ANSWERS, HINTS & SOLUTIONS
FULL TEST – II
(Paper – 2)
ANSWERS KEY
PHYSICS CHEMISTRY MATHEMATICS
Q. No. ANSWER ANSWER ANSWER
1. C A C 2. C D C 3. B A C 4. B B D 5. B B B 6. D B B 7. A C A 8. C A C 9. B A C 10. A D B 11. B C D 12. C A B 1. (A) → (r) (B) → (p) (C) → (q), (s) (D) → (q), (s) (A) → (p, q, t) (B) → (p, r, s) (C) → (p, q, t) (D) → (q, t) (A) → (s) (B) → (t) (C) → (p) (D)→ (q) 2. (A) → (s) (B) → (q) (C) → (r) (D) → (q) (A) → (q, r) (B) → (q, r) (C) → (r, s) (D) → (p, t) (A) → (r) (B) → (r) (C) → (p) (D)→ (r) 1. 5 3 3 2. 4 7 5 3. 2 3 9 4. 4 5 0 5. 4 4 3
ALL INDIA TEST SERIES
FIITJEE
JEE(Advanced)-2013
From Long Term Classroom Programs and Medium / Short Classroom
Program 4 in Top 10, 10 in Top 20, 43 in Top
100, 75 in Top 200, 159 in Top 500 Ranks & 3542 t
o
ta
l
s
e
le
c
ti
o
n
s
i
n
I
IT
-J
E
E
2
0
1
2
P
P
h
h
y
y
s
s
i
i
c
c
s
s
PART – I
SECTION – A 3. t 2CR q 2CV 1 e= − − ; t 2 2CR b W =2CV 1 e − − 4. LC = 4/200 = 0.025. Gravitational potential inside a shell = constant
6. In nuclear, nuclear force play major role. They are charge independent forces.
SECTION – C
1. Current is 2Ω = 0.1 A current in circuit = 0.2 A
equivalent resistance about AB = 5Ω e = I R = 1 volts and also e = B A V e 1 V 5m / s B 2 0.1 ∴ = = = × A .
2. Force on the rod F – Fm = ma B F (B V E) ma r − A A − = 2 2 B E B V mdV F r r dt + A − A = solving 2 2 0 0 B F mV sin( t) V cos t r = − ω ω + A ω B E r − A
Power expended by force is
2 2
2 2 2
0 0 0
B B E
FV mV sin( t)cos( t) V cos t V cos t
r r
= − ω ω ω + A ω − A ω
Its average over the cycle is 2 2 2 0 B V 2r A .
3. Let A1 and A2 be the cross–sectional area of the pipe at points P and Q respectively.
Let v1 and v2 be the velocities of oil at the points P and Q respectively.
By conservation of mass, Q = A1v1 = A2v2 1 2 1 1 2 A v (v ) 4v A ⇒ = =
2 2 1 1 2 2 1 1 P v P v 2 2 + ρ = + ρ 2 2 2 1 1 2 1 P P (v v ) 2 = + ρ − 3 1 1 P 0.9[16 256] 10 2 = + × − × 3 3 280 10 ( 108) 10 = × + − × P2 = 172 × 103 Nm–2 P2 = 172 (KN m–2). 4.
(
) (
)
0 0 net 0 0 0 0 P v W 2P v P v 4 π = − −(
)
0 0 0 0 net 0 0 P v P v W P v 4 4 4 π = − = − π ; Put π = 3.14(
) (
)
net 0.86 0 0 0 0 W P v 0.22 P v 4 = = Now, 0 0 1 0 0 2 0 0 3 P v T R 4P v T R 2P v T R = = = Thus,[
]
[
]
[
]
1 2 2 1 2 3 3 2 3 1 1 3 3R U 1 T T 2 3R U 1 T T 2 3R U 1 T T 2 → → → ∆ = × − ∆ = × − ∆ = × −( )(
) (
)(
) (
)(
)
1 2 0 0 0 0 0 0 Q→ 4.5 P v 1.22 P v 5.72 P v ∆ = + =(
)
2 3 0 0 0 0 Q → 3P v 0 3 P v ∆ = − + = −(
) (
)
(
)
3 1 0 0 0 0 0 0 Q → 1.5 P v P v 2.5 P v ∆ = − − = −Thus efficiency Wnet
veheat η = +
(
)
(
) (
0 00 0)
0.22 P v 0.04 5.72 P v η = = Thus efficiency is 4%5. The total momentum of the system in the horizontal direction is conserved. We draw the F.B.D., assuming the displacement of the block to be x1 and x2 in opposite
directions, and the total extension x is given by, x = x1 + x2 m1 kx x1 m2 kx x2 and m1x1 = m2x2 2 1 1 2 1 2 d x m k(x x ) dt ∴ = − + and 2 2 2 2 1 2 d x m k(x x ) dt = − +
after suitably manipulating the equations, we get,
2 2 1 2 2 1 2 d x k(m m ) .x m m dt − + = −
i.e. the frequency 1 2 1 2 k(m m ) 1 2 m m + = π 1 300(2 3) 2 2 3 + = π × 2.5Hz .
