## Tuesday, December 27, 2011

### Apply for All India Pre-Medical / Pre-Dental Entrance Examination-2012 (AIPMT 2012)

Central Board of Secondary Education (CBSE), Delhi has invited applications in the prescribed form for All India Pre-Medical / Pre-Dental Entrance Examination-2012 for admission to 15% of the merit positions for the Medical/Dental Courses of India.

The exams will be conducted as per the following schedule:

1. Preliminary Examination …. 1st April, 2012 (Sunday) 10 AM to 1 PM

2. Final Examination ………… 13th May, 2012 (Sunday) 10 AM to 1 PM

Both examinations will be objective type.

Candidate can apply for the All India Pre-Medical/Pre-Dental Entrance Examination 2012 (AIPMT 2012) only online.

All details can be obtained from the official site www.aipmt.nic.in

You can find many AIPMT previous questions with solution on this site. Click on the label ‘AIPMT’ below this post or try a search for ‘AIPMT’ using the search box provided on this page, to access all relevant posts.

## Wednesday, November 30, 2011

### Two Questions (MCQ) on Rotation of Rigid Bodies

“Learn from yesterday, live for today, hope for tomorrow. The important thing is to not stop questioning.”

– Albert Einstein

Today we will discuss a couple of questions involving rigid body rotation. The first question may appear to be familiar to you as it has appeared in various entrance exam question papers. The second one is not so common but you must certainly work it out yourself before going through the solution given here. (1) A thin straight uniform rod AB (Fig.) of length L and mass M, held vertically with the end A on horizontal floor, is released from rest and is allowed to fall. Assuming that the end A (of the rod) on the floor does not slip, what will be the linear velocity of the end B when it strikes the floor?

(a) (3L/g)

(b) (2L/g)

(c) (3gL)

(d) (2g/L)

(e) (3g/L)

When the rod falls, its gravitational potential energy gets converted into rotational kinetic energy. Therefore we have

MgL/2 = ½ I ω2 …………….. (i)

where I is the moment of inertia of the rod about a normal axis passing through its end and ω is the angular velocity of the rod when it strikes the floor.

[Note that initially the centre of gravity of the rod is at a height L/2 and that’s why the initial potential energy is MgL/2]

The moment of inertia of the rod about the normal axis through its end is given by

I = ML2/3

[The moment of inertia of the rod about an axis throgh its centre and perpenicular to its length is ML2/12. On applying parallel axes theorem, the moment of inertia about a parallel axis through the end is ML2/12 + M (L/2)2 = ML2/3]

Substituting for I in Eq.(i), we have

MgL/2 = ½ (ML2/3) ω2

Therefore ω = √(3g/L)

The linear velocity v of the end B of the rod is given by

v = ωL = √(3gL) (2) A thin straight uniform rod AB (Fig.) of length L and mass M is pivoted at point O, distant L/4 from the end A. The friction at the hinge is negligible and the rod can rotate freely (about the hinge) in a vertical plane. Initially the rod is held horizontally and is released from rest. The linear velocity of the end B of the rod when it momentarily attains vertical position A1B1 is

(a) (3L/4g)

(b) (27gL/14)

(c) (18gL/7)

(d) (3g/4L)

(e) (2gL)

The gravitational potential energy of the rod gets converted into rotational kinetic energy when the rod is release from its horizontal position. The centre of gravity of the rod is lowered through a distance L/4 and hence the decrease in the gravitational potential energy is MgL/4. The gain in rotational kinetic energy is ½ I ω2 where I is the moment of inertia of the rod about the axis of rotation passing through the point O and ω is the angular velocity of the rod when it attains the vertical position. Therefore, from the law of conservation of energy we have

MgL/4 = ½ I ω2 …………….. (i)

Here I = ML2/12 + M (L/4)2 = M [(L2/12) + (L2/16)]

Substituting in Eq.(i) we have

g = [(L/6) + (L/8)] ω2 = (7L/24)ω2

Or, ω = √(24g/7L)

Since the end B of the rod is at distance 3L/4 from the axis of rotation, the linear velocity v of the end B of the rod is given by

v = ω×(3L/4) = [√(24g/7L)] (3L/4)

Or, v = √[(24g×9L2) /(7L×16)] = (27gL/14)

## Saturday, November 19, 2011

### Apply for All India Engineering Entrance Examination 2012 (AIEEE 2012)

It’s time to apply for AIEEE 2012. Applications for the examination are to be submitted online only, from 15-11-2011 to 31-12-2011.

