All India Pre-Medical/Pre-Dental Entrance Examination (AIPMT) 2011 Questions (MCQ) on Circular Motion and Rotation

Today we will discuss three questions on circular motion and rotation, which appeared in AIPMT 2011 question paper. Even though difficult questions can be easily set from this section, these questions are relatively simple. Most question setters ask simpler questions to medical degree aspirants compared to engineering degree aspirants! Here are the questions and their solution:

(1) The instantaneous angular position of a point on a rotating wheel is given by the equation θ(t) = 2t³ – 6t². The torque on the wheel becomes zero at

(1) 2 s

(2) 1 s

(3) 0.2 s

(4) 0.25 s

The torque on the wheel will be zero when the angular acceleration is zero.

Now, angular acceleration α is given by

α = d²θ/dt²

We have dθ/dt = 6t² – 12t and

d²θ/dt² = 12t – 12

Therefore, torque is zero when 12t – 12 = 0.

This gives t = 1 s

(2) A particle moves in a circle of radius 5 cm with constant speed and time period 0.2π s. The acceleration of the particle is

(1) 5 m/s²

(2) 15 m/s²

(3) 25 m/s²

(4) 36 m/s²

The acceleration ‘a’ of the particle is given by

a = ω²R where ω is the angular velocity and R is the radius of the circular path.

Since ω = 2π/T where T is the time period, we have

a = (2π/T)² R = (2π/0.2π)² ×(5×10^–2) m = 5 m/s²

(3) The moment of inertia of a thin uniform rod of mass M and length L about an axis passing through its midpoint and perpendicular to its length is I₀. Its moment of inertia about an axis passing through one of its ends and perpendicular to its length is

(1) I₀ + ML²

(2) I₀ + ML²/2

(3) I₀ + ML²/4

(4) I₀ + 2ML²

By the theorem of parallel axes, the moment of inertia of a body about any axis is equal to the sum of the moment of inertia about a parallel axis through the centre of mass and the product Ma² where M is the mass of the body and a is the separation between the two axes.

Therefore, the moment of inertia of the rod about an axis passing through one of its ends and perpendicular to its length is I₀ + M(L/2)² = I₀ + ML²/4

You will find similar questions (with solution) in this section here.

Geometric Optics - IIT-JEE 2011 Questions on Refraction

“You can chain me, you can torture me, you can even destroy this body, but you will never imprison my mind.”

– Mahatma Gandhi

Today we will discuss two questions from optics which appeared in IIT-JEE 2011 question paper. The first one is single correct answer type multiple choice question where as the second one is integer answer type in which the answer is a single digit integer ranging from 0 to 9.

(1) A light ray traveling in glass medium is incident on glass air interface at an angle of incidence θ. The reflected (R) and transmitted (T) intensities, both as function of θ, are plotted. The correct sketch is

You can easily rule out sketches (A) and (B) since they indicate 100% transmitted intensity at angle of incidence (θ) of 0º.

[Even at normal incidence (θ = 0) the entire incident light is not transmitted. A small portion is reflected back].

Sketch (C) is the correct option since it indicates total reflection (total internal reflection) at an angle of incidence equal to the critical angle for the glass air interface. The entire light is totally reflected abruptly and the transmitted intensity drops abruptly to zero.

[The gradual change in intensity indicated in sketch (D) is ruled out].

(2) Water (with refractive index = 4/3 ) in a tank is 18 cm deep. Oil of refractive index 7/4 lies on water making a convex surface of radius of curvature, 'R’ = 6 cm as shown. Consider oil to act as a thin lens. An object ‘S’ is placed 24 cm above water surface. The location of its image is at ‘x’ cm above the bottom of the tank. Then 'x' is

The law of distances in the case of refraction at an interface between two media of refractive indices n₁ and n₂ is

n₂/v – n₁/u = (n₂ – n₁)/R where ‘v’ is the image distance, ‘u’ is the object distance and ‘R’ is the radius of curvature of the refracting surface. In the case of the refraction at the air-oil interface we have n₂ = 7/4, n₁ = 1, u = – 24 cm and R = 6 cm. so that

7/4v – 1/(– 24) = [(7/4) –1]/6

[Note that we have applied the Cartesian sign convention which you can find here].

Therefore, 7/4v = 3/24 – 1/24 = 1/12

This gives v = 21 cm.

The image formed (at 21 cm from the oil surface) by the refraction at the air-oil interface will act as a virtual object for the refraction at the oil-water interface so that u = 21 cm (positive according to Cartesian sign convention). In this case n₂ = 4/3, n₁ = 7/4 and R = ∞.

[The radius of curvature is infinity since the oil-water interface is plane].

