CBSE ANNUAL PAPER - 2000

PHYSICS

(SET-I)

Time allowed : 3 hours

Maximum Marks :70

General Instructions :

(i) All questions are compulsory.

(ii) Marks for each question are indicated against it.

(iii) Question numbers 1 to 8 are very short - answer questions, carrying 1 mark each. These are to be answered in one or two sentences.

(iv) Questions numbers 9 to 18 are short - answer questions, carrying 2 marks each. Answer to these questions should be around 30 words each.

(v) Questions numbers 19 to 27 are also short - answer questions, each carrying 3 marks each. Answer to these questions should be around 50 words each.

(vi) Questions numbers 28 to 30 are long - answer questions, each carrying 5 marks. Answer to these questions should around 100 words each.

(vii) Use Log Tables, if necessary.

Give a reason to show that microwaves are better carrier of signals for long range transmission than radio waves.

Because of their smaller wavelength, microwaves are not bent by objects of normal dimensions. So microwaves are better suited than radiowaves for sending a beam signal in a particular direction.

How does the energy gap in an intrinsic semiconductor vary, when doped with a pentavalent impurity ?

State the condition in which terminal voltage across a secondary cell is equal to its e.m.f.

In an open circuit, terminal voltage across a secondary cell is equal to its e.m.f.

Q.4.

If the number of turns of a solenoid is doubled, keeping the other factors constant, how does the self-inductance of the solenoid change ?

What is the ratio of solar constant on the surfaces of two planets, whose surface temperatures are in the ratio 1 : 2 ?

The wavelength of light coming from a distant galaxy is red - shifted. Is the galaxy receding or approaching the earth ?

What is the angle of dip at a place where the horizontal and vertical components of earth's magnetic field are equal ?

Briefly explain how the distance of an inferior planet from earth can be determined ?

The angle formed at the earth between the earth-planet direction and the earth - sun direction is called the planet's elongation. This angle is denoted by symbol

r_ps / r_es = sin e

Distance of the planet from the sun is

r_ps = sin e. r_es = sin e . AU

where r_es = IAU = average earth - sun distance.

Image to be inserted.

Using Gauss' law, show that no electric field intensity exists inside a hollow charged conductor .

Imagine a Gaussian surface near the cavity of a charged hollow conductor. Electric field is zero at all points inside a conductor. by Gauss's law, charge enclosed by the Gaussian surface is zero. Consequently, electric field is zero at all points of the cavity.

Four capacitors are connected as shown in the figure given below :

Calculate the equivalent capacitance between the points X and Y.

Eqvivalent capacitance of the parallel combination of 2 mF, 3mF and 10mF capacitors = 2+3+5 + 10mF

Now we have two 10mF capacitors in series.

\ Equivalent capacitance between X and Y = 10 x 10 / 10 + 10 = 5 mF

Draw the graph showing the variation of binding every per nucleon with mass number of different nuclei. State two inferences from this graph.

The energy responsible to keep all the nucleons together inside the nucleus or in other words the energy necessary to break a nucleus into its constituent particles is known as binding energy.

The variation of binding energy per nucleon with the mass number of elements is shown below :

The analysis of the curves shows that

(a) higher is the binding energy per nucleon, more stable is the atom.

(b) combination of lighter nuclei into heavier one corresponds to nuclear fusion. It is because the total binding energy of the products exceeds that of reactants.

In a single slit diffraction experiment, if the width of the slit is doubled, how does the (i) intensity of light and (ii) width of the central maximum change. Give reason for your answer.

(ii) As width of central maximum µ 1/slit width, so width of central maximum is halved.

Draw the logic symbol of a 2-input NAND gate. Write down its truth table.

A ray of light passes through an equilateral glass prism, such that the angle of incidence is equal to the angle of emergence. If the angle of emergence is 3/4 times the angle of the prism, calculate the refractive index of the glass prism.

AS A + d _m = i₁ + i 60⁰ + d_m = 45⁰ + 45⁰

0r d_m = 90⁰ - 60⁰ = 30⁰

m = sin / A + d_m /2 / sin / A/2 = sin / 60⁰ + 30⁰ /2 / sin 60⁰ / 2 = sin 45⁰ / sin 30⁰ = 1.414

A concave mirror and a convex lens are held separately in water. What changes, if any, do you expect in the focal length of either ?

