PAGE
www.aieeepage.com
key AI - IITJEE - Model test - 06
PAGE
www.aieeepage.com
PAGE - A Premier Institution for IIT - JEE & AIEEE at the National Level
1 of 21
Paper - I : Physics - 1 to 50 Solutions 1.d. Just by looking at the equation ,what would decrease
How does one suply energy to the photoconductor?By electromagnetic radiation. If no electron-hole pair is produced, the radiation may not be energetic enough. How can one increase the energy? Again, analogous to the photoelectric effect, the only way to do this is by decreasing the wavelength (of increasing the frequency) of the radiation. this is choice D. Choice A is incorrect because increasing the intensity only increases the number of photons; it does not increase the energy of each photon. Choice B is incorrect because the gaseous ions are there only to create the external electric field. If does not take part in the process of generating the electron-hole pairs.
TL ?” Decreasing d ( the numerator ) or increasing
2.c.
µ or V0 .From this ,we could elimainate choice A .Coices B and C are incorrect because they have todo with the creation of electron - hole pairs in the photocoductor , they do not deal with the movement of these charge carriers . Choice D is correct because in the number of gaseous ions adsorbed on the surface increase, it is analogous to increasing the charge on the plates of a capacitor . The voltage between the plates would also increase since C = Q / V. This voltage is the dark voltage ,and sowe have increased the deniminator in the expression for the transit time ,thus decreasing its value. Solve the force equation for ε0 . The net result is simply the interchanging of F and ε0 . The 4π has no effect since they are dimensionless pure number . ε0 ,therefore, has units of charge squared divided by the quantity newtons times distance suqared . This is exactly choice C One needs to combine the qeuation given in the passage for the capacitance with the general one for capacitors C = Q / V . As given in the passage
PAGE 6.c.
ε−L
di di = Ri ; ε − Ri = L dt dt
www.aieeepage.com
3.d.
i
di = or ∫ ε − Ri 0
ke 0 A ke 0 A Q = d d V The surface charge density , Q / A is therefore
i=
Q ke 0 V 3ε 0 (800) = = = 1600ε 0 × 10 5 C/m 2 A d 1.5 × 10 5
= 1600ε 0 × 10 5 C/m 2 ×
1m 2 10 4 cm 2
0
; −1 log e ( ε − Ri ) = t R L t R −t R ; L L e = ε − Ri
tR − 1 tR L ε − e or i.e., iR = ε − e − L R tR − − tLR L ε −ε −e = e
V2 = ε − Ri =
= 1600ε 0 × 101 C/cm 2
⇒ induced emf decays exponentially. Hence graph (c) is the choice.
= 1600ε 0 C / cm 2 4.b.
dt
∫L
log e ( ε − Ri ) = −
C=
:
t
If light of the correct wavelength is shone on the photoconductor, electron-hole pairs would be generated, as stated in paragraph 1. If no gaseous ions were adsorbed, however, there would be no external electric field, and so the electrons and the holes would not move in opposite directions. they would eventually recombine and annihilate one another. Choice C is incorrect because the dark voltage changes to the light voltage as the holes migrate to the surface and neutralize the ions. If there were no ions adsorbed, the dark voltage would in fact be 0, and there whould be no change to this value even as light were shone. Since the creation of electron-hole pairs requires some minimum energy to raise the electron to another state, it is analogous to the photoelectric effect.
7.b.
V2 cos θ = V1 ;
PAGE
V1 = cos θ . V2
V2
V2
V1 V2cos
m
www.aieeepage.comM
5.d.
8.b.
mu m
mv
m
1 m
mu + 0 = mv1 + mv
PAGE - A Premier Institution for IIT - JEE & AIEEE at the National Level
1
m×0=0
mv 1
v
2 m
v
— (1)
2 of 21
v − v1 e= ; v − v1 = eu, v1 = v − eu . u
10.c.
mu = m(v − eu ) + mv = 2mv − meu ; 2 mv = mu (1 + e) .
1+ e ∴V = u . 2 mu(1+e) 2
2 m
11d.
3 m
V 2
2
3
V
from (1), (2) E = 40 volts In both upward and downward motion function force at every point is identical, work done against friction in 2 cases are same i.e., mgh. ∴ Total work done by the force F is against gravitational force + work done against friction= 2mgh It displacement will increase after every rebound and finally becomes maximum displ. becomes maximum as it comes to rest after several rebounds.
