Electrostatics: Two Questions (MCQ) on Sharing of Charge

(1) Three identical conducting spheres X, Y and Z carrying charges 8 μC, – 4.8 μC and 6.4 μC respectively are kept in contact and then separated from one another. Then, the sphere Y will have approximately

(a) an excess of 3×10¹³ electrons

(b) a deficiency of 3×10¹³ electrons

(c) a deficiency of 6×10¹³ electrons

(d) an excess of 6×10¹³ electrons

(e) a deficiency of 2×10¹³ electrons

The total charge on the spheres is (8 – 4.8 + 6.4) μC = 9.6 μC. Since the spheres are identical, they have the same capacitance and the charge is shared equally by them. Therefore, the charge on each sphere is 3.2 μC.

Since the charge on the sphere Y (and the spheres X and Z) is positive, there will be a deficiency of electrons.

Remembering that the electronic charge is 1.6×10^–19 coulomb, the deficiency of electrons on the sphere is 3.2×10^–6/1.6×10^–19 = 2×10¹³. Therefore the correct option is (e).

(2) Two capacitors C₁ and C₂ of capacitance 2 μF and 3 μF are charged to 50 volt and 40 volt respectively and arranged as shown, with the key K open. On closing the key, the charge flowing through the key will be

(a) 8 μC

(b) 12 μC

(c) 32 μC

(d) 64 μC

(e) 132 μC

You must remember that the total charge is conserved in all situations. In the above question, the total charge is C₁V₁ + C₂V₂ = 100 μC + 120 μC = 220 μC.

When the key K is closed, the capacitors get connected in parallel and the common potential difference between their terminals will be

V = (C₁V₁ + C₂V₂)/(C₁ + C₂) = 220 μC/ 5 μF = 44 volt.

The charge on C₁ after closing the key will be C₁V = 88 μC.

Since the initial charge on C₁ is 100 μC, the charge flowing through the key must be (100–88) μC = 12 μC.

[You will obtain the same result on considering C₂].

Kerala Engineering Entrance (2007) Questions on Oscillations

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

(1) A particle executes S.H.M. with a time period of 16 s. At time t = 2 s, the particle crosses the mean position while at t = 4 s, its velocity is 4 ms^–1. The amplitude of motion in metre is

(a) √2 π

(b) 16√2 π

(c) 24√2 π

(d) 4/π

(e) (32√2)/π

If we start reckoning time from the instant the particle crosses the mean position (y =0 when t =0), the equation of the simple harmonic motion can be written as

y = A sin ωt, where A is the amplitude and ω is the angular frequency.

In this case, at the instant t = 2 s, we have velocity dy/dt = A ωcos ωt = 4 ms^–1.

Since the period T = 16 s, ω= 2π/T = π/8 so that the above equation becomes

A×(π/8) ×cos [(π/8)×2] = 4

Or, A×(π/8)×(1/√2) = 4, from which A = (32√2)/π

(2) In damped oscillations, the amplitude of oscillations is reduced to one-third of its initial value a₀ at the end of 100 oscillations. When the oscillator completes 200 oscillations, its amplitude must be

(a) a₀ /2

(b) a₀ / 6

(c) a₀ /12

(d) a₀ /4

(e) a₀ /9

In the case of damped oscillations, the amplitude decrease exponentially with time so that at time t, the amplitude (A) is given by

A = A₀ exp(– bt) where A₀ is the initial amplitude (at t = 0) and b is the damping constant.

If the time required for 100 oscillations is ‘t’, the time required for 200 oscillations will be 2t. Therefore we have

a₀/3 = a₀ exp(– bt) and

a = a₀ exp(– 2bt) where ‘a’ is the amplitude after 200 oscillations (at time 2t)

From the first equation, exp(– bt) = 1/3. Substituting this in the second equation,

a = a₀×(1/3)² = a₀/9.

