0 log 5
3 –
x 10
= 0.79 – 1 0591 .
0 log 3–3 x 10
∴ log x = 0.656 ≈ 0.66 ; ∴ x = 100.66
2.[B]
Br Et O
Mg
2
→
δ– O
MgBr CH3
δ+ +
→ OH
CH3
3.[C] Br– Changes to Br2
4.[C] Diazo coupling reaction is an ES reaction in which electrophile is diazo component. Greater electro philicity of diazo group makes it more reactive
EWG ERG
NO2
–
, CH3–
, CH3O–
, Me2N–
5.[A,B,C,D]
– 1/2 ] A [
] A [
d = kdt
Integrating both sides, we get –2[A]1/2 = kt + C
When t = 0 the [A] = [A0] So C = – 2[A0]1/2
∴ k =
t
2([A]1/2 – [A]1/2)
After rearrangement [A]1/2 = – 2
kt+ [A0]1/2 y = mx + C
So graph will be
t [A]1/2
[A0]1/2
Slope = –k/2
t ¾ = k
] A [ 0
and t ½ =
k ] A [ ) 1 2 (
2 − 0
6.[A,B,C]
On heating one molecule of carbon dioxide comes out.
Ph
Ph
H Me
HO2C CO2H
H Me −CO2
→∆
Ph
Ph
H Me
H CO2H
H Me
Ph
Ph
H Me
HO2C H
H Me
[A] [B]
Both have plane of symmetry hence optically inactive 7.[A,B,C,D]
8.[A,B,C]
(A) 100 mg of CaCO3 in 1000 ml = 100 g CaCO3 in 106 ml = 100 ppm
(B) 120 mg of MgSO4 in 103 ml = 120 g MgSO4 in 106 ml = 100 g CaCO3 in 106ml = 100 ppm (C) 84 mg of MgCO3 in 103 ml = 84 g MgCO3 in
106 ml = 100 g CaCO3 in 106ml = 100 ppm (D) 111 g of CaCl2 in 103 ml = 100 × 103 g of CaCO3
in 106 ml = 100000 ppm
SOLUTION FOR MOCK TEST
PAPER - II
IIT-JEE (PAPER - II)
9.[D] Neighbouring group participation occurs through two consecutive SN2 substitution with inversion of configuration, thus the net result is retention of configuration.
Column Matching :
10. (A) → q ; (B) → p,r ; (C) → p,s ; (D) → p,s * For exothermic reaction, ∆H = –ve, ↑ in T backward
whereas for endothermic, ↑ in T forward.
* Reactions for which ∆ng = 0, pressure has no effect
Numerical Response type questions : 12. [4]
Volume of both AgNO3 & HCN are equal so concentration is halved.
[AgNO3] = [HCN] = 0.01M
(Since K is very large so almost entirely forward shifted)
Reaction at C–1 and C–5 are enantiotropic face give racemic products (2)
Reaction at C–2 and C–3 it self cause racemization in addition to that these are diastereotopic faces so diastereomers form (4)
Reaction at C–3 (1) Reaction at C–6 (1) Total isomers are (8) 14. [4]
The cell reaction is
Cd + 2H+ (0.2M) → Cd2+ (0.1M) + H2 (0.5 atm)
MATHEMATICS
1.[B] For vertical tangents dθ
. Corresponding to these values of θ, we have at the points given by
t tangent to the curve if the roots of this quadratic equation are equal, for which 4t4 – 4t = 0 ⇒ t = 0 or t = 1 and an equation
Therefore, equation of the circle having AB as diameter is
(x – α) (x – β) + (y – α′) (y – β′) = 0
where y = 1 – cos x. So f is derivable at x = 0 and hence also continuous.
