RS -11- I -21
TARGET – IIT JEE
CHEMISTRY, MATHEMATICS & PHYSICS
HINTS & SOLUTION
O
CHEMISTRY
1.[C]
1
1
1
=
+
λ2
300
760
As above
1
1 
1
= 
−

λ 2  300 760 
CH3COO– + H+
CH3COOH
C
(C – Cx)
0
0
Cx
Cx
760 − 300
1
=
760 × 300
λ2
i = [1 + (y – 1)x] = 1 + x = 1.0 1
∴λ2 = wavelength of the second photon
x = 0.0 1
= 496 nm
[H+] = Cx = 0.01 × 0.01 = 1 × 10–4 M
∴ pH = 4
2.[1]
z=
Thus, (C)
2.[A]
0.12 M → 0.06 M 50 % change T50 = 10 h
=
0.12 M → 0.03 M 75 % change T75 = 20 h
1× 10 5 × 18 × 10 −3
= 0.97
0.6 × 8.314 × 373
Since T75 = 2 × T50 hence order = 1
Note : All the values have been taken in SI units
Hence (A)
(m in kg mol–1, ρ in kg m–3, R in J mol–1 K–1)
3.[B]
4.[B]
3. [7]
5.[A,B,C]
6.[A,B,D]
4. [1]
7.[C,D ]
8.[B,C]
9.[A,B,D]
Column Matching
1.
A → P; B → S; C → R; D → Q
2.
A → T; B → S; C → P; D → Q
Numerical Response :
1. [5]
PV
PV
PVm
Pm
=
=
=
w
ρRT
nRT
wRT
RT
m
Wavelength of the first photon = λ1,
E1 =
hc
λ1
and that of second photon = λ2
E2 =
hc
λ2
Etotal = E1 +E2 = emitted energy
hc
hc
hc
=
+
λ
λ1
λ2
CAREER POINT, CP Tower, Road No.1, IPIA, Kota (Raj.), Ph: 0744-3040000
Cl–CH2COO– + H+
Cl–CH2COOH
van't Hoff factor i = (1 + x)
using the Ostwald's dilution law fo weak
electrolyte
x=
Ka
=
C
1.36 × 10 −3
= 0.37
0.01
∴ i = (1 + x) = 1.37
hence, elevation in b.p.
(∆T)b = Kbm (molality) i
= 0.51 × 0.01 × 1.37
= 0.007º
hence, b.p. of solution
T = T0 + (∆T)b
= 100 + 0.007º = 100.007ºC
Page # 1
5. [5]
⇒ ∑xy = 24
1.8
= 0.01
180
Moles of carbohydrate =
Moles of acetyl chloride =
3.92
= 0.05
78.5
⇒ xyz =
–OH and CH3COCl react in 1 : 1 parts
Number of (OH) group =
2
3
∴ x3 + y3 + z3 – 3xyz = (x + y + z) (∑x2 – ∑xy)
moles of CH 3COCl
=5
moles of carbohydrate
∴ ∑x3 – 2 = 12 (96 – 24)
6.[8]
7. [1]
∑ xy
= 36
xyz
Now ,
⇒ ∑x3 = 866
–
SO42– + R(OH ) 2 → RSO4 + 2OH
re sin in
exchange
OH– +
3[A]
No. of determination corresponding to
1st column = 2
→ H2O
H+
in H 2 SO4
No. of determinants corresponding to
2nd column = 3
25 × N (OH–) = 21.58 × 2 × 10–3 N(H2SO4)
No. of determinants corresponding to
N(OH–) = 1.7264 × 10–3- N
3rd column = 4
N(CaSO4) = N(SO42–) = 1.7264 × 10–3 N
Total No. = 2 × 3 × 4 = 24
1.7264 × 10 −3
M(CaSO4) =
mol L–1
2
from multiplication theorem
= 0.1174 g L–1
So, choice (A) is correct.
4.[A]
= 117.4 ppm
A
8. [4]
E
O
MATHEMATICS
1[B]
tan 3 α cot α =
=
⇒ tan2 α =
=
M
B
3 tan α − tan 3 α
tan α(1 − 3 tan 2 α )
3 − tan 2 α
1 − 3 tan 2 α
AM =
= x (say)
(3x − 1)( x − 3)
(3 x − 1) 2
2
p , MD
3
1
2
2
p , BM = q BM
3
3
=
x −3
3x − 1
= BM2 + DM2
5.[A, D] 2 ( 2 sin
2
x −3 sin x +1)
Since tan2 α is non-negative,
Let 2 ( 2 sin
either x < 1/3
⇒t+
or
C
D
x ≥ 3, so x cannot lie between 1/3 and 3.
2[D] (x + y + z)2 = 144
2
∴ ∑x + 2∑xy = 144
CAREER POINT, CP Tower, Road No.1, IPIA, Kota (Raj.), Ph: 0744-3040000
2
+ 2 3− ( 2 sin
x −3 sin x +1)
2
x −3 sin x +1)
=9
=t
8
=9
t
⇒ t2 – 9t + 8 = 0
⇒ t = 1, 8
⇒ 2sin2x – 3sinx + 1 = 3
Page # 2
or 2sin2x – 3sinx + 1 = 0
⇒ sinx = −
∆AOB =
1
1
, sin x = , sin x = 1
2
2
⇒ | t1 t2 (t1 – t2) | = 20
⇒ | (t1 – t2) | = 5
6.[B, C] Q a > 2
Also
Now
⇒ t1 = ± 4 or ± 1
8ax
>0
a−2
∴x>0
A ≡ (16, 8) or (16, –8)
as a > 2
2
or (1, 2) or (1, –2)
2
x + 15a
8ax
=
a−2
a−2
2
Column Matching :
2
⇒
x – 8ax + 15a = 0
⇒
(A) → (T),
(B) → (S),
x = 3a, 5a
(C) → (P),
(D) → (Q)
∴
x = 9 (given)
(A) domain of f(x) is (1, 3]
⇒
x = 3,
1.
9
5
∴ integers in the domain are 2 and 3
Hence probability = 1/2
But a > 2
7.[C,D]
1 2
| t1 – 2t2 – t22 – 2t1| = 20
2
(B)
∴
a=3
at
a = 3 ⇒ x = 9, 15
x2
→
x
0, 2, 4, 6, 8
∴ Probability =
S1 = 1, S2 = 23
 ⇒ A & B are not true
S 2 − S1 = 22 
4
(C)
9(S2 – S1) = 198
Hence C & D are correct.
2 2
×
0, 4, 6, 6, 4
2
5
3
2
7
=
1
35
2 2 2
8. [B, D]
(D) Rectangles can be of sizes
α α α
|A(α)| = α α α = 0
α α α
1 × 1, 1 × 2, 1 × 3, ………..1 × 8 = 8 numbers
2 × 2, 2 × 3,……….. 2 × 8 = 7 numbers
3 × 3, 3 × 4,………….3 × 8 = 6 numbers
∴ A–1 does not exist.
α

