jee mains test series

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)


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.


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


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


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.


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


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.


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


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


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 :


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


Documents

jee mains test series