PART TEST-II

GZ, GZK & GZS

JEE MAIN PATTERN
Test Date: 10-03-2017
[2] PT-2 (Main) GZ & GZK & GZS_10-03-17

PHYSICS
1. T  kp x d y E z

[T ]  [ ML1T 2 ] x [ ML3 ] y [ ML2T 2 ] z

x yz0 …(1)

 x  3y  2z  0 …(2)

 2 x  2z  1 …(3)

1
xz …(4)
2
1
y
2
1
y
2

3
by equation (2)  x   2z  0
2
3
 x  2z  …(5)
2
Adding (4) & (5)

1
3 z  1, z 
3
1 1 5
x  
2 3 6
 (A)

1
2. Kinetic energy E  mv 2
2

E v 2  v 2
  100  2
 100 = [(1.5)2 – 1] × 100
E v
E
  100  125 %
E
 (D)

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3. [] = L

2 [ L]1/ 2
–2
ML T =
[ L]
5/ 2 2
 = [ M ][ L ][T ]

 (D)

4. Let after t second particle will reach at P again, v (m/s)

 area of v – t curve = 0

1 1 2
 2  8   (t  8)  (t  8)  1  0
2 2

(t  8)2  16 8 t t (s)
P 6
t8  4
-v
t  12s
 (C)

5. v H  u cos   6 , v v  v 2  u2 cos2   8

u sin   8 u sin   8 82

t1  , t2  , t 2  t1   1.6 s
10 10 10

 (D)
1
d 2 1
6. t   hr
um2  u2r 2
4 3 2
2 7

 (B)

7. ucos53  v cos37
3 4
 100   v   v  75m/s
5 5
u=100m/s
v y   v sin37  45m/s 37°
53° 37°
v
uy  u sin53  80m/s

v y  uy  gt  45  80  10t

t  12.5s
 (B)

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8. For collision,

vA sin  = vB sin 60° vB= 10 3 m/s

vA=25m/s
3
25 sin  = 10 3   60°
2
x
3
sin  =
5

or = 37°

 (D)

9. Distance travelled along OE in 2s = 4 × 2 = 8 m

1 2 1  6 2
Distance travelled perpendicular to OE in 2s = at    2 = 6 m
2 2  2

Displacement = 62  82 = 10 m

 (D)

10. In condition (i), 20g  T  20a , N  20a

20g g
T  N  40a  a 
80 4

2g g
Net acceleration = a1  a 2 , 2a  
4 2 2

g g
In condition (ii) 20g  T = 20a , T  40a , a  , a2 
3 3

a1 g / 2 2 3
 
a2 g/3 2 2

 (A)

mg ma
11. T  …(i)
2 2

ma
T cos 60   …(ii)
cos 60

2g
Solving (i) and (ii), acceleration of ring =
9
 (C)

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12. (B)

N2 N cos
N1
N cos
N a1 a2 N
N sin N sin
mg sin mg sin
mg cos mg cos
mg mg

mg sin   N cos   ma1 …(i)

N cos   mg sin   m a 2 …(ii)

From constraint relation,

a1  a 2

 N  mg tan 

at
13. (B)  acc
m1  m 2

at sliding pseudo force on m1 = friction force

m1at 0
  km1g
(m1  m2 )

14. Total length of string is constant.

x2 = 7x1

a 2  7a1

 (D)

dU
15. F  = – 5(2x – 4) at mean position F = 0  x = 2 m,
dx


U min = – 20 J as F  x and k = 10 N/m  T = sec.
5

 (C)

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16. v b = 20 m/s < 5gl

 Particle leaves the circular path.

2
20  gl(2  3 cos)  cos  
3

20 2
Velocity at top =  
gl cos cos =
3 3
m/s

 (B)

17. (B)

v2
ac  k 2 rt 2 or  k 2 rt 2 or v = krt
r
dv
Therefore, tangential acceleration, a   kr
dt
or tangential force, Ft  mat  mkr
Only tangential force does work,
Power = Ft v  ( mkr)(krt ) or Power = mk 2 r 2 t

18. W = area of F – x curve  20  2  10  1  30J

By work-energy theorem Wnet  K .E.

