Electrostatics: MCQ on Charged Spherical Drops

Often questions involving the calculation of the potential of a drop obtained by the combination of a number of identical charged spherical droplets are seen in college entrance question papers. If n identical small drops each of radius r carrying charge q coalesce to form a single large drop of radius R, we have

R = n^1/3 r

[You will get this by equating the volumes: n×(4/3)πr³ = (4/3)πR³]

The potential V on the surface of each small drop is given by

V = (1/4πε₀)(q/r)

The total charge on the larger drop is nq. Therefore, the potential V’ on its surface is

V’ = (1/4πε₀)(nq/R) = (1/4πε₀)(nq/n^1/3r) since R = n^1/3 r

Therefore, V’ = n^2/3V

The electric field E on the surface of each small drop is given by

E = (1/4πε₀)(q/r²)

The electric field E’ on the surface of the large drop is given by

E’ = (1/4πε₀)(nq/R²) = (1/4πε₀)(nq/n^2/3r² ) since R = n^1/3 r

Therefore, E’ = n^1/3E

If you can remember the above expressions for the electric potential and field on the larger drop, multiple choice questions involving them can be answered in no time. But it is always more rewarding to remember the basic things so that you can calculate the required quantities in all situations.

The calculation of V’ in the above form itself appeared as a multiple choice question in Kerala Engineering Entrance 2008 question paper.

Now answer the following multiple choice question involving charged drops. You may try them yourself and then check with the given solution. Here are the questions:

(1) Sixty four identical drops each charged by q coulomb to potential V volt coalesce to form a single large drop. The charge and potential on the large drop are respectively

(a) 64q, 64V

(b) 64q, V

(c) 8q, 16V

(d) 16q, 16V

(e) 64q, 16V

The charge on the large drop is the sum of the charges on the small drops and is equal to 64q.

As shown above, the potential on the large drop is V’ = n^2/3V = 64^2/3V =16V [Option (e)].

(2) Two identical soap bubbles A and B are uniformly charged with the same amount of charge. But the charge on A is positive where as the charge on B is negative. (The electrical interaction between the bubbles is negligible). Because of charging, the excess of pressure inside

(a) both bubbles will increase

(b) both bubbles will decrease

(c) both bubbles will remain unchanged

(d) A will increase and that inside B will decrease

(e) B will increase and that inside A will decrease

Because of the repulsive force between like charges, the bubbles will expand and hence the excess of pressure inside both bubbles will decrease.

You will find many useful posts on different branches of Physics at your level at apphysicsresources.blogspot.com and at physicsplus.blogspot.com. The essential equations to be remembered in the section, ‘Electric field and Potential’ can be found here.

Kerala Medical Entrance (KEAM) 2008 Questions from Newton’s Laws of Motion

The following questions which appeared in Kerala Medical Entrance 2008 question paper are of the type popular among question setters:

(1) Two blocks of masses 7 kg and 5 kg are placed in contact with each other on a smooth surface. If a force of 6 N is applied on the heavier mass, the force on the lighter mass is

(a) 3.5 N

(b) 2.5 N

(c) 7 N

(d) 5 N

(e) 6 N

The common acceleration a of the system of masses is given by

a = Driving force/Total mass moved = 6 N/(7+5) kg = 0.5 ms^–2

The force on the lighter mass = Ma = 5×0.5 = 2.5 N.

(2) A body of mass 60 kg suspended by means of three strings P, Q and R as shown in the figure is in equilibrium. The tension in the string P is

(a)130.9g N

(b) 60g N

(c) 50g N

(d) 103.9g N

(e) 109g N

Let us denote the tensions in the three strings by P, Q and R. The concurrent forces P, Q and R keep the common meeting point of the strings in equilibrium so that they can be represented by the three sides of a triangle taken in order as shown in the adjoining figure. From this right angled triangle,

Q/P = tan 30º where Q = weight of the mass M = Mg =60g newton.

[This equation is often written as W/_H = tan θ where W is the weight and H is the horizontal force and is known as tangent law].

Substituting for Q, we have 60g/P = 1/√3, from which P = 60g×√3 = 103.9g N.

Kerala Engineering Entrance 2008 Questions on One Dimensional Motion

The following questions (on one dimensional kinematics) numbered 1, 2 and 3 appeared in KEAM (Engineering) 2008 question paper:

(1) A particle starts from rest at t = 0 and moves in a straight line with an acceleration as shown below. The velocity of the particle at t = 3 s is

(a) 2 ms^–1

(b) 4 ms^–1

(c) 6 ms^–1

(d) 8 ms^–1

(e) 1 ms^–1

According to the acceleration – time graph shown, the particle has an acceleration of 4 ms^–2 during the first two seconds. Therefore, its velocity (v₂) at the end of 2 seconds is given by

v₂ = v₀ + at = 0 + 4×2 = 8 ms^–1

From 2 second to 3 second the particle has a retardation of 4 ms^–2. Hence its velocity (v₃) at the end of 3 seconds is given by

v₃ = v₂ – at = 8 – 4×1 = 4 ms^–1 [Option (b)].

