**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.

**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.

** (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.

** (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

**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