1
 30   5 v2 , v  2 3 ms1
2

 (C)

2m1u A u 2 gh
19. u A  2 gh , vB   A 
m1  m2 2 2

2 gh
To complete vertical circle, v B2  5 gR   5 gR
4

 h  10R

 (D)

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20. v  s 2 / 3

ds
 s 2 / 3
dt

s t
ds
0 s 2 / 3  0  dt  3 s1/ 3   t

3 3 3 2
s t  ds = t dt
27 9

3 2
v t
9

2
1 1   3t 2  m 6 t 4
W  mv 2  m  
2 2  9  162

 (B)

21. By conservation of linear momentum m 1u1  m 2u 2  m1v 1  m 2 v 2

u1 v  v1 u1  0 2
3u1  2u 2  0  2u1 , u2  , e 2  =
2 u1  u2 u 3
u1  1
2

 (C)

22. Before explosion, particle was moving along x-axis, i.e., it has no y-component of velocity.
Therefore, the centre of mass will not move in y-direction or we can say y com  0.

m1y1  m 2 y 2
Now, y com 
m1  m 2

(m / 4)( 15)  (3m / 4)(y)

Therefore, 0  or y  5cm
(m / 4  3m / 4)

 (A)

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 Area of force-time graph 

23. Impulse   under the specified interval
 

1 1 –3
 (800  200)  2  10 6   800  10  10 6 = 5 × 10 N-s
2 2

 (A)

2h
24. Let height be h, so t 
g

and v  2gh  gt

2ev 2e 2 v
Now Tt   .......
g g

2ev
 t (1  e  e 2  ...)
g

2gt  e   1 e
t  t
g  1  e   1  e 

 (C)

25. According to theorem of parallel axes,

2 2
2  R 2  R 1 21
I M    M(2R)2  M    4MR 2  MR 2  MR 2
5  2 5  2 5 5

 (A)

26. Due to pure translation point P has a


translational velocity v towards x axis and due P v+rsin
r
to pure rotation about its center of mass it has  
C r
a velocity r as shown in the figure.
r cos 
 (B)

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R2  3R 
R 2 (0) 
14  4  R
27. C.M. = 2
=–
R  20
R 2 
16

 (A)

28. T sin   Mr 2 = M(L sin  ) 2

 T  ML 2  ML(2  n)2

Tcos
2
 2 T
T = 0.1  1   2  
  Tsin
Mg
T = 1.6 N

 (B)

29.: mg  (l / 2)  P  (l / 2)  mg(l / 2)

P  mg  mg and P  mg  ma

hence a  g(1  2 )

 (A)

1 2 2h 2  5
30. h at …(i) or a    2.5m / s2
2 t2 4

mg  T  ma …(ii)

  I  RT …(iii)

RT R 2T R2m(g  a) 0.25  2  (10  2.5)

I   [a  R]   1.5kg  m 2
 a a 2.5

 (D)

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CHEMISTRY
31. (A)
meqH  meqH2SO4  meq HCl  meqHNO3

= 50  10  25  12  40  5
= 1000
eqH  1

V  1 lit  N  1 N

32. (A)
h = ho + KE

E E
E=  KE ;  K.E =
2 2

h E
 ho = = .
2 2

h 6.6  10 34 0.6947  10 18

= = = .
2mKE 34 E E
2  9.1  10 
2
33. (A)

8
The oxidation state of M changes from  to + 2, it is change of 0.67, there are three M
3
resent per mole of M3O4 , the net change of electron is 0.67  3  2.0

at wt
 Eq.Wt. of M =
2
34. (A)
Balmer series lines lies in visible region.
35. (C)
0.50 0.20
Mole of A = = 0.0083 ; Mole of B = = 0.0044 ; Total mole = 0.0127
60 45
mole of A
Total pressure = 750 mm ; Partial pressure of A = total moles x total pres-
sure
0.0083 0.0044
= x 750 = 490 mm ; Partial pressure of B = x 750 = 260 mm
0.0127 0.0127

