Students must start practicing the questions from RBSE 12th Physics Model Papers Set 6 with Answers in English Medium provided here.

## RBSE Class 12 Physics Model Paper Set 6 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.

Multiple Choice Questions

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

(i) If \(\oint_{S} \vec{E} \cdot \overrightarrow{d s}\) =0 over a surface, then : [1]

(a) the electric field inside the surface and on it is zero.

(b) the electric field inside the surface is necessarily uniform.

(c) all charges must necessarily be outside the surface.

(d) all of these.

Answer:

(d) all of these.

(ii) The electric potential at a point in free space due to a charge Q Coulomb is Q x 10^{11} V.

The electric field at that point is : [1]

(a) 12πε_{0} Q × 10^{22}Vm^{-1}

(b) 4πε_{0} Q × 10^{22}Vm^{-1}

(c) 12πε_{0} Q × 10^{20}Vm^{-1}

(d) 4πε_{0} Q × 10^{20}Vm^{-1}

Answer:

(b) 4πε_{0} Q × 10^{22}Vm^{-1}

(iii) In the series combination of two or more than two resistances: [1]

(a) the current through each resistance is same.

(b) the voltage through each resistance is same.

(c) neither current nor voltage through each resistance is same.

(d) both current and voltage through each resistance are same.

Answer:

(a) the current through each resistance is same.

(iv) Current flows through uniform, square frames as shown in the figure. In which case is the magnetic field at the center of the frame not zero? [1]

Answer:

(iv) (a)

(v) The polarity of induced emf is given by: [1]

(a) Ampere’s circuital law

(b) Biot-Savart law

(c) Lenz’s law

(d) Fleming’s right-hand rule

Answer:

(c) Lenz’s law

(vi) ………………………….. frequency of light used to start emission of electrons from any metal. [1]

(a) isotopes

(b) isotones

(c) isomers

(d) isobars

Answer:

(c) isomers

(vi) ………………………….. frequency of light used to start emission of electrons from any metal. [1]

(a) medium

(b) low

(c) threshold

(d) radio

Answer:

(d) radio

(viii) Temperature coefficient of resistance of semiconductor is : [1]

(a) zero

(b) constant

(c) positive

(d) negative

Answer:

(d) negative

(ix) The inverter is ……………………… . [1]

(a) NOT gate

(b) OR gate

(c) AND gate

(d) None of these.

Answer:

(a) NOT gate

Question 2.

Fill in the blanks:

(i) Energy stored in charged capacitor is ……………………….. . [1]

Answer:

U = \(\frac{1}{2} \) CV^{2}

(ii) The reciprocal of …………………. of a conductor is known as conductivity. [1]

Answer:

resistivity

(iii) The core of transformer is laminated to prevent ……………………… being produced in the core. [1]

Answer:

eddy current

(iv) In n-type semiconductor, the semiconductor is dopped with …………………… . [1]

Answer:

pentavalent impurity.

Question 3.

Give the answer to the following questions in one line (i to viii):

(i) Show variation of resistivity of Si with temperature in graph. [1]

Answer:

(ii) When a charge q is moving in the presence of electric (E) and magnetic (B) fields which are perpendicular to each other and also perpendicular to the velocity v of the particle, write the relation expressing v in terms of E and B. 1

Answer:

F_{lorentz} = F_{electric} + F_{magentic}

= qE + q(v × B) = q [E + (u × B)]

(iii) The electric current flowing in a wire in the direction from B to A is decreasing. Find out the direction of the induced current in the metallic loop kept near the wire as shown in the figure. [1]

Answer:

According to Lenz’s law, the direction of induced current will oppose the cause of its production. So, the current in loop will induce in such a way that it will support the current flowing in the wire, i.e., in the same direction. So, the direction of current in the loop will be clockwise.