9.[A,D] cot–1 (1 + x2 – x) = tan–1
−
−
− +
) x 1 ( x 1
x 1 x
= tan–1 x + tan–1 (1 – x) I =
∫
01 − + −2 1(1 x x)dx
cot =
∫
01 −1xdx
tan
+
∫
01 − − 1(1 x)dx tan=
∫
01 − +∫
− =∫
−1 0 1 1
0 1
1xdx tan xdx 2 tan xdx
tan
= 2x tan–1
]
−∫
01 + 2 10 dx
x 1
x x 2
= 2tan–1 (1) – log(1 + x2)
]
10= 2(π/4) – log 2 = π/2 – log 2.
10. A → r, s; B → s; C → p;D → p, q
(A) Any point on the line (t, 1– t). The chord with this as mid point T = S1 passes through the point (a, 2a)
⇒ (1 – t)2 = 2a (1 – a) > 0 ⇒ a ∈ (0, 1) Q LR ∈ (0, 4)
(B) POI are (1, 0) and (4, 0) and circle is (x – 1) (x – 4) + y2 + λ y = 0 the length of the tangent from (0, 0) is √4 = 2
(C) The ⊥r tangents from any point to the parabola intersect on the directrix.
(D) 1 – | h | > 0 ⇒ h ∈ (–1, 1).
11. A → q,t; B → s; C → p; D → r
Let X = the number of steps taken in the forward direction, then X ~ B (n, p) with n = 11, p = 0.4.
p1 = P(X = 5) + P(X = 6) = 11C5 p5q6 + 11C6p6q5
= 11C5 (pq)5 = 11C5 (0.24)5 = 11C6 (0.24)5 p3 = P(X = 4) + P(X = 7)
= 11C4 p4 q7 + 11C5 P7 p7 q4 = 11C4 (pq)4 (1 – 3pq) = 11C4 (0.24)4 (0.28)
p0 = 0
and p11 = P(X = 0) + P(X = 11) = (0.4)11 + (0.6)11 Numerical Response type questions : 12. [3]
We have
|2
z
|
9 = (2 + cos θ)2 + sin2 θ = 5 + 4 cos θ (1)
and z
3+ z
3= 4 + 2 cos θ (2)
Eliminating θ from (1) and (2), we get
|2
z
|
9 – 6
+ z 1 z
1 = – 3
⇒ 3 = 2( z + z) – |z|2 13. [2]
p(x) = (2x + 3) (x97 + x96 + … + 1).
= (2x + 3) ( x + 1) (x96 + x94 + … + x2 + 1) Also, x96 + x94 + … + x2 + 1 > ∀ x ∈ R.
14. [3]
Let the required G.P. be 2 , , 1 2 , 1 2
1
b 2 a b a
a + + ……
Its sum is
7 1 1 2
2 2 / 1 1
2 / 1
b a b b
a =
= −
−
−
We can take a = 3, b = 3.
15. [2]
As A2 = O, Ak = O ∀ k ≥ 2.
Thus, (A + I)50 = I + 50A
⇒ (A + I)50 – 50A = I
∴ a = 1, b = 0, c = 0, d = 1 16. [6]
Let b = xi + yj + zk. So a × b = (z + y) i – xj – xk a × b + c = 0 ⇒ z + y + 1 = 0, –x + 1 = 0
⇒ x = 1 a . b = 3 ⇒ y – z = 3
Solving these equations we have y = 1, z = –2.
Thus b = (1, 1, –2). i.e. |b|2 = 6.
17. [4]
We have BC = 2BD, AD = h and OD = h – r.
∴ BC =2 r2−(h−r)2 =2 2hr−h2
⇒ AB = 2hr−h2+h2 = 2hr so that P = 2AB + BC
=2[ 2hr−h2 + 2hr] Also the area of ∆ABC is
∆ = BD × AD =h 2hr−h2 .
A
Note that it is not given that f is a differentiable function we have
f′ (4) =
The given equation can be written as dx
Time taken by pendulum in going from A to B
=
4
T where T = 2π gl
Time taken by pendulum in going from B to C
=
Free body diagram of point AC –
θ1
M′g TA
TC
Horizontal equilibrium : TA cos θ1 = TC
∴ TC =
) ( sin
cos cos Mg
2 1
2 1
θ + θ
θ θ
Tension will be maximum at A and minimum at C.