There are two modes of examination viz., offline examination (Pen & Paper Mode) and online examination (Computer Based Test) with dates as given below:

Offline Examination - 29th April, 2012

Online Examination- 7th May, 2012 to 26th May, 2012

You may visit the site http://aieee.nic.in for details and information updates.

You will find many old AIEEE questions (with solution) on this blog. You can access all of them by trying a search for ‘AIEEE’ making use of the search box on this page or by clicking on the label ‘AIEEE’ below this post. Make use of the ‘older posts’ button to navigate through older posts related to AIEEE.

## Thursday, October 27, 2011

### IIT-JEE 2011 and IIT-JEE 2010 Questions on Doppler Effect

"I am a friend of Plato, I am a friend of Aristotle, but truth is my greater friend."
– Sir Isaac Newton

Questions on Doppler Effect were discussed on this site earlier. You can access them by clicking on the label ‘Doppler effect’ below this post or by trying a search for ‘Doppler effect’ using the search box provided on this page. Questions on Doppler effect appear frequently in Entrance examination question papers. Today we will discuss two questions in this section Here is the first question which appeared in IIT-JEE 2011 question paper as a single answer type multiple choice question:

(1) A police car with a siren of frequency 8 kHz is moving with uniform velocity 36 km/hr towards a tall building which reflects the sound waves. The speed of sound in air is 320 m/s. The frequency of the siren heard by the car driver is

(A) 8.50 kHz

(B) 8.25 kHz

(C) 7.75 kHz

(D) 7.50 kH

The general expression for the apparent frequency (n’) produced due to Doppler effect is

n’ = n(v+w–vL)/(v+w–vS) where ‘v’ is the velocity of sound, ‘w’ is the wind velocity ‘vL’ is the velocity of the listener, ‘vS’ is the velocity of the source of sound and ‘n’ is the actual frequency of the sound emitted by the source. Note that all the velocities in this expression are in the same direction and the source is behind the listener. In other words, the listener is moving away from the source and the source is moving towards the listener.

[It will be helpful to remember that if the source moves towards the listener or the listener moves towards the source, the apparent frequency increases. If they move away, the apparent frequency decreases].

In the present case the source of sound which produces the Doppler effect is the reflected image of the siren. (The car driver will not detect any change in the frequency of the direct sound from the siren since he is moving along with the siren). The reflected image (source) of the siren moves towards the car driver with a speed of 36 km/hr and the car driver (listener) moves towards the reflected image (source) with the same speed.

Thus in the expression for apparent frequency we have ro substitute

n = 8 kHz, vL = 36 km/hr = 36×(5/18) ms–1 = –10 ms–1, w = 0, vS = + 10 ms–1.

Therefore, n’ = 8×(320 + 10)/(320 – 10) = 8×(33/31) kHz = 8.5 kHz.

The following question appeared in IIT-JEE 2010 question paper as an integer type question (in which the answer is a single-digit integer, ranging from 0 to 9):

(2) A stationary source is emitting sound at a fixed frequency f0, which is reflected by two cars approaching the source. The difference between the frequencies of sound reflected from the cars is 1.2% of f0. What is the difference in the speeds of the cars (in km per hour) to the nearest integer? The cars are moving at constant speeds much smaller than the speed of sound which is 330 ms–1.

This is similar to question no.1 above. The difference between the frquencies of the sound reflected from the cars is given by

n1 n2’ = [f0(v+v1)/(v–v1)] – [f0(v+v2)/(v–v2)] where v1 and v2 are the speeds of the two cars.

Therefore (n1 n2’)/f0 = [(v+v1)/(v–v1)] – [(v+v2)/(v–v2)]

The quantity on the left hand side of the above equation is 1.2/100 as given in the question.

Therefore, 0.012 = [(v+v1) (v–v2) (v+v2) (v–v1)] / [(v–v1)(v–v2)]

Or, 0.012 = (2 vv1 2 vv2)/ [(v–v1)(v–v2)] = 2v(v1 –v2)/v2 since (v–v1) (v–v2) ≈ v

[The cars are moving at speeds much smaller than the speed of sound].

