IIT-JEE 2008 Reasoning Type Question on Rotation

The following question appeared in IIT-JEE 2008 question paper under Reasoning Type Questions.

STATEMENT -1

Two cylinders, one hollow (metal) and the other solid (wood) with the same mass and identical dimensions are simultaneously allowed to roll without slipping down an inclined plane from the same height. The hollow cylinder will reach the bottom of the inclined plane first.

and

STATEMENT -2

By the principle of conservation of energy, the total kinetic energies of both the cylinders are identical when they reach the bottom of the incline.

(A) STATEMENT-1 is True, STATEMENT-2 is True; STATEMENT -2 is a correct explanation for Statement-1

(B) STATEMENT -1 is True, STATEMENT -2 is True; STATEMENT -2 is NOT a correct explanation for STATEMENT -1

(C) STATEMENT -1 is True, STATEMENT -2 is False

(D) STATEMENT -1 is False, STATEMENT -2 is True

The acceleration (a) of a body of mass M rolling down an inclined plane is given by

a = g sinθ/[1 + (I/MR²)]

where θ is the angle of the plane (with respect to the horizontal), I is the moment of inertia (about the axis of rolling) and R is the radius of the body.

Since the moments of inertia of solid cylinder and hollow cylinder are respectively MR²/2 and MR² the acceleration a is greater for the solid cylinder. Therefore, the solid cylinder will reach the bottom of the inclined plane first.

Since the cylinders have the same mass and are at the same height they have the same initial gravitational potential energy Mgh. This potential energy gets converted into translational and rotational kinetic energies (obeying the law of conservation of energy) when the cylinders roll down the incline and the total kinetic energies of both the cylinders are identical when they reach the bottom of the incline.

The correct option is (D).

Let us consider an ordinary type of multiple choice question now:

A, B and C are three equal point masses (m each) rigidly connected by massless rods of length L forming an equilateral triangle as shown in the adjoining figure. The system is first rotated with constant angular velocity ω about an axis perpendicular to the plane of the triangle and passing through A. Next it is rotated with the same constant angular velocity ω about the side AB of the triangle. If K₁ and K₂ are the kinetic energies of the system in the first and the second cases respectively, the ratio K₁/K₂ is

(a) 2/3

(b) 4/3

(c) 8/3

(d) 10/3

(e) 16/3

The rotational kinetic energy (K) is given by

K = ½ Iω²

Since the angular velocity is the same in the two cases, the ratio of kinetic energies must be equal to the ratio of moments of inertia.

Thereore, K₁/K₂ = I₁/I₂ = 2mL²/m(Lsin60º)²

[Note that in the second case the moment of inertia of the system is due to the single mass at C which is at distance Lsin60º from the axis AB].

Thus K₁/K₂ = 2/(√3/2)² = 8/3.

AIPMT 2008 Question on Feed back Amplifier

The following question on negative feed back amplifier is simple even though Plus Two students will normally be unprepared to answer it:

The voltage gain of an amplifier with 9% negative feed back is 10. The voltage gain without feed back will be

(1) 10

(2) 1.25

(3) 100

(4) 90

If the voltage gain without feed back is A_v and the feed back factor (fraction of output voltage fed back to the input) is β, the voltage gain (A_fb) with feed back is given by

A_fb = A_v/(1– βA_v)

In the case of negative feed back the sign of the feed back factor β is negative so that the voltage gain with feed back is given by

A_fb = A_v/(1+ βA_v)

Since A_fb = 10 and the magnitude of β is 9% = 0.09, we have on substituting,

10 = A_v/(1+ 0.09A_v)

This gives A_v = 100.

Here is another question on feed back amplifiers:

Pick out the wrong statement:

When negative feed back is applied in a transistor amplifier

(1) its voltage gain is decreased

(2) its band width is decreased

(3) distortion produced by the amplifier is decreased

(4) the transistor current gain is unchanged

The second option alone is incorrect. The band width of a negative feed back amplifier will be greater than that of the same amplifier without the feed back. Since there is a reduction in the voltage gain consequent on the negative feed back, the 3 dB down frequency on the lower side will be shifted towards lower frequency and the 3 dB down frequency on the upper side will be shifted towards higher frequency.

All India Engineering/Architecture Entrance Examination 2009 (AIEEE 2009)

Application Form and the Information Bulletin in respect of the All India Engineering/Architecture Entrance Examination 2009 (AIEEE 2009) to be conducted on 26-4-2009 are being distributed from 5.12.2008 and will continue till 5.1.2009. Candidates can apply for AIEEE 2009 either on the prescribed Application Form or make application ‘Online’. Visit the site http://aieee.nic.in immediately for details. Apply for the exam without delay.

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