∴ number of complete oscillations in 1 minute 60 24 2.5
C
C
h
h
e
e
m
m
i
i
s
s
t
t
r
r
y
y
PART – II
SECTION – A
1. Gem-dihalides on substitution by OH produces carbonyl compound, which will reacts with
2,4-DNP but 1,2-dichloroethane will give vic-diol.
2. 2 2 3 4 MnO Fe 8H Mn Fe 0.1 M 0.01M 0.09 M − + + + ⊕→ + + + 2 4 2 2 0 0 MnO R.P.H R.P. Mn H E − E + + >
So permanganate electrode behave as cathode. Ecell = 1.51 – 0 = 1.51.
Using Nernst’s equation :- (n = 5)
4 2 8 4 MnO 2 Mn MnO H 0.059 E 1.51 log 5 Mn − + − + + = + = 1.51 + 0.11 = 1.4 V
3. Amine inversion will change the configuration of N only and not that of carbon.
5. Compound (A) will first undergo Cannizzaro reaction and the product B then will undergo esterification.
6. 3 600 0.2 4
( )
500 0.3 0 Moles 760 RT 760 RT NH HCl NH Cl s × × × × + → at room temperature. 150 120 760RT 760RT 30 120 0 760RT 760RTNow total moles = 30 100 1
760RT 760RT × + Total P = 130 RT 760RT 1.5 ×
× (Total volume = 1. 5 Litre) P in torr = 130 86.66 torr
1.5 =
SECTION – C
M
M
a
a
t
t
h
h
e
e
m
m
a
a
t
t
i
i
c
c
s
s
PART – III
SECTION – A 1. 3 1 1 1 1 f(x) x x 2 4 2 2 = − + − + 3 / 4 1/ 4 I=∫
g(x) dx let x t 1 2 = + ⇒ 1/ 4 1/ 4 1 I g t dt 2 − =∫
+ 3 1 1 1 f t t t 2 4 2 + = + + ⇒ 3 3 3 1 1 1 1 1 f f t t t t t 2 4 4 4 2 + = + + + + ⇒ g t 1 p(t) 1 2 2 + = + where p(–t) + p(t) = 0 ⇒ 1/ 4 1/ 4 1 1 I dt 2 4 − =∫
= 2. Required area is(
2)
0 1 2 ex ln x 1 1 (x 1) dx α α + − − − ∫
∫
−1 −1 1 3.(
)
n r 1 n n r t n r r r t r n n n n r 1 t o r 1 1 1 lim C C 3 lim C 4 3 5 5 − →∞ →∞ = = = ⋅ ⋅ = ⋅ − ∑
∑
∑
=(
)
n n n r n r n n r r n n n n r 1 r 1 1 1 lim C 4 C 3 lim 5 4 1 5 5 →∞ →∞ = = − = − = ∑
∑
. 4. In triangle OA1A2 2 2 2R 4 1 cos 4 2R 2 π − = = ⇒ R2 = +4 2 2 Also ⇒ ( )2 ( )2 r 1+ − −r 1 = +4 2 2 ⇒ 4r= +4 2 2 ⇒ r 1 1 2 = + A8 A3 R Rπ/4 O A1 A2 Vertex of octagen O1 r (r–1) O 1 R 1 1 A2 5. f 3 2 2 π = π ⇒ f ' 3 g' 2( ) 1 2 π ⋅ π = ⇒ ( )f ' x 25(2) 2x 3( )24 4 sin x 3 = − π + − ⇒ f ' 3 7 2 3 π = ⇒ ( ) 3 g' 2 7 π =
6. Let the age of friends be a, ar, ar2, r > 1 and money distributed is x, xr, xr2 respectively. ⇒ (1 + r + r2)x = (a(1 + r + r2) + a)t
Also after 3 years ar2 + 3 = 2(a + 3)
⇒ a(r2 – 2) = 3
x + 105 = (a + 3)t and xr + 15 = (ar + 3)t (given)
⇒ (1 + r)x + 120 = (a(1 + r) + 6)t ⇒ 2(a + 3)t = r2x – 120 = 2x + 210