3
mu(1 + e) = mV2 +, mV3 ; 2 V3 − V2 =
ymax
u (1 + e) ⋅e ; 2
V2 = V3 − e
PAGE 12.a.
u (1 + e) 2
mv 2 mv 2 ke 2 − = KE = 2 r 2 2r
u (1 + e) eu (1 + e) = V3 − + V3 ; 2 2 1+ e V3 = 2
www.aieeepage.com 1
ke2 nh KE = mv = 3 ; But mv = 2πr 2 2r 2
2
n2h2 ke 2 4π2 ke 2 m = ;r= 4π2 ⋅ 2mr 2r 3 n2 h 2
n
1 n. 1+ e 1 + 2 u =3 ∴Vn = x = 2 2 4 n
9.a.
−do −∂ ke 2 −ke 2 F= = = 4 = dA dr 3r 3 r
In a circuit, current i = i =
1 ke2 1 ke 2 ∴ TE = ⋅ 3 = ⋅ 6 r 6 4π2 RC 2 m
(
E R+r
6 ∴TE α n
Power consumed (P) is the external resistance
13.d.
2
=i R
dP =0 dR
.
With respect to infinity TE of hydrogen atom in ground state = –13.6eV
−13.6 + 27.2 = n2
PAGE ∴ E2 = 1
−13.6 + 27.2 = 23.8eV 22
= 0 ; ( R + r ) ( R + r − 2 R) = 0
∴R = r ∴ Power consumed is a circuit is maximum when
energy is first excited state.
dQ dv dw = + dt dt dt
www.aieeepage.com
the external resistance is equal to the internal resistance
∴ Pmax =
E
2
(R + r )
2
.R =
Current = i =
E
2
( R + R)
2
2 3
of the ground state is taken as zero PE, PE of the atom and ∴ TE increases by 27.2 eV.
This becomes maximum when
(R + r)4
3
2
∴ PE = 2(TE ) = −2 × 13.6 = −27.2eV
2
E2 E = P = R R (R + r)2 R+r
( R + r 2 )1 − 2 R( R + r )
m3
)
(n h )
=
2
E = 100 ...(1) 4R
E E E = = = 10 ...(2) R + r R + R 2R
14.b.
C=
T⋅
VT = constant;
dv = 0 (PV = RT ) dt
C = CV + P ⋅
PAGE - A Premier Institution for IIT - JEE & AIEEE at the National Level
dv −V = CV + P dt T
= CV − R
3 of 21
15.d.
λ= λ1 =
16.c.
hc hc = ; 2E − E E
A
3 ⋅ hc hc = = 3λ . 2E − E E
F C D
R = 2 h(2 − h) . As the value of h lies between 2 and 1.5; R1 is maximum for h=1.5m.
17.a.
From A → B . V decreasting and temperature increases, means pressure should increase and
22.a.
∆ Q = ∆ U + ∆ W = ∆ U + 2 R log e
n
v2 ∆ T = ∆ U + 2R v1
2V0 ⋅∆T V0
Path different (for bright finge) =
for point A ;
1 1 3 1 + = ; v 20 40 v = 40 − 0 A B
C
2V 0 = ∆ U BC + 2 R | n2 ⋅ ⋅ (300 − 500) 2V 0
www.aieeepage.com
∆ QC → D = ∆ U CD + 2 R | n2 ⋅ (0 − 300) = 0 − 600 R / n
∴
∴ A1 B1 =
dQ dv + dw dv dv = = + p⋅ dt dt dt dt
20.a.
R R α VR = + 2 αV r −1 2 αV
T0
∫ dT = −
T
U 3K
R
∫ r ⋅ dr ; T = T0 + r
v × CD u
24 × 10 = 8cm ; 30
A1 D1 = 40 − 24 = 16m ; area of A B C D = 16 + 1
1
1
1
PAGE 24.d.
The energy omitted by this shell =
4 3 dT πr ⋅ U = − K 4πr 2 3 dr
21.c.
C1 D1 =
R R R(2 + r − 1) R(r + 1) + = = 2( r − 1) 2(r − 1) r −1 2
1 11 3 + = ; v 30 40
v = OD1 = 24cm ; C 1 D1 =
dv R dv ; = = CV . dt 2α V dt
C=
v v 40 × AB = × AB = × 10 = 20cm n n 20
for point D1:
RT dt α α =αV ;T = V2 ; = 2V V R dv R
C = CV + P ⋅
D D1
Heat capacity of gas
But P α V ;
C1
10m
2
∴ ∆ Q = ∆ Q A→ B + ∆ QB →C + ∆ QC → D + ∆ QD → A = 400 R / n 2
=C =
20m
A
D
∆ QD → A = ∆ QB →C 19.d.