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36. (A)

 a 
 P  2  V  RT
 V 

a
or PV   RT
V

PV a
or  1
RT RTV

PV  a 
or  1  Z
RT  RTV 

37. (B)

h 6.625  1034  1  1012 6.625

P h/   P      10 32
2 14 25
 25  10

h 6.626  1034 25
x     102  0.02 m
4P 6.626 4  3.14
 1032  4  3.14
25

38. (D)
Hardness = 200 PPm = 200 g CaCO3 in 106g water

mCaCO3 200 0.02

(A)  6

mH2O 10 100

nCaCO3 200 / 100 2  18

(B)  6
  3.6  105
mH2 O 10 / 18 106

(C) mCaCO  200g in 106 g ~ ml

3

2g in 104 ml
0.2 g in 1 lit

0.2
(D) 1 kg water = 100 g ; mCaCO3   0.002
100

eqCaCO3  0.002  2  0.004 eq

 4 meq.

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39. (B)
V.f of K 2 Cr2O7  6 & v.f. of KMnO 4  5

So, K 2Cr2O7 will oxidise more Fe2 

40. (A)
wtFeO a

wtFe2O3 b

nFeO a / 72

nFe2O3 b / 160

nFe a / 72  2  b / 160 7 / 56
 
no b / 72  3  b / 160 2.5 / 16
a : b = 9 : 10
41. (A) 42. (C) 43. (B) 44. (C)
45. (C) 46. (B) 47. (D) 48. (A)
49. (D) 50. (C)
51. (D)
150ºC 200ºC 
MgSO4 .7H2O  MgSO4 .H2O  MgSO4   MgO  SO3
52. (B)
1
NaNO3 (S) 
 NaNO2  O2
2

4LiNO3   2Li2O  2NO2 (g)  O2 (g)

MgCO3   MgO  CO2

Mg(OH)2   MgO  H2O
53. (B)
Ammoniated solution of sodium is highly paramagnetic blue coloured solution which act as
strong R.A.
54. (C)
Factual
55. (A)
Electronegativity values of 2nd and 3rd period are as following.

Li Be B C N O F
1 1.5 2 2.5 3 3.5 4
Na Mg Al Si P S Cl
0.9 1.2 1.5 1.8 2.1 2.44 3

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56. (D)
57. (D)
Ea
Li  e  Li2 exothermic
2S' 2S

58. (A)
acidic strength  O.S of metal (for oxides)
59. (C)
(i)  Hy  Number of +ve charge

1
(ii) for isoelectronic, r 
Atomic number
60. (A)
4KO2  2CO2  2K 2CO3  3O2

MATHEMATICS
61. (B)
x 2  2(k  1)x  k 2  k  7  0.
for both roots negative
2(k  1)  0  k  1......(1)

D  0  4(k  1)2  4(k 2  k  7)  0  k  6......(2)

Also, k 2  k  7  0
True  k......(3)
62. (A)
Centre of circle S = 0 lies on radical axis of two given circle
radical axis is x = 0
For smallest radius of S = 0
Centre should be intersection point of radical axis and line joining their centres
So, it is (0, 0).
radius = length of tangent drawn from (0, 0) to each circle  16  4
63. (B)
sin x – 3 sin 2x + sin 3x – cos x – 3 cos 2x + cos 3x
 2 sin 2x cos x – 3 sin 2x = 2 cos 2x cos x – 3 cos 2x
 (2cos x – 3) sin 2x = cos 2x (2cos x – 3)
2 cos x - 3  0  tan 2x = 1
n
 x = 2  8
Hence (b) is the correct answer.

