Multiple Choice Questions on Metre Bridge

“If I have seen a little further it is by standing on the shoulders of Giants.”

– Sir Isaac Newton

The metre bridge is basically a Wheatstone bridge, the condition of balance of which is conveniently used in measuring unknown resistances. Today we shall discuss a few questions (MCQ) involving metre bridge:

(1) Resistances R₁ and R₂ are connected respectively in the left gap and right gap of a metre bridge. If the balance point is located at 55 cm, the ratio R₂/R₁ is

(a) 4/5

(b) 5/4

(c) 9/11

(d) 11/9

(e) 6/7

The distance of the balance point from the left end of the wire in the metre bridge is 55 cm. Therefore the distance of the balance point from the right end of the wire is 45 cm. Thus we have

            R₁/R₂ = 55/45

Or, R₂/R₁ = 45/55 = 9/11

(2) A 15 Ω resistance is connected in the left gap and an unknown resistance less than 15 Ω is connected in the right gap of a metre bridge. When the resistances are interchanged, the balance point is found to shift by 20 cm. The unknown resistance is

(a) 5 Ω

(b) 6 Ω

(c) 8 Ω

(d) 10 Ω

(e) 12 Ω

Initially the balance point will be at J₁ as indicated in the figure. On interchanging the resistances the balance point will shift to J₂ so that the length of the bridge wire between J₁ and J₂ is 20 cm. The balance points J₁ and J₂ must be equidistant from the mid point (50 cm mark) of the wire so that J₁ is at 60 cm and J₂ is at 40 cm.

Therefore we have

            R/X =  60/40 = 6/4

Or, 15/X = 6/4

This gives X = 10 Ω

(3) Resistances 4 Ω and 6 Ω are connected across the left gap and right gap respectively of a metre bridge. When a 2 Ω resistance is connected in series with the 4 Ω resistance in the leftt gap, the shift in the balance point is

(a) 10 cm

(b) 15 cm

(c) 20 cm

(d) 25 cm

(e) 30 cm

Initially when the left gap and right gap contain 4 Ω and 6 Ω respectively, the condition of balance is

            4/6 = L/(100 – L) where L is the balancing length

This gives L = 40 cm.

When a 2 Ω resistance is connected in series with the 4 Ω resistance in the leftt gap, the balancing length becomes 50 cm (since the gaps contain equal resistances).

Therefore the shift in the balance point is (50 cm – 40 cm) = 10 cm.

 

Questions Involving Charging of Capacitors

“Science distinguishes a Man of Honour from one of those Athletic Brutes whom undeservedly we call Heroes.”

– John Dryden

Today we will discuss a few multiple choice questions involving charging of capacitors. These questions will serve as practice questions for various entrance examinations including the National Eligibility Cum Entrance Test (NEET) for admission to MBBS and BDS Courses and the Joint Entrance Examination (JEE) for Admission to Undergraduate Engineering Programmes in IITs, NITs and other Technical Institutions. Here are the questions:

(1) An initially uncharged capacitor of capacitance 100 μF is charged at a steady rate of 20 μC/s. What is the time (in seconds) required for the potential difference across the capacitor to build up to 12 V?  

(a) 240 s

(b) 120 s

(c) 60 s

(d) 30 s

The final charge Q on the capacitor is given by

            Q = CV = 100 μF×12 V = 1200 μC

Since the charging rate is 20 μC/s, the time t required for the potential difference across the capacitor to build up to 12 V is given by

            t = 1200 μC/20 μC/s = 60 s

(2) A battery of emf V volt is connected to a combination of four identical capacitors (each of capacitance C farad) arranged as shown in the adjoining figure. What is the amount of energy  supplied by the battery to the capacitor combination?  

(a) CV²/4

(b) CV²/2

(c) CV²/3

(d) 2CV²/3

The effective capacitance connected across the battery is C + (C/3) = 4C/3

The energy E of the capacitor combination on getting charged to V volt is given by

            E = ½ × (4C/3) V² = 2CV²/3

(3) A parallel plate capacitor with air as dielectric has capacitance C. It is charged fully using a battery of emf V. The battery is then disconnected and the separation between the plates of the capacitor is doubled. What is the final energy stored in the capacitor?