(iv) Show the variation of photoelectric current with collector plate potential for different frequencies but same intensity of incident radiation. [1]

Answer:

The variation of photoelectric current with collector plate potential for different plate frequencies is shown as below:

(v) In photoelectric effect, why should the photoelectric current increase as the intensity of monochromatic radiation incident on a photosensitive surface is increased? Explain. [1]

Answer:

The photoelectric effect is a’ one-photon-one electron phenomenon. Therefore, when the intensity of radiation incident on the surface increases, the number of photons per unit area unit time increases (since the intensity of incident radiation number of photons). Hence, the photoelectrons. ejected will be large, which in turn, will contribute to the increase in photoelectric current.

(vi) When four hydrogen nuclei combine to form a helium nucleus estimate the amount of energy in MeV released in this process of fusion (Neglect the masses of electrons and neutrons). [1]

Given:

(i) mass of ^{11}H = 1.007825 u

(ii) mass of helium nucleus = 4.002603u, 1u = 931 MeV/c^{2} [1]

Answer:

Energy released = Δm x 931 MeV

E.m= Δm(^{11}H)-m(^{42}He)

Energy released

Q = {4.m(^{11}H) — m(^{42}He)} × 931] MeV

=[4 × 1.007825— 4002603] × 931 MeV

= 26.72MeV

(vii) Why cannot we take one slab of p-type semiconductor and physically join it to another slab of n-type semiconductor to get p-n junction? [1]

Answer:

In this way, continuous contact cannot be produced at atomic level and junction will behave as a discontinuity for the flowing charge carrier.

(viii) What is the function of a photodiode? [1]

Answer:

A photodiode is a special purpose p-n junction diode fabricated with a transparent window to allow light to fall on the diode.

Section-B

Question 4.

Find out the expression for the potential energy of a system of three charges q_{1}q_{2} and q_{3} located at r_{1}, r_{2} and r_{3} with respect to the common origin O. [1½]

Answer:

Let three-point charges q_{1},q_{2} and q_{3} have position vectors r_{1}, r_{2,} and r_{3}, respectively.

Potential energy of the charges q_{1} and q_{2},

U_{12} = \(\frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{q_{1} q_{2}}{\left|\mathbf{r}_{12}\right|}=\frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{q_{1} q_{2}}{\left|\mathbf{r}_{2}-\mathbf{r}_{1}\right|}\)

Similarly, U_{23} = \(\frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{q_{2} q_{3}}{\left|\mathbf{r}_{3}-\mathbf{r}_{2}\right|}\)

⇒ U_{31} = \(\frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{q_{3} q_{1}}{\left|\mathbf{r}_{3}-\mathbf{r}_{2}\right|}\)

∴ Net potential energy of the system,

U = U_{12} + U_{23} + U_{31}

Question 5.

Two parallel plate capacitors X and Y have the same area of plates and same separation between them, X has air between the plates while Y contains a dielectric medium of ε_{r} = 4.[1½]

(i) Calculate the capacitance of each capacitor, if equivalent capacitance of the combination is 4 µF.

(ii) Calculate the potential difference between the plates of X and Y.

(iii) Estimate the ratio of electrostatic energy stored in X and Y. [1½]

Answer:

Let the capacitance of X be C and capacitance of Y = ε_{r}C = 4C

[ε_{r}=4]

(a) Equwalent capacitance = \(\frac{C \times 4 C}{C+4 C}\)

(X and Y are in series)

= \(\frac{4 C^{2}}{5 C} \Rightarrow \frac{4 C}{5}\)

and it is given that 4C/5= 4µF

So, 4C = 20 µF = capacitance of Y

Capacitance of X = C = 20/4 = 5µF

(ii) Charge stored in the capacitor is given by

q=CV=\(\frac{4 C}{5}\) x 15= \(\frac{4 \times 5}{5}\) x 15=60µC

q = CV = \(\frac{4 C}{5}\) x 15 = \(\frac{4 \times 5}{5}\) x 15 = 60µC

Now, let the potential difference between plates of capacitors X and Y are V_{x} and V_{y} respectively.

