Ijso Solutions 12-11-10

IJSO Question (WithansAns.) Date : 12-11-10 12-11-10 1.

Sol.

2.

Sol.

The number of ordered pairs (a, b) of positive integers such that a + b = 90 and their greatest common divisior is 6 equals. (A) 15 (B) 14 (C*) 8 (D) 10 Given a, b are positive integers such that a + b = 90 and their HCF is 6. Let a = 6k1 and b = 6k2 then 6k1 + 6k2 = 90 ! k1 + k2 = 15 So, k1 + k2 = 15 for which we have (1, 14), (2, 13), (4, ( 4, 11), (7, 8) But the value of k 1 and k2 can be interchanged. The radius of a circular wire is 0.5 m and the current is 10 amp. W hat is the magnitude of magnetic field at the centre of the circular wire ? (A*) 12.57 × 10 –6 T (B) 12.57 × 10+5 T (C) 12.57 10 –4 T (D) 12.57 × 10 –3 T Radius of the coil, r = 0.5 m Current, I = 10A B= B=

3.

%0 I

2r

22 4# 10 &7 10 =4× 7 2 0.5

×10 –6

88 10 &6 = 12.56 × 10 –6 T 7

A basic lining is given to a furnace by using (A*) Calcined dolomite (B) Lime stone

(C) Haematite

(D) Silica

Sol.

Dolomite on calcination gives CaO,.MgO which provides basic lining in furnace.

4.

Livestock is  – (A) all human beings (B) all wild and domestic animals (C*) all domestic useful animals (D) all the plants and animals