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Spectrometer Experiment Viva Questions

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Spectrometer Experiment Viva Questions with Answers

(i) Why is a spectrometer so called?

Ans: It is an instrument to study the spectra of different sources of light.

(ii) What is meant by monochromatic source of light?

Ans: It is a source that emits waves of the same colour or wave length.

(iii) Why do we use sodium vapour lamp in the laboratory as source of light?

Ans: It emits nearly monochromatic light (yellow) of wavelength 5890 Å.

(iv) What is the function of a colimator?

Ans: It is to produce parallel rays

Tangent Galvanometer Viva Questions

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Tangent Galvanometer Experiment Viva Questions

(1) What is meant by a tangent galvanometer?

 

A tangent galvanometer is a moving magnet type galvanometer and is used for measuring strength of electric current.

 

(2) What is tangent law?

 

If Bh and B be two magnetic fields acting perpendicular to each other, and their resultant makes an angle θ with Bh,

tan θ = (B/Bh);

Therefore, B = Bh tan θ

 

(3) What is magnetic meridian?

 

Magnetic meridian is a vertical plane passing through the magnetic axis (axial line) of a freely suspended or pivoted magnetic needle.

 

(4) Why is the pointer long and made of aluminium?

 

Aluminium makes pointer light. Moreover, aluminium is non-magnetic.

 

(5) What is the function of the plane mirror at the bottom of the compass box?

 

It helps in reading the position of the pointer on the scale without error of parallax.

 

(6) Define reduction factor of T.G.

 

Reduction factor of T.G. is defined as that current which when passed through the galvanometer required to produce a deflection of 45° when set in magnetic meridian.

 

(7) What are the two perpendicular magnetic fields in T.G. experiment?

 

The horizontal component of earth's magnetic field and the field produced by the current carrying circular coil.

 

(8) Why the deflection in tangent galvanometer is preferred to be nearly 45°?

 

An error in the measurement of θ will affect the result as tangent of θ. When θ is nearly 45° the variation in tan θ is very small and hence a small error in measurement of θ will cause negligible error.


(9) Aim of Tangent Galvanometer Experiment

To determine the reduction factor of a tangent galvanometer and hence to calculate the horizontal intensity at the place.

(10) Apparatus of Tangent Galvanometer Experiment

Ans: A tangent galvanometer (T.G), accumulator, rheostat, commutator, ammeter etc.
T.G. consists of a circular coil of insulated copper wire mounted vertically on a horizontal base provided with three levelling screws. A graduated compass box is mounted horizontally at the centre of the coil. It is graduated into four quadrant from 0° to 90°. Coils of different number of turns can be used by making connections to the appropriate terminals on the base.

(11) Theory of Tangent Galvanometer Experiment

Ans: If B, the magnetic field at the centre of a circular coil of radius r and number of turns n carrying a current i is perpendicular to Bh, the earth horizontal intensity at the place, then according to the tangent law, B = Bh tan θ; where θ is the mean deflection in the TG.

Therefore, μo/4π x 2πni/r = Bhtan θ

i.e., 10-7 x 2πni/r = Bhtan θ
i = (107 x Bhr/2πn) x tan θ ;  i = K tan θ ; where K is called the reduction factor of TG.

Therefore, K = i/tan θ

But K = (107 x Bhr)/2πn

Therefore, Bh = (2πnK/r) x 10-7

(12) Procedure of Tangent Galvanometer Experiment

Ans: 

Preliminary adjustments

(a) The leveling screws are adjusted till the plane of the coil is vertical. (b) The plane of the coil is brought in the magnetic meridian by making its plane parallel to the magnetic needle in the compass box. (c) Without disturbing the coil, the compass box alone is rotated to make the pointer to read 0 - 0.

To find the reduction factor K of TG

A lead accumulator, rheostat, ammeter and the T.G. are connected in series. The T.G. is connected through a commutator. The commutator is put in one direction and the rheostat is adjusted to get a convenient reading (between 30° and 60°) in T.G. The ammeter reading i is noted. The readings θ1 and θ2 at both ends of the pointer are noted. The current is reversed with the help of the commutator. Again the readings θ3 and θ4 of the pointer are taken. The average of the four readings gives θ, the deflection of T.G. The reduction factor K of the T.G is calculated from the equation.