so, V_{x} = \(\frac{q}{C_{x}}=\frac{60}{5}\) =12V

and V_{y} = \(\frac{q}{C_{y}}=\frac{60}{20}\) =3V

(iii) Electrostatic energy stored in capacitance

X(E_{x}) = \(\frac{1}{2}\) CV^{2}_{x} ………………………… (i)

Similarly for Y, E_{y}, = \(\frac{1}{2}\)4CV_{y2} ……………………… (ii)

From Eqs. (i) and (ii), we get

Ratio = \(\frac{E_{x}}{E_{y}}=\frac{2}{\frac{1}{2} 4 C V_{y}^{2}}=\frac{V_{x}^{2}}{4 V_{y}^{2}}\)

= \(\frac{12 \times 12}{4 \times 3 \times 3}\) = 4:1

Question 6.

Derive an expression for drift velocity of electrons in a conductor. Hence, deduce Ohm’s law.[1½]

Answer:

(i) Let υ_{d} be the drift velocity.

Electric field produced inside the wire is

E=V/l …………………………… (i)

Force on an electron, a = — Ee

Acceleration of each electron a= -Ee/m

[ from Newton’s law, a = F/m]

where m is mass of electron.

Velocity created due to this acceleration = \(\frac{E e}{m} \tau\)

where τ is the time span between two consecutive collisions. This ultimately becomes the drift velocity in steady-state.

So, υ_{d} = \(\frac{E e}{m} \tau=\frac{e}{m} \tau \times \frac{V}{l}\) [from Eq. (i)]

We know that current in the conductor i = neAυ_{d} (where, n is number of free electrons in a conductor per unit volume)

i = neA × \(\frac{e}{m} \tau \frac{V}{l}\)

⇒ i = \(\frac{n e^{2} A \tau V}{m l}\)

= i=V/R [ R=ml/ne^{2}Aτ]

i ∝ V

This is Ohm’s law.

Question 7.

Resistance of a wire at 22°C is 7.5 Ω and at 175°C is 15.7Ω Calculate the value of temperature coefficient of resistance of the substance of the resistor.[1½]

Answer:

Given t_{1} = 22°C; R_{1} = 7.5Ω;

t_{2} = 175°C; R_{2} = 15.7 Ω; α =?

The value of α,

α = \(\frac{R_{2}-R_{1}}{R_{1} t_{2}-R_{2} t_{2}}\)

∴ α = \(\frac{15.7-7.5}{7.5 \times 175-15.7 \times 22}\)

= \(\frac{8.2}{1312.5-345.4}=\frac{8.2}{967.1}\)

or α = 0.00847°C^{-1}

Or α = 8.47 × 10^{-3}°C^{-1}

Question 8.

Define the term self-inductance of a solenoid. Obtain the expression for the self-induction of a long solenoid. [1½]

Answer:

When current in a coil changes, the magnetic flux linked with the coil also changes, and hence emf is induced in the coil. This phenomenon is known as self-induction. According to Lenz’s law, the induced emf opposes the change in magnetic flux. Self Inductance of a Long Solenoid. Let us consider a solenoid of N turns, A cross-sectional area, and length l Let I be the current that flows through it. Magnetic field produced inside the solenoid,

B=µ_{0}nl

where, n = \(\frac{N}{l}\)

Magnetic flux linked with solenoid,

Φ_{B} = N(BA)

Φ_{B} = \(\frac{\mu_{0} N I}{l} \times N A\)

⇒ Φ_{B} = \(\frac{\mu_{0} N^{2} A}{l} I\)

If L is coefficient of the self inductance of solenoid, then

L= \(\frac{\mu_{0} N^{2} A}{l}\)

Question 9.

A wheel with 100 metallic spokes each 0.5 m long is rotated with a speed of 120 rev/min in a plane normal to the horizontal component of the Earth’s magnetic field. If the resultant magnetic field at that place is 4 x 10^{-4} T and the angle of dip at the place is 30°, find the emf induced between the axle and the rim of the wheel.[1½]

Answer:

Spokes = 100, length = 0.5 m and

ω=120 re/min

B_{e} =4 x 10^{-4}T, θ = 30°

= 543 × 10^{-4} V

Question 10.