3.[C] Maximum expansion in spring is given by kx2max
2
1 = v20 2 1µ
[µ = Reduced mass]
⇒ xmax = k µ . v0 =
k 3
m 2 v0
4.[B] Force diagram of block for the view shown
view
N N
θ
mg cos α
N N
180°– θ
mg cos α
⇒ N =
) 2 / sin(
2 cos mg
θ α
∴ Net friction up the plane = 2 µN = µmg
) 2 / sin(
cos θ α
∴ a = g
θ µ α
α sin( /2) – cos
sin 5.[A,D]
S1 θ
S2
2λ
P
δ = 2 π–
λ
2π(2λ sin θ)
⇒ δ = 2
π– 4π sin θ For maxima, δ = nπ
where n = 0, ± 1, ± 2 . . . 2
π– 4π sin θ = nπ
sin θ = 4 2 n 1−
n = 0, sin θ = 8 1
n = ± 1, sin θ =
8 , 3 8 1 +
−
n = ± 2, sin θ = 8
−3, 8 5
n = ± 3, sin θ = 8
−5, 8 7
n = 4, sin θ = 8
−7
6.[A,B,C,D]
a = m F =
m qE=
m ) x ( qα−β
a = 0 at x = β α
Force on the particle is zero at x = β α
So, mean position of particle is at x = β α
dx
vdv= ( x) m
q α−β
Solve for v =
α−βx 2 m
qx 2
v = 0 at x = 0 and x = β 2α
x = β
2α with mean position at β
= α
x .Therefore amplitude of particle is
β α.
Maximum acceleration of particle is at extreme position (at x = 0 or x =
β
2α) and αmax = m qα
.
7.[B,D] As v = 2t and let radius of circular path is r then, aT =
dt
dv = 2, ar = r v2 =
r t 4 2 Therefore,
a = a2T+a2r
a = 2
4
r t 4+16
∴ (B) and (D) are correct.
8.[A,B,D]
A = 0.04 m ω = 25π k = 5π
∴ Position of antinodes is given by cos (5πx) = 1
⇒ x = 0, 20 cm ....
v = k
ω = 5 m/s vmax = Aω = π m/s
9.[A,C] Temperature gradient at distance 'r'
dr dθ
= K 1 × 2
x 1 π
[Where x = cross-sectional radius at distance 'r']
⇒ Temperature increases with increase in 'r'.
10. (A) → (S) ; (B) →(P) ; (C) → (Q) ; (D) → (R)
11. (A) → (P,S) ; (B) → (Q,R) ; (C) → (Q,R,S) ; (D) → (P,Q,R) For process PVn = constant Molar heat capacity of gas
C = R
− −
−
γ n 1
1 1
1 ... (i)
Here, γ = 7/5 PV = nRT = n1
V t tan cos
−
= Constant ×
− n 1 1
P
For n = 1 : Temperature with increase in volume work done positive Hence heat is absorbed by system.
For n = 2
1 : Temperature and volume decrease with increase in pressure
∴ Work done negative Hence heat is rejected For n=
5
6: Temperature increases with decrease in volume work done negative
For n = 2 : Temperature increase with increase in pressure work done negative Hence heat is absorbed.
Numerical Response type questions : 12. [4]
M,R 4M, 2R
6R – x
V = 0 E = 0 x
x2
m 4 .
G = 2
) x – R 6 (
GM
2 (6 R – x) = x x = 4 R
= 4000 km = 4 × 103 km.
13. [1]
Temperature of sun is given by
T =
m
b
λ [b : wein constant]
= 6127 K
∴ Intensity at earth = d2
4 u π = 24
d 4
AT π σ
[d = 1.5 × 1011 m, T = 6127 K, A = 4π r2, r = 7 × 108 m]
= 1740.1 W/m2 = 1.740 × 103 14. [1]
Temperature of gas will increase with increase in volume and becomes maximum at ‘C’
∴ Tmax = nR PV =
314 . 4 8 1
10 4 10
200 3 –3
×
×
×
×
= 384.89 K
= 112ºC = 1.12 × 102 ≈ 1
15. [3]
S1
S2
S3
P
2 d θ1 3
θ2
S2P – S1P = d sin θ1 = d.