K = i/tan θ

The experiment is repeated for different values of i by adjusting the rheostat. The average value of K is calculated.

To calculate the horizontal intensity Bh at the place

The radius r of the coil is determined by measuring its circumference.

r = Circumference/ 2π

The number of turns n of the coil is noted. The horizontal intensity at the place is calculated by the equation,

Bh = 2πnK x 10-7/r

Mapping of Magnetic Field Experiment

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Experiment 1:


Aim: To map the magnetic lines of force surrounding a bar magnet placed in the magnetic meridian with its north pole pointing magnetic north of the earth and, hence to determine the moment of the magnet.

Apparatus: The given bar magnet, compass needle, brass fixing pins, drawing board, drawing paper etc.,

Procedure

(1) To draw the magnetic meridian

A sheet of drawing paper is fixed on a drawing board. A straight line is drawn symmetrically on the paper. The compass needle is placed on the line. The drawing board is slowly turned till the ling becomes parallel to the magnetic needle in the compass box. Since the magnetic needle always comes to rest along the magnetic north-south direction, the line is along the magnetic meridian. A dotted line is drawn perpendicular to the magnetic meridian. The outline of the drawing board is drawn on the table with a chalk. Hereafter the board should not be disturbed. A short straight line with an arrow head is drawn at the upper right hand corner of the paper to indicate the direction of the magnetic north of the earth.

(2) To draw the lines of force

The bar magnet is placed on the paper with its axis in the magnetic meridian; its north pole pointing earth's magnetic north. The dotted line should become the equatorial line of the magnet. The outline of the magnet is drawn on the paper. The compass needle is placed near the north pole of the magnet and a dot is put against the north pole of the needle. The compass box is moved-so that its south pole is against dot. Another dot is put against the north pole of the needle and the process is repeated to get a number of dots. The dots are joined with a smooth line. This is a line of force. A number of such lines of force are also drawn. The general pattern is shown in the figure. Arrow marks are put on the lines to show the directions of the lines of force. It is found that there are two points, N1 and N2, symmetrically situated on either side of the magnet on the equatorial line where there are no lines of force. The compass needle placed at these points will remain in any direction. These are null points. The distance 2d between the two null points is measured and d, the distance of a null point from the centre of the bar magnet is found out. The length 2l of the magnet is measured using a metre scale. From this, the half the length l of the magnet is determined. At null points, the horizontal intensity Bh of the earth's magnetic field and the field due to the bar magnet are equal and opposite.

i.e., Bh = 10-7 x   m/(d2 +l2)3/2

Therefore, m = Bh(d2 + l2)3/2 x 107;

The moment m of the bar magnet is calculated assuming the value of Bh.

Note:-The null points may be marked conveniently as follows:-

The compass box is placed on the equatorial line of the magnet. It is kept near the magnet. Since the field due to the bar magnet is greater than that of the earth, the magnetic needle in the compass box becomes parallel to the bar magnet with its north pole pointing the south pole of the bar magnet, which is towards the earth's magnetic south. The compass box is moved slowly along the equatorial line away from the bar magnet. When it just crosses the null point, the magnetic needle turns so that its north pole points the magnetic north pole. This is because the earth's horizontal field Bh becomes greater than that due to the bar magnet. The position of the compass box is marked. Now, the compass box is moved along the equatorial line towards the magnet. When it just crosses the null point, the magnetic needle turns as the field due to the bar magnet becomes greater. The position of the compass box is marked again. The midpoint between the two positions of the compass box marked on the drawing paper is marked as N1, the null point. In a similar way the null point N2 is noted along the equatorial line on the other side of the bar magnet.

Experiment 2:


Aim: To map the magnetic lines of force surrounding a bar magnet placed in the magnetic meridian with its north pole pointing south and, hence to determine the moment of the magnet.

Apparatus: The bar magnet, compass needle, drawing board, drawing paper, brass fixing pins etc.

