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Vidyamandir Classes
Solutions to Test Series - 2/Paper - 1/IITJEE - 2014
[CHEMISTRY]
1.(A) Covalent bonds are formed by orbital overlapping. It must take place is some direction.
Ionic bonds are electrostatic attraction in the lattice, which can be in any direction
2.(D) Hence pH =
3.(A)
XeF2 linear
4.(C) V1 = Volume of NaOH in mixture.
V1 + V2 = 30 V2 = volume of HCl in mixture.
V2 = 10
V1 = 20
Moles of NaOH in 25 ml =
Moles in 500 ml = 40 m moles
gms = 1.6 gm
5.(B) LiCl, RbCl, BeCl2, MgCl2
Least ionic BeCl2
Most ionic RbCl
6.(D)
7.(B)
8.(D)
9.(A)
10.(B)1/3rd neutralization
Salt = 1/3 Base = 2/3
=
p(OH) = 4.4
pH = 9.6
VMC/JEE-2014/Solutions 1 Test Series-2/ACEG/Paper-1
FAJAN’S RULE
Vidyamandir Classes
11.(D)
II, III are chain isomers
I, II are chain isomers
I, III are position isomers
12.(C)
For 21.6 g B 2 moles
3 moles H2 will be consumed 67.2L
13.(D)
14.(D)
15.(A) (i)
(ii)
(iii)
We have to calculate
gives us the desired reaction
.
16.(A)
x = 464 KJ/mole.
17.(D) pH is maximum when (H+) is least.
CH3COONa is basic salt
Hence pH is max.
18.(C)
On dilution [salt] and [acid] remains the same.
Hence pH remains the same (neglecting minor changes)
19.(C)
VMC/JEE-2014/Solutions 2 Test Series-2/ACEG/Paper-1
Vidyamandir Classes
20.(A) Equivalent weight = 21.(C) 3 lone pairs around I
22.(C) After balancing
2 moles of mole of
Hence 4.5 moles of H2O2 will consume 9 moles of K3[Fe(CN)6]
23.(B) MgSO4
Assume density of water = 1 Kg/lt.
So 106 ml = 103 kg
Hence 6 gm of MgSO4 in 103kg of water
of CaCO2 in 103 kg of water 5 gm in 103 kg of water
24.(D)
25.(C)
26.(B)
PV = Pb + RT
(Compare with y = mx + c)
slope
27.(B)
sp2 sp2 sp2 sp2 sp
28.(C) For 100 KW reactor
105 J/s
In one second
105J will be used.
produces one molecule.
Molecules of H2 produced = =
Moles =
Hence volume of
224 ml
29.(A)
30.(A) Statement 1 is correct
Statement 2 is also correct and reason also.
VMC/JEE-2014/Solutions 3 Test Series-2/ACEG/Paper-1
Vidyamandir Classes
Solutions to Test Series - 2/Paper - 1/IITJEE - 2014
[PHYSICS]
31.(D) 32.(B) Total distance = (5 × 1) + (10 × 1) + (15 × 1) = 30 m.
33.(C)
34.(C) Components of the velocities of the block along the strings are equal
35.(B) Let T1 and T2 are tensions in AB and BC
At B: At C:
and
36.(C)
37.(C) 38.(A)
39.(B) To lift the block 2 kg, tension in the string is 2 g = 20 N
Let x: elongation in the spring for which the tension is 20 N.
kx = T 40x = 20 x = 0.5 m
For 5 kg: loss in GPE = gain in KE + gain in EPE
40.(B) Minimum horizontal force required to apply on 4 kg block to slide it on 5 kg block
= .
For F = 6 kg both are blocks are moving together with same acceleration
Force of static friction f = 5a = 10/3 N.
Displacement of the slab in the given interval
= work done = f s = 10/3 × 5/3 = 50/9 = 5.55 J
41.(A) Let B falls by x then elongation in the spring is 2x, loss in GPE of B = gain in EPE of spring
VMC/JEE-2014/Solutions 4 Test Series-2/ACEG/Paper-1
T1
T2
Mg
T2
mg
T2
Vidyamandir Classes
42.(D) Velocity is maximum when net force (on both the blocks) is zero.
Let T: tension in the string connecting A & x, elongation in spring at this instant
mg = 2T and T = kx1 block B falls by x1/2
loss in GPE of A = gain in KE of A & B + gain in EPE
V: velocity of A
velocity of B = V/2
43.(A) Let : angular velocity of rod just before impact & V: velocity of ball just after impact
.
