Physicsplus

Kerala Engineering Entrance 2007 Questions on Communication Systems

2012-01-28T15:19:00.002+05:30

“I have no special talents. I am only passionately curious.”

– Albert Einstein

Questions on communication systems at the level expected from you are generally simple and interesting. Today we will discuss three questions (MCQ) on communication systems which appeared in Kerala Engineering Entrance 2007 question paper.

(1) The time variations of signals are given as in (A), (B) and (C).

Point out true statement from the following

(a) A, B and C are analog signals

(b) A and B are analog, but C is digital signal

(d) A and C are analog, but B is digital signal

(e) A, B and C are digital signals (e)

Digital signals will have two discrete levels only, corresponding to the zero level and one level as shown in graph (B). Analog signals have various instantaneous values, as shown in graphs (A) and (C). The correct option therefore is (d).

(2) A photo detector used to detect the wave length of 1700 nm, has energy gap of about

(a) 0.073 eV

(b) 1.2 eV

(d) 1.16 eV

(e) 0.73 eV

The energy gap of a photo detector should be equal to or less than the energy of the photon which it is intended to detect. The product of the energy of a photon in electron volt and the wave length in Angstrom is 12400. Therefore, the energy of the 1700 nm photon is 12400/ 17000 electron volt = 0.73 eV. So, the correct option is (e). Option (a) is not suitable since a semiconductor with a gap as low as 0.073 eV (if at all available) will be unreliable due to the breaking of bonds by thermal excitation.

[You can use the expression hc/λ for calculating the energy of the photon in joule and then convert it into electron volts by dividing it by the electronic charge of 1.6×10^–19joule. But it will be more difficult and time consuming].

(3) The optical fibres have an inner core of refractive index n₁ and a cladding of refractive index n₂ such that

(a) n₁ = n₂ (b) n₁ ≤ n₂ (c) n₁ < n₂ (d) n₁ > n₂ (e) n₁ ≥ n₂

The cladding has to be rarer than the core so as to totally reflect the beam of light entering the optical fibre. So, the correct option is (d).

Apply for Kerala Engineering Agricultural Medical Entrance Examinations 2012 (KEAM 2012)

2012-01-20T13:31:00.002+05:30

"Strength does not come from physical capacity. It comes from an indomitable will."

– Mahatma Gandhi

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 2012-13.

(a) Medical (i) MBBS (ii) BDS (iii) BHMS (iv) BAMS (v) BSMS

(b) Agriculture (i) BSc. Hons. (Agriculture) (ii) BSc. Hons. (Forestry)

(d) Fisheries BFSc.

(e) Engineering B.Tech. [including B.Tech. (Agri. Engg.), B.Tech. (Food Engg.) Courses

under the Kerala Agricultural University and B.Tech. (Dairy Science & Tech.) under the Kerala Veterinary & Animal Sciences University]

(f) Architecture B.Arch.

Dates of Exam:

Engineering Entrance Examination (For Engineering courses except Architecture)

23.04.2012 Monday 10.00 A.M. to 12.30 P.M. Paper-I : Physics & Chemistry.

24.04.2012 Tuesday10.00 A.M. to 12.30 P.M. Paper-II: Mathematics.

Medical Entrance Examination (For Medical, Agriculture, Veterinary and Fisheries courses)

25.04.2012 Wenesday10.00 A.M. to 12.30 P.M. Paper-I : Chemistry & Physics.

26.04.2012 Thursday10.00 A.M. to 12.30 P.M. Paper-II: Biology.

All candidates are required to apply online through the website www.cee.kerala.gov.in for the Entrance Examination for admission to Medical, Agriculture, Veterinary, Fsheries, Engineering and Architecture Courses.

All details regarding KEAM 2012 can be obtained from the website http://www.cee-kerala.org

Hurry up! The Print out of your Application and other relevant documents to be submitted with the Application, are to be sent to the Commissioner for Entrance Examinations so as to reach him before 5 p.m. on 15.02.2012 (Wednesday).