Procedure:
As in Experiment 1, the magnetic meridian is drawn on a drawing paper fixed on the drawing board. The bar magnet is placed on the meridian with its north pole pointing earth's magnetic south. The lines of force are drawn as explained above. The general pattern is shown in the figure. The null point N1 and N2 are obtained on the axial line of the-magnet. The distance 2d between the null points and hence d is found out. The length 2l of the magnet is determined using a meter scale. From this half the length 1 of the magnet is calculated. The moment of the magnet is calculated from the equation,

Bh = 10-7 x 2md/(d2l2)2

Therefore, m = Bh(d2l2)2 x 107/2d

Mapping of Magnetic Field Viva Questions with Answers

(i) What is a magnetic line of force?

Ans: It is a line in a magnetic field which indicates the direction along which the magnetic force acts. It is a continuous curve in a magnetic field such that the tangent at any point on it gives the direction of the resultant field at that point.

(ii) What is a neutral point in a magnetic field?

Ans: It is a point where the resultant magnetic field is zero.

(iii) No two lines of force intersect. Why?

Ans: If they intersect at any point it means that the magnetic field at that point are in two directions which is impossible.

(iv) What is the expression for the magnetic field at a point on the axis of a bar magnet?

Ans: B = (10-7 x 2md)T / (d2 — l2)2

(v) When a bar magnet is placed in the magnetic meridian with north pole pointing north we get two null points symmetrically situated. why?

Ans: There are two points on either side of the magnet on the equatorial line where the fields are equal.

(vi) What is the nature of the lines of force in a uniform magnetic field?

Ans: The lines of force are parallel.

(vii) Can you have a magnet with one pole?

Ans: No. A magnet has always two poles. Even if it is broken each piece becomes a complete magnet.

(viii) What is a magnetic dipole?

Ans: It is a current loop of moment  = i Am

(ix) What is magnetic moment of a magnet?

Ans: It is numerically equal to the torque acting on it when it is placed at right angles to unit uniform field.

(x) What is magnetic equator?

Ans: It is the line joining points where the dip is zero.

PN Junction Diode Viva Questions with Answers

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The PN Junction Diode Viva Questions with Answers

(i) What is a pn junction?

A PN-junction is said to be condition when an N-type material is combined together with a P-type material creating a semiconductor diode.

(ii) Give the symbol for a semiconductor diode.

The 'P' side of a pn junction diode is always positive terminal and is called as anode. Other side which is negative is called as cathode. The symbol is shown in figure.

(iii) What is meant by forward biasing of a pn junction?

If the anode is connected to positive terminal of a battery and cathode to the negative terminal, the set up is called forward bias.

(iv) What is meant by reverse biasing of a pn junction?

When the positive of the battery is connected to the N side and negative of the battery is connected to the P side, then the diode is called to be reverse biased.

(v) What is forward resistance of a diode?

A diode offers an extremely small resistance (not equal to zero) when forward biased and it is called a forward resistance of a diode.

(vi) Distinguish between static and dynamic forward resistances of a diode.

When the diode is forward biased, it creates a small resistance in the circuit. The static resistance is defined as the ratio of DC voltage generated in the diode to the DC current runs through it. Dynamic resistance otherwise called as AC resistance is the reciprocal of the slope of the tangent of the characteristics curve.

ie., Dynamic resistance = change in voltage/resulting change in current = ΔV/ΔI

(vii) What is meant by a semiconductor?


Semiconductors arc substances having resistivity or conductivity lying between conductors and insulators.


(viii) Give some properties of a semiconductor?


i. It has covalent bonding

ii. It is crystalline in nature

iii. It has a negative temperature coefficient of resistance.

iv. Its conductivity increases with addition of impurities.


(ix) What is meant by doping?


Doping is the process of addition of a desirable impurity to a pure semiconductor in order to increase its conductivity.


(x) What is meant by forbidden gap?


The band separating the valence hand and the conduction band is called forbidden gap.


(xi) What is meant by a junction diode?


When a p-type semiconductor is fixed with n-type semiconductor, it forms a junction diode.


(xii) What is depletion region in p-n junction?