Focal of concave mirror remains unchanged when immersed in water. Focal length of a convex lens in water will increase because the refractive index of glass with respect to water is less than refractive index of glass with respect to air.

A rectangular coil of N turns and area of cross - section A, is held in a time-varying magnetic field given by B = B) sin w, with the plane of the coil normal to the magnetic field. Deduce an expression for the e.m.f. induced in the coil.

f= NBA = NB₀ A sin wt

Draw the graph showing variation of thermo e.m.f. of a thermo-couple with the temperature difference of its junction. How does its neutral temperature vary with the temperature of the cold junction ?

Thermocouple is the arrangement of two different metal wires joined end to end so as to form a closed circuit.

The Sb-Bi thermocouple produces greater emf than Cu-Fe thermocouple for the same temperature difference hot and cold functions.

Application of thermocouples are

(i) In the generation of thermoelectricity.

(ii) In the measurement of temperature over a wide range.

Derive an expression for the electric potential at a point along with axial line of an electric dipole.

= q / 4pe₀ [ 1 / r-a - 1/r+a] = q/4pe . 2a / r² - a²

or v = 1/4pe₀ . p/r² - a² [ q x 2a = p]

For a <<r, V = 1/4pe₀ . P/r²

A copper voltmeter is in series with a heater coil of resistance 0.1 ohm. A steady current flows in the circuit for 20 minutes, and a mass of 0.99 gm of copper is deposited at the cathode. If the electro - chemical equivalent of copper is 0,00033 gm/coulomb, calculate the heat generated in the coil.

As m = zl t

l = m/zt = 0.99 / 0.00033 x 1200 = 2.5 A.

H = I² RT + (2.5)² x 0.1 x 1200 = 750 J.

State Huygen's postulates of wave theory. Sketch the wavefront emerging from a

(i) point source

(ii) linear source of light like a slit.

Huygen's postulates of wave theory are as follows :

(i) Every point of the medium situated on the wave front acts a new wave-source from which waves, called secondary wavelets, originate.

(ii) The secondary wavelets travel in the medium in all directions with the speed of light.

(iii) The envelope of the secondary wavelets in the forward direction at any instant gives the new wavefront at that instant.

State the condition of total internal reflection of light to take place at an interface separating two transparent media. Hence derive the expression for the critical angle in terms of the speeds of light in the two media.

Essential conditions for total internal reflection are

(i) Light should travel from a denser to a rarer medium.

(ii) The angle of incidence should be greater than the critical angle for the two media.

Expression of critical angle : -

When a ray of light travels from a denser to a rarer medium it deviates away from the normal. The angle of incidence in denser medium for which the angle of refraction is 900, is called critical angle. At this situation.

rarer denser = 1 /Sin C

But rarer denser = Speed of light in rarer medium / Speed of light in denser medium

Hence = Speed of light in rare medium / speed of light in denser medium = 1 / Sin C

State the dependence of work function on the Kinetic energy of electrons emitted in a photocell. If the intensity of incident radiation is doubled, what changes occur in the stopping potential and the photoelectric current.

Energy of incident photon = Maximum K.E. of photoelectron = work function

hn = 1/2 mv² _max + W_o

E_k = 1/2 mv² _max = hn - W_o

Higher is the work function of the metal, the lesser will be the kinetic energy of the emitted electron.

If the intensity of incident radiation is doubled, stopping potential remains unchanged and the photoelectric current gets doubled.

A short bar magnet of magnetic moment 0.9 joule / tesla, is placed with its axis at 450 to a uniform magnetic field. If it experience a torque of 0.063 joule, (i) calculate the magnitude of the magnetic field and (ii) what orientation of the bar magnet corresponds to the stable equilibrium in the magnetic field ?

(i) Magnetic field, B = p / m sin = 0.063 / 0.9 x sin 300 = 0.99 T

(ii) For stable equilibrium, the magnetic moment of the bar magnet should be in the direction of the magnetic field.

A conductor of length 'l' is connected to a d.c. source of potential 'V'. If the length of the conductor is tripled, by stretching it, keeping 'v' constant, explain how do the following factors vary in the conductor :

(i) Drift speed of electrons, (ii) Resistance and (iii) Resistivity.