* * * * * * * * * * * * * * *

To access earlier KEAM questions discussed on this site, type in ‘Kerala’ in the search box at the top left of this page and strike the enter key or click on the label ‘Kerala’ below this post.

Optics: Karnataka CET Questions (MCQ) on Lenses

2012-01-17T13:07:00.002+05:30

Today we will discuss two questions (MCQ) involving lenses. The first one appeared in Karnataka CET 2011 question paper and the second one appeared in Karnataka CET 2010 question paper.

(1) Two thin lenses have a combined power of + 9 D. When they are separated by a distance of 20 cm, their equivalent power becomes + 27/5 D. Their individual powers (in dioptre) are ……..

(1) 4, 5

(2) 3, 6

(3) 2, 7

(4) 1, 8

We have P₁ + P₂ = 9 where P₁ and P₂ are the powers of the two lenses.

[Note that combined power means 1/f₁ + 1/f₂which is the effective power when the lenses are in contact. Power is the reciprocal of focal length and when the lenses are in contact, we have 1/F = 1/f₁ + 1/f₂ where f₁ and f₂are the focal lengths of the lenses and F is the combined focal length].

When the lenses are separate by a distance d, the power becomes P₁ + P₂– d P₁P₂

[This follows from the expression, 1/F = 1/f₁ + 1/f₂– d/f₁f₂].

Therefore, we have

P₁ + P₂– d P₁P₂ = 27/5

Or, 9 – (0.2× P₁P₂)= 27/5

[You have to substitute the separation in metre since the power is the reciprocal of focal length in metre].

From the above equation, 18/5 = 0.2× P₁P₂

Or, P₁P₂ = 18

Evidently the values given in option (2) satisfy the above condition so that we have P₁ = 3 and P₂ = 6

(2) The focal length of a plano-convex lens is ‘f’ and its refractive index is 1.5. It is kept over a glass plate with its curved surface touching the glass plate. The gap between the lens and the glass plate is filled by a liquid. As a result, the effective focal length of the combination becomes 2f. Then the refractive index of the liquid is ……

(1) 1.5

(2) 2

(3) 1.25

(4) 1.33

In the case of a lens placed in air or vacuum, lens maker’s equation is

1/f = (n – 1)(1/R₁ – 1/R₂) where n is the refractive index of the lens.

[General form of lens maker’s equation is 1/f = (n₂/n₁ – 1)(1/R₁ – 1/R₂) where ‘f’ is the focal length of the lens, R₁ and R₂ are the radii of curvature of its faces, n₂ is its refractive index and n₁ is the refractive index of the medium in which the lens is placed].

In the case of the plano-convex lens, the equation becomes

1/f = (n – 1)(1/R₁– 1/∞) = (n – 1)(1/R₁)

Since n = 1.5, R₁ = 0.5f ……..(i)

The focal length of the combination (F) of the plano convex lens (of focal length f) and the plano-concave liquid lens (of focal length f_ℓ)is given by

1/F = 1/f + 1/f_ℓ

Therefore, 1/2f = 1/f + 1/f_ℓ

This gives f_ℓ = –2f………(ii)

But from lens maker’s equation, the focal length of the plano-concave liquid lens is related to the refractive index (n_ℓ) of the material (liquid) as

1/f_ℓ = (n_ℓ – 1)(–1/R₁– 1/∞) = (n_ℓ – 1)(– 1/R₁)

[Note that the radius of the concave surface, which the liquid lens presents to the incident light, has to be negative in accordance with the Cartesian sign convention. Click on the label ‘Cartesian sign convention’ below this post to learn more about this convention].

Therefore, 1/(–2f) = – (n_ℓ – 1)/0.5f , on substituting for R₁ and f_ℓ from equations (i) and (ii).

This gives 2 n_ℓ = 2.5 so that n_ℓ = 1.25

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

2011-12-27T11:10:00.001+05:30

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

2. Final Examination ………… 13^th 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.

Two Questions (MCQ) on Rotation of Rigid Bodies

2011-11-30T18:45:00.005+05:30

“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)

(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)

(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)

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

2011-11-19T15:13:00.001+05:30

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.