1 1 1 + = ; v m f
PAGE 23.b.
∆ QA→ B = ∆ U A→ B + 2 R | n 2 ⋅ (500 − 0) = 1000 R ln 2
∆ QB →C
F2
d ⋅d xd 2 d2 d2 = = = nλ; n = D D 2D 2λD
from B → C ; V α T ∴ pressure should be constant. 18.b.
F1
B
8 + 20 = 224cm 2 . 2
amplitude of incident and reflected wave are a1, a2 respectively amplitude of antinode = a1+a2, amplitude of node = a1– a2 given.
a1 + a2 3 a22 1 = ; a1 = 5a2 ; 2 = a1 − a2 2 a1 25
U (R2 − r 2 ) 6K
% of energy transmitted = www.aieeepage.com a
Since foca lengths of the 2 parts will increase, the parallel rays on the region AB and that on the region CD will focus away from F towards right. But the rays passing through the region BC will still focus at F.
2
2 1 − 2 ×100 = 96% . a 1
25.c.
Let the source emitted I pulse when it is at A. Then time taken by the pulse to reach observer = t1 =
PAGE - A Premier Institution for IIT - JEE & AIEEE at the National Level
d v
4 of 21
A
Source
B
O
1 a T2 2
−(2 − y ) ∴ ax =
Fy =
Observer
Let the second pulse be emitted when it is at B. ∴ time taken by pulse to reach B = T
ax = −1, a y = −3 : a = ax 2 + a y 2 = 10 29.c.
1 T = f
Path difference between light rays reflected=
µ t + µ t = 2µ t from 2 surfaces of coating. ∴2µ t =
= time taken by source to reach B (=T)
1 2 at 2
−dv Fy = −(2 x − 1) ; ay = = −(2 x − 1) dy m
At (2,1) :
d
∴ AB =
Fx = −(2 − y ) m
1 2 (∵ u = 0) BO = d − at 2
λ (for maximum) 2
PAGE
1 1 2 d − at 2 at d 2 1, T + ;t1 = v ;t2 − t1 = V − 2 2 v Frequency heard by observer = ∴
⇒t =
30.d.
λ = 10−7 m. =10-5cm . 4µ
n1, n2 be the refractive indices of 2 lenses when lenses are immersed in water.
1 n1 1 n2 2 −2 = − 1 and = − 1 F2 nw R F1 nw R (Say converging) (Say diverging) Combined focal length =
www.aieeepage.com
1 t2 − t1 26.b.
2
2vf . 2vf − a
F=
1 q2 q2 − + mgx = 4πF0 r0 − x r0 1 2 kr ; x = 0.1m . 2
27.a.
31.a.
Volume of elemental shell = Charge at shell =
1 1 1 2 n −n = + = 1 2 F F1 F2 R nw
n1 − n2 =
R nw 20 4 1 5 ⋅ = × × = 2 F 2 3 24 9
Path diff. =
λ 2λ λ λ × = , ∴ Phase diff. = λ 8 4 8
Intensity at this point
(2πx 2 ) dx
dQ = ρ ⋅ 2πx 2 dx
= I = 2 I + 2 I ⋅ I ⋅ cos
Potential energy
= dU =
x
−q ρ U = ρdU = 2ε 0
R
∫ x ⋅ dx = 0
PAGE 33.b.
2
0
−q ρ R 2 −q ρ R 2 ∆U = U α − U = 0 − = 4 ε0 n ε0
− dv = dx
V + V0 V − V0 m 10 = n − n ; u = 2.5 s V + VS V + VS
From ;
B → A chose the path containing the cells
V − V = 8 − 3 + 10 = 15V . = q ρR www.aieeepage.com 4ε
Work done =
U = 2 x − y ; Fx =
= 4 I (= I C )
I 2 I ( 2 + 1) = = 0.853 IC 4I
32.b.
R
dx
28.a.
Intensity at this centre
1 ( − q ) ( dQ) q ρ ⋅ = x.dx 4πε0 2ε 0 x
π = 2 I ( 2 + 1) 4
A
34.b.
