Students must start practicing the questions from RBSE 12th Physics Model Papers Set 4 with Answers in English Medium provided here.

## RBSE Class 12 Physics Model Paper Set 4 with Answers in English

Time . 2 Hours 45 Min.

Max Marks: 56

General Instruction to the Examinees:

- Candidate must write first his/her Roll No. on the question paper compulsorily.
- All the questions are compulsory.
- Write the answer to each question in the given answer book only.
- For questions having more than one part the answers to those parts are to be written together in continuity.

Section – A

Question 1.

Write the Correct Answer from multiple choice question 1 (i to ix) and write in given answer book:

(i) Four metal conductors having different shapes:

Spherical, Cylindrical, Pear, Lightning conductor are mounted on insulating stands and charged. The one which is suited to retain the charges for a longer time is: [1]

(a) Spherical

(b) Cylindrical

(c) Pear

(d) Lightning conductor

Answer:

(a) Spherical

(ii) Two conducting spheres of radii r_{1} and r_{2} are equally charged. The ratio of their potential is: [1]

(a) \(\frac{r_{1}}{r_{2}}\)

(b) \(\frac{r_{2}^{2}}{r_{1}^{2}}\)

(c) \(\frac{r_{2}}{r_{1}}\)

(d) \(\frac{r_{1}^{2}}{r_{2}^{2}}\)

Answer:

(c) \(\frac{r_{2}}{r_{1}}\)

(iii) If voltage across a bulb rated 220 V, 100 W drops by 2.5% of its rated value, the percentage of the rated value by which the power would decrease is: [1]

(a) 20%

(b) 2.5%

(c) 5%

(d) 10%

Answer:

(a) 20%

(iv) The magnetic force \(\vec{F}\) on a current carrying conductor of length l in on external magnetic field \(\vec{B}\) is given by: [1]

(a) \(\frac{I \times \vec{B}}{\vec{l}}\)

(b) \(\frac{\vec{l} \times \vec{B}}{I}\)

(c) I\((\vec{l} \times \vec{B})\)

(d) I^{2}\(\vec{l} \times \vec{B}\)

Answer:

(c) I\((\vec{l} \times \vec{B})\)

(v) Two similar circular co-axial loops carry equal currents in the same direction. If the loops be brought nearer, the currents in loops. [1]

(a) Decreases

(b) Increases

(c) Remains same

(d) Different in each loop

Answer:

(a) Decreases

(vi) The photoelectric cut off voltage in a certain experiment is 1.5 V. The maximum kinetic energy of photoelectrons emitted is: [1]

(a) 2.4 V

(b) 1.5 eV

(c) 3.1 eV

(d) 4.5 eV

Answer:

(b) 1.5 eV

(vii) After two hours, one sixteenth of the starting amount of a certain radioactive isotope remained undecayed. The half life of the isotope is: [1]

(a) 15 minutes

(b) 30 minutes

(c) 45 minutes

(d) one hours

Answer:

(b) 30 minutes

(viii) Which statement is correct: [1]

(a) TV-type germanium is negatively charged and P-type germanium is positively charged

(b) Both TV-type and P-type germanium are neutral

(c) TV-type germanium is positively charged and P-type germanium is negatively charged

(d) Both TV-type and P-type germanium are negatively charged

Answer:

(b) Both TV-type and P-type germanium are neutral

(ix) Which of the following gates will have an output of: [1]

Answer:

Question 2.

Fill in the blanks:

(i) The direction of the electric field is always ……………………….. to the plane of the equipotential pacing surface. [1]

(ii) Resistance R is ………………………. proportional to its length. [1]

(iii) Induced emf is directly proportional to rate of chagne of ………………………… . [1]

(iv) A photodiode is a p – n junction diode arrange in ……………………….. biasing. [1]

Answer:

(i) perpendicular

(ii) directly

(iii) magnetic flux

(iv) reverse.

Question 3.

Give the answer of the following questions in one line:

(i) I-V graph for a metallic wire at two different temperatures T_{1} and T_{2} is as shown in the figure below. Which of the two temperature is lower and why? [1]

Answer:

Since, slope of 1 > slope of 2.

