HINTS & SOLUTIONS (Y (YEAR EAR-2 -20 013 13-1 -14) 4)_ _ NSEJ S (ST (STAGE AGE-I) -I) ANSWER KEY Ques.
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
A n s
a
d
b
b
d
b
d
c
c
a
b
d
b
b
b
a
b
a
c
a
Ques.
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
A n s
c
a
c
d
c
b
c
b
d
b
c
a
a
d
c
d
b
b
c
b
Ques.
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
A n s
d
c
a
c
b
a
d
a
c
** **
c
b
c
c
b
a
b
d
c
c
Ques.
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77
78
79
80
A n s
d
c
d
c
c
c
c
a
c
d
a
a
d
b
a
c
b
d
a
d
1.
4.
Pea is a kharif crop and it cannot be grown in summur.
5.
Food was distrubted and fed equally. equall y.
6.
Grasshopper Grasshopper has not stored food f or future use.
= 180 – 2 × 45 = 90
H
7. 2.
41
P.E. = K.E. mgh = 100 mg (20sin) = 100 (1) (10) (20sin ) = 100 sin = 1/2 = 30º
= 6.40 ... 5 41 = 5 6.401
= 32.01
3.
Pt Cl2 (NH (NH3)2 OS x–2=0 x = +2 coordination number = no. of ligands that that form coordinate bond with with central metal atom. atom . coordination number = 4
0 m 2
8.
3
3
2 way
171
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3
3
3
12. 12.
0.025 KOH solution [OH – ] = 25 × 10 –3 M pOH = –log [25 × 10 –3] pOH = – log(5) 2 + 3log10 pOH = (–2 × 0.6990) + 3 log 10 = 1 pOH = –1.39 + 3 pOH = 1.602 pH + pOH = 14 pH + 1.602 = 14 pH = 14 – 1.602 pH = 12.398 = 12.40
14. 14.
xy2 = a3, yz2 = b 3 , zx2 = c 3
3
on multiplying x3y3z3 = a3b3c3 xyz = abc 2 w ay
9.
xy 2
2 w ay
Total 6 ways
yz2
O > N > S > Br [Decreasing order of EN]
xy
10. 10.
11.
F = P( A) F = gh( A)
b3 a3 b3
a3
xy =
sides of an equilateral triangle are
a3
=
=
z2
..(i)
b
z2
3
...(ii)
substitute substitute in (i)
3x + 3y – 1 ,4x 2 + y – 5 ,4x + 2y 4x + 2y = 3x + 3y – 1
z
3
y=x+1
a3 b
3
= abc
4x2 + y – 5 = 4x + 2y 2y 4x2 – 4x – y = 5
z = 3
4x2 – 4x 4x – x – 1 = 5
abc b 3 a3
4x2 – 5x – 6 = 0 x = 2, x = – 3/4 ( not possible possible ) 3
z =
so x = 2, y = 3 side of triangle (a) = 14 Area of triangle triangle =
=
3 4
142
= 84.868 It is nearer to 85
3 4
a2
15. 15.
b 4c a2
nC12H22O11 =
3.4 342
= 0.0099 1 mole C H O = 22 × N H-atoms 12 22 11 1 A 0.0099 mole C12O22O11 = 22 × 6.023 × 1023× 0.0099 = 1.3 × 1023 16. 16.
(a) Mountain streems has most fresh water and diatoms grow well in fresh f resh water & it sewage canal has more pollution so the diatoms will grow less, Lake water has less diversity because because of because because the of little pollution.
172
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17. 17.
Because of of W ater pollution different varities of diatom will not grow
18. 18.
Different Diff erent species species have diff erent structure. structure.
19. 19.
On Surface of earth g s =
GM R2
GM At height R, gh =
( h = R)
(R R )2
23. 23.
given by n2 closest integer to , n is even 48
GM gh = 4R 2
g =
No. of integer sided triangle with perimeter n i s
GM GM – 4R 2 R 2
1 1 = – 3 g 4 4
g = gs
(n 3)2 closest integer to , n is odd 48
s
m=
Percentage change in g is
g gs
x100= –
3 × 100 %= – 75 % 4
n =
(decreased by 75%) 20. 20.
m – n= 7 –5 = 2
1 4
and b = 3 then
1 a
24. 24.