Now A (α) = 3 α 2
α 2

2
2
2
2
8 × 8 = 1 numbers
α
2
α
2
α
2
α
2
Total rectangles = 1 + 2 + 3 + …. + 8 = 36
2
α 
α2 

∴ Required probability
=
2
A (a) = 3A(α ) ⇒ A (1) = 3A(1)
⇒ A3(1) = 3A2(1) ⇒ A3(1) = 9A(1)
9.[A, B, C, D] A(t1) & B(t2) ∠AVB = 90°
2.
8
2
=
36 9
(A) → S, T,
(B) → R,
(C) → S,
(D) → P,Q
2 2
= – 1 ⇒ t1 t2 = – 4
.
t1 t 2
f(x) = (x + 1)3 – 3α(αx + 1) + 2α2
f′(x) = 3(x + 1 – α) (x + 1 – α)
Q g(x) = f(x).f ′(x)
CAREER POINT, CP Tower, Road No.1, IPIA, Kota (Raj.), Ph: 0744-3040000
Q f′(x) = 0 for two distinct values of x so
Page # 3
g(x) has atleast three roots
f(α – 1) –α(α – 1) (α – 3/2);
f(–1 –α) = α(α + 3)(2α – 1)
(A)
Q g(x) has atleast three roots
So A∈φ
(B)
so
2.[1]
Equation represents parabola.
3.[3]
Rank is 2710
4.[4]
apply
B can be anything.
R1 → R1 – R2
R2 → R2 – R3
g(x) = 0 has five solution then
1
−1
0
0
1
− 1 = 1([z] + 1 + [y] ) + 1⋅([x])
[ x] [ y ] [ z ] + 1
f(x) should have three solution
1–α
= [x] + [y] + [z] + 1
–1–α
Maximum value = 2 + 0 + 1 + 1 = 4
α∈ (–∞, –3) ∪ (1/2, 1) ∪ (3/2, ∞)
(C)
Let X = x
5.[3]
So f(–1 – α) f(1 – α) × 0.
1
1
+
lny lnz
.
1
1
+
y lnz lnx
 1
 1
1 
1 
⇒ ln X = ln x 
+ (ln y) 
 ln y + ln z 
 ln z + ln x 