I_{1}/V_{1} > I_{2}/V_{2}

⇒ V_{2}/I_{1} > V_{1}/I_{1}

∴ R_{2} > R_{1}

V/I = R

Also, we know that resistance is directly proportional to the temperature.

Therefore, T_{2} > T_{1}

(ii) Write the relation for the force acting on a charged particle q moving with velocity υ in the presence Of a magnetic field B. [1]

Answer:

When a charged particle q moves with velocity υ in a uniform magnetic field B, then the force acting on it is given by

F = q(υ × B)

(iii) The motion of copper plate is damped, when it is allowed to oscillate between the two poles of a magnet. What is the cause this damping? [1]

Answer:

As, the copper plates oscillate in the magnetic field between the two poles of the magnet, there is a continuous change of magnetic flux linked with the plate. Due to this, eddy currents are set up in copper plate which try to oppose the motion of the plate (according to Lenz’s law) and finally bring it to rest.

(iv) Define intensity of radiation in photon picture of light. [1]

Answer:

The intensity of radiation is defined as the rate of emitted energy from unit surface area through unit solid angle. Its SI unit is w/m^{2}sr^{-1}.

(v) The stopping potential depends upon what? [1]

Answer:

The stopping potential depends only on the frequency of incident light.

(vi) Define the activity of a given radioactive substance. Write its SI unit. [1]

Answer:

The rate of disintegration in a radioactive substance is known as its activity. SI unit is becqueral (Bq).

(vii) Two nuclei have mass numbers in the ratio 1 : 2. What is the ratio of their nuclear densities? [1]

Answer:

∴ f = \(\frac{m}{\frac{4}{3} \pi R_{o}^{3}}\)

Thus, f is independent of mass number.

∴ The ratio of density will be 1 : 1.

(viii) What happens to the width of depletion layer of a p-n junction when it is

(i) forward biased?

(ii) reverse biased? [1]

Answer:

(i) Width of depletion layer decreases in forward bias.

(ii) Width of depletion layer increases in reverse bias.

Section – B

Question 4.

A spherical capacitor consist of two concentric spherical conductors, held in position by suitable insulating supports. Show that the capacitance of this spherical capacitor is given by C = \(\frac{4 \pi \varepsilon_{0} r_{1} r_{2}}{r_{1}-r_{2}}\) where r_{1} and r_{2} are the radii of outer and inner spheres respectively. [1½]

Answer:

As is clear from figure +Q charge spreads uniformly on inner surface of outer sphere of radius rr. The induced charge – Q spreads uniformly on the outer surface of inner sphere of radius r_{2}. The outer surface of outer sphere is earthed. Due to electrostatic shielding, E = 0 for r < r_{2} and E = 0, for r> r_{1}.

In the space between the two spheres, electric intensity E exists as shown. Potential difference beween the two spheres,

V = \(\frac{Q}{4 \pi \varepsilon_{0} r_{2}}-\frac{Q}{4 \pi \varepsilon_{0} r_{1}}\)

= \(\frac{Q}{4 \pi \varepsilon_{0}}\left[\frac{1}{r_{2}}-\frac{1}{r_{1}}\right]\)

= \(\frac{Q}{4 \pi \varepsilon_{0}} \frac{\left(r_{1}-r_{2}\right)}{r_{1} r_{2}}\)

As C = \(\frac{Q}{V}\)

∴ C = \(\frac{Q 4 \pi \varepsilon_{0} r_{1} r_{2}}{Q\left(r_{1}-r_{2}\right)}\) = \(\frac{4 \pi \varepsilon_{0} r_{1} r_{2}}{r_{1}-r_{2}}\)

Question 5.

A spherical capacitor has an inner sphere of radius 12 cm and an outer sphere of radius 13 cm. The outer sphere is earthed and the inner sphere is given a charge of 2.5 μC. The space between the concentric spheres is filled with a liquid of dielectrc constant 32.

(a) Determine the capacitance of the capacitor.

(b) What is the potential of the inner sphere?

