**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.