‘MOCK JEE’ SOLUTION / PAGE - 1
PART I : PHYSICS
SECTION – 1 : (Only One Option Correct Type)
Sol.1.
(C)
KE = ½ mv2 + ½ I2 = ½ mv2 + ½ I(v/R)2
By conservation of energy as rolling motion is taking place
mgh = ½ mv2 + ½ I(v/R)2 = mg (v2/g)
2
 I = R2 (2mv 2  mv 2 ) = mR2
v
Sol.2.
(A)
Mg 2R  Mg 4R 

Sol.3.
(C)
Sol.4.
(C)
24g
11R
P sin = W sin 
P=
1 3
1

 MR 2   2  M 4R 2 2
2  2
2

l
2
W sin 2

2sin
For in plane oscillation
I=

T = 2

Sol.5.
(A)
MVCM = mv
Sol.6.
(C)
mvR = mv'R + I'
(C)
(3 / 2)MR 2
3R
= 2
MgR
2g
VCM =
m
v
M
v ' 
mvR = mv'R + mR2  
R

Sol.7.
MR 2
3MR 2
+ MR2 =
2
2
= v/2
Considering O as origin
ml 3
2 =0
8m
7mx 
Sol.8.
(B)
Sol.9.
(C)

x=
3l
14
 2M  m  g  T   2M  m  acm
 T   2M  m  g
Sol.10.
(B)
T  mg cos 
mv 2
l
T  2 mg
mgl cos  
cos  
mv 02
1
mv 2 
2
2
1
4
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‘MOCK JEE’ SOLUTION / PAGE - 2
SECTION – 2 : (One or More Than One Option Correct Type)
Sol.11.
(ABD) In CM frame the blocks will perform SHM as
VCM is constant hence both will can have
maximum velocity v0/2 towards right in CM frame.
(vB)max = v0 in ground frame
In CM frame (vA) min = –v0/2

(vA)min = 0
in ground frame
i.e. A move only in right direction
In CM frame vB = v0/2 left in the case when spring is at natural length

Sol.12.
(BD)
vB = 0 in ground frame at that instant.
When they have equal linear velocities 
(B)
h 0 relative motion so force of friction will disappear angular momentum no
conserved as there will be external force on the centre of disk (axle)
There is loss of rotational Kinetic Energy
Sol.13.
L
(AC)
mg
L 1 2
 ml 
2 3
3g
2L
(A)

(C)
Initial acceleration of right end rod is
3g
3g
L 
and difficult points would have
2L
2
different acceleration.
Sol.14.
(ABCD)
B to C
 5 sec
A to D
 15 sec
B to B to C  25 sec
Sol.15.
(ABCD)
 fmax 4 kg
 40 
1
6
 20 N ,  fmax 5 kg  50 
 30 N ,
2
10
 fmax 2 kg
 20 
4
 8N
10
In all the cases shown A,B,C,D there is at least an incident where the force exerted by
friction > frictional force.
(A)
13
2
 500 
 300  65  6  71 N
100
100
(B)
8
3
 500 
 300  40  9  49 N
100
100
(C)
1
8
 500 
 300  5  24  19 N
100
100
(D)
2
 1


 500 
 300   5  6  11 N
100
 100

So, in all cases it system is not in equilibrium.
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‘MOCK JEE’ SOLUTION / PAGE - 3
SECTION – 3 : (One Integer Value Correct Type)
Sol.16.
(5 N)
Force = Rate of change of momentum
Force 
K

Sol.17.
Change in momentum 2m 2  m  2gh


Time
t
t
2  0.5  2  10  4.05
0.002
 4500
K
4500

5
900
900
(0003 J)
As no external force on the system and energy is conserved
mAv A  mBv B
A
 2 VA   6  0.5 
 VA 
3
2
Energy in Compressed Spring 
Sol.18.
(8)
y
y
(5)
m  0   m  0   m 10   m 10   m  20 
5m
(5)
40
8
5
 20g  5  50 2 p
PE of A