When 'l' is tripled, drift speed becomes one - third of the original one

(ii) R = p l/a

When 'l' tripled, resistance becomes 3 times the original one.

(iii) Resistivity remains unchanged as it does not depend on the length 'l'.

A proton and an alpha particle of the same velocity enter in turn a region of uniform magnetic field acting in a plane perpendicular to the magnetic field. Deduce the ratio of the radii of the circular paths described by the particles. Explain why the kinetic energy of the particle after emerging from the magnetic field remains unaltered.

Now ma = 4 m_p and q_a = 2q_p

\ r_p/r_a = m_p/4m_p x 2q_p/q_b = 1/2 = 1:2

State the postulates of Bohr's model of hydrogen atom. The electron, in a given Bohr orbit has a total energy of -1.5eV. Calculate its (i) kinetic energy, (ii) potential energy and (iii) wavelength of light emitted, when the electron makes a transition to the ground state. ( Ground state energy = -13.6 eV)

Bohr's Postulates (1) An atom consists of a small and massive central core called nucleus around which electrons revolve. The centripetal force required is provided by electrostatic force of attraction between electron and the nucleus i.e.

mv² / r = KZe² / r²

(2) The electrons are permitted only in those orbits in which the angular momentum of electron is an integral multiple of h / 2

L = m vr = nh/2p where n = 1,2,3, ....

(3) An atom can emit or absorb radiation in the form of discrete energy only when an electron jumps from higher energy level to lower energy level i.e.,

Numerical : Total energy, E =- 1.5eV

(i) Kinetic energy, T = -E= -(-15) + 1.5 eV

(ii) Potential energy, V =- 2T = -2 x 1.5 =-3.0 eV.

(iii) D E = hn = hc/ l

\ l = hc/DE

But D E = -1.5 -(-13.6) = 12.1 ev = 12.1 x 1.6 = 10^-19 J

\ l = 6.6x10^-34x3x10⁸ / 12.1x10^-19 = 1.0227 x 10-7 m = 1022.7 A

For a given a.c. circuit, distinguish between resistance, reactance and impedance. An a.c. source of frequency 50 hertz is connected to a 50 mH inductor and a bulb. The bulb glows with some brightness. Calculate the capacitance of the capacitor to be connected in series with the circuit, so that the bulb glows with maximum brightness.

Resistance : It is the opposition to the flow of current by a pure resistor.

R = V/I

Reactance : It is the resistance offered by a pure inductor or capacitor to the flow of a.c.

X1 or X1 = Veff / Ieff

Impedance : It is the combined resistance offered by LR - or CR - or LCR circuit to the flow of a.c.

Z = Veff / Ieff

Numerical : Here f = 50 Hz, L = 50 mH = 0.05 H,

X_L = X_c or 2pfl = 1/2pfc

or C = 1/4p²f²L = 1/4x9.87x(50)² x 0.05

= 0.1 x 10^-4F

Formula to be inserted.

Drawing a labelled circuit diagram, explain the working principle of a common - emitter transistor amplifier. State the phase relation between input and output signals.

An npn transistor used as a common emitter amplifier is shown in figure.

The emitter circuit is forward biased while the collector emitter circuit is reverse biased;. The input signal Vi is superimposoeod on the forward bias. The load RL is connected between the collector and emitter as shown in figure.

When the signal is fed to the input circuit, its positive half cycle increases the forward bias of the circuit which in turn, increases the emitter current and hence the collector current. The increase in collector current increases the potential drop across RL which makes the output voltage less positive or more negative. This means that as the input signal goes through its positive cycle, the output signal goes through a negative half cycle.

The negative half cycle of the input signal opposes the forward bias. This decreases the emitter current and hence the collector current. The decrease in collector current decreases the potential drop across RL which makes the output voltage more positive or less negative. As the input signal goes through its negative half cycle, the output signal goes through its positive half cycle.

Voltage gain : The ratio of the output voltage to input voltage is called the voltage gain of the amplifier.

Voltage gain = V output / V input = D V_CE / DV_BE = DIC/DIB X R output / R input

= Current gain x resistance gain.