If we did all things we are capable of, we would literally astound ourselves.

– Thomas A. Edison

Wednesday, January 30, 2008

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).

Monday, January 28, 2008

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

Wednesday, January 09, 2008

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/Qseries = 1/Q1 +1/Q2.

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 Qseries through the series combination is therefore given by

1/Qseries = 1/V + 1/2V from which Qseries = 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 vterminal, we can equate the magnitude of viscous force to the apparent weight of the rain drop so that

6πrηvterminal = (4/3)πr3(ρσ)g with usual notations.

Therefore, vterminal 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)

6πrηvterminal = Wapp

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

vterminal α Wapp

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

v’terminal α 2Wapp.

From the above we obtain v’terminal = 2 vterminal = 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.