Why does bluish colour predominate in a clear sky? [1½]

Answer:

The blue colour of the sky is due to the scattering of sunlight by the molecules of the atmosphere. According to Rayleigh’s law of scattering, the intensity of scattered light, I ∝ \(\frac{1}{\lambda^{4}}\) blue colour having the short wavelength in the visible spectrum scatter the most. When we look at the sky, the scattered light enters our eyes and this light contains blue colour in a large proportion, so the sky appears blue.

Question 11.

When monochromatic light travels from one medium to another its wavelength changes but frequency remains the same. Explain.[1½]

Answer:

If υ_{1} and υ_{2} denote the velocity of light in medium 1 and medium 2 respectively and λ _{1} and λ_{2} denote the wavelength of light in medium 1 and medium 2. Thus.

\(\frac{\lambda_{1}}{\lambda_{2}}=\frac{v_{1}}{v_{2}}\)

Or \(\frac{v_{1}}{\lambda_{1}}=\frac{v_{2}}{\lambda_{2}}\)

The above equation implies that when a wave gets refracted into denser medium (υ_{1} > υ_{2}). the wavelength and

the speed of propagation decreases but the frequency ν = \(\frac{v}{\lambda}\) remains the same.

Question 12.

A converging lens has a focal length of 20 cm in air. It is made of.a material of refractive index 1.6.

It is immersed in a liquid of refractive index 1.3. Calculate its new focal length.[1½]

Answer:

Given, f_{1} = + 20 cm

_{a} n_{g} = 16, _{a}n_{w} = 13

⇒ _{w} n_{g} = \(\frac{{ }_{a} n_{g}}{n_{w}}=\frac{1.6}{13} \)

Using lens maker’s formula water) for converging lens,

\(\frac{1}{f_{2}}=\left({ }_{w} n_{g}-1\right)\left(\frac{1}{R_{1}}+\frac{1}{R_{2}}\right)\) …………….. (i)

In air , \(\frac{1}{f_{1}}=\left({ }_{a} n_{g}-1\right)\left(\frac{1}{R_{1}}+\frac{1}{R_{2}}\right)\)

Dividing Eq. (ii) by Eq. (i), we get

\(\frac{f_{2}}{f_{1}}=\frac{\left({ }_{a} n_{g}-1\right)}{\left({ }_{w} n_{g}-1\right)}=\frac{(16-1)}{\left(\frac{16}{13}-1\right)}=\frac{0.6 \times 13}{03}\)

\(\frac{f_{2}}{f_{1}}\) = 26

New focal length,

f_{2} = 2.6 x f_{1} = 2.6 x 20

f_{2} = 52cm

Question 13.

Define : (a) Spherical mirror formula (b) Refraction of light (c) Refractive index.[1½]

Answer:

(a) Spherical Mirror Formula:

This give relation between object distance υ, image distance ν, and the focal length f of a spherical mirror.

\(\frac{1}{u}+\frac{1}{v}=\frac{1}{f}\)

(b) Refraction of Light: It is the phenomenon of bending of light from its straight path when it passes at an

angle from one transparent medium to another.

(c) Refractive Index: Refractive index of a met him for a light of given wavelength may be defined as the ratio of the speed of light in vacuum to its speed in that medium

n = \(\frac{\text { Velocity of light in vacuum }}{\text { Velocity of light in medium }}=\frac{c}{v} \)

Question 14.

Deduce the expression, N = N_{0}e^{-λt} for the law of radioactive decay.[1½]

Answer:

Let the number of atoms present at any given time t is N and at the time t + Δt this number decreases N – ΔN to its value then, the rate of decay of atoms is – \(\frac{\Delta N}{\Delta t}\) Therefore, according to the law of Rutherford and Soddy.