(c) Compare the capacitance of this capacitor with that of an isolated sphere of radius 12 cm. Explain why the latter is much smaller? [1½]

Answer:

Here, r_{α} 12cm = 12 × 10^{-2} m,

r_{b} = 13cm = 13 × 10^{-2} m,

q = 2.5μC = 2.5 × 10^{-6} C,ε_{r} = 32

(a) C = ? As C = 4πε_{0}\(\frac{\varepsilon_{r} r_{a} r_{b}}{r_{b}-r_{a}}\)

= \(\frac{1}{9 \times 10^{9}}\)\(\frac{32 \times 12 \times 10^{-2} \times 13 \times 10^{-2}}{(13-12) 10^{-2}}\)

= 5.5 × 10^{-9} farad

(b) V = ?

V = \(\frac{q}{C}\) = \(\frac{2.5 \times 10^{-6}}{5.5 \times 10^{-9}}\)

= 4.5 × 10^{2} volt

(e) Capacity of an isolated sphere of radius

R= 12 × 10^{-2} mis

C’ = 4πε_{0}R = \(\frac{1}{9 \times 10^{9}}\) × 12 × 10^{-2}

= 1.33 × 10_{-2} farad

The capacity of an isolated sphere is much smaller because in a capacitor, outer sphere is earthed, potential difference decreases and capacitance increases.

Question 6.

Derive an expression for the current density of a conductor in terms of the drift speed of electrons. [1½]

Answer:

Let potential difference V is applied across a conductor of length l and hence an electric field E produced inside the conductor.

∴ E = \(\frac{V}{l}\) ………………. (i)

Let n = number density of free electrons

A = cross-sectional area of conductor

e = electrons charge.

∴ Number of free electrons present in length l of conductor = nAl

Total charge contained in length l which can contribute in current,

q = (nAl)e …………….. (i)

The time taken by free electron to cross the length l of conductor is

t = l/υ_{d} ………………… (ii)

where, υ_{d} = drift speed of electron.

∴ Current through the conductor [from Eqs. (i) (ii)]

I = q/t

I = \(\frac{(n A l) e}{t}=\frac{(n A l) e}{\left(l / v_{d}\right)}\)

= neAυ_{d}

∴ Current density

(J) = \(\frac{I}{A}=\frac{n e A v_{d}}{A}\)

= neυ_{d}

∴ = neυ_{d} ,i.e., J ∝ υ_{d} Thus, current density of conductor is proportional to drift speed.

Question 7.

The network PQRS, shown in the circuit diagram, has the batteries of 4 V and 5 V and negligible internal resistance. A milliammeter of 20 Ω resistance is connected between P and R. Calculate the reading in the milliammeter. [1½]

Answer:

The given diagram is shown below Applying Kirchhoffs second law to the loop PRSP,

-I_{3} × 20 – I_{2} × 200 + 5 = 0

4I_{3} + 40I_{2} = 1 ……………. (i)

For loop PRQP,

-20I_{3} – 60I_{1} + 4 = 0

5I_{3} + 15I_{1} = 1 …………….. (ii)

Applying Kirchhoffs first law,

On solving, we get

I_{3} = \(\frac{11}{172}\) A = \(\frac{11000}{172}\)mA

I_{2} = \(\frac{4000}{215}\)mA, I_{1} = \(\frac{39000}{860}\) mA

∴ The reading in the milliammeter will be \(\frac{11000}{172}\) mA

Question 8.

Drive an expression for the mutual inductance of two long coaxial solenoids of same length wound one over the other. [1½]

Answer:

Mutual Inductance between Two co-axial Solenoids

Let, in solenoid S_{1} current flow is I_{1}, due to which magnetic flux induced which is linked with the coil S_{2}. The magnetic flux (Φ_{21}) is proportional to the current I_{1} then :

Φ_{21} ∝ I_{1} ……………. (1)

∴ Φ_{21} = M_{21}I_{1} …………….. (2)

Where M_{21} = coefficient of mutual inductance Magnetic field due to current in solenoid S_{1}

B_{1} = μ_{0} n_{1}I_{1} = \(\frac{\mu_{0} N_{1}}{l}\)I_{1} ……………. (3)

Since S_{1} solenoid is wounded on S_{2} solenoid, therefore magnetic flux linked with each turns of solenoid ‘S_{2}’