(3) For a simple pendulum, the graph between T² and L is

(a) a straight line passing through the origin

(b) parabola

(c) circle

(d) ellipse

(e) hyperbola

The period (T) is given by T = 2 π√(L/g) with usual notations so that

T² = (4π²/g)L

This is in the form y = mx, which represents a straight line passing through the origin. So, the correct option is (a).

Kerala Engineering Architecture Medical (KEAM) 2008 Entrance Examinations

The Commissioner for Entrance Examinations, Govt. of Kerala, has invited applications for the Entrance Examinations for admission to the following Degree Courses in various Professional Colleges in the State for 2008-09.

(a) Medical: (i) MBBS (ii) BDS (iii) B.Pharm. (iv) B.Sc. (Nursing) (v) B.Sc. (MLT)

(vi) BAMS (vii) B.H.M.S (viii) B.S.M.S (Siddha) (ix) B.Sc. Nursing (Ayurveda)

(x) B.Pharm. (Ayurveda) and (xi) B.P.T (Physiotherapy).

(b) Agriculture: (i) B.Sc. Hons. (Agriculture) (ii) B.F.Sc. (Fisheries) (iii) B.Sc. Hons. (Forestry)

(c) Veterinary: B.V.Sc. & AH

(d) Engineering: B.Tech. [including B.Tech. (Agricultural Engg.) / B.Tech. (Dairy Sc. & Tech.) courses under the Kerala Agricultural University]

(e) Architecture: B.Arch.

The last date for the receipt of completed applications by the Commissioner for Entrance Examinations is 29-2-2008 Friday (before 5 PM).

For full details you may click here

Multiple Choice Questions on Viscosity

As promised in the last post, we will discuss some multiple choice questions on viscosity. Consider the following MCQ):

The rate of steady volume flow of water through a capillary tube of radius ‘r’ and length ‘L’ under a pressure difference P between its ends is V. This tube is connected in series with another tube of radius 2r and length 8L. Then the rate of volume flow through the series combination under the same pressure difference between the ends of the combination is

(a) V (b) 2V/3 (c) 3V (d) V/2 (e) V/4

We have 1/Q_series = 1/Q₁ +1/Q_2.

Here Q₁ = V and Q₂ = 2V since Q α r⁴/L (When the radius is doubled, the rate of flow becomes 16 times; when the length is made 8 fold, the rate of flow becomes one eighths so that the rate of flow with the second tube is twice that with the first tube).

The net rate of flow Q_series through the series combination is therefore given by

1/Q_series = 1/V + 1/2V from which Q_series = V×2V/(V+2V) = 2V/3

[If there is sufficient water column, the acceleration will finally become zero and the sphere will then move with constant (terminal) velocity].

You may be remembering that rain drops arriving near the earth’s suface move down with constant (terminal) velocity. The terminal velocty will depend on the radius of the drop. Here is a question on this:

Two rain drops of radii ‘r’ and √r arriving at the ground will have their terminal velocities in the ratio

(a) 1:1

(b) r:1

(c) 2:1

(d) 1:2

(e) √r:1

Since the rain drops move with terminal velocity v_terminal, we can equate the magnitude of viscous force to the apparent weight of the rain drop so that

6πrηv_terminal = (4/3)πr³(ρ–σ)g with usual notations.

Therefore, v_terminal α r².

The ratio of terminal velocities in the present case is (r)²:(√r)² = r:1

Here is another MCQ involving terminal velocity:

A spherical glass bead released at the top of a column of castor oil in a tall jar is found to attain a terminal velocity of 2 cm s^–1. If it is pulled up with a force equal in magnitude to three times its apparent weight in castor oil, its terminal velocity will be

(a) 4 cm s^–1

(b) 6 cm s^–1

(c) 3 cm s^–1

(d) 2 cm s^–1

(e) 1 cm s^–1

We have (from Stokes formula)

6πrηv_terminal = W_app

where W_app is the apparent weight, which tries to move the sphere downwards. When terminal velocity is attained, the opposing viscous force (6πrηv_terminal) becomes equal in magnitude to the apparent weight. Since 6πrη is constant, the terminal velocity v_terminal is directly proportional to the force trying to move the sphere. Therefore,

v_terminal α W_app

When the sphere is pulled up with a force equal in magnitude to 3 times the apparent weight, the net force on it is 2W_app and hence we have

v’_terminal α 2W_app.

