and B:

(i) Series combination of C₁ and C₆ giving a capacitance C/2.

(ii) Series combination of C₂ and C₅ giving a capacitance C/2.

(iii) Capacitor C₄ of value C.

The parallel combined value of the abve three connections is (C/2) + (C/2) + C = 2C.

(2) In a circuit, capacitors C₁ and C₂ (in series) are connected between two points A and B (Fig). If the potentials at A and B are V₁ and V₂ respectively, what is the potential at the junction point P between C₁ and C₂?

(a) C₁(V₁+V₂)/(C₁+C₂)

(b) (V₁ + V₂)/2

(c) V₁ – V₂

(d) (C₁V₂ + C₂V₁)/ (C₁ + C₂)

(e) (C₁V₁ + C₂V₂)/ (C₁ + C₂)

Since the capacitors are in series, they carry the same charge. If the potential at the junction point P is ‘V’, the charges on the capacitors C₁ and C₂ are respectively C₁(V – V₁) and C₂(V₂ – V).

Therefore, we have

C₁(V – V₁) = C₂(V₂ – V), from which

V = (C₁V₁ + C₂V₂)/ (C₁ + C₂).

You can find more posts on electrostatics by clicking on the label ELECTROSTATICS below this post or on the left side of this page.

Three KEAM 2007 Multiple Choice Questions from Heat and Thermodynamics

Four questions from Heat and Thermodynamics appeared in the Kerala Engineering Entrance 2007 question paper. One question was quite simple, meant for testing your knowledge of the temperature at which water has minimum volume, which most of you know, is 4º C. Here are the remaining three questions:

(1) A closed gas cylinder is divided into two parts by a piston held tight. The pressure and volume of gas in two parts respectively are (P, 5V) and (10 P, V). If now the piston is left free and the system undergoes isothermal process, then the volume of the gas in two parts respectively are

(a) 2 V, 4 V (b) 3 V, 3 V (c) 5 V, V (d) 4 V, 2 V (e) 2.5 V, 3.5 V

Since the temperature is constant (isothermal process), we can apply Boyle’s law. The piston will move towards the lower pressure side until the pressures on the two sides are the same. If this common pressure is P’, we have

P×5V + 10P×V = P’×6V, since the total volume is 6 V.

From this, P’ = (15/6) P.

Now, applying Boyle’s law to the gas on the two sides separarely, we have

P×5V = (15/6) P×V₁ and

10P×V = (15/6) P×V₂

where V₁ and V₂ are the volumes of the gas in the two parts respectively.

From the above equations, we obtain V₁ = 2V and V₂ = 4V [Option (a)].

(2) A Carnot engine with sink’s temperature at 17º C has 50% efficiency. By how much should its source temperature be changed to increase its efficiency to 60%?

(a) 225 K (b) 128º C (c) 580 K (d) 145 K (e) 145º C

We have efficiency η = (T₁ – T₂)/T₁ where T₁ and T₂ are the temperatures of the source and sink respectively.

Here, η = 0.5 and T₂ = 17º C = 290 K. On substituting these values, we obtain

1– (290/T₁) = 0.5, from which T₁ = 580 K.

When the efficiency is 60%, we have

1– (290/T₁^') = 0.6, from which T₁^' = 725 K.

Therefore, the change in source temperature = 725 – 580 = 145 K.

(3) Two moles of oxygen is mixed with eight moles of helium. The effective specific heat of the mixture at constant volume is

(a) 1.3 R (b) 1.4 R (c) 1.7 R (d) 1.9 R (e) 1.2 R

Oxygen is diatomic and hence its molar specific heat at constant volume is (5/2) R (corresponding to 5 degrees of freedom).

Helium is mono-atomic and hence its molar specific heat at constant volume is (3/2) R (corresponding to 3 degrees of freedom).

Therefore, the total quantity of heat required to raise the temperature of two moles of oxygen and eight moles of helium through 1 K is

2×(5/2)R + 8×(3/2)R = 17 R.