= B_{1} × Area of cross section

Φ_{21} = \(\frac{\mu_{0} N_{1} I_{1}}{l}\)A …………….. (4)

Total flux linked with the solenoid ‘S_{2}’

Φ_{21} = Φ_{21} × Total number of turns in ‘S_{2}’

= \(\frac{\mu_{0} N_{1} I_{1} A N_{2}}{l}\) = \(\frac{\mu_{0} N_{1} N_{2} I_{1} A}{l}\) ……………… (5)

From the equation (2) and (5),

M_{2}I_{1} = \(\frac{\mu_{0} N_{1} N_{2} I_{1} A}{l}\)

M_{21} = μ_{0} \(\frac{N_{1} N_{2}}{l}\)A ……………. (6)

M = M_{12} = M_{21} = \(\frac{\mu_{0} N_{1} N_{2}}{l}\)A

Question 9.

A rectangular coil of dimensions 0.10 m × 0.50 m has 2000 coils of wire. This coil is rotated at the rate of 2100 rotations per minute along the axis parallel to the long side of the coil with the magnetic field intensity of 0.1 T. What would be the maximum induced emf of the coil? What would be the instantaneous value of emf when the coil makes an angle of 30° with the field? [1½]

Answer:

Given : B = 0.1T;

A = 0.10 × 0.50 m^{2} = 0.05m^{2};

N = 2000;

Angular frequency,

n = \(\frac{2100}{60} \frac{\text { rotations }}{\text { second }}\) = 35 Hz

∴ Angular velocity,

ω = 2πn = 2 × \(\frac{22}{7}\) × 35 = 220 rads^{-1}

Thus, the maximum value of induced emf

ε_{0} = BANω

= 0.1 × 0.05 × 2000 × 220

= 2200V

The instantaneous value of emf

ε = ε_{0} sinωt = ε_{0} sinθ

The plane of the coil makes an angle of 30° with the magnetic field. Thus, the angle between the normal drawn on the plane of the coil and the field,

θ = 90° – 30° = 60°

∴ ε = 2200 sin 60°= 2200 × \(\frac{\sqrt{3}}{2}\)

= 1100 × 1.732 = 1905.2 V

Question 10.

(a) The angle subtended at the eye by an object is equal to the angle subtended at the eye by the virtual image produced by a magnifying glass. In what sense then does a magnifying glass provide angular magnification?

(b) In viewing through a magnifying glass, one usually positions one’s eyes very close to the lens. Does angular magnification change if the eye is moved back?

(c) Magnifying power of a simple microscope is inversely proportional to the focal length of the lens. What then stops us from using a convex lens of smaller and smaller focal length and achieving greater and greater magnifying power? [1½]

Answer:

(a) When viewing the object without magnifying lens, the distance of the object should not be less than the distance of near point (i.e., 25 cm). But, when viewing by magnifying lens, the object can be placed comparatively at some lens distance so that the final image becomes forming at least distance of distinct vision.

(b) While viewing by the magnifying lens, the observer keeps his eyes close to the lens because by this action the visual angle subtended by the image at eye increases and the object appears of larger size. On displacing the eye away from the lens, magnification is changed because by this action the visual angle subtended by the image at eye will become less than that at the lens. Therefore the magnification will be reduced.

(c) If we reduce the focal length of the lens, it will become thicker and more over the process of preparing such lenses is not easy.

Question 11.

How does the power of a convex lens vary, if the incident red light is replaced by violet light? [1½]

Answer:

According to Lens Maker’s formula

P = \(\frac{1}{f}\) = (μ – 1) (\(\frac{1}{R_{1}}-\frac{1}{R_{2}}\))

μ_{violet} > μ_{red}

∴ Power of the lens will be increased.

Question 12.

An object of 3 cm height is placed at a distance of 60 cm from a convex mirror of focal length 30 cm. Find the nature, position and size of the image formed. [1½]

Answer:

Given, length of object O = + 3 cm

u = – 60 cm, f = + 30 cm

1/f = 1/υ + 1/u [mirror formula]

1/30 = 1/υ + 1/(—60)

or \(\frac{1}{v}=\frac{1}{30}+\frac{1}{60}=\frac{2+1}{60}\) ⇒ υ = 20 cm

∴ \(\frac{I}{O}=-\frac{v}{u}\) ⇒ \(\frac{I}{(+3)}=-\frac{(+20)}{(-60)}\)

I = 1 cm

So, the virtual, erect and diminished image will be formed on the other side of the mirror.