(a) ½ CV²

(b) CV²

(c) 3CV²/2

(d) 2 CV²

(e) CV²/4

The charge Q on the capacitor is unchanged since the battery is disconnected. The charge Q is given by

            Q = CV ------------- (i)

When the separation between the plates is doubled, the capacitance is halved (C/2).

[Note that the capacitance of a parallel plate air capacitor is given by C = ε₀A/d where A is the plate area and d is the separation between the plates.

The energy E stored in the capacitor is given by

            E = Q²/2C₁ where C₁ = C/2

Substituting for Q from Eq. (i), we have

            E = C²V²/(2C/2) = CV², as given in option (b).

[Note that the increased energy of the capacitor is the result of the external work done for increasing the separation between the plates, overcoming the electrostatic attractive force between the plates]

(4) In the above question, suppose the battery remains connected to the capacitor while doubling the separation. What will be the final energy stored in the capacitor>

(a) ½ CV²

(b) CV²

(c) 3CV²/2

(d) 2 CV²

(e) CV²/4

In this case charges can flow through the wires connecting the battery to the capacitor. The capacitor will hold a smaller amount of charge (let us say, Q₁) since the capacitance is decreased from C to C/2. We have\       

            Q₁ = (C/2) V

The final energy (E₁) stored in the capacitor is given by

            E₁ = Q₁²/2C₁ = [(C/2)V]²/(2C/2) = CV²/4

Joint Entrance Examination (JEE) 2013 for Admission to Undergraduate Engineering Programmes in IITs, NITs and other Technical Institutions

Central Board of Secondary Education has notified that admission to Undergraduate Engineering Programmes in IITs, NITs and other Centrally Funded Technical Institutions will be based on a Joint Entrance Examination (JEE) with effect from 2013.

JEE will be conducted in two parts – JEE (Main) and JEE (Advanced).

JEE (Main) will replace AIEEE and JEE (Advanced) will replace IIT-JEE, with the important criterion that only the top 150,000 candidates performing in JEE (Main) will qualify for appearing in JEE (Advanced). Admission to IITs, IT-BHU and ISM Dhanbad will be based only on category-wise All India Rank in JEE (Advanced), subject to the additional condition that such candidates are in the top 20 percentile of successful candiates in Class XII/equivalent examination conducted by their Boards.

For admission to Undergraduate Engineering Programmes in NITs and other Centrally Funded Technical Institutions the rank list will be prepared based on 40% weightage to school board marks in Class XII and 60% weightage to JEE (Main) marks.

JEE (Main) 2013 Paper I (for admission to B.E./B.Tech courses) which replaces Paper I of earlier AIEEE, will be of 3 hours, consisting of objective type questions from Physics, Chemistry and Mathematics.

Paper II for admission to B.Arch/B.Planning courses also will be of 3 hours and will consist of Mathematics, Aptitude Test and Drawing Test as per past practice of AIEEE.

The JEE (Main) (Paper I of earlier AIEEE) for B.E./B.Tech will be held in two modes, viz., offline and online (CBT). The offline examination for JEE (Main) will be conducted on 7^th April, 2013 and the online examinations will be conducted thereafter in April, 2013.

For details you may visit http://jeemain.nic.in 

CBSE to conduct National Eligibility Cum Entrance Test (NEET) 2013 for admission to MBBS and BDS Courses

“The pursuit of truth and beauty is a sphere of activity in which we are permitted to remain children all our lives.”

– Albert Einstein

Most of the students aspiring for admission to MBBS and BDS courses in India will be happy about the recent notification that admission to MBBS and BDS Courses in the institutions approved by the Medical Council of India and Dental Council of India are subject to merit position of candidates in the National Eligibility Cum Entrance Test (Under-Graduate), 2013.

The Central Board of Secondary Education (C.B.S.E.) will conduct the National Eligibility cum Entrance Test (NEET) for admission to MBBS and BDS Courses on Sunday, the 5^th May 2013.

The syllabus for the National Eligibility cum Entrance Test (NEET) is available on the Medical Council of India website www.mciindia.org

Lorentz force – An IIT-JEE 2012 Multiple Correct Answer(s) Type Question

“Making the simple complicated is commonplace; making the complicated simple, awesomely simple, that’s creativity.”