Depletion layer is a thin layer formed amid p and n regions of junction diode devoid of free electrons and the holes.


(xiii) Is Ohm's law obeyed in a semiconductor or not?


In a semiconductor, Ohm's law is valid only for low electric fields.


(xiv) What is a rectifier?


Rectifier is a device which converts AC into DC.


(xv) Aim of the PN Junction Diode Experiment

To draw the static current-voltage (I — V) characteristic of a junction diode in forward bias and hence to calculate its ac and dc forward resistances.

(xvi) Apparatus of the PN Junction Diode Experiment

The given junction diode (ex: IN 4007), a 2-volt battery, voltmeter (a low range voltmeter reading upto 1 V with 0.1 V per divisions), milliammeter, key, etc.

(xvii) Procedure of the PN Junction Diode Experiment

Connections are made as shown in the figure. A 2 volt battery is connected to a rheostat Rh. The rheostat is used as a potential divider arrangement. The voltmeter V is connected across the diode. The milliammeter m A is used to measure the diode current. The potential difference  across the diode is increased from zero in steps of 0.1 V till the current increases steeply and becomes 25 to 30 mA. In each step the ammeter reading is noted. A mph is plotted with ammeter readings in mA along the Y-axis and the voltmeter readings in volts along X-axis. This gives the forward characteristic of the diode.

(a) To find static (dc) forward resistance (Rf):

Static forward resistance is the ratio of the direct voltage (V) applied at any point P on the straight portion of the characteristic to the direct current (I) corresponding to the point P. The voltage (V) and current (I) corresponding to the point P on the straight portion (i.e., beyond the knee voltage) are noted. The static forward resistance, 

R= V/I.

(b) To find dynamic (ac) forward resistance (rf): 

It is defined as the ratio of a small change in the forward bias voltage to the corresponding change in the forward current 

rf = (dV/dI)P 

The reciprocal of the slope of the graph is determined at the point P. This gives the dynamic resistance of the diode in forward bias at the point P.

Refractive Index of a Glass Slab Viva Questions

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Refractive Index of a Glass Slab Using Travelling Microscope Viva Questions and Answers

(i) What is meant by the refractive index of the material of a transparent medium?

Ans: Refractive index is called as the measure of the bending of a light ray when passing from one medium to another.

(ii) Give the relation for the refractive index of a medium in terms of velocity of light.

Ans: n = c/v, where n is refractive index, c is velocity of light and v is velocity of substance.

(iii) What is Snell's law of refraction?

Ans: Snell's law is a formula that defines the relationship between angles of incidence and refraction. The mathematical formula is given below.

n1sinθ1 = n2sinθ2

(iv) Give an equation for the refractive index of a glass slab in terms of real and apparent thicknesses of a glass slab.

(v) What is meant by shift in the position of image produce by a glass slab?

(vi) Aim of the refractive index of a glass slab Experiment?

Ans: To determine the refractive index of the material of a glass slab.

(vii) Apparatus of the refractive index of a glass slab Experiment?

Ans: The glass slab, a travelling microscope, chalk powder, etc.

(viii) Theory of the refractive index of a glass slab Experiment?

Ans: The refractive index of the material of the glass slab is given by,
μ = Real thickness of the glass slab/ Apparent thickness of the glass slab

(ix) Procedure of the refractive index of a glass slab Experiment?

Ans: A few small grains of chalk power is sprinkled on the horizontal base of the travelling microscope. The microscope is kept vertically and adjusted to get a well focussed image of the chalk powder. The reading R1 of the microscope is taken on its vertical scale. The glass slab is placed over the chalk powder. The microscope is raised and adjusted to get the image of the chalk powder. The microscopic reading R2 is again taken. The chalk powder is sprinkled over the glass slab. The microscope is raised further to get the chalk powder focussed. The reading R3 of the microscope is noted. The real thickness of the glass slab,

t = R3 - R1

The apparent thickness of the glass slab,

t' = R3 - R2

The refractive index of the material of the glass slab is calculated by the equation,

μ = Real thickness /Apparent thickness = t/ t' = (R3 - R1)/(R3 - R2).