Since the number of moles in the mixture is 10, the effective molar specific heat of the mixture is (17/10)R = 1.7 R.

Solution to MCQ on Friction

In the post dated 9^th August 2007, two questions on friction were given without solution. These questions with solution are given below as promised in the post:

(1) A block of mass ‘M’ equal to 1 kg is resting on a frictionless horizontal surface. Another block of mass ‘m’ equal to 0.2 kg is resting on the top surface of the 1 kg block. The coefficient of friction between the two blocks is 0.2. The motion of the 1 kg kg block is controlled by a light spring of force constant 24 Nm^–1 arranged as shown in the figure. What is the maximum amplitude of oscillation of the blocks (together) so that the upper block does not slip relative to the lower block? (g = 10 ms^–2)

(a) 2 cm (b) 5 cm (c) 10 cm (d) 20 cm (e)50cm

The figure is not given here. You will find it with the post dated 9^th August 2007 by moving a little down on this page.

If ‘x’ is the maximum possible amplitude, the maximum restoring force exeretred by the spring is Kx and the maximum acceleration of the two blocks (while moving together) is Kx/(M+m).

The force experienced by the upper block while moving with the above acceleration is mKx/(M+m). The upper block will just start slipping when this force exceeds the frictional force μmg where ‘μ’ is the coefficient of friction so that in the limiting case, we have

mKx/(M+m) = μmg.

Therefore, x = μg(M+m)/K = 0.2×10(1+0.2)/24 = 0.1 m = 10 cm.

(2) A metallic cube takes ‘t’ seconds to slide down a smooth inclined plane of angle 45º. When the inclined plane is made rough, the block takes nt seconds to slide down. The coefficient of friction between the block and the rough incline is

(a) 1/n (b) 1 – (1/n) (c) 1 – (1/n²) (d) (1/n²) (e) 1 + (1/n²)

In the case of the smooth inclined plane, the acceleration of the block down the plane is gsinθ where ‘θ’ is the angle of the plane (equal to 45º). If ‘s’ is the length of the plane, we have

s = 0 + ½ (gsinθ)t²

[We have used the equation of linear motion, s = ut + ½ at²]

In the case of the rough inclined plane, since the net force acting on the block is (mgsinθ – μmgcosθ), the acceleration of the block down the plane is g(sinθ – μcosθ).

Therefore, we have

s = 0 + ½ g(sinθ – μcosθ) (nt)²

From the above equations, we obtain

sinθ(n² – 1) = μn²cosθ

Therefore, μ = [(n² – 1) /n²] tanθ.

Since θ = 45º, tanθ =1 and hence μ = (n² – 1)/n² = 1 – (1/n²)

Two Practice Problems (MCQ) on Friction

Posts on friction were published on this site on 12^th August 2006, 7^th May 2007, 23^rd May 2007 and 1^st June 2007. You can access these posts easily by clicking on the label ‘friction’ below this post or on the side of this page. It will be beneficial for you to go through those posts before trying the practice problems given below:

(1) A block of mass ‘M’ equal to 1 kg is resting on a frictionless horizontal surface. Another block of mass ‘m’ equal to 0.2 kg is resting on the top surface of the 1 kg block. The coefficient of friction between the two blocks is 0.2. The motion of the 1 kg block is controlled by a light spring of force constant 24 Nm^–1 arranged as shown in the figure. What is the maximum possible amplitude of oscillation of the blocks (together) so that the upper block does not slip relative to the lower block? (g = 10 ms^–2)

(a) 2 cm (b) 5 cm (c) 10 cm (d) 20 cm (e)50cm

(2) A metallic cube takes ‘t’ seconds to slide down a smooth inclined plane of angle 45º. When the inclined plane is made rough, the block takes nt seconds to slide down. The coefficient of friction between the block and the rough incline is

(a) 1/n (b) 1 – (1/n) (c) 1 – (1/n²) (d) (1/n²) (e) 1 + (1/n²)

Try to solve the above problems. I’ll be back with the solution shortly.