 
PE of B  20g  7.5  75 3 q

Sol.20.
1
1
mAv A2  mB v B2  3 J
2
2
m1y1  m2 y 2  m3 y 3  m4 y 4  m5 y 5
m1  m2  m3  m4  m5
y 
Sol.19.
B

pq  5
v 0 sin 
v
g
10  5
5




2
R 2v 0 sin  cos  2v cos  2  10  4 8
g
 8  8 
5
5
8
PART II : CHEMISTRY
SECTION – 1 : (Only One Option Correct Type)
Sol.21.
Sol.22.
(D)
(A) (r max) 3s > (rmax)2s > (rmax)1s
Sol.23.
Sol.24.
(C)
(C)
Sol.25
(A)
On y-axis
x = 0 or P  0
In low pressure region, real gas behaves ideally.




PM = dRT
RT
P
=
d
M
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‘MOCK JEE’ SOLUTION / PAGE - 4
8
7
Sol.26.
2
4
3
5
O
(A)
6
Sol.29
(C)
Sol.30.
(D)
1
Sol.27. (A)
Sol.28. (C)
As molecular force increases, london dispersion force increases so solubility of inert gases
in water increases
HF < HCl < HBr HI (order of acidic strength)
On moving down the group, bond length increases so bond becomes weaker and hydracids
becomes stronger
SECTION – 2 : (One or More Than One Option Correct Type)
Sol.31.
(ABC) The apparent O.S. of S in H2SO5 = + 8
The maximum O.S. of S = + 6
The no. of peroxo linkages in H2SO5 =
86
2 =1
The apparent O.S. of S in H2S2O8 = + 7
The maximum O.S. of S = + 6
The no. of peroxo linkages per S atom in H2S2O8 =
76
1
=
2
2
 in H2S2O8 the no. of peroxo linkage is one.
The apparent O.S. of Cr in CrO5 = + 10
The maximum O.S. of Cr = + 6
The no. of peroxo linkages in CrO5 =
10  6
=2
2
In FeO42  the O.S. of Fe is + 6
Sol.32.
(ACD)
rvap.
 O2
2

32
32
4
  
,
Mvap.
3
M
 
vap.
 V.D = 9  Density =
Now
Sol.33.
Z=
(BCD) nRT = 1 × 0.0821 ×
Mvap. = 18
18
gram/litre = 0.8035 gram/litre
22.4
PV
P
PM
18P


Z=
W
nRT
dRT
dRT
RT
MV
1
=1
0.0821
(A) PV = nRT
lnP + lnV = ln nRT = 0
(B) log 10P + log10V = 0
(C) PM = dRT

P RT
= constant = 0.01

d
M
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‘MOCK JEE’ SOLUTION / PAGE - 5
(D) PV = nRT
 PV2 = (nRT)V
PV2 = (1) V
Sol.34.
Sol.35.
(ABCD)
(ABC)
SECTION – 3 : (One Integer Value Correct Type)
Sol.36.
(6)
Volume strength of H2O2 = 5.6 × Normality
= 5.6 × 1.08 = 6.048
Sol.37.
(2)
Only B2 and Cl2 are paramagnetic.
Sol.38.
(9)
Angular momentum for p-orbital e s
–
= (   1)
h
=
2
2
h

2
h
2
So we have to find out total no. of electrons in p-orbitals of phosphorous.
15P
 1s2 2s2 2p6 3s2 3p3
–
Total p-orbital e = 9
Sol.39.
(7)
NaCN + HCl  HCN + NaCl,
[HCN] =
0.2  50
= 0.1 M
100
H3O+ + CN–
HCN + H2O
For the acids and it's conjugate base
KaKb = 10–14
Hence Ka for HCN =
(H3O+] = (CN–] =
Sol.40.
1.0  10 –14
= 5 × 10–10
2  10–5
K aC  5  10 –10  0.1
(2 ml) Molarity of H2SO4 solution =
= 7 × 10–6
1.25  39.2  10
=5M
98
No. of m moles of H 2SO4 needed to react with 15
millimoles of base = 10
5 × V = 10  v = 2 ml
So
PART III : MATHEMATICS
Sol.41.
(C)
SECTION – 1 : (Only One Option Correct Type)
3x, 4y, 5z are in G.P.
 16y2 = 15xz ....(1) and y =
2xz
xz
....(2)
using (2) in (1)
16 × 4x2z 2 = 15(x + z)2xz
( x  z )2
64
=
;
xz
15
x z
64
+
+2=
;
z x
15
 m = 34 and n = 15