If at the time Δt the nucleus ΔN will be disintegrated then the rate of disintegration,

\(\frac{-d N}{d t} \propto N\)

\(\frac{-d N}{d t}=\lambda N\)

or dN = -λNdt ……………………. (i)

Now, arranging equation (1) again,

Integrating both the sides, we have

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

In N – ln N_{0} = – λt

Or ln \(\frac{N}{N_{0}}\) = – λt

Taking the exponential on both the sides, we get

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

or

\(N=N_{0} e^{-\lambda t} \) ………………………………… (2)

Question 15.

The half-life of radium is 1600 year, with a sample of radium, how much time it will take to disintegrate 75% of its initial mass? [1½]

Answer:

Given, T = 1600 year, t =?

m = m_{0} – \(\frac{75}{100}\) m_{0} = \(\frac{25}{100}\) m_{0}

m = \(m_{0}\left(\frac{1}{2}\right)^{n} \)

∴ \(\frac{25}{100} m_{0}=m_{0}\left(\frac{1}{2}\right)^{n}\)

⇒ \(\frac{25}{100}=\left(\frac{1}{2}\right)^{n}\)

⇒ \(\frac{1}{4}=\left(\frac{1}{2}\right)^{n}\)

⇒ \(\left(\frac{1}{2}\right)^{2}=\left(\frac{1}{2}\right)^{n}\)

∴ n = 2

So, total time t = nT = 2 x 1600 = 3200 year

Section-C

Question 16.

Draw a schematic sketch of a cyclotron. State its working principle. Describe briefly, now it is used to accelerate charged particles? [3]

Or

State Biot-Savart’s law expressing it in the vector form, use it to obtain the expression for the magnetic field at an axial point distanced from the centre of a circular coil of radius a carrying current I. Also, find the ratio of the magnitudes of the magnetic field of this coil at the centre and at an axial point for which d = a\(\sqrt{3}\).[3]

Answer:

Cyclotron

The cyclotron was invented by E.O. Lawrence and M.S. Livingston.

Principle:

Cyclotron works on the principle that a charged particle moving normal to a magnetic field experiences magnetic Lorentz force due to which the particle moves in a circular path. Cyclotron uses the fact that the frequency of revolution of the charged particle in a magnetic field is independent of its energy. Here, the magnetic field moves the charged particles in a circular path and electric field increases their energy in each frequency.

Construction:

Cyclotron consists of two-semicircular disc-like metal containers D_{1} and D_{2}, which are called dees. The cyclotron uses both electric and magnetic fields in combination to increase the energy of charged particles. As the fields are perpendicular to each other they are called ‘crossed fields’.

Inside the metal boxes the particle is shielded and is not acted on by the electric field. The magnetic field, however, acts on the particle and makes it go around in a circular path inside a dee. Every time the particle moves from one dee to another. It is acted upon by the electric field. The whole assembly is evacuated to minimise collisions between the ions and the air molecules. A high-frequency alternating voltage is applied to the dees. Magnetic field acts perpendicular to the plane of dees.

Working: A radio frequency oscillator generates approximately 10^{4} V voltage and 10^{6} Hz frequency alternating voltage. Positive ions or positively charged particles (as protons) are released at the center P. They move in a semi-circular path in one of the dees and arrive in the gap between the dees in a time interval T/2; where T is the period of revolution.

As shown in the figure when any charged particle enters the dees from the ion source then due to the perpendicular magnetic field it does motion in a circular path. Suppose at any instant t, D_{1} is positive and D_{2} is negative and positive ion completes its semi-circular path in D_{1} and moves towards D_{2} and its energy increases by qV. The time taken by the particle to do motion in D_{1} is T/2. The alternating voltage in a cyclotron is so adjusted that as any charged particle completes its semicircular path in a dee; the polar effect of the alternating voltage changes. The particle also does motion in a circular path in D_{2} due to the perpendicular magnetic field.