Question 13.

A fluorescent tube of 40 watt provides more light than that of filament bulb of 40 watt. Why? [1½]

Answer:

Filament bulb is a thermal source of light from which ultraviolet and infrared radiation are emitted besides.

While inside the fluorescent tube ultraviolet radiation is emitted with light due to gas discharge. Ultraviolet radiation is modified into visible light by photo luminescent. So fluorescent tube provides more light than filament bulb of same weightage.

Question 14.

Derive the expression for the law of radioactive decay of a given sample having initially N_{0} decaying to the number N present at any subsequent time t.

Plot a graph showing the variation of the number of nuclei versus the time lapsed. Mark a point on the plot in terms of T_{1/2} value the number present N = N_{0}/16. [1½]

Answer:

Let N be the number of undecayed nuclei in the sample at time t and ΔN nuclei undergoing decay in time Δt.

Then \(\frac{-\Delta N}{\Delta t}\) ∝ N, \(\frac{-\Delta N}{\Delta t}\) = λN

where, λ is disintegration constant. The rate of change in N in time Δt → 0, can be expressed as

\(\frac{d N}{N}\) = – λdt

On integrating both sides

\(\int_{N_{0}}^{N} \frac{d N}{N}=-\int_{0}^{t} \lambda d t\)

where, N_{0} is in initial undecayed nuclei.

In \(\frac{N}{N_{0}}\) = -λt, N = N_{0}e^{-λt}

Question 15.

Given the mass of iron nucleus as 55.85 u and A = 56. Find the nuclear density. [1½]

Answer:

Nuclear density, ρ = \(\frac{3}{4} \frac{m}{\pi R^{3}}\)

\(\frac{3}{4}\) × \(\frac{9.27 \times 10^{-26}}{3.14 \times\left(1.2 \times 10^{-15}\right)^{3}}\) × \(\frac{1}{56}\)

\(\frac{27.81 \times 10^{-26}}{1215.41 \times 10^{-45}}\)

= 0.02288 × 10^{9}

= 2.288 × 10^{-17} kg m^{-3}

Question 16.

A long straight wire of circular cross section of radius ‘R’ carries a steady current ‘I’. The current is uniformly distributed across the cross section. Apply Ampere’s circuital law to calculate the magnetic field at a point P in the region for (i) r < R and (ii) r > R. [3]

Or

(a) Define the current sensitivity of a galvanometer.

(b) The coil area of a galvanometer is 1.6 × 10^{4} m^{2}. It consists of 200 turns of a wire and is in a magnetic field of 0.2 T. The restoring torque constant of the suspension fibre is 10^{-6} Nm per degree. Calculate the maximum current that can be measured by the galvanometer. If the scale can accomodate 30° deflection. [3]

Answer:

Consider and infinite long thick wire of radius R with asix XY. Let I be the current flowing through the wire.

When the point P lies outside the wire:

Let r be the perpendicular distance of point P from the axis of the cylinder,

where r > R. Here \(\vec{B}\) and \(d \vec{l}\) are acting in the same direction. Applying Ampere’s circuital law, we have

\(\oint \vec{B} \cdot d \vec{l}\) = μ_{0}I

\(\oint B \cdot d l \cos \theta\) = μ_{0}I

⇒ \(B \int_{o}^{2 \pi r} d l\) = μ_{0}I [ cos 0 = 1]

⇒ B.2πr = μ_{0}I

B = \(\frac{\mu_{0} I}{2 \pi r}\)

∴ B ∝ \(\frac{1}{r}\)

When the point P lies inside the wire: here r ≤ R. We have two possibilities. According to ampere circuital law,

(i) Whenever the current flows only through the surface of the wire, B = 0

B = 0 as current in the closed path will be zero.

(ii) Whenever in the case when current is uniformaly distributed through the cross section of conductor, current through the closed path will be :

I’ – Current per unit area of the wire × area of the circle of radius R

Question 17.