Density(d) =
= 4
which satisfy the condition a < b <
mass volume
= 0.7979 kg
moleculer mass of solute = 27 × 12 + 28 + 2 × 80 + 5 × 16 + 32 = 624
1 a
No. of moles of solute =
1 1 x=a+ – b b a
22. 22.
256 162 = , closest integer = 5 48 48
given condition condit ion is possible only when 0 < a < 1 take a =
21. 21.
324 (15 3)2 = , closest closest integer integer = 7 48 48
x=a+
1 1 – b – a b
Molality(m) =
x=4+
1 1 – 3– 4 3
=
x = 1–
1 12
25. 25.
100 624
= 0.16025 moles
No. of moles of solute mass of solvent in kg
0.16025 = 0.201 mol kg-1 0.7979
R50 > R100
x>0 Standard Standard enthalpy enthalpy of f ormation is 0 kJ/mole f or Na(s).
2 V () So in Parallel Parallel P = P 1 R R
Initially acc. is +ve and constant constant so s = ut
() in series P = I2 R P R So 50 W bulb will glow more.
+
1 a t2 and then particle decelerates so s = 2 0
ut -
So 100W bulb will glow more
1 a0t2 2
173
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26. 26.
abc 3
let at 4 hours x m in angle is 120º
11
(b + 3)
(a + c)
max va value
max value
12
18
min value
min value
3
1
x =
1 1 1
18
7
21 2
17
6
3 1 3
16
5
4 1 4
15
4
5
14
3
12
31. 31.
8
11
11
8
10
10
9
9
9
8
8
8
7
7
7
6
6
6
5
5
5
4
4
4
3
3
3
1
1
12
=
M2 = 10 10 –3 M2V2 = M3V3 10 –3 × 10 = M 3 × 100 M3 =
10 – 3 10 100
M3 = 10 –4 pH = – log [10 –4] pH = 4
50.01 103 1.01325 105
V=
255 1000
30. 30.
Full course of Antibiotics should be taken to cure the diseases. At 4 o’ clock angle between between min and hour
L
T = 288 K R = 0.0821 L atm K –1 mol –1 m = 0.5755 g mRT PV
M =
M=
1.01325 10 5 1000 0.5755 0.0821 288 50.01 10 3 255
M = 108 g form ula of compound is SF 4. The molecular formula 32.
(a) zees.
33. 33.
P A = PB
Humans didnot evolv e from chimpan-
1.6 × g × 26. 6 =
B 28. 28.
atm
V = 255 mL
M1V1 = M2V2 0.1 × 1 = M 2 × 100 0.1 1 M2 = 100
mRT M
P = 50.01 KP a = 50.01 × 10 3 Pa
81 27. 27.
Following PV = nRT PV =
12
480 11
7 min = 43 min 38 sec] 11
= 43
No. of ways a c b 3
7
11 xº – 30º 4 = 120º 2
so angle is
34. 34.
=
B × g × 50
1.6 26.6 = 0.8512 50
a,b are two positive real numbers a2
b2 ab
hand = 120º
a b
= 6
b
= 6 a
174
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Let x =
42. 42.
a
x3 – 3ax 2 + 3ax 3ax – a = 0 a3 – 3a3 + 3a2 – a = 0
b
– 2a3 + 3a2 – a = 0 x+
1
a(2a2 – 3a + 1) = 0
=6
x
a = 0 or 2a or 2a2 – 2a – a + 1 = 0
x – 6x + 1 = 0 2
x=
x=
2a (a – 1) – 1 (a – 1) = 0
6 32
(a – 1) (2a – 1) = 0
2 1
a = 0, 1, 1/2 number of values v alues of a is 2 as at a = 1/2 equation
64 2 2
43. 43.
x = 32 2
does not have all roots real. Molecular formula = C4H8 Isomers
x = 3 + 2 2 or x = 3 - 2 2 x = 3 + 2 1.41 or x = 3 – 2 1.414 x = 3 + 2. 2.82 or x = 3 - 2.828 x = 5.82 or x = 0.172 (not in option) So x =
Total 5 isomers
a lies between 5 and 6 b
35. 35.
and
Position of Cl is different both are position isomers.
44. 44.
(c) Iodine is important for the actvation of thyroxin hormone.