In this way a charged particle does two times motion in a magnetic field in one cycle and its energy increases by 2qV. As energy increases, the radius of the circular path also increases, hence the time period is kept constant.

The frequency of the applied voltage is so adjusted that the polarity of the dees is reversed in the same time that it takes the ions to complete one half of the revolution. The requirement is called resonance condition. The ions are repeatedly accelerated across the dees until they have the required energy to have a radius approximately that of the dees. They are then deflected by a magnetic field and leave out the system via an exit slit.

Question 17.

Define power of a lens. Write its units. Deduce the relation \(\frac{1}{f}=\frac{1}{f_{1}}+\frac{1}{f_{2}}\) for two thin lenses kept in contact coaxially.[3]

Or

(a) With the help of a ray diagram show how a concave mirror is used to obtain an eract and magnified image of an object.

(b) Using the above ray diagram, obtain the mirror formula and the expression for linear magnification. [3

Answer:

“Power of a lens can be defined as the deviation produced by the lens in the rays coming parallel to principal axis at a unit distance from the principal axis.” S.I. unit of power of lens is Dioptre.

Two lenses L_{1}, L_{2} of respective focal lengths f_{1} and f_{2} are kept in contact.

A point object O is situated at a distance u in front of the combination and the final image is forming at I.

The intermediate image I’, formed by first lens, behaves as virtual image for the second lens.

From the lens formula for first lens,

\(\frac{1}{v^{\prime}}-\frac{1}{u}=\frac{1}{f_{1}}\) …………………….. (1)

For second lens, \(\frac{1}{v}-\frac{1}{v^{\prime}}=\frac{1}{f_{2}}\) ………………………… (2)

On adding equation (1) and (2), we get,

\(\frac{1}{v^{\prime}}-\frac{1}{u}+\frac{1}{v}-\frac{1}{v^{\prime}}=\frac{1}{f_{1}}+\frac{1}{f_{2}}\)

Or \(\frac{1}{v}-\frac{1}{u}=\frac{1}{f_{1}}+\frac{1}{f_{2}}\) …………………. (3)

If the focal length of combination of lenses be f, then

\(\frac{1}{f}=\frac{1}{f_{1}}+\frac{1}{f_{2}} \)

Question 18.

Monochromatic light of wavelength 632.8 nm is produced by a helium-neon leaser. The power emitted is 9.42 mW.

(a) Find the energy and momentum of each photon in the light beam.

(b) How many photons per second, on the average, arrive at a target irradiated by this beam? (Assume the beam to have uniform cross-section which is less than the target area), and

(c) How fast does a hydrogen atom have to travel in order to have the same momentum as that of the photon? [3]

Or

The maximum kinetic energy of photoelectrons emitted from a photocathode is 0.46 eV when a light of 6000 Å wavelength is incident on it. If the wavelength of the incident light is reduced to 5000 A, then the energy increases to 0.875 eV.

Determine :

(i) Work function of photocathode surface;

(ii) Threshold wavelength, and

(iii) Planck’s constant. [3]

Answer:

Given, λ=632.8nm=632.8 × 10^{9}

m,P =9.42mW=9.42 x 10^{-3} W

(a) Energy of each photon,

E = \(\frac{h c}{\lambda}=\frac{6.63 \times 10^{-34} \times 3 \times 10^{8}}{632.8 \times 10^{-9}}\)

= 3.14 x 10^{-19}

Momentum of each photon,

p = \(\frac{h}{\lambda}=\frac{6.63 \times 10^{-34}}{632.8 \times 10^{-9}}\)

= 1.05 x 10^{-27} kg. ms^{-1}

(b) Number of photons arriving per second at the target,

N = \(\frac{P}{E}=\frac{9.42 \times 10^{-3} \mathrm{~W}}{3.14 \times 10^{-19}}\)

= 3 × 10^{16} Photons/second

(c) Let u be the velocity of the hydrogen atom.

