CBSE ANNUAL PAPER - 1998

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.

Name the physical quantity whose SI unit is coulomb/volt.

Write the frequency limit of visible range of electromagnetic spectrum in KHz.

Frequency limit of visible light is from 4 x 10¹¹ kHz to 8 x 10¹¹ kHz.

How does the conductance of a semi-conducting material change with rise in temperature ?

The conductance increases when the temperature of a semi-conducting material increases.

of these, name the pairs of vectors which are always at right angles to each other.

Two wires A and B are of same metal, have the same area of cross section and have their lengths in the ratio 2:1. What will be the ratio of currents flowing through them respectively when the same potential differences is applied across length of each of them ?

Calculate the rms value of the alternating current shown in the figure.

The image of an object formed by a lens on the screen is not in sharp focus.Suggest a method to get clear focusing of the image on the screen without disturbing the position of the object, the lens or the screen.

Sharp image can be obtained by either removing paraxial rays or marginal rays by using suitable stoppers.

Two protons A and B are placed between two parallel plates having a potential difference V as shown in the figure.

Figure to be pasted.

Will these protons experience equal or unequal force ?

Define the terms 'threshold frequency' and 'stopping potential' for photo-electricity effect.Show graphically how the stopping potential, for a given metal, varies with frequency of the incident radiation's. Mark threshold frequency on this graph.

Threshold frequency : The threshold frequency of a metal is the minimum value of the frequency of incident light below which the emission of photoelectrons stops completely.

Stopping potential : It is the minimum value of the negative potential which should be applied to the anode in a photo cell so that the photoelectric currents becomes zero. Figure shows the variation of stopping potential (V₀) with frequency (v) of the incident radiation for a given metal.

Draw labelled diagram of a leclanche cell. Write the function of charcoal powder and maganese dioxide used in its porous pot.

1. the distance between the coils is increased ?

Justify your answer in each case.

(i) When the distance between the two coils is decreased, the mutual inductance decreases because lesser flux passes from primary to secondary coil.

(ii) Mutual inductance, M = m N₁N₂A/l. Clearly when the number of turns N1 and N2 in the two coils are decreased, the mutual inductance decreases.

Define the terms magnetic inclination and horizontal component of Earth's magnetic field at a place. Establish the relationship between the two with the help of a diagram.

Magnetic inclination (d) : It is the angle which the direction of total intensity of earth's magnetic field makes with the horizontal.

Horizontal component of earth's magnetic field (H) : It is the component of the earth's magnetic field along the horizontal direction.

As shown in the figure, if d is the angle of inclination and H is the horizontal component of earth's total magnetic field, then H/B = Cos d or H = B cos d This is the required relation between d and H.

An electron in an atom revolves around the nucleus in an orbit of radius 0.53 ⁰A. Calculate the equivalent magnetic moment if the frequency of revolution of electron is 6.8 X 10⁹ MHz.

Write the function of base region of a transistor. Why is this region made thin and slightly doped ?

Base controls the flow of electrons or holes between emitter and collector.

The base is thin and slightly doped, so that it may contain smaller number of majority charge carriers. This reduces the recombination rate of free electrons and holes in the base region.

Derive an expression for the energy stored in a charged parallel plate capacitor with air as the medium between its plates.

Work done in charging a capacitor. This work done is stored as its electrical potential energy. Suppose a capacitor is charged with charge q snow that potential difference between its plates is V = q/C Work done to increase the charge by an amount dq is dW = Vd_q = (q/C) d_q Total work done to charge the capacitor from O to Q is W = ^Q|_o q/C d_q = 1/C [q²/2]^Q_o = Q²/2C

Energy of the capacitor, U = 1/2 Q²/2C = 1/2 QV [ C = Q/Y]

In the diagram given below for the stationary orbits of the hydrogen atm, mark the transitions representing the Balmer and Lyman series.

Transitions representing the Lyman and Balmer scries of a hydrogen atom are show in the above figure.

The given figure shows an inductor L and resistor R connected in parallel to a battery B through a switch S. The resistance of R is the same as that of the coil that makes L. Two identical bulbs, Q and P are put in each arm of the circuit as shown.

