Multiple Choice Questions (MCQ) on Newton’s Laws of Motion

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

A shell at rest at the origin explodes into three fragments of masses 1 kg, 2 kg and ‘m’ kg. The 1 kg and 2 kg pieces fly off with speeds of 5 ms^–1 along the X- axis and 6 ms^–1 along the Y-axis respectively. If the m kg piece flies off with speed 6.5 ms^–1, the total mass of the shell must be

(a) 4 kg (b) 5 kg (c) 3.5 kg (d) 4.5 kg (e) 5.5 kg

The initial momentum of the shell is zero. Explosion of the shell involves internal forces only. Since there are no external forces, the total momentum of all the fragments of the shell taken together must be zero.

The momentum of the 1 kg fragment is 5 kg ms^–1 and that of the 2 kg fragment is 12 kg ms^–1. Since these are at right angles, the net momentum of these two fragments taken together is √(5² + 12²) = 13 kg ms^–1. The momentum of the m kg fragment must be equal and opposite to this (to make the net momentum of the three fragments zero).

Therefore, m×6.5 = 13, from which m = 2 kg.

The mass of the shell is therefore 1 + 2 + 2 = 5 kg.

The following question is simple, but many among you will have certain doubts about it:

A uniform rope of length ‘L’ is resting on a smooth horizontal surface. If it is pulled at one end by applying a horizontal force ‘F’ parallel to its length, what will be the tension in the rope at a distance ‘d’ from the other end?
(a) F (b) F/d (c) FL/d (d) Fd (e) Fd/L

When you work out questions on bodies connected by strings, you will assume that the string is light and therefore the tension in a segment of the string connecting two bodies will be the same everywhere. The acceleration of the system in such cases will be determined by the masses of the connected bodies only.

In the above problem, there are no bodies connected to the rope and hence the acceleration ‘a’ of the rope is determined by the mass of the rope only.

Therefore, a = F/ mL where ‘m’ is the mass per unit length of the rope.

The tension ‘T’ in the rope at the distance ‘d’ from the other end is responsible for pulling the portion of length ‘d’ of the rope. Therefore we have

T = mass × acceleration = md × F/mL = Fd/L

The following question on connected bodies is a popular one and you are expected to know how to solve it:

Three blocks of masses m₁, m₂ and m₃ are connected by light inextensible strings and carried by a light frictionless pulley as shown. If (m₁+m₂) > m₃, the tension in the string connecting m₁ and m₂ is

(a) 2m₁m₃g/ (m₁+m₂+m₃)

(b) 2(m₁+m₂)m₃g/(m₁+m₂+m₃)

(c) m₁g

(d) (m₁+m₂)

(e) 2m₁m₂m₃g/ (m₁+m₂+m₃)

Note that the tension in the string connecting m₁ and m₂ is different from the tension in the string connecting m₂ and m₃.

The common acceleration ‘a’ of the system of masses is the ratio of the net driving force to the total mass moved.

Therefore, a = (m₁+m₂ – m₃)g/(m₁+m₂+m₃)

The tension (T) in the string connecting m₁ and m₂ is produced by the apparent weight ( as in the case of motion in a lift) of the mass m₁ which moves downwards with acceleration ‘a’.

Therefore, T = m₁(g – a) = m₁[g – (m₁+m₂ – m₃)g/(m₁+m₂+m₃)]

= m₁g[1– (m₁+m₂ – m₃)/(m₁+m₂+m₃)]

= 2m₁m₃g/ (m₁+m₂+m₃)

You will find more questions (with solution) in this section by clicking on the label ‘Newton’s laws’ below this post or on the side of this page.

Similar multiple choice questions with solution can be found at physicsplus: Multiple Choice Questions on Newton’s Laws of Motion

Two Multiple Choice Questions on Electromagnetic Induction

(1) The initial rate of increase of current when a battery of emf 6V is connected in series with an an inductance of 2H and a resistance of 12 Ω is

(a) 0.5 A s^–1 (b) 1 A s^–1 (c) 3 A s^–1 (d) 3.5 A s^–1 (e) 4.5 A s^–1

Initially the current in the circuit is zero but the rate of increase of current is maximum. There will be no voltage drop (which is equal to R×I) across the resistance and the entire voltage will appear across the inductance. The magnitude of the voltage across the inductance will be L(dI/dt) so that initially we have

L(dI/dt) = V, from which dI/dt = V/L = 6/2 = 3 A s^–1.

[You can calculate the rate of increase of current at instant using the expression for the current in an LR circuit during the rise, given by I = I₀(1– e^–Rt/L) where ‘e’ is the base of the natural logarithm, ‘I’ is the current at the instant ‘t’ and I₀ is the final (maximum) current equal to V/R.

Differentiating the above equation, dI/dt = (I₀R/L)e^–Rt/L.

Initially, since t = 0, dI/dt = I₀R/L = (V/R)×R/L = (6/12)×12/2 = 3 As^–1

The rate of decrease of current at any instant ‘t’ when the current in an LR circuit is switched off is – (I₀R/L)e^–Rt/L since the current during the decay is given by I = I₀ e^–Rt/L]

(2) The emf induced in a coil at time t = 0.5 s when the magnetic flux (Φ) linked with the coil changes as Φ = t² – 4t + 2 weber is

(a) 3V (b) 4V (c) 5V (d) 6.5V (e) 8V

The emf (V) induced is given V = – dΦ/dt = –d[t² – 4t + 2]/dt = –2t + 4.