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

2011-10-27T10:36:00.007+05:30

"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

(D) 7.50 kH

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

n’ = n(v+w–v_L)/(v+w–v_S) where ‘v’ is the velocity of sound, ‘w’ is the wind velocity ‘v_L’ is the velocity of the listener, ‘v_S’ is thevelocity 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, v_L = – 36 km/hr = 36×(5/18) ms^–1 = –10 ms^–1, w = 0, v_S = + 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.

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

2011-10-16T14:22:00.002+05:30

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/

IIT Madras: http://jee.iitm.ac.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.

Multiple Choice Questions on Unusual Combinations of Capacitors

2011-10-02T19:54:00.007+05:30

"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

(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

(d) 4ε₀A/d

(e) 2ε₀A/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

(d) 4ε₀A/5d

(e) 4ε₀A/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.

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

2011-09-22T12:13:00.001+05:30

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= I₂ω₂where 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 = I_CM + Ma² where I_CM 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 i – j + k

(2) 2 i – j – 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 i – j + 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.

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

2011-09-02T17:20:00.005+05:30

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

2011-08-21T14:47:00.008+05:30

“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

2011-07-31T12:28:00.003+05:30

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

(D) 225 Ω

(E) 240 Ω

Power gain A_p is given by

A_p = β_acA_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_ois 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

(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

2011-07-20T11:25:00.002+05:30

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.

(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

(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

(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

2011-07-09T16:46:00.005+05:30

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

(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

2011-06-12T20:25:00.003+05:30

“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

(d) 7 s

(e) 8 s

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

10 t = 48 + ½ at²where 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

(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 v₁and 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

(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].

AIEEE 2011 Question (MCQ) on Magnetic Field due to Current Carrying Conductors

2011-05-18T22:06:00.005+05:30

"Seven Deadly Sins:

Wealth without work

Pleasure without conscience

Science without humanity

Knowledge without character

Politics without principle

Commerce without morality

Worship without sacrifice."

– Mahatma Gandhi

The following multiple choice question included in the All India Engineering/Architecture Entrance Examination (AIEEE) 2011 will be interesting and useful to you:

A current I flows in an infinitely long wire with cross section in the form of a semi-circular ring of radius R. The magnitude of the magnetic induction along its axis is

(1) μ₀I/4πR

(2) μ₀I/π²R

(3) μ₀I/2π²R

(4) μ₀I/2πR

It would have been better if the magnitude of the magnetic induction at the axis was asked for since the words along the axis usually means directed along the axis. [The component of field directed along the axis is zero].

Forget about it. The question setter requires you to calculate the magnetic flux density at points on the axis (of the semicircular ring shaped cross section of the infinitely long wire). In the adjoining figure we have shown the cross section (of the given wire) lying in the XY plane. The length of the wire is along the Z-axis and the current in the wire is supposed to flow along the negative Z-direction. The broad infinitely long wire can be imagined to be made of a large number of infinitely long straight wire strips, each of small width dℓ.

With reference to the figure, we have dℓ = Rdθ.

The magnetic flux density due to the above strip is shown as dB₁ in the figure. It has an X-component dB₁sinθ and a Y-component dB₁cosθ. When we consider a similar strip of the same with dℓ located symmetrically with respect to the Y-axis, we obtain a contribution dB₂to the flux density. The flux density dB₂ has the same magnitude as dB₁. It hasX-component dB₂sinθ and Y-component dB₂cosθ. The X-components of dB₁ and dB₁ are of the same magnitude and direction and they add up. But the Y-components of dB₁ and dB₁ are in opposite directions and have the same magnitude. Therefore they get canceled. The entire conductor therefore produces a resultant magnetic field along the negative X-direction.