Draw a ray diagram to show refraction of a ray of monochromatic light passing through a glass prism. Deduce the expression for the refractive index of glass in terms of angle of prism and angle of minimum deviation. [3]

Or

Draw a labelled ray diagram to obtain the real image formed by an astronomical telescope in normal adjustment position. Define its magnifying power. [3]

Answer:

The angle between the directions of incident ray and emergent ray is known as the angle of deviation.

In ΔO’QR, Exterior angle = δ

and opposite interior angle

∠O’QR = (i_{1} – r_{1})

and ∠O’RQ = (i_{2} – r_{2})

∴ According to Geometry

δ = ∠O’QR + ∠O’RQ

or δ = (i_{1} – r_{1}).+ (i_{2} – r_{2})

or δ = (i_{1} + i_{2}) – (r_{1} + r_{2}) …………… (1)

In ΔAQR,

∠AQR = ∠AQO – ∠OQR

= 90° – r_{1}

Similarly,∠ARQ = 90° – r_{2}

The sum of all three interior angles of a triangle is equal to 180°.

∴ ∠QAR + ∠AQR + ∠ARQ = 180°

or A + (90 – r_{1}) + (90 – r_{2}) = 180°

A – (r_{1} + r_{2}) + 180°= 180°

or A – (r_{1} + r_{2}) = 0

∴ A = (r_{1} + r_{2}) ……………… (2)

From equation (1) and (2),

δ = (i_{1} + i_{1}) – A

or (i_{1} + i_{2}) = (δ + A)

Angle of Minimum Deviation (δ_{m}):

If the graph is plotted between different incident angles and corresponding angles of deviation, the obtained curve will be as shown in figure below.

δ = i + i_{2} – A

i. e., δ remains the same as i and i_{2} are interchanged. Physically it means that the path of the ray in figure shown above can be traced back resulting the same angle of deviation.

“The minimum value of the angle of deviation suffered by a ray on passing through a prism is called the angle of minimum deviation and is denoted by δ_{m}.”

When i_{1} = i_{2}, then r_{1} = r_{2}.

Hence in the state of minimum deviation :

(i) Angle of incidence is equal to angle of emergence.

According to Snell’s law, the refractive index of the substance of the prism.

n = \(\frac{\sin i}{\sin r}\)

In ΔQOR,

∠r + ∠r + ∠O = 180°

or 2r + ∠O = 180°

In AQOR,

∠AQO = ∠ARO = 90°

∴ ∠A + ∠O = 180°

(Because the sum of all four interior angles of a quadrilateral is equal to four right angles.)

On comparing equations (2) and (3),

2r + ∠O = A + ∠O

or 2r = A

∴ r = \(\frac{A}{2}\)

Now in ΔO’QR, the exterior angle,

δ_{m} =(i – r) + (i – r) = 2i – 2r

or δ_{m} = 2i – A [from equation (4)]

or 2i = A + δ_{m}

∴ i = \(\frac{A+\delta_{m}}{2}\) ……………… (5)

Now substituting the values of i and r in equation (1)

n = \(\frac{\sin \left(\frac{A+\delta_{m}}{2}\right)}{\sin \left(\frac{A}{2}\right)}\)

This formula is also known as “Prism Formula”.

Question 18.

(a) A proton and a deuteron are accelerated through the same accelerating potential. Which one of the two has:

(i) greater value of de-Broglie wavelength associated with it, and

(ii) less momentum?

Give reasons to justify your answer.

(b) What is the de-Broglie wavelength associated with an electron accelerated through a potential of 100 volts. [3]

Or

Write three basic properties of photons which are used to obtain Einstein’s photoelectric equation. Use this equation to draw a plot of maximum kinetic energy of the electrons emitted versus frequency of incident radiation. [3]

Answer:

(a) De-Broglie wavelength is given by

λ = \(\frac{h}{\sqrt{2 m V_{0} q}}\) λ ∝ \(\frac{1}{\sqrt{m}}\)

[V_{0} and q are same, because proton and deuteron have been accelerated by same potential and have same charge]

Since, mass of proton is less as compared to a deuteron. So, it will have higher value of de-Broglie wavelength associated with it.