When f(1 – α) = 0
or f(–1 – α) = 0
(D)
 1
1 
 (ln z)
+ 
+
 ln x ln y 
g(x) = 0 has at most
three solution if
Now given ln x + ln y + ln z = 0
∴
ln x ln z
=–1
+
ln y ln y
similarly
f(1 – α) . f(–1 –α) > 0
Numerical Response
∴
R.H.S. = – 3
∴
ln X = – 3
X = e–3
→ →
1.[4] a . b = 0 ⇒ sec α . cos 5 α + 1 = 0
6.[2]
⇒ cos 5 α = – cos α
Let α is common root
α3 + a α – 1 = 0 ……(i)
⇒ 5 α = 2nπ ± (π – α)
…..(i)
cos α ⇒ cos α > 0
⇒ α is in I and IV quadrant
2
ln y ln z
=–1 &
+
ln x ln x
ln x ln y
=–1
+
ln z ln z
So α∈(–3, 1/2) ∪ (1, 3/2)
Condition (i)
& α4 + a α2 + 1 = 0 ……(ii)
Multiply equation (i) by α & subtract from (ii)
∴ α = –1
2
(ii) ln (2πα – α ) ⇒ 2πα – α > 0
∴ put in (i)
⇒ α ∈ (0, 2π)
⇒ a = –2
so α =
. z
1
1
+
lnx lny
π π 7 π 11π
,
, ,
6 4 4
6
hence 4 solutions
CAREER POINT, CP Tower, Road No.1, IPIA, Kota (Raj.), Ph: 0744-3040000
Page # 4
7.[4]
eˆ2 =
6 tan α 
–1  tan α 
= tan–1 
 + tan 

2
 8 + 2 tan α 
 4 
eˆ2 = −
2


Q 3 tan α < 1

 16 + 4 tan 2 α


4.[A]
3
a=
2  2   iˆ + ˆj 



3  2  2 
iˆ − ˆj − kˆ
3
2F
where M is mass of rod.
M
3F – T =
Q cot4x – 2(1 + cot2x) + a2 = 0
−
FNet = 3F – F = 2F
= tan–1 (tan α) = α
8.[0]
iˆ + ˆj − kˆ
3 sin 2α 
–1  tan α 
tan–1 
 + tan  4 
5
+
3
cos
2
α




cot4x – 2cot2x + a2 – 2 = 0
M
× x×a
L
T
(cot2x – 1)2 = 3 – a2
3F
x
to have at least one solution 3 – a2 ≥ 0
∴ a ∈ [– 3 , 3 ]
3F − T =
∴ integer values of a = – 1, 0, 1
M
2F
× x×
L
M
∴ sum = 0
x
O
PHYSICS
1.[C]
P1 +
T = 3F −
1
1
ρv12 = P2 +
ρ2v22
2
2
=
TL
= ∆L
AY
1
ρ(v22 – v12)
2
P1 – P 2 =
L
∴ Total extension
1
ρ(16 – 4) = 6ρ
2
1
P1 + ρgh1 + + ρg (h2 – h1) + ρv32
2
2 Fx
L
=
5.[A,D]
3FL FL 2 FL
−
=
AY AY
YA
A
30º (1)
(3)
1
1
6P + ρv32 = ρv42
2
2
30º
60º
(2)
28
B
2.[B]
3.[D]
Fx  dx
0
1
= P2 + ρgh2 + ρv42
2
v4 =