The wire strip of width dℓ cn be imagined to be an ordinary thin straight infinitely long wire carrying current Idℓ/πR since the total current I flows through the semicircular cross section of perimeter πR. Putting dB₁ = dB₂ = dB we have

dB = μ₀ (Idℓ/πR)/2πR = μ₀ (IRdθ/πR)/2πR = μ₀Idθ/2π²R

The X-component of the above field is (μ₀I/2π²R) sinθ dθ

The field due to the entire conductor is B = ₀∫^π[(μ₀I/2π²R) sinθ]dθ

Or, B = μ₀I/π²R since ₀∫^πsinθ dθ = 2

Kerala Engineering Entrance (KEAM) 2011 Question (MCQ) on Zener Diode

2011-05-04T13:02:00.005+05:30

“Everything should be made as simple as possible, but not simpler”

– Albert Einstein

The following multiple choice question on zener diode was included in Kerala Engineering Entrance (KEAM) 2011 question paper. If you identify that R₁ is the current limiting resistor and R₂ is the load resistor in the simple shunt voltage regulator making use of a zener diode, you won’t have any confusion. Here is the question:

In the circuit given the current through the zener diode is

(a) 10 mA

(b) 6.67 mA

(d) 3.33 mA

(e) 0 mA

Since the zener breakdown voltage (across the zener diode) is 10 V, the voltage across the resistor R₂is 10 V. The current through R₂is 10 V/1500 Ω = 0.00667 A = 6.67 mA.

The voltage drop across R₁is 15 V – 10 V = 5 V.

Therefore, the current through R₁is 5 V/500 Ω = 0.01 A = 10 mA

The above current gets divided between the resistor R₂ and the zener diode. Therefore, the current through the zener diode is 10 mA – 6.67 mA = 3.33 mA.

By clicking on the label ‘zener diode’ below this post you can access similar questions on zener diodes posted on this site.

IIT-JEE 2011 Questions (MCQ) on Simple Harmonic Motion

2011-04-28T10:53:00.004+05:30

“Do not worry about your problems with mathematics, I assure you mine are far greater.”

– Albert Einstein

The following two questions on simple harmonic motion were included in the IIT-JEE 2011 question paper. Both questions are simple for those who have mastered the fundamental principles:

(1) A wooden block performs SHM on a frictionless surface with frequency, ν₀. The block carries a charge +Q on its surface. If now a uniform electric field E is switched on as shown, then the SHM of the block will be

(A) of the same frequency and with shifted mean position

(B) of the same frequency and with the same mean position

(D) of changed frequency and with the same mean position

When the uniform electric field is switched on, the positively charged block experiences a constant force in the direction of the electric field and so the mean position of the block is shifted. But the frequency (ν)of oscillation of the block is unchanged since it depends only on the mass m of the block and the force constant k of the spring, in accordance with the relation

ν = (1/2π) [√(k/m)]

The correct option is (A).

(2) A point mass is subjected to two simultaneous sinusoidal displacements in x-direction, x₁(t) = A sin ωt and x₂(t) = A sin(ωt + 2π/3). Adding a third sinusoidal displacement x₃(t) = B sin(ωt + φ) brings the mass to a complete rest. The value of B and φ are

(A) √2 A, 3π/4

(B) A, 4π/3

(D) A, π/3

The particle in the above question is subjected to two simultaneous simple harmonic motions (SHM) of the same frequency and amplitude in the same direction. the resultant motion of the particle is simple harmonic as is evident by adding x₁(t) and x₂(t):

x₁(t) + x₂(t) = A sin ωt + A sin(ωt + 2π/3) = A sin(ωt +π/3)

The amplitude of the resultant simple harmonic motion is A itself but the initial phase of the motion is now π/3.

As the particle remains at rest on adding the third simple harmonic motion, x₃(t) = B sin(ωt + φ) the amplitude (B) of the third SHM must be A itself, but it must be 180º (or, π radian) out of phase. In other words, the initial phase (φ) of the third simple harmonic motion must be π/3 + π = 4π/3

The correct option is (B).

You can find a useful post on simple harmonic motion here.

IIT-JEE 2011 – Paragraph Type Multiple Choice (Single Answer) Questions Involving Phase Space Diagrams

2011-04-15T13:02:00.010+05:30

“The earth provides enough to satisfy every man’s needs, but not every man’s greed.”