Therefore we have

0.012 = 2(v1 –v2)/v from which (v1 –v2) = 0.006×330 ms–1

The difference in the speeds of the cars in km per hour is 0.006×330×(18/5) = 7.128

The answer, which is the nearest integer, is 7.

## Sunday, October 16, 2011

### Joint Entrance Examination 2012 (IIT-JEE 2012) for Admission to IITs

Notification in respect of the Joint Entrance Examination 2012 (IIT-JEE 2012) for Admission to IITs has been issued. This Joint Entrance Examination is for admission to the under graduate programmes in the IITs at Bhubaneswar, Bombay, Delhi, Gandhinagar, Guwahati, Hyderabad, Indore, Kanpur, Kharagpur, Madras, Mandi, Patna, Rajasthan, Roorkee and Ropar as well as at IT-BHU Varanasi and ISM Dhanbad.

Important Dates for IIT JEE 2012:

Online application process: October 31 to December 10, 2011
Sale of off-line application forms: November 11 to December 5, 2011.

Last date for receiving application forms at IITs: Thursday, December 15, 2011

Date of Exam of IIT JEE 2012:

Sunday, April 8, 2012 Paper 1: 09:00 to 12:00 hrs;

Paper 2 : 14:00 to 17:00 hrs

All information and information updates regarding online application and offline application and other important details about IIT JEE 2012 can be obtained from the following websites:

IIT Bombay: http://www.jee.iitb.ac.in
IIT Guwahati: http://www.iitg.ernet.in/jee/
IIT Delhi: http://jee.iitd.ac.in
IIT Kharagpur: http://www.iitkgp.ernet.in/jee
IIT Kanpur: http://www.jee.iitk.ac.in
IIT Roorkee: http://jee.iitr.ernet.in/

You will find many earlier IIT-JEE questions with solution on this blog. Click on the label ‘IIT’ below this post or try a search for ‘IIT’ using the search box provided on this page.

## Sunday, October 02, 2011

### Multiple Choice Questions on Unusual Combinations of Capacitors

"It is unwise to be too sure of one's own wisdom. It is healthy to be reminded that the strongest might weaken and the wisest might err."

– Mahatma Gandhi

Today we will consider a few questions on unconventional combinations of capacitors which occasionally find place in entrance examination question papers conducted for admitting students to medical, engineering and other degree courses.

(1) Four parallel metal plates are arranged in air as shown, with the same separation ‘d’ between neighbouring plates. If the area (of one surface) of each plates is A, the capacitance between the terminals T1 and T2 is

(a) 4ε0A/d

(b) ε0A/d

(c) 2ε0A/d

(d) 3ε0A/d

(e) ε0A/3d

Let us number the plates starting from the top as 1, 2, 3, and 4 respectively. The lower surface of plate No.1 and the upper surface of plate No.2 with air in between makes a capacitor of capacitance ε0A/d.

The lower surface of plate No.2 and the upper surface of plate No.3 with air in between makes another capacitor of capacitance ε0A/d.

The lower surface of plate No.3 and the upper surface of plate No.4 with air in between makes a third capacitor of capacitance ε0A/d.

This arrangement therefore makes three capacitors in parallel. The capacitance value of each capacitor is ε0A/d so that the parallel combined value which appears between the terminals T1 and T2 is 3ε0A/d [Option (d)].

(2) If the connection of metal plates in the above question is modified as shown in the adjoining figure, the capacitance between terminals T1 and T2 is

(a) ε0A/2d

(b) 3ε0A/2d

(c)0A/d

(d) 0A/d

(e) 0A/3d

In this case the capacitor at the centre (having capacitance ε0A/d) appears directly across the terminals T1 and T2. The upper and lower capacitors (each of value ε0A/d) are connected in series and this series combination having effective capacitance ε0A/2d also appears between the terminals T1 and T2. Therefore, the effective capacitance between T1 and T2 is ε0A/d + ε0A/2d = 3ε0A/2d [Option (b)].

(3) If a fifth plate is added to the combination shown in question No.1, as shown in the figure, what is the effective capacitance between T1 and T2?