Then, according to the condition,

m_{H} u = momentum of photon

u = \(\frac{p}{m_{H}}\)

(m_{H} is mass of hydrogen atom)

= \(\frac{1.05 \times 10^{-27} \mathrm{~kg} \mathrm{~m} / \mathrm{s}}{1.673 \times 10^{-27} \mathrm{~kg}}\)

= 0.63 ms^{-1}

Section-D

Question 19.

(a) Show that the electric field at the surface of a charged conductor is given by E = \(\frac{\sigma}{\varepsilon_{\mathbf{0}}} \hat{\mathbf{n}}\), where o is the surface charge density and is a unit vector normal’

to the surface in the outward direction. [4]

(b) A spherical conducting shell of inner radius and outer radius R_{2} has a charge Q. A charge q is placed at the centre of the shell.

(i) What is the surface charge density on the (a) inner surface, (b) outer surface of the shell?

(ii) Write the expression for the electric field at a point to x > R_{2} from the centre of the shell.

Or

State Gauss’ law in electrostatics. A cube with each side a is kept in an electric field given by E = Cx \(\hat{\mathbf{i}}\) as shown in the figure, where C is a positive dimensional constant. Find out

(i) the electric flux through the cube.

(ii) the net charge inside the cube. [4]

Answer:

Let q charge be uniformly distributed over the spherical shell of radius r.

∴ Surface charge density on spherical shell

σ = \(\frac{q}{4 \pi r^{2}}\) ……………………. (i)

Electric field intensity on the surface of spherical shell

E = \(\frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{q}{r^{2}} \hat{\mathbf{n}}\)

[E acts along radially outward and along \(\hat{\mathbf{n}}\) ]

E = \(\frac{\left(q / 4 \pi r^{2}\right)}{\varepsilon_{0}} \hat{\mathbf{n}} \Rightarrow \mathbf{E}=\frac{\sigma}{\varepsilon_{0}} \hat{\mathbf{n}}\)

(b) Here, two points are important.

1. Charge resides on the outer surface of spherical conductor (skin effect).

2. Equal change of opposite nature induces in the surface of conductor nearer to source charge.

(i)

(a) Charge produced on the inner surface due to induction = — q

∴ Surface charge density of inner surface = \(\frac{-q}{4 \pi R_{1}^{2}}\)

(b) When charge -q is induced on inner walls, then equal charge +q is produced at outer surface.

Charge on outer surface = q + Q

∴Surface charge density of outer surface = \(\frac{q+Q}{4 \pi R_{2}^{2}} \)

(ii) electric field intensity at point P at a distance x (x> R_{2})

\(\frac{1}{4 \pi \varepsilon_{0}} \times \frac{(q+Q)}{x^{2}}\)

[along CP and away from spherical shell]

Whole charge is assumed to be concentrated at the centre.

Question 20.

Draw the circuit arrangement for studying the V-I characteristics of a p-n junction diode in forward.

Briefly explain how the typical V -I characteristics of a diode are obtained and draw these characteristics. [4]

Or

(a) Distinguish between a metal and an insulator on the basis of energy bond diagram.

(b) Mention the important considerations required while fabricating a p-n junction to be used as a light-emitting diode (LED). [4]

Answer:

(i) Forward Bias Characteristics For the V-I characteristics of a forward biasing P-N junction diode the experimental circuit arrangement is shown in the

figure.

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. which is also called as the P-N junction diode forward bias characteristic curve.

In forward biasing for every less value of voltage. (for Ge is 0.3 V and for Si is 0.7 V) the forward current is very less. The reason for this is that the value of applied voltage is less than the potential barrier due to which the flow of the current is not uniform. This behavior of the diode is represented by the OA part of the curve. When the value of applied potential is more increased the current in the diode increases exponentially.

This behaviour is shown by part AB. The potential at which the current’s value increase very rapidly is called the diode’s knee voltage or cut in voltage. For germanium diode its value is approximately 0.3 V and for silicon diode its value is 0.7V.

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