– Mahatma Gandhi

The practice of asking two or three multiple choice questions (often single answer type) based on a given paragraph, is resorted to in many entrance examinations. The following three questions[(1), (2) and (3)] which appeared in IIT-JEE 2011 question paper are relatively simple:

Paragraph for questions (1), (2) and (3)

Phase space diagrams are useful tools in analyzing all kinds of dynamical problems. They are especially useful in studying the changes in motion as initial position and momentum are changed. Here we consider some simple dynamical systems in one–dimension. For such system, phase space is a plane is which position is plotted along horizontal axis and momentum is plotted along vertical axis. The phase space diagram is x (t) vs p(t) curve in this plane. The arrow on the curve indicates the time flow. For example, the phase space diagram for a particle moving with constant velocity is a straight line as shown in the figure. We use the sign convention in which position or momentum upwards (or to right) is positive and downwards (or to left) is negative.

(1) The phase space diagram for a ball thrown vertically up from the ground is

The arrow on the diagram shows the flow of time. Initially the ball is on the ground and its velocity and momentum have maximum positive values (since it moves upwards). At the highest point of its trajectory the displacement of the ball has maximum positive value but its momentum is zero. Finally when the ball just hits the ground, its displacement is zero but its momentum has maximum negative value(since it is moving downwards).

The correct option is obviously (D).

(2) The phase space diagram for simple harmonic motion is a circle centered at the origin. In the figure, the two circles represent the same oscillator but for different initial conditions, and E₁ and E₂ are the total mechanical energies respectively. Then

(A) E₁ = √2 E₂

(B) E₁ = 2 E₂

(D) E₁ = 16 E₂

When the total energy of the simple harmonic motion (SHM) is E₁, the amplitude is 2a and when the total energy is E₂, the amplitude is a, as is evident from the phase space diagrams. The total energy E of a particle of mass m in simple harmonic motion of amplitude A and angular frequency ω is given by

E = ½ mω²A²

So when the amplitude is doubled, the total energy is quadrupled.

Thus E₁ = 4 E₂ as given in option (C).

[The phase space diagram given in the above question represents an undamped simple harmonic motion since respective amplitudes appropriate to the initial conditions remain constant. If the simple harmonic motion is a damped one, the curve will be a spiral, proceeding inwards].

(3) Consider the spring-mass system, with the mass submerged in water, as shown in the figure. The phase space diagram for one cycle of this system is

The correct option is (B) on account of the following reasons:

(i) The amplitude of oscillations must decrease with time [which is not the case in diagrams (C) and (D)] as the viscous forces in water damps the motion of the system.

(ii) When the displacement is maximum positive value (corresponding to maximum positive position co-ordinate), the momentum is zero and is going to increase in magnitude in the negative direction [unlike in the case of diagram (A)].

All India Pre-Medical/Pre-Dental Entrance Examination (AIPMT) 2011 Questions on Magnetic Force

2011-04-09T16:19:00.006+05:30

“I have no special talents. I am only passionately curious.”

– Albert Einstein

Two questions involving magnetic force on current carrying conductors and moving charges were included in the AIPMT 2011 (Preliminary) question paper. Here are those questions with solution:

(1) A current carrying closed loop in the form of a right angled isosceles triangle ABC is placed in a uniform magnetic field acting along AB. If the magnetic force on the arm BC is F, the force on the arm AC is

(1) √2 F

(2) –√2 F

(3) – F

(4) F

Since the current loop is placed in a uniform magnetic field, the net force acting on the loop is zero.

[Note that a current loop in a uniform magnetic field will experience a torque but no net force. If the magnetic field is non-uniform there will be a torque as well as a net force].

Since the arm AB of the loop is parallel to the magnetic field, the magnetic force on AB is zero. Since the magnetic force on the arm BC is F (as given in the question), the magnetic force on the arm Ac must be – F (so as to make the net force on the loop zero).

(2) A uniform electric field and a uniform magnetic field are acting along the same direction in a certain region. If an electron is projected in the region such that its velocity is pointed along the direction of the fields, then the electron

(1) will turn towards left of the direction of motion

(2) will turn towards right of the direction of motion

(3) speed will decrease

(4) speed will increase

The magnetic force will be zero since the electron is projected parallel to the field. The electric field will exert a force opposite to the direction of motion of the electron and hence its speed will decrease [Option (3)].