From the above we obtain v’_terminal = 2 v_terminal = 2×2 = 4 cm s^–1

Now, consider the following MCQ:

Two horizontal capillary tubes of lengths L and 3L having radii r and 2r are connected in series and a liquid is flowing slowly and steadily through this combination. If P₁ is the pressure difference between the ends of the first tube and P₂ is that between the ends of the second tube, then P₁/P₂ is

(a) 9/5

(b) 1/6

(c) 12/5

(d) 8/3

(e) 16/3

Since the tubes are in series, the rate of flow (πPr⁴/8Lη) through them is the same. Therefore we have,

πP₁r⁴/ 8Lη = πP₂ (2r)⁴/ 8(3L)η, from which P₁/P₂ = 16/3.

Viscosity – Equations to be Remembered

The essential things you must remember in the section, ‘viscosity’ are given below:

1. Viscous force between two layers of a fluid = ηAdv/dx where η is the coefficient of viscosity, A is the common area of the layers and dv/dx is the velocity gradient.

2. Poiseuille’s formula for the volume (V) of a liquid flowing through a capillary tube of radius 'r’ in a time ‘t’ under a pressure difference ‘P’ between the ends of the tube is

V = πPr⁴t / 8Lη

where L is the length of the tube and η is the coefficient of viscosity of the liquid.

If the liquid flows through a horizontal capillary tube under a constant hydrostatic pressure produced by a height ‘h’ of liquid column, V = πhρgr⁴t / 8Lη where ρ is the density of the liquid.

It will be better to remember the rate of flow (which is the volume flowing per second) as

Q = V/t = πPr⁴/ 8Lη

If Q₁ and Q₂ are the rates of flow through two tubes (of radii r₁, r₂ and lengths L₁, L₂) under a given pressure head P, then the rate of flow (Q_series) under the same pressure head P when the tubes are connected in series is given by the reciprocal relation,

1/Q_series = 1/Q₁ +1/Q₂.

[You can easily prove this by combining the equations Q₁ = πPr₁⁴/ 8L₁η, Q₂ = πPr₂⁴/ 8L₂η, P = P₁+P₂ (where P₁ and P₂ are the pressures between the ends of the two tubes when they are in series) and Q = πP₁r₁⁴/ 8L₁η = πP₂ r₂⁴/ 8L₂η. Do this as an exercise]

If there are many tubes in series, the above relation gets modified as

1/Q_series = 1/Q₁ +1/Q₂ + +1/Q₃ +1/Q₄ +…etc.

3. Reynold’s number, R = vρr/ η where ‘v’ is the velocity of the liquid of density ρ and viscosity (coefficient) η through a tube of radius ‘r’.

Note that R is dimensionless and that the flow will be streamlined only if R is less than 2000 (approximately). It therefore follows that stream lined flow is more likely in the case of liquids of small density and large viscosity.

4. Stokes formula for the viscous force (F) on a sphere of radius ‘r’ moving with a velocity ‘v’ through a fluid having coefficient of viscosity η is

F = 6πrηv

If the sphere moves with terminal velocity v_terminal as is the case when it moves down under gravity through a column of viscous medium, we can equate the magnitude of viscous force to the apparent weight of the sphere so that

6πrηv_terminal = (4/3)πr³(ρ–σ)g where ρ is the density of the material of the sphere and σ is the density of the viscous medium.

Note that the terminal velocity is directly proportional to the radius of the sphere.

In the next post we will discus some typical multiple choice questions on viscosity.

Merry Christmas!