About axis e = 1/3
44.(C)
About centre of mass: TR = fR
And to avoid sliding Ans. (C)
45.(D)
= Ans. (D)
46.(D) .
47.(C) Acceleration of point 1 = acceleration of point 2
48.(D) Frictional torque on the ring is
Angular retardation = .
49.(C) About the triangular edge and m = 8 kg
VMC/JEE-2014/Solutions 5 Test Series-2/ACEG/Paper-1
NT
f
mg cosmg sin
NT
mg cosmg sin
a
1
22m/s2
4m/s2
Vidyamandir Classes
50.(B)
51.(D)
52.(A) P = 8.31 × 105 N/m2
53.(C) = constant.
54.(C) for edge length
in first case in second case
55.(B)
56.(C)
57.(B) isochoric with increasing pressure.
58.(A) ; parabola passing through origin opening upwards.
59.(A)
=
Work done = inc. in GPE
= =
60.(C) Net acceleration is vector addition of tangential & radian acceleration.
VMC/JEE-2014/Solutions 6 Test Series-2/ACEG/Paper-1
d
Vidyamandir Classes
Solutions to Test Series - 2/Paper - 1/IITJEE - 2014[MATHEMATICS]
61.(D)
62.(C) Equality holds if
63.(A)
64.(C)
argument =
principal argument is
65.(A) , n = 3, 4, 5, . . .
: : . . . . . . . .
66.(A)
So sum
VMC/JEE-2014/Solutions 7 Test Series-2/ACEG/Paper-1
Vidyamandir Classes
67.(D) We have, for all x
for all x and and
Thus, the triplets are , where .
Hence, there are infinitely many triplets.
68.(B) Line is equally inclined to axes as slope is 1. As is equidistant from the points
and (3, 4),
69.(B) The lines PQ and PR will be inclined at an angle of with the line .
Therefore, the lines are and
Find the equation of pair of lines by multiplying these two equations.
70.(A)
Now,
71.(B)
72.(A)
73.(B)
So,
i.e., f (x) is continuous at x = 0.
VMC/JEE-2014/Solutions 8 Test Series-2/ACEG/Paper-1
Vidyamandir Classes
Now,
and
So, is not differentiable at x = 0.
74.(A) Let
for all if has two distinct real roots and and are negative.
and
These conditions give
75.(C)
= .
76.(D) Given that
Squaring both sides, we get :
Trick : Put so that
77.(A) Put
VMC/JEE-2014/Solutions 9 Test Series-2/ACEG/Paper-1
Vidyamandir Classes
So,
78.(B) Graph of and Graph of
Graph of g (x) = is :
graph of [ | g (x) |] is :
range of is {0, 1}
79.(D)
, where
80.(B) [Apply L.H. rule]
(As f (x) is continuous)
Let
For
81.(C)
z =
Its argument is
VMC/JEE-2014/Solutions 10 Test Series-2/ACEG/Paper-1
1 1
1 1O
1 1
O
1
Vidyamandir Classes
82.(C)
.
83.(B)
84.(A)
85.(B) Let
Determinant
But for
So determinant < 0, a > 0 for g (x)
Hence
86.(A) Given O(0, 0) is the orthocenter. Let A(h, k) be the third vertex, ,
and the other two vertices. Then the slope of the line through
A and O is k/h, while the line through B and C has the slope . By the
property of the orthocenter, these two lines must be perpendicular, so we
have
. . . .(i)
Also, Slope of AB
Slope of OC
. . . .(ii)
Solving it with (i) we get :
VMC/JEE-2014/Solutions 11 Test Series-2/ACEG/Paper-1
Vidyamandir Classes
87.(B) i.e. m 1 + m 2 = 0
where m1 and m 2 are the slopes of the tangents from P(h, k).
If is a tangent from P(h, k) then
i.e.
m1 and m2 are the roots of this equations such that m1 + m2 = 0
i.e. locus of P(h, k) is xy = 0
88.(B) and
For
L.H.S.
is a solution
For x = 0 L.H.S.
x = 0 is a solution
sum of the solutions =
89.(A)
90.(C)
Possible values of k are
VMC/JEE-2014/Solutions 12 Test Series-2/ACEG/Paper-1
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