x
z
34
+
=
z
x
15

m + n = 34 + 15 = 49
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‘MOCK JEE’ SOLUTION / PAGE - 6
Sol.42.
(D)
Given
P = x + y cos z°
Also
P = 10 + b + c
Now using sine law
....(1)
10
b
c
=
=
sin 60
sin50
sin70
 b=
10 sin50
sin60
b+c=
10
(sin 50° + sin 70°)
sin 60
and c =
10 sin70
sin 60
10
[2 sin 60° · cos 10°] = 20 cos 10°
sin 60
=
 P = 10 + 20 cos 10°
 x = 10;
y = 20;
z = 10
x + y + z = 40
Sol.43.
(B)
O O;
B B B;
number of ways =
RRRR
7!
7!
7!
+
+
3! 4!
4!2!
2! 3! 2!
OO
B B
R  R
= 35 + 105 + 210 = 350
Sol.44.
(C)
r · 100Cr
r ·100!
( r  1)!(101  r )!
=
·
= 101 – r
100
Cr 1
r !(100  r )!
100!
100
 S=
 (101  r ) = 100 + 99 + 98 + ....... + 1 = 5050
r 1
Sol.45.
(A)
=
0 1 1
1
2 2 1 = 2;
2
a 0 1
0
0
1
 2 1 1 =4
a 1 1
 |a+2|=4
1 [–2 – a] = 4
a=2
or
a=–6
sum of the abscissa = – 4
Sol.46.
(C)
n
C (21/ 3 )n  6 (31/3 )6
1
 n 6 1/3 6 1/3 n  6 or 61  64.6n / 3  6n / 3 4
6
Cn  6 (2 ) (3 )

Sol.47.
(B)
n
 4  1  n = 9.
3
circle is x2 + y2 = 1 and P = (x1, y1)

x12  y 12 = 1
2
 (PA)

2
= (x1 –
 (PA)
2
1)2
2
2

1
3
1


+ y +  x1   +  y 1 
+  x1   +


2
2 
2



2
1
= 3( x12  y12 ) + 1 +

3
 y1 

2 

2
1
3
1
3
+
+
+
=3+3=6
4
4
4
4
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‘MOCK JEE’ SOLUTION / PAGE - 7
Sol.48.
(A)
Put y  2 sin x in
y  5 x 2  2 x  3  2sin x  5 x 2  2 x  3
 5 x 2  2x  3  2sin x  0
x
.....(i)
2  4  20(3  2sin x )
.
10
It is clear that number of intersection point is zero, because 1  sin x  1 and in all the
values roots becomes imaginary.
Sol.49.
(A)
sin  =
x
l
....(1);
1 + cos 2 =
2 cos2 =
l=
Sol.50.
(D)
cos 2 =
also
6x
x
6
;
x
6
lsin
{substituting x = l sin  from (1) }
3
sin cos2 
(sin 23x – sin2 x) = sin 3x · cos x – sin x · cos 3x
sin 4x · sin 2x = sin 2x
 sin 2x = 0 or
 {x = 0,
sin 4x = 1
  5
,
,
, }
2 8
8

(D)
SECTION – 2 : (One or More Than One Option Correct Type)
Sol.51.
(ABD)
S1 : x2 + y2 = 4
and S2 = x2 + y2 – 2x – 4y + 4 = 0
centre: (0, 0); radius = 2
(A)
centre : (1, 2); radius = 1
d = distance between centres =

r1 + r 2 = 3
5
| r1 – r2 | = 1

| r1 – r2 | < d < r1 + r2

these 2 circles are intersecting.

number of common tangents is 2. 
(B)
P(h, k) power of point P is same w.r.t. these two circles.