45. 45.
EK = hf – On comparing this with y = mx + c Slope m = h & intercept c = –
36. 36.
46. 46.
Area of lawn Area of pavement
(d) Biodiversity means variaties species species in a particular area and different species has different diff erent gens and and different diff erent ecosystem
r 22 25 2 2 r 1 r 2 24
r 1 r 2
38. 38.
ATQ ATQ x2 + y2 +z2 = 1000 (x + y + z)2 = 2500 x2 + y2+ z2 + 2(xy + yz + zx) = 2500 2(xy + yz + zx) = 1500 xy + yz + zx = 750
39. 39.
Ag is less less reactive than hydrogen hydrogen Ag + dil. HCl No reaction
40. 40.
(b) Both the explanation explanati on for the squirt of horned lizard are correct.
41. 41.
r 1
7
r = 5 2 perimeter outer peremeter inner
=
r 1 r 1 7 = = r 2 r 2 5
47. 47.
Boiling can remov e only temporary hardness.
48. 48.
Companion cell and sieve tube t ube originate same mother cell so they are associated with each other.
Q = mS 96 = m × 0.8 × 6 m=
96 0.8 6
= 20 g 51. 51.
4PCl3 t=0 1 t = equi. 1–x
P4 + 6Cl2 0 x
0 6x 175
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KC =
KC =
59. 59.
[P4 ][Cl2 ]6
concept of chemistry when ph changed it
[PCl3 ] 4
caused loss of tertiary structure.
(6x )6 x (1 – x )
It is a biochemistry question and according according to
60. 60.
Nucleus controls controls all the cellular activ ities.
4
52. 52.
Placenta is layer that provides prov ides food, respiration and elemination of the waste of the foetes.
53. 53.
altitude = 350 × 6 = 2100 m. = 2.1 km. 62. 62.
54. 54.
r A = P 1 100
2
r 96800 = P 1 100
2
...(1)
r compound half yearly, y early, 97240 = P 1 200 .
P lies an y =
4
... (2)
After solving we get r = 10% 55. 55.
Polyalkanes are not not inflammable. infl ammable.
56. 56.
Receptor receive the message that is taken by sensory sensory neuron to the CNS and the reply of t he message message is given giv en by motor neuron to the effector.
57. 57.
Body B & C should have same kind of charge while body A may either have opposite charge or be neutral.
58. 58.
Equation formed is : 13x + 35y = 1000 x y both should be integer So, x = 15, y = 23 1st integer value which satisfy the equation next, x = 50, y = 10 So maximum of x + y = 60, is means 60 max. items can be purchased and minimum minim um of x + y = 38
(x + y)max = 60 = m (x + y)min = 38 = n
m + n = 98
x 3
= 50 K =
h 3
h = 3K
P (3K,K) Now OAP is isoscles if (i) OP = PA (ii) PA = OA (iii) OA = OP Coee (i) OP = PA OP2 = PA PA2 (3K – 0)2 + (K – 0)2 = (3K – 5)2 + (K – 0) 2 9k2 + k 2 = 9k2 – 30k + 25 + k 2 30K = 25 K=
25 30
K =
5 5 5 so P , 2 6 6
Case (ii) PA = OA OA PA2 = OA2 (3K – 5)2 + (K – O) 2 = 52 9K2 – 30 K + 25 + K 2 = 25 10K2 – 30K = 0 10K ( K – 3) = 0 K = 0 / K = 3 P (0 (0,0) ,0) {n {not po possible ible}} P (9 (9,3) ,3) Case (iii) OA2 = OP2 52 = (3k – O)2 + (K – O) 2 25 = 9k2 + k2 10k2 = 25 5 K2 = 2 5 K=± 2
5 5 P 3 2, 2 5 5 – 3 ,– P 2 2 total 4 points are there 176
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68. 68.
63. 63.
– [A] = 2[C]
64. 64.
Ovum Ov um has X-chromosome and sperms has Xchromosome so the child child will be girl.