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64. (B)

a2  2a  0

6a  a2  8
& 0
a2  6a  5

 a  ( , 2]  [0,1)  (2,4)  (5, )

65. (B)
A

Another fixed point will be image of P (1,2)

x+y–5=0

with respect to x + y – 5 = 0, (3,4)

P (1,2)
66. (D)
Base AC = 4
Corresponding maximum height = 2

1
So, Maximum Area = 4  2  4 sq.unit
2
So, possible Area = 1, 2, 3, 4
Corresponding to area 1 , 2 and 3
Number of vertex for B are = 12
and for Area = 4
Number of vertex B are 2
So, number of vertex = 12 + 2 = 14
67. (D)

 x 2  2x  n  0

2  4  4n
x  1  1  n
2
for integer roots, n + 1 = perfect square of any integer
n = 8, 15,....,80, 99.
Total no. of values of n = 8
68. (C)
L1  4x – 7y + 10 = 0 & L2  7x + 4y = 15
are perpendicular
Hence, triangle is right angled. Orthocentre is point of intersection of L1 & L2
i.e., (1, 2)

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69. (B)
Equation of AB A
P(5,5)
H
5x + 5y – 9 = 0

Orthocentre of  PAB is H
B
H will be image of (0,0) with respect to 5x + 5y – 9 = 0

9 9
i.e.  , 
5 5
70. (A)
C’  reflection of C in x-axis  (5, –3)

C’(5, – 3)

B
• y=0

A(1, 2) C(5, 3)

 equation of AB (or AC’)

 5
(y – 2) =   4  (x – 1)
 
 5x + 4y = 13
71. (D)
1 1 1 1
, , ,
a H1 H2 b are in A.P..
We know that 2A  a  b and G 2  ab
1 1 1
2   .
H1 a H2

1 1 1
Similarly, 2   
H2 b H1
On adding and solving we get,
 1 1   1 1  1 1
2     
 H1 H2   H1 H2  a b
1 1 a  b 2A
   2.
H1 H2 ab G

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72. (A)
From sin x + sin2x = 1, we get sin x = cos2x, Now, the given expression is equal to
cos6x (cos6x + 3cos4x + 3cos2x + 1) – 1 = cos6x (cos2x + 1)3 – 1
= sin3x (sinx + 1)3 – 1 = (sin2x + sinx)3 – 1 = 1 –1 = 0
73. (A)
A.M .  G.M .
1 1
a  a  ...  an1  2an
 1 2 n
 (a1,a2,....an 12an )  (2c) n
n
 Minimum value of
a1  a2  .....  an1  2an  n(2c)1/n.

74. (D)

1
ab  log4 5.log5 6  log4 6  log 6
2 2

1
ab  (1  log2 3)  2ab  1  log2 3
2

1
 log3 2  .
2ab  1
75. (A)

(2 sin2 91  1)(2 sin2 92  1)...(2 sin2 180  1)

In this product there exists a factor

(2 sin2 135  1) which is equal to zero.

 The product of all terms is zero.

76. (C)

 sin x
The given equation can be written as  2 cos x  1 + sin x = 2 cos2x = 2(1 –
cos x
sin2x)
 2 sin2x + sin x – 1 = 0  (1 + sin x)(2 sin x – 1) = 0  sin x = 1 or 1/2
Now sin x = 1  tan x and sec x not defined. Sin x = 1/2  x = /6 or 5/6.
 The required number of solution is 2.

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77. (A)

1 1 1
   ....  
(1  a)(2  a) (2  a)(3  a) (3  a)(4  a)

nth term of series

1 1 1
Tn   
(n  a)(n  1  a) n  a n  1  a

1 1 1 1
T1   ;T2   , T3  1  1
1 a 2  a 2a 3a 3a 4a
..........................
..........................

1 1 1 1
Tn1   ,Tn  
n  1 a n  a n  a n  1 a

 S n  T1  T2  T3  ........  Tn

1 1 n
  
1  a n  1  a (1  a)(n  1  a)

1
Sn   S  Sn,
 1 a
(1  a)  1   
 n n
when n  

1
 S  .
(1  a)
78. (B)

3
cos(A  B)  cos(B  C)  cos(C  A) 
2

 2cos(A  B)  2 cos(B  C)  2 cos(C  A)  3  0

  sin A  sinB  sin C 2  (cos A  cosB  cosC)2  0

 cos A + cos B + cos C = 0, sin A + sin B +sin C = 0
 cos 2A + cos 2B + cos 2C = 0

1 3
Also, cos2A + cos2B + cos2C  (1  cos 2A  1  cos 2B  1  cos 2C)  .
2 2

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79. (B)
Clearly AB is common chord.