Therefore we have

4/3v₁ – 7/(4×21) = [(4/3) – (7/4)]/∞ where v₁ is the distance of the final image (from the water surface).

Or, 4/3v₁ – 7/(4×21) = 0

This gives v₁ = 16 cm.

The water column in the tank has a height of 18 cm. Therefore the final image is 2 cm above the bottom of the tank.

Therefore, x = 2

Kerala Engineering Entrance (KEAM -Engineering) 2011 Questions (MCQ) on Transistor Amplifier and Oscillator

Today we will discuss two multiple choice questions involving transistors which appeared in Kerala Engineering Entrance (KEAM -Engineering) 2011 question paper. These questions are simple as is the case with questions in this section at the level expected in the case of 12^th grade (Plus Two) students.

(1) In a common emitter transistor amplifier, the output resistance is 500 KΩ and the current gain β = 49. If the power gain of the amplifier is 5×10⁶, the input resistance is

(A) 325 Ω

(B) 165 Ω

(C) 198 Ω

(D) 225 Ω

(E) 240 Ω

Power gain A_p is given by

A_p = β_ac A_v where β_ac is the current gain and A_v is the voltage gain (for small a.c. signals).

Since A_v = β_acR_o/R_i where R_o is the output resistance and R_i is the input resistance, we have

A_p = β_ac²R_o/R_i

Therefore R_i = β_ac²R_o/A_p = 49²×(500×10³) /(5×10⁶) = 240 Ω, very nearly.

(2) A transistor oscillator is (i) an amplifier with positive feedback (ii) an amplifier with reduced gain (iii) the one in which dc supply energy is converted into ac output energy. Then

(A) all (i), (ii) and (iii) are correct

(B) only (i) and (ii) are correct

(C) only (i) and (iii) are correct

(D) only (ii) and (iii) are correct

(E) only (ii) is correct

Statements (i) and (iii) are correct where as statement (ii) is incorect. So the correct option is (c).

Multiple Choice Questions on Magnetic Force on Moving Charges

Questions involving magnetic force on moving charges are included in most of the medical, engineering and other degree entrance examinations. Here are some simple questions which may easily tempt you to commit mistakes:

(1) The magnetic Lorentz force equation is F = q v×B. In this equation

(a) F, v and B must be mutually perpendicular.

(b) F must be perpendicular to v but not necessarily to B.

(c) F must be perpendicular to B but not necessarily to v.

(d) v must be perpendicular to both F and B

(e) F must be perpendicular to both B and v.

The equation F = q v×B gives the magnetic force F on a charge ‘q’ when it moves with velocity v in a magnetic field B. The angle between the velocity v and the field B can be any value, but the magnetic force F is at right angles to both v and B. So the correct option is (e).

[The vector product v×B which gives the force F indeed demands that F is at right angles to both v and B].

(2) A charged particle moving in the north east direction at right angles to a magnetic field experiences a force vertically upwards. This charged particle will not experience any force if it moves towards

(a) south west direction

(b) north

(c) east

(d) west

(e) north west

Since the magnetic force is vertical, the magnetic field must be horizontal.

Since the charged particle is moving in the north east direction at right angles to the magnetic field, it follows that the magnetic field must be directed either north west or south east (Fig.). So it will not experience any force if it moves towards north west or south east. The correct option is (e).

[You can use Fleming’s left hand rule to obtain the directions easily. See yourself that if the magnetic field in the above question is along the north west direction, the charge on the particle must be positive to obtain a vertically upward magnetic force. If the magnetic field in the above question is along the south east direction, the charge on the particle must be negative].

(3) A proton traveling vertically downwards experiences a southward force due to a magnetic field directed at right angles to its path.. An electron traveling northward in the same magnetic field will experience a magnetic force directed

(a) downwards

(b) upwards

(c) towards east

(d) towards west

(e) towards south east

Since the proton (which is positively charged) experiences a southward force while traveling vertically downwards, the perpendicular magnetic field must be acting towards the east.

[As required by Fleming’s left hand rule, hold the fore-finger, middle finger and thumb of your left hand in mutually perpendicular directions, with the middle finger pointing downwards (in the present case) and the thumb pointing southwards. The fore-finger then points towards the east].

If the proton were to move northward in this magnetic field, it would experience a downward magnetic force. Since the electron is negatively charged, it will experience an upward magnetic force [Option (b)]..

Karnataka CET Multiple Choice Questions on Combination of Capacitors

The following question appeared in Karnataka Common Entrance Test (CET) 2010 question paper. Even though at the first reading it may appear somewhat difficult for you on seeing the circuit, a careful look at the circuit will assure you that it’s simple.