On substituting for t (=0.5 s), V = 3 volt

KEAM (Medical) 2007 Questions on Electromagnetic Induction

The following questions which appeared in Kerala Medical Entrance 2007 question paper are simple, as is usually the case as far as the questions on electromagnetic induction are concerned:

(1) A wire of length 50 cm moves with a velocity of 300 m/minute perpendicular to a magnetic field. If the emf induced in the wire is 2V, the magnitude of the magnetic field in tesla is

(a) 2 (b) 5 (c) 0.4 (d) 2.5 (e) 0.8

The emf (V) induced in a wire of length ‘L’ metre moving with velocity ‘v’ metre/second perpendicular to a magnetic field ‘B’ tesla is given by

V = BLv from which B = V/Lv

Be careful to convert the given velocity in to meter/second.

Thus we have B = 2/(0.5×5) = 0.8 tesla

(2) Whenever a magnet is moved either towards or away from a conducting coil, an emf is induced, the magnitude of which is independent of

(a) the strength of the magnetic field

(b) the speed with which the magnet is moved

(c) the number of turns in the coil

(d) the resistance of the coil

(e) the area of cross section of the coil

The magnetic flux linked with a circuit (and the emf induced) is independent of the resistance of the circuit. So the correct option is (d)

However, remember that the induced charge flowing in a circuit depends on the resistance as induced charge = (change of flux)/resistance

You will find more multiple choice questions with solution at physicsplus

A Multiple Choice Question on Lenz’s Law

Two plane coils A and B are arranged coaxially as shown. Using an external battery, an increasing or decreasing clockwise current (as judged by viewing along the axis in the direction marked by the arrow) can be made to flow through coil A. The directions of the induced current through coil B when the current in coil A is (i) increased and (ii) decreased are

(a) clockwise in cases (i) and (ii)

(b) anticlockwise in cases (i) and (ii)

(c) anticlockwise in case (i) and clockwise in case (ii)

(d) clockwise in case (i) and anticlockwise in case (ii)

(e) unpredictable

This is a question based on Lenz’s law in electromagnetic induction, according to which the induced current has to oppose the magnetic flux change which produces the induced current.

When the clockwise current in coil A increases as in case (i), there is an increasing magnetic flux through coil B. The direction of the magnetic field lines (due to the current in coil A) through coil B is along the direction of view marked by the arrow. The induced current in coil B should therefore produce a magnetic field in the opposite direction. Evidently, this will be the case if the sense of induced current in coil B is anticlockwise.

When the current in coil A is decreasing as in case (ii), the flux through coil B is decresing. The direction of the magnetic field lines (due to the current in coil A) is same as in case (i). But the induced current in coil B has to oppose the decrease of the flux. This can be done by producing a magnetic field in the same direction as that produced by coil A. Evidently, the sense of the induced current through coil B must be clockwise for this. So, the correct option is (c).

[Instead of the above lengthy argument, you can have a relatively shorter argument like this:

Coil A has to move away from coil B when the flux linked with coil B increases because of the increasing current in coil A. The required repulsive force between the coils can be produced only if they carry unlike currents. So, the induced current in B in case (i) is anticlockwise. In case (ii) the decrement in the flux linked with coil B can be opposed if coil B moves towards coil A. The attractive force required for this can be produced if the coils carry like currents (currents in the same direction). So, the induced current in B is clockwise in case (ii)].

IIT-JEE 2007 Assertion-Reason Type MCQ on Kinetic Energy of Gas Molecules

Assertion-Reason Type Multiple Choice Questions are often interesting to answer, as is the case with the following one which appeared in IIT-JEE 2007 question paper:

STATEMENT-1

The total translational kinetic energy of all the molecules of a given mass of an ideal gas is 1.5 times the product of its pressure and volume

because

STATEMENT-2

The molecules of a gas collide with each other and the velocities of the molecules change due to the collision

(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. To prove this you may use the expression for the translational kinetic energy (E) of a gas molecule, given by

E = (3/2) kT

where ‘k’ is Boltzman’s constant and ‘T’ is the absolute (Kelvin) temperature.

[Remember that the translational kinetic energy of any type of molecule is (3/2)kT since the energy per degree of freedom is ½ kT and in our three dimensional space, there can be three translational degrees of freedom only]

The total translational kinetic energy of all the molecules of the given mass of gas is (3/2) kT×nV where ‘V’ is the volume of the gas and ‘n’ is the number of molecules per unit volume.

But, the pressure (P) of the gas is given by

P = nkT .

Therefore, total translational kinetic energy = (3/2)PV = 1.5 PV.

Statement -2 is true. In fact this follows from the postulates of kinetic theory. But, the pressure exerted by a gas is not because of the collision between the gas molecules, but is because of the collision of gas molecules with the walls of the container. So, statement-2 is not a correct explanation for statement-1.

The correct option therefore is (B).