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Physics-1997-Set-II

Q1) Is the ratio of number of holes and the number of conduction electrons in a n-type extrinsic semiconductors more than, less than or equal to 1? (Marks 1)
Ans1) nh/ne in N type < 1 as holes are minority charge carriers & electrons are majority charge carriers.

Q9) Use mirror formula to show that a convex mirror always produces a virtual image independent of the location of the object. (Marks 2)

Ans9) According to mirror formula
1 / f = 1 / v + 1/u
For convex mirror f > 0 (being +) & u < 0 (being - ve)
1 / v = 1 / f - 1 / u = 1 / |f| + 1 / |u|
v is always +ve i.e. , image will be behind the mirror. Thus convex mirror always gives virtual & erect image.

Q10) An astronomical telescope of magnifying power 10 consists of two thin lenses 55 cm apart. Calculate the focal lengths of the lenses. (Marks 2)
Ans10) m = 10, L = 55 cm
m = fo/fe or fo = mfe
L = fo + fe = mfe + fe = (m + 1) fe
55 = (10 + 1) fe or fe = 5 cm
fo = 10 fe = 50 cm.

Q11) An -particle and a proton are accelerated through same potential differences. Calculate ratio of linear momentum acquired by the two. (Marks 2)
Ans11) Now qV = 1/2 mv2
For particle
(2e)V = 1/2 mv2 – (i)
For proton
(e)V = 1/2 mpvp2 – (ii)
Dividing (i) by (ii)
2 = mv2/mpvp2
m/mp = 4
=> v2/vp2 = 2 x 1/4 =1/2
or v/vp = 1/2
The ratio of linear momentum is
P/Pp = mv/mpvp = 4 x 1/2 = 22

Physics-1997-Set-III

Q1) Horizontal component and vertical component of Earth’s magnetic field at a place are equal. what is the value of angle of dip at this place? (Marks 1)
Ans1) BV = BH
tan = BV/BH = 1
or = 45o

Q5) Is the ratio of number of holes and the number of conduction electrons in a p-type extrinsic conductor more than, less than or equal to 1? (Marks 1)
Ans5) The ratio of number of holes to the number of conduction electrons in a p-type extrinsic conductor more than one as holes are the majority charge carriers.

Q9) What percentage of a given mass of a radioactive substance will be left undecayed after four half-life periods? Justify your answer. (Marks 2)
Ans9) N/No = (1/2)n, where n is no. of half-lives.
n = 4
… N = No(1/2)4
or N/No = (1/2)4 = 1/16
or N/No = 1/16 x 100% = 6.25%

Q12) Use mirror formula to show that for an object lying between F and 2F of a concave mirror, a real image is formed beyond 2F.
(Marks 2)

Ans12) According to mirror formula
1/f = 1/v + 1/u
For a concave mirror f is -ve , u is -ve
… -1/|f| = 1/v – 1/|u|
or 1/v = 1/|u| – 1/|f| – (i)
object lies between F and 2F
So |f| < |u| < |2f|
or 1/|f| > 1/|u| > 1/|2f| – (ii)
or 0 > 1/|u| – 1/|f| > -1/|2f|
using Equation (i)
0 > 1/v > -1/|2f|
1/v < 0 or v < 0 i.e. v is -ve, hence a real image is formed.
Also 1/v > -1/|2f|
or v < -|2f| or |v| > |2f| i.e. image is formed beyond 2F.

Q13) An astronomical telescope of magnifying power 7 consists of two thin lenses 40 cm apart. Calculate the focal length of the lenses. (Marks 2)
Ans13) 7 = fo/fe or fo = 7fe
Also L = fo + fe
40 = 7fe + fe = 8fe
or fe = 5 cm
& fo = 7fe = 35 cm.

Physics-1998-Set-I

Q1) Name the physical quantity whose SI unit is Coulomb/Volt. (Marks 1)
Ans1) Capacitance

Q2) Write the frequency limit of visible range of electromagnetic spectrum in kHz. (Marks 1)
Ans2) 8 x 1011 HHz to 4 x 1011 KHz

Q3) How does the conductance of a semi-conducting material change with rise in temperature? (Marks 1)
Ans3) Increases.

