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).

Multiple Choice Questions on Newton’s Laws of Motion

You will have to apply Newton’s Laws of Motion in different branches of Physics, but you will find questions specifically meant for checking your understanding of these laws in AP Physics Examination, Graduate Record Examination (GRE) and Medical and Engineering Entrance Examinations.

Consider the following MCQ based on impulse:
The force ‘F’ acting on a particle of mass ‘m’ is indicated by a force-time graph (Fig.). The momentum received by the particle during the time from zero to 8 s is

(a) 24 Ns (b) 20Ns (c) 12Ns (d) 6Ns (e) zero

The area under the force-time graph gives the impulse imparted to the particle. Impulse is a vector quantity and so you must consider its sign while adding the areas. The impulse received from zero to 2 seconds is positive and is equal to the area of the triangle, which is 6 Ns. The impulse received during the time from 2 s to 4 s is the area of the rectangle, which is – 6 Ns. The impulse received during the time from 4 s to 8 s is the area of the larger rectangle, which is 12 Ns.

Hense the net impulse received during the time from zero to 8 seconds is 6 –6 +12 = 12 Ns.

Since the impulse is equal to the change of momentum, the correct option is (c).

Now consider the following question:

A boy caught a ball of mass 200g moving with a speed of 30 ms^–1. If the catching process be completed in 0.1 s, the force of impact exerted by the ball on the hands of the boy is

(a) 60 N (b) 40 N (c) 30 N (d) 20 N (e) 10 N

We have force F = dp/dt where ‘dp’ is the change in linear momentum during the time dt.

The initial momentum of the ball is 0.2×30 = 6 kgms^–1 and the final momentum is zero so that dp = 6.

Therefore, F = 6/0.1 = 60 N.

The following MCQ which appeared in Karnataka CET 2002 question paper is meant for checking your grasp of the law of conservation of linear momentum:

A projectile is moving at 20 ms^–1at its highest point, where it breaks into two equal parts due to an internal explosion. One part moves vertically up at 30 ms^–1 with respect to the ground. Then the other part will move at

(a) 30 ms^–1 (b) 50 ms^–1 (c) 10√3 ms^–1 (d) 20 ms^–1

If the mass of the projectile is ‘m’, the mass of each fragment after the explosion is m/2. The momentum of the part which moved upwards is (m/2)×30 =15m. This is shown as vector OA in the figure.

The momentum of the projectile at the highest point of its path is 20m and its direction is horizontal. This is shown as vector OC.

The momentum of the other part (after the explosion) is (m/2)×v where ‘v’ is its velocity. This is shown as vector OB in the figure.

Since the momentum is conserved, the total final momentum, which is the vector sum of the momentum vectors OA and OB must be equal to the initial momentum, represented by the vector OC.

Evidently, OB = √(OA² + OC²).

Or, mv/2 = √[(15m)² + (20m)²] = 25m.

This gives v = 50 ms^–1.

Solution to Practice Questions on Electromagnetic Induction

In the post dated 24^th July 2007, two questions on electromagnetic induction were given without solution. The questions with solution are given below as promised in the post:

(1) A search coil in the form of a plane circular copper loop of 10 turns and diameter 2 cm is held within a magnetic field of 20000 gauss, with the plane of the loop perpendicular to the magnetic field. The resistance of the search coil is 0.1 Ω. If the search coil is withdrawn from the magnetic field in a time span of 0.1s, the total charge (in coulomb) and the average current (in ampere) flowing during this time are respectively

(a) 0.1π, 0.5π (b) 0.2π, 2π (c) 0.2π, π

(d) 0.1π, π (e) 0.02π, 0.2π

The charge ‘Q’ induced in the search coil is given by

Q = Flux change/Resistance = nAB/R where ‘B’ is the magnetic flux density and ‘n’, ‘A’ and ‘R’ are respectively the number of turns, area and the resistance of the search coil.

Therefore, Q = [10× π(0.01)²×2]/0.1 = 0.02 π coulomb.

The average current ‘I’ in the coil is given by

I = Q/t = 0.02 π/0.1 = 0.2 π ampere.

(2) A triangular copper loop ABC is moving in its own plane with velocity ‘v’ in a uniform magnetic field acting perpendicular to the plane of the loop. If the direction of motion is perpendicular to the side AB of the loop, an electric field is induced

(a) in AB only

(b) in AC as well as BC, but not in AB

(c) in BC only

(d) in AB, AC and BC

(e) in no side of the loop

There is no change in the magnetic flux linked with the loop and hence the net emf acting in the whole loop is zero. But, since all the conductors AB. AC and BC are moving in a magnetic field, cutting the magnetic field lines, the free charge carriers in them experience magnetic force (Lorentz force) and get shifted towards their ends . So, there are electric fields in all these conductors (generally). In other words, there are induced voltages across AB, AC and BC, but the net voltage in the whole loop is zero since the voltage across AB is equal and opposite to the net voltage across the series combination of AC and BC.

The correct option is (d).

A special case is the possibility of the triangle to be right angled at A or B. If the triangle is right angled at A, the electric field induced across AC is zero. If the triangle is right angled at B, the electric field induced across BC is zero. But the most suitable choice in the above question is still (d).