3 1 A r=3
sin     (1,1)
3 2 2
 
 P(4,4)
4 B

1
AB  2r cos   2  3  3 2
2
80. (C)

 6 2
1  cos 1
 12  4
cot 
24  6 2
sin
12 4

4 6  2 6 2
   6  2 2 3  2  3  4  6
6 2 6 2
81. (B)
Since the circle does not touch or intersect the coordinates axes, the absolute values of x
and y coordinates of the centre are greater than the radius of the circle. Coordinates of the
centre of the circle are (3, 5) and the radius 9  25  p

3  9  25  p  p  25 ..........(1)
So that
5  9  25  p  p  9 ..........(2)
And the point (1, 4) lies inside the circle  1  16  6  10  4  p  0  p  29 ..........(3)
From (1), (2), (3) we get 25 < p < 29.
82. (B)
Image of A say A  w.r.t x – 2y + 1 = 0 lies on BC

x 1 y  2 (1  4  1) 4 9 2
Here,   2   A   , 
1 2 1  22 5 5 5
9 2
 Equation of BC joining A    ,  and B (2, 1) is
5 5
2
1
5 3
y 1 (x  2)   x  2 
9 1
2
5
3x – y – 5 = 0  a  b  3 1 2

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83. (B)
S = 1 + 4x + 7x2 + 10x3 + ..........
x.S = x + 4x2 + 7x3 + ..................
Subtract
S (1 – x) = 1 + 3x + 3x2 + 3x3 + ........
 1 
S (1 – x) = 1 + 3x  , |x| < 1
 1– x 
1  2x
S =
(1– x)2
1  2x 35
Given 
(1– x)2 16
1 19
 16 + 32x = 35 + 35x2 – 70x  (5x – 1) (7x – 19) = 0  x = ,
5 7
1
But |x| < 1  x =
5
84. (A)

1 2r  3  (2r  1)
Tr = 
(2r  1) (2r  1) (2r  3) 4(2r  1) (2r  1) (2r  3)

1 1 1  1 1 1 
   =   
=
4  (2r  1) (2r  1) (2r  1) (2r  3)  4  (2r  1) (2r  3) (2r  1) (2r  1) 
1
= -  Vr  Vr 1 
4
n
1 1 1 1 1 1 1 
 T  4
r 1
r [Vn - V0] =     =  
4  (2n  1) (2n  3) 3  4  3 (2n  1) (2n  3) 
85. (C)
Given, ,  are roots of equation
x2 – 2x + 3 = 0
 2  2  3  0 ...(i)

and 2  2  3  0 ...(ii)

  2  2  3   3  2 2  3

 P  (2 2  3 )  3 2  5  2    2  2  2  3  2  1 . [Using (i)]

Similarly, we can have Q = 2.

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86. (B)

Putting H  2ab
ab

Ha Hb 2(H2  ab)

 
H  a H  b (H  a)(H  b)

 2[ ( H  a)( H  b)  H 2  H (a  b)  ab

2ab
 H2  (a  b)  ab  H 2  ab].
( a  b)

1 1
Trick : Let a  1, H  and b  , then
2 3

H a Hb 3 / 2 5 / 6
    2.
H  a H  b 1/ 2 1/ 6
87. (D)

1 x2 1 x2 1 1 x2
 | x | x   . x  0  x [1, 1]  {0}
x x x x
88. (B)
Point I and I1 divide AD in the same ratio internally & externally respectively
89. (C)

(x  a)(x  8)  2

 x  a  2 & x  8  1  a  5 or x  a  2 & x  8  1  a  11

or x  a  1 & x  8  2  a  5 or x  a  1 & x  8  2  a  11

90. (D)

f(x,y)  (x  3)2  9  (y  2)2  4

which is minimum when x = 1 & y = 0

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