(a) ε0A/4d

(b) ε0A/5d

(c) 0A/d

(d) 0A/5d

(e) 0A/d

Once you answer question No.1, this question becomes quite simple. You can easily see that there are four identical capacitors (each having capacitance ε0A/d) connected in parallel. Therefore, the effective capacitance between T1 and T2 is 4ε0A/d [Option (e)].

You can find a few more questions (with solution) in this section here.

## Thursday, September 22, 2011

### All India Pre-Medical/Pre-Dental Entrance Examination (AIPMT) 2009 Questions on Rotational Motion and Centre of Mass

A few questions on circular motion and rotation, which appeared in AIPMT 2011 question paper were discussed in the last post. A few more questions on rotational motion and centre of mass with their solution are given below. These questions were included in the All India Pre-Medical/Pre-Dental Entrance Examination (Preliminary) 2009 question paper:

(1) A thin circular ring of mass M and radius R is rotating in a horizontal plane about an axis perpendicular to its plane with a constant angular velocity ω. If two objects each of mass m be attached gently to the opposite ends of a diameter of the ring, the ring will then rotate with an angular velocity

(1) ωM /(M + 2m)

(2) ω(M + 2m) /M

(3) ωM /(M + m)

(4) ω(M – 2m) /(M + 2m)

This question is a popular one and has appeared in various entrance question papers. Its answer is based on the law of conservation of angular momentum:

I1ω1 = I2ω2 where I1 and I2 are the initial and final moments of inertia and ω1 and ω2 are the initial and final angular velocities respectively.

Therefore we have

MR2ω = (MR2 + 2 mR2) ω2

Therefore, ω2 = ωM /(M + 2m)

(2) Four identical thin rods each of mass M and length L, form a square frame. Moment of inertia of this frame about an axis through the centre of the square and perpendicular to its plane is

(1) (2/3) ML2

(2) (13/3) ML2

(3) (1/8) ML2

(4) (4/3) ML2

We have to use the theorem of parallel axes: I = ICM + Ma2 where ICM is the the moment of inertia about an axis through the centre of mass of the rod and I is the moment of inertia about a parallel axis at distance a.

The moment of inertia of a thin rod of mass M and length L about an axis through the mid point of the rod and perpendicular to its length is ML2/12. The moment of inertia of the rod about a parallel axis at a distance L/2 (which is the distance of the centre of the frame from each ro.is ML2/12 + M (L/2)2 = ML2/3

You must remember that moment of inertia is a scalar quantity. Since there are four rods in the square frame, the moment of inertia of the entire frame is four times the moment of inertia of one rod. Therefore the answer is (4/3) ML2

(3) If F is the force acting on a particle having position vector r and τ be the torque of this force about the origin, then

(1) r.τ > 0 and F.τ < 0

(2) r.τ = 0 and F.τ = 0

(3) r.τ = 0 and F.τ ≠ 0

(4) r.τ ≠ 0 and F.τ = 0

This question may appear to be a difficult one at the first glance. But this is a very simple question if you remember that the torque τ is the ‘cross product’ (vector product) of r and F:

τ = r×F

The ‘cross product’ of two vectors is a vector perpendicular to both individual vectors. The torque vector is thus perpendicular to both r and F. Therefore the ‘dot product’ (scalar product) of r and τ as well as that of F and τ is zero [Option (2)].

(4) Two bodies of mass 1 kg and 3 kg have position vectors i + 2 j + k and –3 i – 2 j + k espectively. The centre of mass of this system has a position vector

(1) –2 ij + k

(2) 2 ij – 2 k

(3) – i + j + k

(4) –2 i + 2 k

If two point masses m1 and m2 have position vectors r1 and r2 their centre of mass has position vector R given by

R = (m1 r1 + m2 r2) /( m1 + m2)

Here we have m1 = 1, m2 = 3, r1 = i + 2 j + k and r2 = –3 i – 2 j + k

Substituting, R = [1(i + 2 j + k) + 3(–3 i – 2 j + k)] /(1+3)

Or, R = (– 8 i – 4 j + 4 k) /4 = –2 ij + k

You can access all questions in this section by clicking on the label ‘rotation’ below this post. Make use of the buttons ‘older posts’ and ‘newer posts’ at the end of the page to navigate through all the posts.