(2) Two cars A and B are moving with same speed of 45 km/hr along same direction. If a third car C coming from the opposite direction with a speed of 36 km/hr meets two cars in an interval of 5 minutes, the distance of separation of two cars A and B should be (in km)

(a) 6.75

(b) 7.25

(c) 5.55

(d) 8.35

(e) 4.75

The relative velocity of car A with respect to car B (and that of car B with respect to A) is zero since they have the same speed in the same direction. The relative velocity of car C with respect to A and B is 45 + 36 = 81 km/hr. Since the car C takes a time of 5 minutes to cover the distance between A and B, the separation between A and B is 81×(5/60) km = 6.75 km.

(3) An object is dropped from rest. Its v - t graph is

The correct option is (a) since the velocity ‘v’ is directly proportional to the time t in accordance with the equation v = v₀ + at where v₀ is the initial velocity which is zero and a is the acceleration which is the constant acceleration due to gravity. The graph should evidently pass through the origin.

Suppose you were asked to draw the v – t graph in the case of a ball projected vertically upwards with a velocity u. If you are asked to draw the graph from the instant the ball leaves your hand to the instant you catch it while returning, you can do it as shown, ignoring the air resistance and the time taken for the velocity to reduce to zero on hitting your hand.

Questions involving Viscosity and Buoyancy

The following simple question involving viscous force and force of buoyancy is meant for checking whether you have a good understanding of basic points:

A wooden ball of relative density 0.5 is released from the bottom of a still water reservoir. Its acceleration while moving up will

(a) go on decreasing initially

(b) go on increasing initially

(c) remain constant at g/2 (in magnitude)

(d) remain constant at g (in magnitude)

(e) be zero throughout

The net force acting on the ball at the moment of releasing is its apparent weight which is upwards. [The apparent weight = Real weight – up thrust = Vρg – Vσg. Since the density of wood (ρ) is less than that of water (σ), the apparent weight is negative which means it is directed upwards]

The sphere therefore moves up with an acceleration. But, when the velocity (v) increases from zero, the opposing viscous force (6πrηv) also increases thereby reducing the net upward force. The upward acceleration therefore goes on decreasing. So, the correct option is (a).

The following question which appeared in Kerala Engineering Entrance 2008 question paper also involves viscous force and force of buoyancy; but the latter is negligible:

Eight drops of a liquid of density ρ and each of radius ‘a’ are falling through air with a constant velocity of 3.75 cm s^–1. When the eight drops coalesce to form a single drop, the terminal velocity of the new drop will be

(a) 1.5×10^–2 ms^–1

(b) 2.4×10^–2 ms^–1

(c) 0.75×10^–2 ms^–1

(d) 25×10^–2 ms^–1

(e) 15×10^–2 ms^–1

The viscous force acting on a sphere of radius ‘a’ is 6πahv where h is the coefficient of viscosity of the fluid and ‘v’ is the velocity of the sphere. Terminal velocity is attained when the opposing viscous force is equal in magnitude to the apparent weight of the sphere. Therefore,

6πahv = (4/3)πa³(ρ – σ)g where ρ is the density of the sphere, σ is the density of the fluid and ‘g’ is the acceleration due to gravity.

The above equation shows that the terminal velocity ‘v’ is directly proportional to the square of the radius of the sphere.

In the above problem, eight identical drops coalesce to form a single drop. The new drop thus formed has eight times the volume of each small drop. The radius ‘R’ of the new drop is given by

(4/3)πR³ = 8×(4/3)πa³

Therefore, R =2a

Since the radius of the new drop is twice that of each small drop, the terminal velocity of the new drop must become four times. The correct option therefore is 15×10^–2 ms^–1.

Two Multiple Choice Questions involving Kinetic Energy

Check whether you can find the answers to the following two questions in a couple of minutes:

(1) When a running boy increases his speed by 2 ms^–1, his kinetic energy becomes three times the original value. His original speed is (in ms^–1)

(a) √3 +1

(b) √3 –1

(c) √3

(d) 2√3

(e) 2

If his original speed is ‘v’ we can write

(½)mv²×3 = (½)m(v+2)² where ‘m’ is his mass.

This gives v√3 = (v+2) so that v(√3 –1) =2.

Therefore, v = 2 /(√3 –1) = √3 +1 on multiplying the numerator and denominator by (√3 –1).

(2) A particle of mass ‘m’ at rest is acted on by a force ‘F’ for a time‘t’. Its kinetic energy after the time t is

(a) Ft²/2m

(b) F²t/2m

(c) F²t²/2m

(d) F²t²/3m

(e) F²t²/m

The impulse received by the particle during the time t is Ft. But impulse is equal to the change of momentum. Since the initial momentum of the particle is zero, Ft is the final momentum of the particle. The kinetic energy of the particle after the time t is therefore equal to F²t²/2m (remembering that KE = p²/2m where p is the momentum).