# 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.