h2  k 2  4 =
(A) is correct
h 2  k 2  2h  4k  4
– 4 = – 2h – 4k + 4
2h + 4k – 8 = 0

x + 2y – 4 = 0
(C)
y intercept of S1 is
2
4 = 4
y intercept of S2 is
2
44= 0

sum of y-intercept = 4
(D)
2(0 + 0) = – 4 + 4

circle is orthogonal.
(B) is correct
(C) is incorrect

(D) is correct
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‘MOCK JEE’ SOLUTION / PAGE - 8
Sol.52.
(ABD)
Any point on the line x + y = 1 can be taken as (t, 1– t).
Equation of the chord, with this as mid-point is T = S1
y (1 – t) – 2a (x + t) = (1 – t)2 – 4at, it passes through (a, 2a).
So t2 – 2t + 2a2 – 2a + 1 = 0,
this should have 2 distinct real roots so discriminant > 0, we get a2 – a < 0 



 0 < a < 1, so length of latus rectum < 4 



 latus rectum  4.
Sol.53.
(AB)
Line
y = 1 + c(x + 3) is tangent to circle x2 + y2 = 1



p=r

1  3c

1 c2
=1
(1 + 3c)2 = 1 + c 2
9c2 + 6c = c 2
8c2 = – 6c

Sol.54.
(ABC)
Sol.55.
(ACD)
c = 0 or
B U L B U L;
c = – 3/4
number of ways =
6!
2!·2!·2!
(A) 2 Apples can be distributed in 3 people in 4C2 way
and
4 Mangoes in 6C2 ways

Total ways = 6C2 · 4C2 =

(A) is correct
O O O O Ø Ø
6!
4!
6!
·
=
2!·4! 2!·2!
2!·2!·2!
(B) 6 books in 3 bundles, two in each bundle =

O O Ø Ø
6!
2!·2!·2!·3!
(B) is incorrect
(C) Tr + 1 in (x + y + z)6 is 6C r(x + y) 6 – r · zr
put
T3 =
6C
2(x +
r=2
y) 4 · z2
= 6C2 · z2 (4Cp · x4 – p · yp)
put
p=2
=
6C
2
· z2 · 4C2·x2y2)
= 6C2· 4C2 x2y2z2
=
6!
2!·2!·2!

(C) is correct
(D) 6 prizes in 3 children, two to each =





6!·3!
6!
=

2!·2!·2!·3! 2!·2!·2!
(D) is correct]
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‘MOCK JEE’ SOLUTION / PAGE - 9
Sol.56.
(0)
SECTION – 3 : (One Integer Value Correct Type)
2
2
2
2
2
Locus will be director circle x + y = a + b < 2c
which do not intersect with xy = c2
Sol.57.
(3)
2x 
3 
  1

2
9 
  
n
n
For T6 to be N.G.T.
( n  1)
5
2x
9
6
2x
1
9
when x = 3
5
( n  1)2
6
5
25  2n + 2  30
23
 n  14
2
 n = 12, 13, 14
Sol.58.
(B)
Normal to x2 = 4y is x = my –2m –m 3
passes through (1, 2)  1 = 2m –2m – m 3
m 3 = –1 or m = –1
Sol.59.
(2)
Let the point P be (h, k)
3
equation of normal k = mh – 2m – m
m 3 + m (2 – h) + k = 0
m 1m 2m 3 = – k  m 3 = 
k
[ m 1m 2 =  ]

3
k
 k
2
2
2
3
      (2 – h) + k = 0 k =  h – 2 + 


2
2
2
locus of (h, k) is y =  x – 2  + 
3
2
2 = 4 & – 22 + 3 = 0
compare with y = 4x we get
2
Sol.60.
(4)
Let the ellipse be
and
x
y
+ 2 = 1 whose area = ab
2
a
b
x2
y2
+
= 1 whose area = bc
b2
c2
Required area = ab – bc
2
 = 2
2
2
2
2
2
= b (a – c)
2
b = a (1– e ) and c = b (1 – e )
 c2 = a2 (1 – e2)2  c = a (1 – e2)  a – c = ae2
so required area = b (ae2) = abe2
2
2
= 9 ×   = 4 sq. units.
3
END OF TEST
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CH 15 - icivil-Hu

FIMECC Factory in Brief

Ottobock - Start Junior - Order Form

w = x2 + 2y2 + 3z2 - xy

Lessons 17