Solutions of IJSO Daily Practice Test- 3 (13-10-10)
Specimen copy of Resonance Distance Learning program for IJSO Plus. Visit @ www.edushoppee.com to purchase.Full description
nsejs-Paper-2014-15
Practice Problems for IJSO Stage-1Full description
Bio IJSO DPP
NSEJS-Paper-2012-13
Practice Problems for IJSO Stage-1Full description
NSEJS Paper 2008-09
NSEJS-Paper- 2010-11
NSEJS-Paper-2013-14
NSEJS-Paper-2011-12
IJSO Last Year PapersFull description
PRACTICE PROBLEMS FOR IJSO STAGE-1Full description
Practice Problems for IJSO Stage -1Full description
IJSO Solutions 18-10-10
Question Paper 2007 (MCQ) |International Junior Science Olympiad|Full description
IJSO QUESTION WITHANSWER (25-10-10) 1.
Two parallel lines are one unit apart. apar t. A circle of radius 2 touches one of the lines and cuts the other line. The area of the circular cap between the two parallel lines can be written in from of
3.
(i) Zn
a – b 3 . The sum (a + b) of the two inte3 gers a and b equal to : (A) 3 (B) 4 (C*) 5 (D) 6 C D1 1
A 2
Sol.
The values of standard oxidation potentials of following reactions are given below.
(ii) Fe
Fe
(iii) Cu
(iv) Ag
B 2
O
Sol.
In triangle AOD 1 2 = 60º AOB = 2 = 120º
Zn2+ + 2e – ; Eº = 0.762 V
2+
+ 2e – ; Eº = 0.440 V
Cu2+ + 2e – ; Eº = 0.345 V Ag1+ + e – ; Eº = –0.800 V
The following that can be easily reduced according to the above reactions is (A) Fe. (B*) Ag. (C) Zn. (D) Cu. As the oxidation potential of Ag is lowest among the given elements, it would be reduced most readily.
cos =
so net balanced equation
2 1 2 area of segment ACB = r sin2 r2 – 360 º 2
=
120 º 2 1 2 r sin120 r – 360 º 2
4 – 3
a 3 = 3 – b 3
The work done in taking a charge q once round another charge Q along a circular path of radius R (Q as centre) : (A*) Zero
(C)
Qq
R
4.
The linkage that holds monosaccharide units together in a polysaccharide is called : (A) Peptide linkage (B*) Glycoside linkage (C) Ester linkage (D) Ionic linkage