Two Questions (MCQ) on Electric Potential

(1) An infinite number of point charges each equal to +Q coulomb are arranged at random around a point P such that the distances of the charges from the point P are 1m, 2m, 4m, 8m, 16m,…….etc... The electric potential at P is

(a) zero (b) infinite (c) negligibly small

(d) Q/2πε₀ (e) Q/4πε₀

Note that the electric potential is a scalar quantity. Therefore, the direction of the location of the charge does not matter and the potentials simply add up. The resultant potential (V) at P is given by

V = (Q/4πε₀) × [(1/1) + (1/2) + (1/4) + (1/8) + ………]

The infinite series within the square bracket yields a value equal to 2 so that V = Q/2πε_0.

(2) A thin spherical conducting shell of radius R has a charge +Q. Another point charge –q is placed at the centre of the shell. The electrostatic potential at a point P distant R/2 from the centre of the shell is

(a) (Q/4πε₀R) – (2q/4πε₀R) (b) (q/4πε₀R) – (2Q/4πε₀R)

(c) – (2q/4πε₀R) (d) (Q/4πε₀R) (e) zero
The electrostatic potential at any point within the shell due to the charge Q on the shell is constant and is equal to Q/4πε₀R.

The potential at distance R/2 due to the charge –q placed at the centre of the shell is – q/4πε₀(R/2) = – 2q/4πε₀R.

Therefore, the net potential (V) at the point P distant R/2 from the centre of the shell is given by

V = (Q/4πε₀R) – (2q/4πε₀R), given in option (a).

[Note that positive charges will produce positive potential where as negative charges will produce negative potential].

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

You will find similar multiple choice questions with solution at physicsplus also.

Two KEAM (Engineering) 2007 Multiple Choice Questions on Gravitation

The following questions which appeared in KEAM 2007 question paper are typical and of the type repeatedly asked in various entrance examinations:

(1) The change in potential energy when a body of mass ‘m’ is raised to a height nR from earth’s surface is (R = radius of the earth)

(a) mgR(n/n–1) (b) mgR (c) mgR(n/n+1)

(d) mgR(n²/n² +1) (e) mgR/n

This question appears often in entrance question papers. There may be slight change in the wording (as for example, “what is the work done in lifting a body of mass m, from the earth’s surface, through a height nR?”).

Gravitational potential energy of a mass ‘m’ at a height ‘h’ is given by

U = – GMm/(R+h) where M is the mass of the earth and G is the Gravitational constant.

Since h = nR, the change in potential energy is

– GMm/(R+nR) – (–GMm/R) = (GMm/R)[1 – 1/(1+n)] = (GMm/R)[n/(1+n)]

Since g = GM/R², the change in potential energy becomes mgRn/(n+1), given in option (c).

(2) The escape velocity of a body on the surface of the earth is 11.2 km/s. If the mass of the earth is doubled and its radius is halved, the escape velocity becomes

(a) 5.6 km/s (b) 11.2 km/s (c) 22.4 km/s

(d) 44.8 km/s (e) 67.2 km/s

The escape velocity (v_e) of a body on the surface of the earth is given by

v_e = √(2GM/R)

Therefore we have √(2GM/R) = 11.2 km/s

If the mass (M) of the earth is doubled and its radius (R) is halved, the escape velocity becomes √(2×4GM/R) = 2×√(2GM/R) = 22.4 km/s.

Two Multiple Choice Questions on Work and Energy

The following MCQ which appeared in KEAM (Engineering) 2007 question paper is worth noting:

A particle is released from a height S. At a certain height its kinetic energy is three times its potential energy. The height and speed of the particle at that instant are respectively

(a) S/4, 3gS/2 (b) S/4, [√(3gS)]/2 (c) S/2, [√(3gS)]/2

(d) S/4, √(3gS/2) (e) S/3, √(3gS/2)

Suppose the given conditions occur at the instant when the particle is at height ‘h’.

Potential energy at the height h = mgh.

Since the initial energy is completely potential and is equal to mgS, the kinetic energy at the height ‘h’ is equal to the loss of potential energy and is equal to mg(S–h).

Since the K.E. is three times the P.E., we have

mg(S–h) = 3mgh, from which h = S/4.

But, kinetic energy = ½ mv².

Therefore, ½ mv² = mg(S–h) = mg[S– (S/4)] = 3 mgS/4 so that

v =√(3gS/2)

Now, consider the following MCQ which is simple. But simple qiestions may cheat you some times. Here is the question:

A uniform chain of mass M and length L is placed on a table with 20% of its length overhanging from the edge of the table. The minimum work that has to be done to place the hanging portion on the table is

(a) MgL/5 (b) MgL/10 (c) MgL/20

(d) MgL/25 (e) MgL/50

The hanging portion has mass M/5 and therefore weight Mg/5. The length of the hanging portion is L/5 and hence the centre of gravity of the hanging portion is at a depth L/10 from the surface of the table. The entire weight of the hanging portion can be imagined to act through its centre of gravity. When you pull the hanging portion up to keep it on the table, you are moving the point of application of the force (equal to the magnitude of the the weight Mg/5) through a distance L/10 along the direction of the force.

Therefore, the work done is (Mg/5)×(L/10) = MgL/50

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