– Charles Mingus    

IIT-JEE Multiple Correct Answer(s) Type Questions are designed to have four choices (A), (B), (C) and (D) out of which ONE or MORE are correct. The following question appeared in IIT-JEE 2012 question paper. The question is meant for checking your understanding of Lorentz forces.

Consider the motion of a positive point charge in a region where there are simultaneous uniform electric and magnetic fields E = E₀ ĵ and B = B₀ ĵ. At time t = 0, this charge has velocity v in the x-y plane, making an angle θ with the x-axis. Which of the following option(s) is (are) correct for t > 0?

(A) If θ = 0°, the charge moves in a circular path in the x-z plane.

(B) If θ = 0°, the charge undergoes helical motion with constant pitch along the y-axis.

(C) If θ = 10°, the charge undergoes helical motion with its pitch increasing with time, along the y-axis.

(D) If θ = 90°, the charge undergoes linear but accelerated motion along the y-axis.

Note that the electric and magnetic fields are along the positive y-direction.

If θ = 0, the charged particle is moving at right angles to the electric and magnetic fields (Fig.). The magnetic force (qv×B₀) will make the particle move along a circle and the electric force (qE₀) will push it along the positive y-direction. The path of the particle will therefore be a helix of increasing pitch.

If θ = 10°,the magnetic field will make the particle move along the positive y-direction, following a helical path of constant pitch; but since the electric force (qE₀) pushes it along the positive y-direction, the helical path has increasing pitch.

If θ = 90°, the charged particle is moving parallel to the electric and magnetic fields. The magnetic field has no action on the motion since there is no magnetic force. But the electric force (qE₀) will push the particle along the positive y-direction. The particle is thus in accelerated linear motion. 

Therefore options (C) and (D) are correct.

Multiple Choice Questions on Electrostatics

“Example isn't another way to teach, it is the only way to teach.”

– Albert Einstein

Today we will discuss a few questions from electrostatics. You will find many questions (with solution) in this section discussed earlier on this site. You can access all those questions by clicking on the label ‘electrostatics’ below this post.

(1) A 6 μF capacitor is connected in series with a 2 μF capacitor. The 6 μF capacitor can withstand a maximum voltage of 3 kV where as the 2 μF capacitor can withstand a maximum voltage of 6 kV. The maximum voltage that the parallel combination can withstand is

(a) 2 kV

(b) 3 kV

(c) 6 kV

(d) 8 kV

(e) 12 kV

We have capacitors C₁ and C₂ (let us say) having values 6 μF and 2 μF. If the maximum voltage that the parallel combination can withstand is V_max, the voltage V₁ across the 6 μF capacitor on applying this voltage across the series combination is given by

            V₁ = V_max C₂/(C₁ + C₂)

[The charge Q on each capacitor on connecting the voltage V_max across the series combination is given by Q = C₁C₂ V_max/(C₁ + C₂), remembering that the effective capacitance of the series combination is C₁C₂/(C₁ + C₂)].

Therefore, V₁ = V_max×2/(6+2) = V_max/4

Since C₁ can withstand a maximum voltage of 3 kV we have V_max/4 = 3 kV

This gives V_max = 12 kV.

[Do not jump to a conclusion at this stage, You have to check whether C₂ will be intact on applying the above 12 kV across the series combination].

The voltage V₂ across the 2 μF capacitor on applying this voltage V_max across the series combination is given by

            V₂ = V_max C₁/(C₁ + C₂)

Threfore V₂ = V_max×6/(6+2) = 6 V_max/8

Since C₂ can withstand a maximum voltage of 6 kV we have 6 V_max/8 = 6 kV

This gives V_max = 8 kV.

This value of V_max being lower than that obtained (12 kV) on considering the 6 μF capacitor, the correct option is 8 kV.

(2) A uniform electric field of intensity E newton/coulomb directed along the positive x-direction exists in a region of space (Fig.). The x- direction is horizontal. A, B, C and D are points at the corners of a square of side a, with AB and CD  parallel to the x-direction. If the electric potential at point A is V volt, what is the potential (in volt) at the point D?