Two Practice Questions (MCQ) on Electromagnetic Induction

Try to work out the following questions on electromagnetic induction:
(1) A search coil in the form of a plane circular copper loop of 10 turns and diameter 2 cm is held within a magnetic field of 20000 gauss, with the plane of the loop perpendicular to the magnetic field. The resistance of the search coil is 0.1 Ω. If the search coil is withdrawn from the magnetic field in a time span of 0.1s, the total charge (in coulomb) and the average current (in ampere) flowing during this time are respectively
(a) 0.1π, 0.5π (b) 0.2π, 2π (c) 0.2π, π
(d) 0.1π, π (e) 0.02π, 0.2π

(2) A triangular copper loop ABC is moving in its own plane with velocity ‘v’ in a uniform magnetic field acting perpendicular to the plane of the loop. If the direction of motion is perpendicular to the side AB of the loop, an electric field is induced
(a) in AB only
(b) in AC as well as BC, but not in AB
(c) in BC only
(d) in AB, AC and BC
(e) in no side of the loop
If you have mastered the basic points in electromagnetic induction, you will be able to answer the above questions in a couple of minutes. Try. I’ll be back shortly with the solution.

IIT-JEE 2007 Assertion-Reason Type MCQ on Lenz’s Law

The following Assertion-Reason Type MCQ which appeared in IIT-JEE 2007 question paper is meant for gauging your grasp of Lenz’s law. Here is the question:

STATEMENT-1

A vertical iron rod has a coil of wire wound over it at the bottom end. The rod goes through a conducting ring as shown in the figure. The ring can float at a certain height above the coil

because

Statement-2

In the above situation, a current is induced in the ring which interacts with the horizontal component of the magnetic field to produce an average force in the upward direction.

(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

Statement-1 is true. But statement-2 false[Option (c)].

When the magnetic flux linked with the ring increases because of the increase of current in the coil, the ring will be repelled in accordance with Lenz’s law, so as to oppose the increase of flux through the ring. When the current through the coil decreases, the flux linked with the ring decreases. In this case, the ring will be attracted towards the coil, so as to oppose the decrease of flux through the ring. Since the coil is fed with an alternating current, these actions are repeated successively and the ring can therefore float above the coil.

The ring floats because of Lenz’s law. Basically it is the result of the interaction of the magnetic field produced by the induced current with the magnetic field produced by the coil, but the magnetic field produced by the coil is along the vertical (through the iron rod) and is not horizontal. The vertical field produces the flux and the change of flux through the ring and this is responsible for the floating of the ring.

Statement-2 is therefore false.

KEAM (Engineering) 2007 Question on Polarisation

In the post dated 3^rd May 2007, four questions on Optics which appeared in Kerala Engineering Entrance 2007 test paper were discussed. Here is an additional question which appeared in the same paper:

If the polarising angle of a piece of glass for green light is 54.74º, then the angle of minimum deviation for an equilateral prism made of same glass is (given tan 54.74º = 1.414)

(a) 45º (b) 54.74º (c) 60º (d) 90º (e) 30º

Brewster’s law gives you the refractive index ‘n’ in terms of the polarising angle ‘θ’:

n = tan θ

Therefore, the refractive index of the given equilateral prism is 1.414, which is √2

Since n = sin[(A+D)/2] /sin(A/2) where ‘A’ is the angle of the prism and ‘D’ is the minimum deviation, we have

√2 = sin[(60º +D)/2]/sin 30º

This gives sin(60º +D)/2 = 1/√2 so that D = 30º

Now that you have worked out a problem on polarisation, note the following points:

(i) Transverse waves only can be polarized (Longitudinal waves cannot be polarised).

(ii) The proof for the transverse nature of light is the exhibition of polarisation by light.

Solution to the Question on Two Dimensional Motion

In the post dated 10^th July 2007, a questions on two dimensional motion was given without solution. The question with solution is given below as promised in the post:

A steel sphere A projected up with a velocity of 10 ms^–1 at an angle of 60º with the horizontal, collides elastically with an identical steel sphere B located at the highest point of the trajectory of A. The sphere B is the bob of a simple pendulum. After the collision, if the sphere B just moves along a vertical circle, what is the length of the pendulum? ( Take g = 10 ms^–2)

(a) 50 cm (b) 55 cm (c) 60 cm (d) 65 cm (e) 70 cm

At the highest point of the trajectory, the velocity of the projectile is horizontal and is equal to ucosθ = 10 cos 60° = 5 ms^–1. Since the collision is elastic and the spheres are identical, the sphere A transfers its entire momentum to sphere B. Sphere B therefore moves with the same horizontal velocity of 5 ms^–1.

If the sphere is to just move along a vertical circle, the minimum velocity required at the lowest point of the circle is √(5gR) where ‘R’ is the radius of the circle (which is the length of the pendulum here). Therefore, we have

√(5gR) = 5 ms^–1.

On substituting for the acceleration due to gravity, g (=10 ms^–1), R = 0.5 m = 50 cm.

Now, consider the following MCQ:

A ball rolling along a horizontal floor with a velocity of 7 ms^–1 reaches the edge of a long stair case from where it moves down as a projectile. If the width and the height of the steps of the stair case are 25 cm and 10 cm respectively, the ball will strike the edge of the n^th step from the top, where n is equal to

(a) 10 (b) 12 (c) 14 (d) 16 (e) 18

Since the horizontal velocity is unchanged throughout the trajectory of the ball, we have

7 t = 0.25 n where ‘t’ is the time taken to hit the edge of the n^th step after leaving the top of the stair case.

Therefore, t = 0.25 n / 7

The vertical fall of the ball during this time is ½ g t² and we have

½ g t² = 0.1 n

Substituting for t and g, ½ ×9.8×0.0625 n²/49 = 0.1n

Therefore, n = (49×0.1)/(4.9×0.0625) =16