Q4) The force experienced by a particle of charge e moving with velocity in a magnetic field is given by. = e( x ) Of these, name the pairs of vectors which are always at right angles to each other. (Marks 1)
Ans4) = e( x )
| and |

Q5) 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 difference is applied across length of each of them?
(Marks 1)

Ans5) Resistance length
RA/RB = 1/2
Current is inversely proportional to resistance
… IA/IB = 1/2

Q6) Calculate the rms value of the alternating current shown in the figure. (Marks 1)

Ans6) Irms = i2 = ((22 + (-2)2 +22)/3) = (12/3) = 2A

Q7) 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. (Marks 1)
Ans7) The reason for such image is spherical aberration. Clearer focusing can be got by using the other central position of the lens & blocking the inter portion or using other proton & blocking central portion.

Q8) Two points A and B are placed between two parallel plates having a potential difference V as shown in the figure.

Will these protons experience equal or unequal force? (Marks 1)
Ans8) Both A and B will experience equal force as field between the plates is uniform.

Q9) Define the terms ‘threshold frequency’ and ’stopping potential’ for photo-electric effect. Show graphically how the stopping potential, for a given metal, varies with frequency of the incident radiations. Mark threshold frequency on this graph. (Marks 2)
Ans9) Threshold frequency: It is the maximum frequency of the incident radiation below which no emission of photoelectrons takes place.
Stopping potential : It is the maximum negative potential required to stop the fastest photoelectrons i e, photoelectric current becomes zero.

Physics-1998-Set-II

(Only those uestionshave been given here, which are different from set I )

Q6) Calculate rms value of alternating current in diagram.(Marks 1)
Ans6) Irms = I2 = (1.52 + (-1.5)2 + 1.52) = 1.5A

Q7) Two wires A and B are of the same metal and of the same length have their areas of cross-section in the ratio of 2 : 1. if the same potential difference is applied across each wire in turn, what will be the ratio of the currents flowing in A and B? (Marks 1)
Ans7) Resistance is inversely proportional to area
RA/RB = AB/AA = 1/2
Now current is inversely proportional to resistance
… IA/IB = RB/RA = 2/1

Q14) An electron in an atoms revolves around the nucleus in an orbit of radius 0.53 A. Calculate the equivalent magnetic moment if the frequency of revolution of electrons is 1010 MHz. (Marks 2)
Ans14) Given r = 0.5A = 0.5 x 10-10m
= 1010 MHz = 1016 Hz
e = 1.6 x 10-19c
magnetic moment = eA = er2
= 1016 x 1.6 x 10-19 x 3.14 x (.5 x 10-10)2
= 1.256 x 10-23 Am2

Q15) Describe the use of a vibration magnetometer for comparing magnetic moments of two bar magnets of same size and mass.
(Marks 2)

Ans15) Let T1 & T2 be the time periods of vibration of two magnets at a place. Then
T1 = 2(I/M1H)
T2 = 2(I/M2)
or T1/T2 = (M2/M1)
or M2/M1 = T22/T12

Q17) Draw a labelled diagram of Daniel cell. What is the function of copper sulphate and sulphuric acid solutions used in it? (Marks 2)
Ans17) Daniel cell

Copper Sulphate acts as depolarizer
Sulphuric acid is electrolyte

Q22) Draw a ray diagram to show formation of image by a refracting type astronomical telescope. On what factors does its (i) magnifying power and (ii) resolving power depend ? (Marks 3)
Ans22) Refracting type Astronomical telescope:

Magnifying power depends on focal length of objective lens & eye lens.
Resolving power = D/1.22 . So it depends on
D : Diameter or the aperture of objective lens
: Wavelength of light used.

Physics-1998-Set-III

(Only those questions have been given, which are different from Set I)

Q2) Draw the energy band diagram for a p-type extrinsic semi-conductor. (Marks 1)
Ans2) Energy band diagram of p-type semi conductor

Q8) Two wires A and B of the same length and of the same metal have their areas of cross-section in the ratio of 1 : 4 respectively. If the same potential difference is applied across the terminal ends of each wire, what will be the ratio of the current flowing through A and B.
(Marks 1)

Ans8) RA/RB = AB/AA = 4 : 1
As current is inversely proportional to resistance
IA/IB = RB/RA = 1/4

Q14) State the laws of photoelectric emission. (Marks 2)
Ans14) Laws of photoelectric emission :
(i) For a give metal & frequency of incident radiation, the no of photoelectrons ejected per second is directly proportional to the intensity of incident light.
(ii) For a given metal there exists a certain min frequency of the incident radiation called threshold frequency below which no emission of photoelectrons takes place.
(iii) Above the threshold frequency the max kinetic energy of the emitted photoelectrons is independent of the intensity of the incident light but depends upon the frequency of the incident light.
(iv) The photoelectric emission is an instantaneous process.

