Solutions of IJSO Daily Practice Test- 3 (13-10-10)
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ISJO QUESTION WITHANSWER (16-10-10)
1.
The diagonals of a rhombus are 12 and 24. The radius of the circle inscribed in the rhombus, is 12
(A*)
5 6
(B)
5
(C) 6 5 (D) not possible as a circle in a rhombus can not be inscribed Sol.
D
C
6 A
12 B
65 AB =
6 2 12 2 = 6 5 Let h is the height of rhombus
Area of Rhombus =
1 × 12 × 24 2
1 × 12 × 24 = 6 5 ×h 2 h=
12 24 1 × 6 5 2
h = 2r =
r= 5.
24 5
12 5
The focal length of a concave mirror is f and the distance from the object to the principal focus is a. The magnitude of magnification obtained will be : (A) (f + a)/f
Sol.
(B*) f /a
Focal length of concave mirror = – f Distance of object from principal focus = a So distance of object from pole, u = –(f + a)
–
1 1 1 = f v u
1 1 1 1 1 1 f v f a v f a f
1 f f a f ( f a) v v f ( f a) a
magnification, m = –
v u
f ( f a) m = – m = – f/a a( f a ) magnitude of magnification , m = f/a
(C)
f / a
(D) f2 /a2
3.
The type of bond present in CCl4 is : (A) Ionic bond.
(B*) Covalent bond.
(C) Electr Electrova ovalen lentt bon bond. d. (D) Dative Dative bon bond. d. Sol.
CCl4 is made of one carbon atom and 4 chlorine atoms.The carbon atom has 4 outermost electrons which shares share s with 4 valence electrons of four chlorine atoms(each chlorine atom has one valence electron).Thus, CCl4 has four single covalent bonds.
4.
A red and tall dominant character hybrid plant is crossed with recessive white dwarf plant (RrTt × rr tt). What W hat will be the ratio of respective four combinations : red tall, red dwarf, white tall and white dwarf plants in the next generation gener ation ? (A) 9 : 3 : 3 : 1 (B) 15 : 1 : 0 : 0 (C) 9 : 3 : 4 : 0