(a) V

(b) V – aE

(c) V – √2 aE

(d) V + √2 aE

(e) V + aE

Since the electric field acts along the positive x-direction, the potential decreases as we move along the positive x-direction.

[Remember that the electric field is the negative gradient of potential].

While moving from A to D the x-coordinate increases by ‘a’ and hence the potential decreases by aE. Therefore, the potential at D is V – aE [Option (b)].

(3) In the above question what is the potential difference between points A and C?

(a) V

(b) V – aE

(c) V + aE

(d) aE

(e) Zero

Since the electric field acts along the x-direction, the potential will change only if the x-coordinate changes. Points A and C have the same x-coordinates and hence they are at the same potential. Therefore, the potential difference between points A and C is zero.

(4) Two small identical spheres are charged equally and suspended in air by strings of equal length. The strings make a small angle θ with each other (Fig.). When the spheres are immersed in oil of density 800 kg m^–3 the angle between the strings is found to be unaltered. If the density of the material of the spheres is 1200 kg m^–3, what is the dielectric constant of the oil?

(a) 1.5

(b) 2.5

(c) 3

(d) 3.5

(e) 4

The repulsive electrostatic force F  between the spheres in air is given by

             F = (1/4πε₀) (q²/d²) where ε₀ is the permittivity of free space (and air, very nearly), qis the charge on each sphere and d is the distance between the spheres.

            When the spheres are in the oil the electrostatic force F₁ between the spheres is given by

            F₁ = (1/4πε₀K) (q²/d²) where K is the dielectric constant of the oil.

The real weight W of each sphere is given by

            W = Vρg where V is the volume, ρ is the density of the material of the sphere and  g is the acceleration due to gravity.

The apparent weight W₁ of each sphere when immersed in oil is given by

            W₁ = Vρg – Vσg where σ is the density of the oil.

[Note that Vσg is the upthrust or the force of buoyancy due to the oil]

When the spheres are in air, we have (Fig.)

     tan α = F /W = (1/4πε₀) (q²/d²)/ Vρg………………..(i)

When the spheres are in oil, we have

tan α = F₁ /W₁ = (1/4πε₀K) (q²/d²) / (Vρg – Vσg)……(ii)

Dividing Eq. (i) by Eq. (ii) we have

            1 = K(ρ – σ) / ρ

Therefore, K = ρ/(ρ – σ) = 1200/400 = 3

Kerala Engineering Entrance 2012 (KEAM Engg. 2012) Questions on Surface Tension

“It doesn't matter how beautiful your theory is; it doesn't matter how smart you are. If it doesn't agree with experiment, it's wrong.”

– Richard Feynman

Two questions on surface tension were included in the KEAM (Engineering) 2012 question paper. Here are the questions with solution:

(1) If two capillary tubes of radii r₁ and r₂ in the ratio 1 : 2 are dipped vertically in water, then the ratio of capillary rises in the respective tubes is

(a) 1 : 4

(b) 4 : 1

(c) 1 : 2

(d) 2 : 1

(e) 1 : √2

The capillary rise h due to surface tension is relate to the surface tension S as

            S = hrρg/2 cosθ

where r is the radius of the capillary tube, ρ is the density of the liquid, g is the acceleration due to gravity and θ is the angle of contact.

Evidently h is inversely proportional to r.

Therefore, h₁/h₂ = r₂/r₁ = 2 : 1

(2) If the excess pressure inside a soap bubble of radius r₁ in air is equal to the excess pressure inside air bubble of radius r₂ inside the soap solution, then r₁ : r₂ is         

(a) 2 : 1

(b) 1 : 2

(c) 1 : 4

(d) √2 : 1

(e) 1 : √2

The excess pressure inside a soap bubble in air is 4S/r where as the excess pressure inside an air bubble in the soap solution is 2S/r where S is the surface tension (of soap solution) and r is the radius of the bubble.

[In the case of the air bubble in the soap solution there is one liquid surfaoe only and that is why the excess pressure is 2S/r and not 4S/r. Remember that in the case of a soap bubble in air there are two liquid surfaces]. 

As given in the question, we have

            4S/r₁ = 2S/r₂

This gives r₁/r₂ = 2

Or, r₁ : r₂ = 2 : 1