Q16) How does the self-inductance of a coil change when :
(i) the number of turns in the coil is decreased ?
(ii) an iron rod is introduced into it?
Justify your answer in each case. (Marks 2)

Ans16) (i) Self inductance is directly proportional to the square of the no of turns of the coil, hence with the decrease in the no of turns of the coil, self inductance decreases.
(ii) When iron is introduced in the coil, self inductance increases as the permeability of self iron is about 1000 times

Physics-1999-Set-I

Q1) Draw an equipotential surface in a uniform electric field. (Marks 1)
Ans1)

Equipotential surfaces are parallel planes perpendicular to the field lines.

Q2) If a wire is stretched to double its original length without less of mass, how will the resistivity of the wire be influenced? (Marks 1)
Ans2) The resistivity of the wire will remain same as it is the resistance of wire of unit length & unit cross-sectional area.

Q3) Why do magnetic lines of force prefer to pass through iron than through air? (Marks 1)
Ans3) The magnetic lines of force prefer to pass through iron then through air as iron has very high magnetic permeability.

Q4) What is the power factor of an LCR series circuit at resonance? (Marks 1)
Ans4) Power factor cos = R/(R2 + (L – 1/c)2)
At resonance L = 1/c
… cos = R/(R2 + 0) = 1

Q5) Why is the transmission of signals using ground waves restricted to frequencies upto 1500 kHz? (Marks 1)
Ans5) The transmission of signals using ground wave is restricted to frequencies upto 1500 kHz because higher frequencies are damped by interaction with matter.

Q6) The polarising angle of a medium is 60o. What is the refractive index of the medium? (Marks 1)
Ans6) = tan ip
ip = 60o
… = tan 60o = 3

Physics-1999-Set-II

Q1) Why is the transmission of signals using sky waves, restricted to frequencies greater than 30 MHz. (Marks 1)
Ans1) The transmission of signals using sky waves is restricted to frequencies greater then 30 MHz since lower frequencies are highly damped by interaction with matter.

Q8) How does magnetic induction of a paramagnetic substance vary with temperature. (Marks 1)
Ans8) The magnetic induction of a paramagnetic substance decreases with the increase in temperature.

Q9) Draw a labelled ray diagram to show the image formation in an astronomical telescope for a distant object. (Marks 2)
Ans9) Ray diagram of image formation by Astronomical telescope.

A”B” : Final image
O : Objective
E : Eye piece
d : least distance of distinct image

Q11) The half-life of a radioactive sample in 30 seconds calculate (i) the decay constant & (ii) time taken for the sample to become 1/4th of its initial value (Marks 2)
Ans11) (i) T1/2 = 30 s
= 0.693/T1/2 = 0.693/30 = 0.0231 s-1
(ii) In 30 s, substance is reduced to half & in next 30s, it will be reduced to 1/4th
Hence time taken for substance to reduce to 1/4th of its initial value = 60 s

Physics-1999-Set-III

(Only those questions have been answered, which are different from Set I and Set II)

Q6) What is the difference between voltage across an inductor & a capacitor in an a.c. circuit. (Marks 1)
Ans6) Voltage leads the current by /2 in an a.c circuit containing inductor & voltage lags the current by /2 in an a.c circuit containing capacitor.

Q7) Why are microwaves used in RADAR. (Marks 1)
Ans7) Microwaves are used in RADAR because they can be transmitted as beam signals in a particular direction. They donot spread or bend around the corners of an obstacle coming in their way.

Q9) What is self-induction? State any two factors on which the self inductance of a long solenoid depends. (Marks 2)
Ans9) Self induction is the property of a coil by virtue of which it opposes any change in the strength of current flowing through it by inducing an emf in itself.

Q10) Light from a distant galaxy having wavelength 6000 Ao is found to be shifted toward red by 60 Ao. Calculate the velocity of recession of the galaxy. (Marks 2)
Ans10) Given = 6000 Ao
d = 60 Ao
Now d/ = V/C
or V = dC/ = 60/6000 x 3 x 108 = 3 x 106 m/s

Q12) Draw a labelled ray diagram to show the image of an object in a compound microscope. (Marks 2)
Ans12) Ray diagram of image formation by compound microscope :

O : Objective
E : Eyepiece
AB : Object
A”B” : Final image
d : Least distance of distinct vision.