When S is closed, which of the two bulbs will light up earlier? Justify your answer.

The bulb P lights up earlier than the bulb Q because the induced e.m.f. across L opposes the growth of current in the bulb Q.

Two point electric charges of values q and 2q are kept at a distance d apart from each other in air. A third charge Q is to be kept along the same line in such a way that the net force acting on q an d2q is Zero. Calculate the position of charge Q in terms of q and d.

Explain, with the help of a circuit diagram, the use of potentiometer for determination of internal resistance of a primary cell. Derive the necessary mathematical expression.

The resistance offered by electrolyte inside the cell circuit is called internal resistance of cell.

Let there be a cell of emf E, internal resistance r, used to draw current in a circuit of resistance R. then.

I = E/R+r

IR + Ir = E

Sketch the wave-fronts corresponding to converging rays. Verify Snell's law of refraction using Huygen's wave theory.

Wavefronts corresponding to converging to conveying rays are shown in the figure.

Let the surface AB represent a surface of separation of two media in which velocities of light are C1 and C2 respectively.

Let the wave front PQ be incident at an angle i to the surface AB. By the time Q reached surface AB at S, the disturbance at P reaches R such that RS is the refracted wavefront.

The time taken by a ray to travel from T to U is

An electric dipole is held in a uniform electric field.

(i) Show that no translatory force acts on it.

(ii) Derive an expression for the torque acting on it.

(i) Consider an electric dipole consisting of charges +Q AND _q and of length 2a placed in a uniform electric field E making an angle q with it.

Force an charge =q at B = =q E ( in the direction of E)
Force on charge -q at B = - q E ( in the opposite direction of E)
Total translatory force on the dipolse = + Q E -Q E = ).

(ii) The two forces acting on the dipole form a couple. The torque exerted by the couple is

t = Either force x Perpendicular distance between the two forces.

= QE x AN = qE X 2a sin q

t = pE sin q

where p = q x 2a, is the electric dipole moment.

Answer to be inserted.

Derive an expression for the width of the central maxima for diffraction of light at a single slit. How does the width change with increase in width of the slit.

Suppose a parallel beam of light falls on rectangular slit AB. According to Huygen's hypothesis, each point in the slit becomes a source of secondary wavelets. These wavelets are initially in same phase and spread out in all directions.

Cental Maximum : - Let C be the centre of slit AB. The wavelets which fall on the lens parallel to CO meet at point O in same phase because their optical paths between the slit and the point O are equal. The wavelets add up constructively and give rise to a central maximum at O.

Positions of secondary maxima and minima : Consider point P on the screen at which wavelets travelling in a direction making an angle with CO are brought to focus by a convex lens. The path difference between the extreme rays starting from A and B is

p = BN + AB SIN q = d = SIN q

where d is width of slit AB.

capacitor of capacitance 100 mF and a coil of resistance 50 and inductance 0.5 H are connected in series with a 110 V, 50 Hz source. Calculate the rms value of the current in the circuit.

Draw a labelled ray diagram to show the image formation in a reflecting type telescope. Write its two advantages over a refracting type telescope. On what factors does its resolving power depend ?

(ii) Image is brighter compared to that in a refracting type of telescope. The resolving power of the telescope depends on the aperture of the objective and wavelength of light used.

Define the terms 'solar constant' and 'solar luminosity'. Explain how their knowledge helps us to calculate the surface temperature of the sun. Derive the necessary mathematical expression.

Solar Constant : It is defined as the amount of radiant energy received per second per unit area of a perfect black body placed on the earth with its surface perpendicular to the direction of the radiation from the sun.

Solar luminosity : The total amount of energy radiated by the sun in all directions per second measured in watt is called the solar luminosity. It is denoted by L.

Let S be the solar constant and r be the radius of the sun.

According to Stenfan's law, energy radiated by the sun per unit area per unit time.

E = sT⁴, s is Stefan's constant.