Two Multiple Choice Questions on Concurrent Forces

The following questions appeared in KEAM 2007 (Engineering) question paper. Questions of this type are often found in various Medical and Engineering Entrance examinations and other similar examinations such as AP Physics Exam and GRE.

(1) A mass of 6kg is suspended by a rope of length 2m from a ceiling. A force of 50 N in the horizontal direction is applied at the mid point of the rope. The angle made by the rope with the vertical, in equilibrium is

(a) 50º (b) 60º (c) 30º (d) 40º (e) 45º

This can be easily worked out using tangent law in Statics dealing with concurrent forces. With reference to the figure, we have

tanθ = F/W

where F is the horizontal force, W is the weight (in newton) suspended and θ is the angle made by the rope with the vertical. [ Note that the tangent law follows from the law of triangle of forces].

In the present case, F = 50 N and W= mg.

Even if you take the value of ‘g’ to be approximately equal to 10 ms^–2, you will arrive at the answer. Thus W = 6×10 = 60 N so that tanθ = 50/60 = 5/6.

Even though you are not allowed to use physical and mathematical tables at the examination, you can arrive at the correct option as 40º since you know that tan 30º = 1/√3 and tan 45º = 1.

(2) The sum of the magnitudes of two forces acting at a point is 18 N and the magnitude of their resultant is 12 N. If the resultant is at 90º with the smaller force, the magnitude of the forces in newton are

(a) 6, 12 (b) 11, 7 (c) 5, 13 (d) 14, 4 (e) 10, 8

If P and Q are the two forces, we have P + Q = 18 N so that P = 18 – Q.

Since the resultant R (12 N) is at right angles to the smaller force P (fig.) we can write

Q = √(P² + 12²) = √[(18–Q)² + 12²], from which Q = 13 N.

Therefore, P = 18–13 = 5 N.

The correct option therefore is (c).

IIT-JEE 2007 Questions on Newton’s Laws of Motion

The following MCQ which appeared in the IIT-JEE question paper is interesting:

Two particles of mass 'm' each are tied at the ends of a light string of length 2a. The whole system is kept on a frictionless horizontal surface with the string held tight so that each mass is at a distance 'a' from the centre P (as shown in the figure). Now, the mid point of the string is pulled vertically upwards with a small but constant force F. As a result, the particles move towards each other on the surface. The magnitude of acceleration, when the separation between them becomes 2x is
(a) Fa/2m√(a²–x²) (b) Fx/2m√(a²–x²)
(c)Fx/2ma (d) F√(a²–x²) /2mx

With reference to the figure, F = 2T sinθ so that T = F/2sinθ where 'T' is the tension in the string

The horizontal force on each particle is T cosθ = Fcosθ/2sinθ = F/2tanθ.

Therefore, the magnitude of acceleration of each particle is given by

acceleration = force/mass = F/(2m tanθ) = F/[2m√(a²–x²)/x] = Fx/2m√(a²–x²)

The following Assertion-Reason Type Multiple Choice Question also appeared in the IIT-JEE 2007 question paper:

STATEMENT-1

A cloth covers a table. Some dishes are kept on it. The cloth can be pulled out without dislodging the dishes from the table

because

STATEMENT-2

For every action there is an equal and opposite reaction.

(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 simple question which you can answer in no time if you have understood the property of inertia. In fact, by repetition this question has become a ‘knowledge’ type even though it is supposed to be an ‘understanding and application’ type.

If you pull the cloth out abruptly in the horizontal direction, the dishes will remain intact on the table. Because of inertia, the dishes cannot move sideways immediately and will touch the bare surface of the table where they will rest as before.

Statement-2 is Newton’s third law which is not the reason for statement-1.

So, the correct option is (b).