Q16) Briefly describe the method to determine the surface temp of sun. (Marks 2)
Ans16) Surface temperature of the sum can be determined by the knowledge of solar constant S. Solar constant is defined as the amount of radiant energy received per second by a unit area of a perfectly black body surface held at right angles to the direction of the sun rays at mean distance of earth from the sun. Its value is 1388 Wm-2
Consider sun to be a black body at temperature T and radius R at the centre of a hollow sphere of radius r
According to Stefan’s law E = T4
Total energy radiated per unit time by the sun
= 4R2E = 4R2T4
Solar luminousity = 4r2S
… 4R2T4 = 4r2S
or T = (r2S/R2)1/4
substituting the value of r, R, S &
T = [((1.496 x 1011)2 1288)/(6.928 x 108)2(5.735 x 10-8)]1/4
or T = 5800 K = 5,527oC

Physics-2000-Set-I

Q1) Give a reason to show that microwaves are better carriers of signals for long range transmission than radio waves. (Marks 1)
Ans1) Microwaves have smaller wavelength due to which they can be transmitted as beam signals in a particular direction, better than radiowaves because microwaves do not spread or bend around the corners of any obstacle coming in their way.

Q2) How does the energy gap in an intrinsic semiconductor vary, when doped with a pentavalent impurity ? (Marks 1)
Ans2) The energy energy gap in semiconductor decreases when doped with a petavalent impurity.

Q3) State the condition in which terminal voltage across a secondary cell is equal to its e.m.f (Marks 1)
Ans3) The terminal voltage across a secondary cell is equal to its e.m.f. when no current is drawn from it i.e., it is in an open circuit.

Q4) Draw an equipotential surface in a uniform electric field.
(Marks 1)

Ans4)
(Equipotential surfaces are parallel planes at 90o to field lines)

Q5) If the number of turns of a solenoid is doubled, keeping the other factors constant, how does the self-inductance of the solenoid change ? (Marks 1)
Ans5) The self inductance becomes four times on doubling the no of turns of solenoid since L = oN2A/e , L N2

Q6) What is the ration of solar constants on the surfaces of two planets, whose surface temperatures are in the ratio 1 :2 ? (Marks 1)
Ans6) Solar constant (Temp)4
S1/S2 = (T1/T2)4 = (1/2)4 = !/16
or S1 : S2 = 1 : 16

Q7) The wavelength of light coming from a distant galaxy is red-shifted. Is the galaxy receding or approaching the earth ? (Marks 1)
Ans7) The red shift indicates that the galaxy is receding

Q8) What is the angle of dip at a place where the horizontal and vertical components of earth’s magnetic field are equal ? (Marks 1)
Ans8) tan = BV/BH = 1
= 45o

Physic-2000-Set-II

Q1) Draw an equipotential surface for a point charge Q > 0.
(Marks 1)

Ans1) Equipotential surface are the surface of concentric spheres with the charge at their centre.

Q2) If the ratio of the horizontal component of earth’s magnetic field to the resultant magnetic field at a place is 1/2, what is the angle of dip at that place ? (Marks 1)
Ans2) Cos = BH/B = 1/2
= 45o

Q3) How does the energy gap in an intrinsic semiconductor vary, when doped with a pentavalent impurity ? (Marks 1)
Ans3) Decreases

Q5) Why is the transmission of signals using ground waves, restricted only to a frequency of 1500 kHz ? (Marks 1)
Ans5) The transmission of signals using ground waves is restricted only to a frequency of 1500 kHz because it becomes weaker as the frequency increases as they are damped by interaction with matter.

Q9) Four capacitors are connected as shown in the figure given below :

Calculate the equivalent capacitance between the points X and Y. (Marks 2)
Ans9)

The equivalent circuit is

C1, C2, C3 are in parallel. There equivalent capacitance = C’ =C1 + C2 + C3 = 20 + 30 + 50 = 100 F
This equivalent capacitance is in series with C4.
Net capacitance ‘C’ is given by
1/C = 1/C’ + 1/C4 = 1/100 + 1/100 = 1/50
or C = 50 F

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