(ii) de-Broglie wavelength is given by

λ = \(\frac{h}{p}\) P = \(\frac{h}{\lambda}\)

As λ_{d} < λ_{P}

So P_{d}

(b) λ = \(\frac{h}{p}=\frac{1.227}{\sqrt{V}}\) nm

= \(\frac{1.227}{\sqrt{100}}\) nm = 0.123 nm

Question 19.

(a) Use Gauss’ law to derive the expression for the electric field (E) due to a straight uniformly charged infinite line of charge density λ C/m.

(b) Draw a graph to show the variation of E with perpendicular distance r from the line of charge.

(c) Find the work done in bringing a charge q from perpendicular distance r_{1} to

r_{2}(r_{2}>r_{1}). [4]

Or

Obtain the formula for the electric field due to a long thin wire of uniform linear charge density λ without using Gauss’s law. [4]

Answer:

Electric Field Intensity due to an Infinite Long Line Charge from Gauss’s Law

Let the linear charge density is λ and the intensity of electric field at point P is to be determined. To apply Gauss’s law, we have to decide the direction of electric field at P. For this purpose let us consider two small elements of length dl at A and B at equal distance from O as shown in the figure. The intensities of electric field due to these elements at P, dE_{1} and dE_{2} are equal in magnitude. If the electric fields are resolved in two normal components, then the components along OP provide the electric field \(\vec{E}\) along OP and normal components dE_{1} sin θ and dE_{2} sinθ cancel out each other. Similar will be the result for other pairs of length elements considered.

Thus the direction of electric field at P will be in OP direction i.e., normal to linear charge.

Calculation of Electric Field: For this purpose, we draw a cylindrical Gaussian surface of radius r and axis coinciding the linear charge AB. Point P lies on the curved surface of the cylinder. The enclosed charge Σq = λl,

where l is the length of the cylinder. Therefore from Gauss’s law,

Now equating equation (1) & (2), we have,

E 2πrl = \(\frac{\lambda l}{\varepsilon_{0}}\)

∴ E = \(\frac{1}{2 \pi \varepsilon_{0}} \frac{\lambda}{r}\)

or E = \(\frac{1}{4 \pi \varepsilon_{0}} \frac{2 \lambda}{r}\) ……………. (3)

In vector form

\(\vec{E}=\frac{1}{4 \pi \varepsilon_{0}} \frac{2 \lambda}{r} \hat{r}\)

where r̂ = unit vector in direction OP i.e., in perpendicular to linear charge.

From equation (3),

E ∝ \(\frac{1}{r}\)

(b) electric field (E) due to the linear charge is inversely proportional to the distance (r) from the linear change. The variation of electric field (E) with distance (r) is shown in figure.

Question 20.

(i) Explain with the help of suitable diagram, the two processes which occur during the formations of a p-n junction diode. Hence, define the terms (i) depletion region and (ii) potential barrier.

(ii) Draw a circuit diagram of a p-n junction diode under forward bias and explain its working. [4]

Or

(i) Distinguish between re-type and p-type semiconductors on the basis of energy band diagrams.

(ii) With what considerations in view, a photodiode is fibricated? [4]

Answer:

(i)

As the P-N junction starts, due to the concentration gradient across P-and N-sides, holes diffuse form P-side to N side (P →N) and electrons diffuse form N-side to P-side (N → P). This motion of charge gives rise to diffusion current across the junction. As the holes continue to diffuse,a layer of negative charge (or negative space-charge region) on the P-side of the junction is developed. This space-charge region of either side of the junction together is known as depletion region.

The loss of electrons for the N-region and the gain of electron by the P-region causes a difference of potential across the junction of two regions. The polarity of this potential is such as to oppose further flow of carriers so that a condition of equilibrium exists.

This potential is known as potential barrier.

(ii) Here the applied voltage V on the diode is changed by the potential divider arrangement which can be easily read by the voltmeter connected in parallel to the diode. The current I flowing in the diode related to various values of voltages are noted by a milliammeter. In this way, the curve obtained for different values of V and I is shown in the figure. (14.17), which is also called as the P-N junction diode forward bias characteristic curve.

## Leave a Reply