∫  3F − L  AY
On mirror B light fall normally, thus light retrace
and reflect at mirror A. Total no. of reflection is 3.
eˆ2 = eˆ1 − 2(eˆ1.Nˆ ) Nˆ
eˆ1 =
iˆ + ˆj − kˆ
3
iˆ + ˆj
Nˆ =
2
Deviation of light is π.
6.[A,B,D] When block was floating upthrust is equal to
weight of block. When it is pressed upthrust
become more than weight and force required to
press block = upthrust – mg.
CAREER POINT, CP Tower, Road No.1, IPIA, Kota (Raj.), Ph: 0744-3040000
Page # 5
Work done by force F will be equal to work done
against (upthrust – mg). It mean work done is
equal to work done against upthrust – loss of
potential energy of block.
Option-(D) : Volume and temperature is increasing
V = aT – b
PV = nRT
P(aT – b) = nRT
7.[A,B]
nRT
nR
=
b
aT − b
a−
T
P=
8.[A,C]
9.[A,B,D] En =
− 13.6Z 2
as T will increasing P will decrease.
eV/atom
n2
For Z = 2, E1 = – 54.4 eV
40.8 eV is the difference between two energy
levels n = 2 and n = 1. Also from ground state the
electron cannot fall and hence cannot emit
photon.
– E = k and E =
1.
2.
A
12Ω
1.[4]
24Ω
C1
12Ω
C2
24Ω
B
∠AC1B = 120º
U
2
AC1 = AC2
A → P,Q,S;
B → P,R,S;
C → P,S,T;
D → P,Q,R,S,T
A → Q,R,T ;
B→P;
C → R,S ;
D → Q,R,S
= C1C2 = radius
resistances of four sections are 24, 12, 12 and
24Ω equivalent resistance = R across AB
1
1
1 1
1
=
+ + +
R 24 12 12 24
Option-(A) : P is decreasing and volume is increasing
⇒ R = 4Ω
P = – aV + b
power =
PV = nRT
(– aV + b) V = nRT
T=–
60º
60º60º
2.[2]
20 2 400
=
= 100 watt
R
4
v
2
aV
bV
+
nR
nR
20 2 45º
u
T/V graph is parabola
∴ T first increase and then decrease.
Option-(B) : Pressure is increasing and temperature is
constant.
energy conservation
40 =
PV = nRT
v2 = 400
(aT + b) V = nRT
v = 20
CAREER POINT, CP Tower, Road No.1, IPIA, Kota (Raj.), Ph: 0744-3040000
1
1
mu 2 = m × 10 × 20 + mv 2
2
2
v 2 sin( 2 × 45º )
10
P = aT + b
bV
b
T=
=
nRT − aV nRT
−a
V
as T is increasing V will increase.
20
A
Option-(C) : Pressure and temperature is increasing
bV = nRT – aTV
B
1 2
1
u = 200 + × 400
2
2
u2 = 400 + 400
u=
800 = 20 2
Page # 6
5.[3]
mg sin 37º
3.[3]
Mg sin 37º
λ
= 0.5
2
mg cos 37º
mg
37º
λ = 1Å
C
A
We can say
Mg cos 37º
E=
Mg
p2
2m
37º
h
 
E= λ
2m
As system is in static equilibrium,
moment of forces acting on the system is zero.
2
Moment about C = 0
Mg sin 37º × R + mg sin 37º × R
6.[3]
Plane make angle θ with horizontal
R
= mg cos 37º× R
m = 3 kg
Considering normal forces
170g
⇒ N Mg cos 37º + mg cos 37º
θ
= 32 Newton
Considering force along the plane
F = Mg sin 37º + mg sin 37º
µmin =
4.[1]
8
15
R = 170 g cosθ
= 24 Newton
F ≤ µN ∴ µ ≥
tan θ =
F
,
N
24 3
=
32 4
R = 1500 N
Force on friction on A = 1500 × 0.2
= 300 N
Force on friction on B = 1500 × 0.4
= 600N
For cylinder A process is isobance while for B,
it is isochoric
For A, QA = nCp ∆TA
For B, QB = nCv ∆TB
Given QA = QB
n Cp ∆TA = n Cv ∆TB
∆TB Cp
=
= 1.4 = 1
∆TA Cv
Considering the two blocks as a system the net
force parallel to the plane
= 2 × 170 g sin θ – 300 – 600
= 1600 – 900
= 700 N
acceleration =
700 35
m/s2
=
340 17
170g sin θ – 300 – P = 170 ×
35
17
⇒ P = 150 N (P is pull on bar)
CAREER POINT, CP Tower, Road No.1, IPIA, Kota (Raj.), Ph: 0744-3040000
Page # 7
7.[2]
m2
m1
θ
l
x
x + y + x 2 + l 2 = constant
1
dx dy
dx
+
+
× 2x
=0
dt dt 2 x 2 + l 2
dt
dx
= velocity of m2 = v
dt
dy
= velocity of m1 = – 3
dt
v −3−
v
=0
2
v = 2m/s
8.[3]
Magnetic induction of origin is due to one semi
finite wire and two quarter circle of radii R and
2R.
CAREER POINT, CP Tower, Road No.1, IPIA, Kota (Raj.), Ph: 0744-3040000
Page # 8