Total energy radiated by the sun per unit time. = 4pr² E = 4 pr² sT⁴ 4 pr² sT⁴ = 4 pr² S

or T⁴ = R²S/r²s

or T = [R2S/r²s]^1/4

An object is kept in front of a concave mirror of focal length 15cm. The image formed is three times the size of the object. Calculate two possible distances of the object from the mirror.

v/u = -3 or v = -3u

As i/u + 1/v = 1/f

\ i/u - 1/3u = 1/f or 3-1/3u = 1/f

or u = 2f/3 = 2x(-15) / 3 = -10 cm.

(ii) When the image formed is real

v/u = +3 or v = 3u

\ 1/3u +1/u = 1/f or 1+3 / 3u = 1/f

or u = 4f/3 = 4x(-15)/3 = -20cm.

A voltmeter V of resistance 400 W is used to measure the potential difference across a 100 W resistor in the circuit shown here. (a) What will be the reading on the voltmeter? (b) Calculate the potential difference across 100 W resistor before the voltmeter is connected.

= 400 x 100/400+100 = 80W

(a) Total resistance is the circuit, R = 80 + 200 = 280 W

Current in the circuit, I = V/R = 84/280 = 3/10 A

Voltmeter reading = 80 x 3/10 = 24 V.

(b) Total resistance before the voltmeter is connected

= 100 + 200 = 300 W

Current in the circuit, I = 84/300 = 7/25 A

P.D. across 100 W resistor = 100 x 7/25 = 28 W

Derive a mathematical expression for the force per unit length on each of the two straight parallel metallic conductors carrying current in the same direction and kept near each other. Why do such current carrying conductors attract each other.

OR

Derive a mathematical expression for the force acting on a current carrying straight conductor kept in a magnetic field. State the rule used to determine the direction of this force.

Draw a labelled diagram of Thomson's experimental set-up to determine e/m of electrons. Explain, by deriving the necessary mathematical expression, how e/m of electrons can be determined by this method.

(a) Thomson set up for the determination of e/m of electrons in the figure below

Now potential energy of electrons at the cathode = eV

Kinetic energy of an electron at the anode = 1/2mv²

By law of conservation of energy, 1/2 mv² = eV

So the velocity of electrons, n = (2eV/m)^1/2

b) E = V/d

\ e/m = E² / 2VB² = V² / 2VB² d² = V/2B² d²

As d is fixed , therefore,

B² µ V or B µ (V)^1/2

If V is doubled, then

B'/B' = (v)^1/2 / (v)^1/2 = (2v)^1/2 / (v) ^1/2 = (2)^1/2

or B' = (2B)^1/2

So magnetic field should be increased to (2) ^1/2 B to keep the electron beam undeflected.

Theory : Let E be the intensity of electric field then the electron will experience a force F1 in upward direction

F1 = Ee

Let B be the intensity of magnetic field. The electron moving with velocity n will experience a force Ben in the direction opposite to that in electric field.

F2 = Ben

If the electric and magnetic fields are so adjusted that the electron beam undergoes no deflection, then

Ben =Ee

Or n = E/B

If V be the accelerating voltage applied between the cathode and anode, then

Ve = 1/2 Mn²

or e/m = 1/2 n²/V

= 1/2 E² / B² V

Knowing the value of E, B and V, the value ^e/m can be determined.

Define the terms 'potential barrier' and 'depletion region' for a p-n junction. Explain, with the help of a circuit diagram, the use of a p-n diode as a full wave rectifier. Draw the input and output wave-forms.

Potential Barrier : This is the potential difference established across the p-n junction due to diffusion of electrons from n-region to p-region and holes from p-region to n-region.

Depletion region : It is a thin layer formed between p - and n-sections which is devoid of free electrons and holes and has only immobile ions.

A full wave rectifier uses two diodes as shown in Fig.

Figure to be inserted.

Let the point A of transformer be positive during first half of the input cycle. The voltage decreases as we move from A to C. Obviously the point A is at higher potential with respect to B. the anode of diode D₁ is positive with respect to cathode and it conducts. The direction of current flow is shown by solid arrow. The diode D₂ is reversed biased so does not conduct. During the second half cycle let C be at positive potential relative to A, then diode D₂ will conduct and not the diode D₁. The direction of current is shown by dotted line. It is clear from the figure that during both halves the current through load resistance ^RL is unidirectional. Since there is an output during both halves of the input signal, the output also is continuous. The input and output signals are shown below.