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IJSO QUESTION WITHANSWER (23-10-10) 1.
Sol.
2.
The equation 8x + 4 = 4x + 2x + 2 has (A) no real solutions (B) one real solution (C*) two real solutions (D) three real solutions 8x + 4 = 4x + 2x + 2 23x + 4 = 22x + 4 × 2x Let 2x = y y3 + 4 = y2 + 4y y3 – y2 – 4y + 4 = 0 y2 (y – 1) – 4(y – 1) = 0 (y2 – 4)(y – 1) = 0 y = 2, –2, 1 2x = 2, x=1 2x = – 2, that is not possible 2x = 1 x=0 two real solution for x
B2 =
0I2
2
2r 2r Net magnetic field at centre 0I11 I B = B 1 – B2 = – 0 2 2 2r 2r 2r 2r 0
B= 3.
4 r 2
(I11 – I22 ) = 0
In the balanced chemical reaction a, b, c and d respectively correspond to O13 – a1 – bH cH2O d 2
Sol.
(A*) 5,6,3,6 (B) 5,3,6,3 (C) 3,5,3,6 (D) 5,6,5,5 First remove a, b, c and d and balance the equation by ion electron method. O13 – 1 – H H2O 2
on splitting above equation into half reactions. Oxidation half reaction :
Any two ends of a circular conducting wire are connected by a cell. The magnetic field at the centre O is:
I
–
I2
2I –1 I2 + 2e –
- - - - (i)
Reduction half reaction IO
1 –
IO
1 –
3
3
(A)
2 2 o (i1 i2 )
2r (C*) zero
(B)
I
2
I
2
+ 3H2O
O13 – 6H 5e – 2 3H2O - - - - (ii)
o (i1 i2 )
2r (D) None of these
2I – I2 + 2e –
---
- (i) × 5
Sol.
i
1 –
O3 6H 5e 2 3H2O
2
i
2O13 –
1
O
R1 R2
I1 2 1 but I so 2 I2 1 R I11 = I22 ...(i) Magnetic field at centre due to current I1 0I1 . ( here N = 1 ) 1 O B2 = 2r 2r 2r Magnetic field due to current I2 1
---
12H 12 2 6H2O
so net balanced equation O13 – 51 – 6H 6 2 3H2O
Then the resistance will will be in the ratio of the and 2
1 –
10
- (ii) × 2
Let the length 1 and 2 1
–
4.
Sickle cell anaemia is disease in which patient suffering from a disorder of respiration and transportation system.This disease is known as (A) Metabolic disorder (B*) Genetic disorder (C) Degenerative disorder (D) Pathogenic disorder