All capacitors used in the diagram are identical and each is of capacitance C. Then the effective capacitance between the points A and B is

(a) 1.5 C

(b) 6 C

(c) C

(d) 3 C

The first three capacitors are in parallel, giving an effective capacitance 3 C. The last three capacitors also are in parallel, giving an effective capacitance 3 C.. These two parallel combinations are connected in series. Therefore, the effective capacitance between the points A and B is 3 C/2 = 1.5 C.

The following question appeared in the Karnataka CET 2008 question paper:

How many 6 μF, 200 V condensers are needed to make a condenser of 18 μF, 600 V?

(1) 9

(2) 18

(3) 3

(4) 27

Since the voltage rating of each 6 μF capacitor is 200 V, you should connect 3 capacitors in series to obtain the required voltage rating of 600 V. But then the effective capacitance of the series combination of these three capacitors will be 2 μF (6/3 = 2). To obtain the required capacitance of 18 μF, you need to connect nine such series combinations in parallel. So the total number of capacitors required is 9×3 = 27.

The following question also appeared in the Karnataka CET 2008 question paper:

The total energy stored in the condenser system shown in the figure will be

(1) 2 μJ

(2) 4 μJ

(3) 8 μJ

(4) 16 μJ

The series combination of 6 μF and 3 μF gives an effective capacitance of (6×3)/(6+3) = 2 μF. Since this is in parallel with a 2 μF capacitor, the effective capacitance C across the 2 volt battery is 4 μF. The energy stored in the system of capacitors is ½ CV² = (½)×(4×10^–6) ×2² = 8×10^–6 J = 8 μJ.

You will find a few more multiple choice questions (with solution) in this section here.

Kerala Engineering Entrance (KEAM) 2011 Questions on Kinematics in One Dimension

“Common sense is the collection of prejudices acquired by age eighteen.”

– Albert Einstein

Today we will discuss three questions (MCQ) on one dimensional kinematics. These questions were included in the KEAM (Engineering) question paper. They are of the type often seen in similar entrance tests. Here are the questions with their solution:

(1) A bus begins to move with an acceleration of 1 ms^–2. A man who is 48 m behind the bus starts running at 10 ms^–1 to catch the bus. The man will be able to catch the bus after

(a) 6 s

(b) 5 s

(c) 3 s

(d) 7 s

(e) 8 s

If the man can catch the bus after t seconds we have

10 t = 48 + ½ at² where a is the acceleration of the bus (a = 1 ms^–2).

[10 t is the distance covered by the man in time t and ½ at² is the distance covered by the bus in the same time. Since the man is 48 m behind the bus, he has to cover an additional distance of 48 m].

Since a = 1 ms^–2 the above equation reduces to

t² – 20 t + 96 = 0

Therefore t = [–(–20) ±√{20² – (4×1×96}] /(2×1)

Or, t = [20 ±√16] /2 = 12 s or 8 s

The smaller time 8 s is the answer.

(2) A car moves a distance of 200m. It covers first half of the distance at speed 60 km h^–1 and the second half at speed v. If the average speed is 40 km h^–1, the value of v is

(a) 30 km h^–1

(b) 13 km h^–1

(c) 60 km h^–1

(d) 40 km h^–1

(e) 20 km h^–1

The total distance moved is 200 m = 0.2 km

The total time (in hours) taken by the car is (0.1/60) + (0.1/v)

Since the average velocity is 40 km h^–1, we have

40×[(0.1/60) + (0.1/v)] = 0.2

Or, 4/v = 0.2 – (4/60) = 8/60 from which v = 240/8 = 30 km h^–1

[If v₁ and v₂ are the velocities while traversing the two halves of the path length s, the average velocity v_av is given by

v_av = s/[(s/2v₁) + (s/2v₂)]

Or v_av = 2v₁v₂/( v₁+v₂)

Since this final equation contains velocities only you can make substitutions without confusion (no need of conversion of metre into kilometre when distances are to be handled).

Therefore, 40 = (2×60×v)/(60+v) from which v = 30 km h^–1]

(3) A particle is moving with constant acceleration from A to B in a straight line AB. If u and v are the velocities at A and B respectively then its velocity at the mid point C will be

(a) [(u² + v²)/2u]²

(b) (u + v)/2

(c) (v – u)/2

(d) √[(u² + v²)/2]

(e) √(v² – u²)/2

If ‘s’ is the length of the path AB, the velocity of the particle changes from u to v when it moves through the distance ‘s’. Therefore we have,

v² – u² = 2as from which the acceleration, a = (v² – u²)/2s

If’ ’v_c ’ is the velocity of the particle at the mid point C of the path AB, we have v_c² = u² + 2a(s/2). Substituting for the acceleration ‘a’ from the above equation, v_c = √[(u² + v²)/2].