All India Pre-Medical/Pre-Dental Entrance Examination (Preliminary) 2010 (AIPMT 2010) Multiple Choice Questions on Centre of Mass

2011-04-01T12:01:00.001+05:30

The following questions (MCQ) on centre of mass were included in the AIPMT 2010 question paper:

(1) A man of 50 kg mass is standing in a gravity free space at a height of 10 m above the floor. He throws a stone of 0.5 kg mass downwards with a speed 2 ms^–1. When the stone reaches the floor, the distance of the man above the floor will be

(1) 20 m

(2) 9.9 m

(3) 10.1 m

(4) 10 m

You can work out this question either by using the concept of centre of mass or by applying the law of conservation of momentum.

The position of the centre of mass of the man-stone system will be unchanged as there are no external forces. If x is the distance of the man from the centre of mass when the stone reaches the floor, we have

50×x = 0.5×10 from which x = 0.1 m

As the centre of mass of the system is initially at a distance of 10 m above the floor, the distance of the man when the stone reaches the floor will be 10 + x = 10.1 m.

[You can apply the law of conservation of momentum and obtain the ‘recoil speed’ v of the man from the equation, 50×v = 0.5×2, from which v = 0.02 ms^–1.

The time taken by the stone to reach the floor is 10/2 = 5 s. During this time the man will move up through a distance 0.02×5 = 0.1 m so that when the stone reaches the floor, the man will be at a distance 10 + 0.1 = 10.1 m above the floor].

(2) Two particles which are initially at rest move towards each other under the action of their internal attraction. If their speeds are v and 2v at any instant, then the speed of centre of mass of the system will be

(1) v

(2) 2 v

(3) zero

(4) 1.5 v

Since the particles are initially at rest, the speed of the centre of mass of the system is zero. Since there are no external forces on the system, the centre of mass will continue to be at rest even though the particles are moving under the action of their internal attraction. Therefore the correct option is (3).

By trying a search for ‘centre of mass’ using the search box provided on this page, you can access all posts on centre of mass on this site.

You will find some additional posts on centre of mass at AP Physics Resources.

Kerala Engineering Entrance [KEAM (Engineering)] 2007 Question on Percentage Error

2011-03-28T21:43:00.001+05:30

In continuation of the previous post (dated 23^rd March 2011) related to errors in measurements, I give you today the following question which appeared in KEAM (Engineering) 2007 question paper:

The value of two resistors are R₁ = (6 ± 0.3) kΩ and R₂ =(10 ± 0.2) kΩ. The percentage error in the equivalent resistance when they are connected in parallel is

(a) 5.125 % (b) 2 % (c) 3.125 %

(d) 7 % (e) 10.125 %

When resistances R₁ and R₂ are connected in parallel the equivalent resistance R is given by

R = R₁R₂/(R₁+ R₂)

The fractional error in R is ∆R/R, which you can find by taking logarithm and differentiating:

∆R/R = (∆R₁/R₁) + (∆R₂/R₂) + ∆(R₁ + R₂)/(R₁+ R₂)

[Here ∆R₁ is the error in R₁, ∆R₂ is the error in R₂ and ∆(R₁ + R₂) is the error in (R₁ + R₂). On differentiation the third term is negative but we have taken the sign of the third term as positive since we have to consider the worst case in which the maximum possible error is obtained].

Therefore, ∆R/R = (0.3/6) + (0.2/10) + (0.5/16)

Percentage error in R = percentage error in R₁ + percentage error in R₂ + percentage error in (R₁+ R₂) = (0.3/6)×100 + (0.2/10)×100 + (0.5/16)×100 = 5 % + 2 % + 3.125 % = 10.125 %.

As an extension of the above problem consider the following question:

In the above question what will be the percentage error in the heat produced per second when a current of (4 ± 0.1) ampere is passed through the parallel combination of R₁and R₂?