D 2 / d =
D 2 d2 =
3 λ
S3P – S2P = d sin θ2 = d D
2 / d 3 =
D 2
d 3 2 = λ
120°
I0
4I0
IResultant = I0 + 4I0 + 2 I0 4I0 cos120°
= 3I0
16. [4]
Let, a = side of cube P = Impulse imparted
P a
A B
∴ After hitting, v0 =
m
P and ω0 = I 2 Pa
[I = Moment of inertial about axis passing through centre of mass]
For just toppling I 20
21 ω = mga
−
2 1 2 1
(Applying energy conservation between situation A and B)
⇒ P = a I 2ω0
= 4 kg m/s
17. [1]
Let rate of production = R
∴ dt
dN= R−λN
dt
dN + λN = R
eλt dt
dN+ λNeλt = Reλt
dt ) Ne ( d λt
= Reλt
Neλt = λ Reλt + C
At t = 0, N = 0 ⇒ C =
−Rλ
∴ N = λ
R(1– e–λt)
At equilibrium quantity N = λ
R for t → ∞
∴ 2λ R =
λ
R(1 – e–λt)
⇒ e–λt = 2 1
t = λ
2
ln = T1/2 = 100 years = 1 × 102
18. [2]
eV1 =
1
hc λ – φ eV2 =
2
hc λ – φ
e(V2 – V1) = hc
λ λ
λ
− λ
2 1
2 1
V2 – V1 = e
hc
λ λ
λ
− λ
2 1
2 1
= –19
8 34
–
10 6 . 1
10 3 10 6 . 6
×
×
×
× × –5
10 6
100
× =
32
66× 10–34+ 8 + 2 + 19 + 5
= 16
33 = 2.0625 volt ≈2 volt
19. [2]
A l
Wgas = −
{
Wgravity+Wexternalpressure}
= mgl + P0lA [m = mass of Hg pallet]
= 2.136 J
∴ ∆Q = ∆W = 2.136 J ≈ 2
PHYSICS
1.[A] Initially relative velocity between coin and lift is zero.
h = ut + 2
1 gt2 here u = 0
t = g h 2 =
8 . 9
45 . 2×2
= 2 1 sec.
2.[C] Mass of liquid inside the capillary = πr2 h d = (πrh d). r since, hr = constant
∴ mass of liquid inside α r
3.[D] 2π r = nλ or r =
π λ 2 n
⇒ r2 – r 1 = π λ 2 2 –
π λ 2 =
π λ 2 4.[C] KE = hυ + φ …..(i)
2KE = hυ' + φ ……(ii) or 2 (hυ + φ) = hυ' + φ or υ' = 2υ +
h
φ ⇒ υ' > 2υ
5.[C]
6.[B]
7.[D]
8.[A] E = 2 1 Ka2
E = 2
1 mw2a2
So a2 t2 = hence (A) constant 9.[C] M = M1M2
M = 24.2×20 = 484 = 22g
10.[D] Due to introduction of glass slab, number of fring between two points remains constant because if n number of fring crosses point P, then same number of fring also crosses point O in same sence.
11.[A] Energy stored in capacitor is, 2
1 CV2 = 3 J On connecting this capacitor to an uncharged capacitor since charge distributes equally, hence both capacitors are of same capacity Now common potential, V' =
C C
) 0 ( C CV
+
+ =
2 V
Total energy stored in two capacitors is – U' =
2
1 CV'2 + 2
1 CV'2 = C
2 2 V
= 4
1 CV2 = 2
3 = 1.5 J
12.[C] Current, I = R V =
5 . 0
30 = 60 A Total no. of free e– s, N = nAl
and linear momentum of each e¯s, P = mvα
∴ Total momentum of all free e¯s, P = (nAl) (mυα)
But I = neAυα , so nAυα = e I