Deodar and pinus belong to gymnosperms. m = 2kg
69. 69.
m = 3kg
u1 = 4m/s
u2 = 1m/s m1 + m2
N = 20N
after collision 2kg
65. 65.
v
2kg
F = 20N
v =0 x = 10m
f = 6N
m1 u1 + m2 u2 = (m1 + m2) v 2 × 4 – 3× 1 = 5v v = 1 m/s
Initially, u = 0 a=
F f 20 6 = = 7m/s2 2 2
Initial K.E. K1 =
V2 = u2 + 2aS V2 = 02 + 2 × 7 × 4 V2 = 56 V=
+
3 = 17.5 J 2
56 m/s Final K.E. K2 =
1 1 mv 2 = × 2 × 56 = 56J 2 2
K.E. =
x3 = a + 1 x+ x2
b x
... ........... (i)
= a
b
= a x x2 – ax + b = 0 (i) – (ii) x x3 – a – 1 = 0 x3 – ax2 + bx = 0 – – ____ ______ ____ ____ ____ ____ ____ ___ _ 2 ax – bx – a – 1 = 0 (iii) – (ii) a ax2 – bx – a – 1 = 0 ax2 – a2x + ab = 0 – + – ____ ______ ____ ____ ____ ____ ____ ____ __ 2 (a – b)x = ab + a + 1 x=
67. 67.
given (ab)2 = (bc) 4 = (ca)x = abc = k (Let) ab = k1/2 bc = k1/4 ca = k1/x abc = k now. abc = k 2
.......(ii)
2 2
(abc)2 = k
1 1 1 2 4 x
1 1 1 2 4 x
1 1 1
= k
2
k
2=
.......... ....(iii iii)
b
Cu2+(aq) + M(s) Cu(s) + M2+(aq) EºCell = Eºcathode – Eºanode
1 2
1 4
2 4 x
1 x
4 5 Alpha (due to high high mass) x=
71. 71. 72. 72.
Cell Nucleus Chromosomes DNA Protein
73. 73.
Current due to electrons I1
ab a 1 a2
1 5 (2 +3) (1) 2 = J 2 2
Loss in energy = 17.5 – 2.5 = 15 J 70. 70.
66. 66.
1 1 2(4)2 + 3(1)2 = 16 2 2
2 1016 1.6 10 19 = 2 Current due to protons I 2
0.75 = 0.34 – EºM2
2 1016 1.6 10 19 = 2
EºM2 = 0.34 – 0.75
So total current I = I 1 + I2 =
EºM2 = – 0.410 V
2 1016 1.6 10 19 = 3.2 × 10 –3
= 3.2 mA 177
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74.
x = 1–
+
1 2
+
1 3
–
1 4
+
1 5
–
1 6
+ .......... –
so, slant area of original cane = rl
1 2012
4r 2) = 9(h2 + r 2) 4(h2 + 4r 4h2 + 16r 2 = 9h2 + 9r 2 5h2 7r 2 = 5h
1 2013
1 1 1 x = 1 3 5 ..... 2013
7r 2 = h2 5
1 1 1 1 ..... 2012 2 4 6
Now 4r
–
1 1 .... 3 1006
12
= K 2
integer closest to K is 11 79. 79.
x =
16 87
116
1 1 1 1 1 1 .... 1 .... 1006 2013 1007 1008 2 3
87r 2 12r 2 2 = K 5 5
16
x =
– 1
16r 2 = K r h2 r 2
7r 2 7r 2 2 2 16 r 2 r = K 16 5 5
1 1 1 1 1 x = 1 2 3 4 ..... 2012 2013
2
7r 2 put h = 5
1 1 1 1 ...... 2012 2 4 6
1 2
h
2
– 2
(i) (i)
16 (h2 + 16r 2) = K2 (h2 + r 2)
1 1 1 1 1 x = 1 2 3 4 ..... 2012 2013
– 1
2r h 2 4r 2 = 3r h 2 r 2
ATQ ATQ
1 1 1 .... 2 3 1006
1 1 1 + + ....+ 1007 1008 2013
75. 75.
Mole =
0.2 =
Given mass Gram molecular mass
W 32
W = 6.4 g 80. 80.
Wuchereria is an organisms that belong to nematoda phylum.
Due to presence of lp shape is triangular pyramidal like NH3. CO32– , NO3 – , SO3 all are sp2 hybridised. 76. 76.
To conserved the environment person should first minimize minim ize the use of plastics
78. 78.
Let radius, height and slant height be r, h and l 178
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