Kvpy-2009 S
Kvpy-2009 S
Kvpy-2009 S
net
x3 + 3x + 2 = 0
PART-I no solution
One Mark Questions x = 1 only one solution
15. Number of integers between 2n and 2n + 1 is 30. As masses are equal for cude and sphere
2n + 1 – 2n – 1 So, ms = vs × density
and term = 2n + 1 me = ve × density
last term = 2n+1 – 1 as ms = me vs × density ve × density
Sn =
2 n1
2n 1 n
[2 + 1 + 2n + 1 – 1]
vs = ve
4 3
2 r s3
3
=
2 n1
2n 1 n
(2 )(1 + 2)
surface area comparison 6s2 > 4r2
2
CHEMISTRY
= (2n – 1)
2 .3 n
31. Silicon is tetravalent, so it forms SiCl4 .
2
Sn = 9 ; 32. NH4Cl undergoes sublimation while NaCl does not.
3 × 2n – 1 × (2n – 1) = 9
2n – 1 × (2n – 1) = 3 Oxidation
2n(2n – 1) = 6
It is possble when n is even. 34. SO2 + 2H2S 3S + 2H2O
Reduction
PHYSICS
17. Given mass of planet A = mA 36. As per Boyle’s law
and mass of planet B = mB PV = constant
mB = 8 mA 1
and P
Gm A GmB V
gA = r 2 and gB = r 2
A B
38. CaC2 + 2H2O Ca(OH)2 + C2 H2
2 Calcium Water Calcium Ethyne
gA mA rB
or .......(i) carbide hydroxide (acetylene)
gB mB rA
mB = 8 mA 39. As we move downwards in a group, atomic radii in-
creases. So the order will be Li < Na < K < Cs.
4 3 4 rB
rB d 8 rA3 d r = 2
3 3 A
40. CH3 – C CH (Propyne) and (cyclo propene)
gA 1 1 are two possible structural isomers of C3H4 .
4
gB 8 2
41. Propionic acid and methyl acetate both have same
gB = 2gA molecular formula (C3H6O2) but different functional
groups, so they are isomers.
25. A = r2
A A 42. N2 + 3H2 2NH3
2
A r Initial
no. of
A A moles 1 0.5 0
% 2 100
A r After
reaction 1 – 0.167 0.5 – 0.5
A 0.334 mole
% 2 0.15 = 0.30%
A 0.334 moles of NH3 = 2.0 × 1023 molecules
43.
235 207
27. 92 U 82 Pb 724 He 4 1e 0
CH2OCOR CH2 – OH
so, n = 4 ( particles)
–1
Alkaline
CHOCOR + 3NaOH CH – OH + 3RCOONa
Hydrolysis
Soap
29 . Bulbs are connected in series and resistance of 100W CH2OCOR CH2 – OH
is greater then that of 200 watt. Triester Glycerol
In series, P = I2 R (Here I is constant)
So, PR 44. Copper cannot displace lead from its solution as it is
So, power of 100W will be greater in the combination. less reactive than lead..
1 x6 r
=
1 x
a/2 a/2
It has 5 roots let 1, 2, 3, 4, 5 they are 6th roots of
unity except unity
Now, P(x12) =1 + x12 + x24 + x36 + x48 + x60
= P(x) Q(x) + R(x)
Here R(x) is a polynomial of maximum degree 4
Put x = 1, 2, .......5
we get Case-III
R(1) = 6, R(2) = 6, R(3) = 6, R(4) = 6, R(5) = 6
i.e. R(x) – 6 = 0 has 6 roots
Which contradict that R(x) is maximum of degree 4 r – a/4
So, it is an identity
R(x) = 6 a/2
30º
r
63. BD b and CE c a/4
A a/2
c D
E a/2
b
k r + a/4
h
B C
a
c sin A b ...(i)
and b sin A c ...(ii)
(i) + (ii)
(c + b) sin A b + c a a
sin A 1 A3 = a r r = 2ra
4 4
A = 90º
Now from (i) and (ii) Hence A1 = A2 = A3
c b and b c
c=b Hence angles are 45º, 45º, 90º PHYSICS
C
u=0
(37 )2 (7)2 – (33 )2 (37 )2 x 2 – (33 )2 h=?
64. cos = =
2.37.7 2.37.x
A 67.
d
37 33 37
Time taken by the photon, t = d/c
B 7 D x C
1 2
h=0×t+ gt
[(37)2 + (7)2 – (33)2] x = 7(31)2 + 7x2 – 7(33)2 2
x = 40
70. After removing charge from P, net force on central 75. On dilution, mili equivalent of the solute remains con-
charge will be : stant.
Initially pH of HCl = 4
Kq1q2 9 10 9 10 5 5 10 5 so normality of HCl = 10–4 N
F=
r2 12 after dilution pH of HCl = 5
F = 4.5 N so normality of HCl will be = 10–5 N
m = 0.5 kg N1V1 = N2V2
so, acceleration, 10–4 × 10 = 10–5 × V
V = 100 mL
F 4.5
a= = 9 m/s2 upwards So, 90 mL of water should be added for this pH change
M 0.5
CHEMISTRY
71. 2NaCl + H2SO4 Na2SO4 + 2HCl(g)
(X)
2HCl + CaCO3 CaCl2 + H2O + CO2
(X) (Y)