(a) 3.5 %

(b) 7 %

(d) 15.125 %

(e) 20.25 %

The heat H produced is given by H = I²R, where I is the current.

The fractional error in the heat produced per second is

∆H/H = 2×(∆I/I) + ∆R/R

Percentage error in H is 2×(∆I/I) ×100 + (∆R/R) ×100

Therefore, percentage error in H = 2×(0.1/4)×100 + 10.125 = 15.125%

Percentage Change and Percentage Error in a Physical Quantity

2011-03-21T17:57:00.003+05:30

“Everything that is really great and inspiring is created by the individual who can labour in freedom.”

– Albert Einstein

The fractional change in a physical quantity Q is (ΔQ)/Q where ΔQ is the actual change in the quantity. The percentage change in the quantity = Fractional change × 100.

Let us consider a few multiple choice questions to make things clear:

(1) A copper wire having resistance 10 Ω is stretched so that its radius decreases by 3%. What is the resistance of this wire after stretching?

(a) 10 Ω (b) 10.3Ω (c) 10.6Ω (d) 11.2Ω (e) 11.8Ω

The resistance R of a wire is given by

R = SL/A where S is the resistivity (specific resistance), L is the length and A is the area of cross section. Since S is constant for a given material, R depends only on L and A. When the radius of the wire is decreased by 3 %, its cross section area A is decreased by 6 % (since A = πr²). The resistance increases by 6% on account of this. When A decreases by 6 %, the length L increases by 6 % since the volume of the wire is unchanged. On account of the 6% increase in length there is an additional increase in resistance by 6%. So the total increase in resistance is 12%. The resistance after stretching = 10 + 1.2 =11.2 Ω [Option (d)]

(2) A nichrome wire has resistance 20 Ω. What will be resistance of another nichrome wire whose length is greater by 5 % and radius is greater by 1%?

(a) 20.2 Ω (b) 20.4 Ω (c) 20.6 Ω (d) 20.8 Ω (e) 21 Ω

There is a 5 % increase in resistance due to the increase in length and a 2 % decrease in resistance due to the increase in radius.

[Note that 1% increase in radius produces 2 % increase in the area of cross section].

The net increase in resistance = 5 % – 2 % = 3 % = 20 × 3/100 = 0.6 Ω. The correct option therefore is (c).

In the above questions, we have considered the changes in certain quantities (length and radius in the above cases). The changes may be positive (increments) or negative (decrements). The final result may increase or decrease because of the individual changes.

(3) A wire has mass m = 0.2 ± 0.002 g, radius r = 0.25 ± 0.005 mm and length L = 6 ± 0.06 cm. The maximum percentage error in the measurement of the density ρ of the material of this wire is

(a) 1 (b) 2 (c) 4 (d) 6 (e) 8

Here the percentage error in mass = (0.002/0.2)×100 = 1 %. Percentage error in radius = (0.005/0.25) × 100 = 2 %. Percentage error in length = (0.06/6)×100 = 1%. Since the density ρ = mass/volume = m/πr²L, the maximum percentage error in density = percentage error in mass + 2×percentage error in radius + percentage error in length = 1+ (2×2) + 1 = 6%.

[The percentage error in r is multiplied by 2 since the power of r is 2. This follows thus:

We have ρ = m/πr²L. On taking logarithms and differentiating, we have

dρ/ρ = dm/m – 0 – (2 dr/r) – dL/L

dρ/ρ, dm/m, dr/r and dL/L are respectively the fractional errors in density, mass, radius and length. Therefore we have

Percentage error in density = Percentage error in mass + 2×percentage error in radius + percentage error in length.

Note that we have taken the signs of all errors as positive since we want to make the errors add up to obtain the maximum percentage error].

Multiple Choice Practice Questions on Electrostatics

2011-03-06T18:17:00.004+05:30

“Live as if you were to die tomorrow. Learn as if you were to live for ever.”