⇒ (r2 – 2)x = 330
Using (r2 – 2)x = 330, a(r2 – 2) = 3 ⇒ x = 110a
⇒ 110a + 105 = (a + 3)t ⇒ 5(22a + 21) = (a + 3)t and 110ar + 15 = (ar + 3)t ⇒ 5(22ar + 3) = (ar + 3)t
⇒ 22a 21 a 3
22ar 3 ar 3
+ = +
+ + ⇒ 5ar = 7a + 6
Where 5ar = 7a + 2(a(r2 – 2)) {Using a(r2 – 2) = 3} ⇒ 2r2 – 5r + 3 = 0 ⇒ r 3
2
= . Hence a = 12 years 7. Given cn = a1 + a2 + a3 + ….. + an
where a1, a2, ….., an are in A.P. with d = 2
and dn = b1 + b2 + b3 + …… + bn
where b1, b2, b3, ….., bn are in A.P. with d = 2
Also (an, cn) lies on y = px2 + qx + r
Now cn = pan2+qan+ ….. r (1)
cn– 1 = pa2n 1− +qan 1− + ….. r (2) ∴ From (1) and (2), we get
cn – cn – 1 = p a
(
n2−an 12−)
+q a(
n−an 1−)
⇒ an =p a(
n+an 1−)(
an−an 1−) (
+q an−an 1−)
(
) (
)
n n n 1 n n 1 a = a −a − p a +a − +q ….. (3)(
an−an 1− =d)
10. y = t(1 – 3t2) ⇒ t = y x ⇒ at A, t = – 1 Slope of tangent at A, m = 2 1 9t 4 6t 3 − = − − . ⇒ Equation of tangent at A ⇒ 3y + 4x + 2 = 0 Since tangent passes through the point B(t1) hence3(t1 – 3t13) + 4(1 – 3t12) + 2 = 0 ⇒ (t1 + 1)2(3t1 – 2) = 0 ⇒ t1 = 2 3 ⇒ B ≡ 1, 2 3 9 − − .
11. Slope of tangents at x = 0 and m1 = 1 3 and m2 = – 1 3 ⇒ tan φ = 3 π. SECTION – B 2. (A) We have 1 1 f(x) lncos x− =
For domain of f(x), ln cos x
(
−1)
>0 ⇒ cos x 1−1 > and cos x−1 ≤ π ⇒ cos π ≤ x ≤ cos 1 ⇒ –1 ≤ x < cos 1Hence number of integers in the domain of f(x) are 2 i.e., –1 and 0
(B) Vector normal to the plane is nG= −ˆi 3 j 2kˆ+ ˆ and vector along the line is VG =2iˆ ˆ+ −j 3kˆ
Now sin x v 2 3 6 7 x v 14 14 14 ⋅ − − θ = = = G G G G Hence cosec θ = 2 (C) We have
(
)
(
)
4 2 2 5 J 3 x tan 3 x dx − − =∫
− − . Put (x + 5) = t, we get ( )(
)
(
( ))
1 2 2 0 J=∫
3− −t 5 tan 3− −t 5 dt =(
)
(
)
1 2 2 0 22 10t t tan 22 10t t dt − + − − + −∫
Now(
)
(
)
1 2 2 2 K 6 6x x tan 6x x 6 dx − − =∫
− + − − . Put (x + 2) = z, we get ( ) ( )(
)
(
( ) ( ))
1 2 2 0 K=∫
6 6 z 2− − + z 2− tan 6 z 2− − z 2− −6 dz(
)
(
)
1 2 2 0 22 10z z tan 22 10z z dz− + − + −∫
Hence (J + K) = 0 SECTION – C 1. A 1, 4, 0 3 − − , 7 B 0, , 1 2 − and 7 C , 0, 8 3 . Hence V 56 18 = 2.(
)
10 r 10 10 r r r 1 1 3· C r· C = +∑
+ = 410 + 10 × 29 ⇒ α = 1 and β = 5 ⇒ k > 4 3. n(
1/ 3) (
n 6 1/ 3)
6 6 x = C 3 − 4−(
) (
6)
x 6 n 1/ 3 1/ 3 n 6 y = C − 3 4− − and x 1 y =12 ⇒ ( ) ( ) n 12 / 3 1 12 − = 12 − ⇒ n = 94. z2 + z3 = 0 5. Let n 2n+1 r r 0 y C = =
