IJSO Solution 20-10-10

IJSO QUESTION WITHANSWER (20-10-10) 1.

All the 3 sides of a right triangle are integers and one side has a length 11 units. Area Area of the triangle in square units lies between (A) 1 and 100 (B) 100 and 200 (C) 200 and 300 (D*) More than 300

Sol.

3.

The total number of structural isomers possible for the hydrocarbon pentane (C5H12) is

Sol.

(A) 2.

(B*) 3.

(C) 4.

(D) 5.

The molecular formula of pentane is C5H12. It has 5 carbon atoms in its chain. So, only two

c

arrangements are possible other than the nor-

a

mal structure. CH3 – CH2 – CH2 – CH2 – CH3

b = 11

n-Pentane

c2 = a2 + b2 c2  – a2 = 11 112 c2  – a2 = 121 (c + a) (c  – a) = 121 × 1

CH – CH  – CH  – CH 3

2

3

CH iso-pentane 3

c + a = 121 c  – a = 1 2c = 122 CH

c = 61 a = 60

CH – C  – CH 3

1 1 ab  – × 60 × 11 = 330 2 2 Current in 4 resistance resistanc e is : (Battery is ideal)

3

CH neo-pentane

Area =

2.

3

3

4.

Binomial nomenclature is given by Carolous linneaus according to this nomenclature each scientific name is consist of two words (A*) Genus and species (B) Order and family

5 A 3 2 (C) A 3 62 3 Req = = 8 2 10 2 ×  i1 = 8 3

(A)

Sol.

( B*)

5 A 6

(C) Family and genus (D) Species and variety

(D ) None o f thes e 

i=

52 10 = 3 3

=

5 A 6