– Mahatma Gandhi

If you grasp the fundamental principles well, you will be able to answer multiple choice questions within the stipulated time. If you are confused about the fundamental principles, you will be tempted to waste your time while dealing with even simple questions just because of some simple distractions. See the following questions:

(1) A battery of emf 12 V is connected in series with two initially uncharged capacitors, each of value 100 μF (Fig.). The capacitors are fully charged. If the total energy spent by the battery for charging the capacitors is E joule, the energy of each capacitor is

(a) E

(b) E/2

(d) E/4

(e) E/8

If the total charge supplied by the battery is Q coulomb, the total energy E supplied by the capacitor is given by

E = VQ joule where V is the emf of the battery (which is 12 volt).

The two identical 100 μF capacitors in series is equivalent to a single capacitor of capacitance C = 50 μF. The energy of this charged capacitor combination is ½ CV² joule = ½ VQ joule since Q = CV.

Therefore, the energy of the charged capacitor combination is ½ E.

Since there are two identical capacitors in the combination, the energy of each capacitor is E/4.

[So the capacitors have gained a total energy of E/2, even though the battery has delivered a total energy E. What happens to the other half of the energy? Well, it is irrecoverably lost as heat generated in the resistance of the circuit].

(2) Charges of equal magnitude are fixed at the diagonally opposite corners A and C of a non conducting square frame ABCD (Fig.). In case (i) the charges at A and C are positive and a third free negative charge q is moved from B to D. In case (ii) the charges at A and C are negative and a third free negative charge q is moved from B to D. In case (iii) the charges at A and C are of opposite sign and a third free negative charge q is moved from B to D. Pick out the correct statement regarding the work done for moving the charge q:

(a) In case (i) the work done is positive

(b) In case (ii) the work done is positive

(d) In case (iii) the work done is positive

(e) In all cases the work done is zero

In case (i) the potentials at B and D are positive, but of equal value. In case (ii) the potentials at B and D are negative, but of equal value. In case (iii) the potentials at B and D are zero. Therefore, in all cases the potential difference between points B and D is zero so that the work done in moving the charge q from B to D is zero.

You will find some useful questions (MCQ) with solution here.

Two Questions on Zener Diode Voltage Regulators

2011-02-08T22:01:00.004+05:30

“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

Zener diodes, as you know, are widely used as reference voltage elements in a variety of voltage regulator circuits. Today I give you a couple of multiple choice questions involving the use of zener diodes in simple voltage regulators:

(1) In the circuit shown, D₁ is a silicon diode which has a voltage drop of 0.7 V while in full conduction under forward bias. The zener diode D₂ has a breakdown voltage of 6.8 V. What is the current through the 450 Ω resistor?

(a) 1 mA

(b) 10 mA

(d) 100 mA

(e) 0 mA

You can use the voltage drop across a forward biased ordinary silicon diode as a reference voltage in voltage regulator circuits. In the circuit shown, a reverse biased zener diode (of breakdown voltage 6.8 V) and a forward biased silicon diode (of voltage drop 0.7 V) in series make a reference voltage of 7.5 V. The output voltage of the circuit is thus 7.5 V.

Since the input voltage to the regulator circuit is 12 V, the voltage drop across the 450 Ω resistor is 12 V – 7.5 V = 4.5 V.

The current through the 450 Ω resistor is (4.5 V)/(450 Ω) = 0.01 A = 10 mA.

(2) The circuit shown in the adjoining figure is the simplest shunt voltage regulator using a zener diode of breakdown voltage 6 V. What is the power dissipated in the zener diode?

(a) 100 mW

(b) 120 mW

(d) 360 mW

(e) 480 mW

Since the input voltage to the regulator circuit is 10 V and the regulated output voltage is 6 V, the voltage drop across the 40 Ω resistor is 10 V – 6 V = 4 V.

Therefore, the current through the 40 Ω resistor (current limiting resistor) is (4 V)/ 40 Ω = 0.1 A = 100 mA. This is the total current flowing into the parallel combination of the zener diode and the100 Ω load resistor.

The current through the load resistor of 100 Ω is (6 V)/(100 Ω) = 0.06 A = 60 mA.

Therefore, the current flowing through the zener diode is 100 mA – 60 mA = 40 mA.

Power dissipated in the zener diode is 6 V×40 mA = 240 mW.