B
E and B are mutually perpendicular
E = Ei ; B = B j and V = Vx i + Vy j + Vz k . Since the charged particle goes undeviated, net force on the charge is zero.
∴ F = qE + q (V × B ) = 0
PAGE - A Premier Institution for IIT - JEE & AIEEE at the National Level
5 of 21
35.b.
j k = EL + i = E L + i( − BVz ) + k ( BVx ) Vx Vy Vz = i ( E − BVz ) + kBVx = 0 O B O
− q0 ⋅ e− t / CR CR
E = BVz and Vx = 0 means component of
i=
velocity along E (which is in x-direction = 0) Due to induction q charge flows to the inner side of 1 plate As A, B are exposed to same medium.
at t = 0 ; i =
Q1 − q = Q4 + Q3 + Q2 + q q=
Q1 + Q2 + Q3 + Q4 2
- (Q2 + q)
A
36.c.
F =−
PAGE B
www.aieeepage.com
1 ε0 kAE 2 − x 2
b
1
∫ Bwxdx = 2 Bw (b
x
1 1 V2 1 U = ε0 kA ⋅ 2 − x = ε0 kA ⋅V 2 ; 2 2 x x 1 1 dv = + ε0 kAV 2 ⋅ 2 ⋅ velocity 2 dt x
Slope of the line =
2 − r = − ; r = 0.4Ω 5
V
dx
l
PAGE
m ; l
mass of element = dm = λ ( dx) l
2
1 mv 2 xv KE of spring = ∫ (λ ⋅ dx ) . = 2 6 l 0
41.c. Gravitational force is the internal force www.aieeepage.com
Equation of discharging is, q = q0 e − t / CR
i=
− a2 )
v . l
λ = mass length ; λ =
∴ Terminal voltage = V = E − ir ; i = 0 ; V = E = 2 Volt
2
When free end which is at distance l moves with velocity ν , then the element located at x moves 1
dv 1 α 2 dt x
2
0
40.c.
with velocity ν = x ⋅
V E= x
38.d.
dx = B ( dx )v = Bwx ⋅ dx
∴ emf between centre and edge of disc =
1 k ε 0 AE 2 2
37.a.
dx
Consider an element (d λ ) at distance x from centre. This element is moving with speed, v = wx . ∴ Induced emf across
Force between & Plates of Condenser =
∴
V
Q4 + Q3 + Q2 + q
- (Q3 + Q2 + q )
dv ; dx
U = ∫ F ⋅ dr = F ⋅ x =
V . R
39.d.
Q3 + Q2 + q
Q2 + q
Q1 - q
CV e − t / CR V − t / CR = ⋅e CR R
V CV . ρd VA k ε A V i= = = 0 = A ρd d ρkε0 ρkε0
Q1 − Q2 − Q3 − Q4 2
∴ Q1 − q =
(q0 = CV )
dq 1 = q0 ⋅ e − t / CR ⋅ − = dt CR
M ⋅ v1 = M ⋅ v2 ; v1 = 2v2 2 from energy conservation :
PAGE - A Premier Institution for IIT - JEE & AIEEE at the National Level
V1
V2 Earth
6 of 21
1 M 2 1 M gh ⋅ v1 = M ⋅ v2 2 = 2 2 2 2
T=
Max. extension in spring during oscillation =
M ⋅g +A k
T=
∴ Max. force acting on M during oscillation of
M g M 1 = Fmax = K 1 + A k for M2 to be at rest,
43.c.
dv a= = −α V ; dt V = V0 e ρ
α
0
0
αt
µ≥
t
dv ∫V v = −α ∫0 dt ; 0
V0 α
Power = F.V. =
mg sin θ ⋅ 2 gl sin θ = 2m 2 g 3l sin 3 θ 45.d.
At any instant, points are situated at the vertices of equilateral triangle. ∴ Relative velocity of approach =
Kt − kt cos 60º
T
∫ ( Kt − kt cos 60º )dt = a ;
Kt 60º
0
Kt
T= 46.c.
= 4a
K.
60º
60º
Kt
700 × 10 −9
= 4143 K
The temperature of the blackbody must be somewhere between these two values, choice (C) is correct. In this problem, you are asked to figure out the relationship between the color of an object and the surface temperature. Since the color of light depends on its wavelength, the real question is : how does wavelength of emission and temperature relate? The passage tells us that the emission spectra of stars are pretty close to the blackbody spectrum. So we can make use of Wien’s displacement law: as the temperature goes up, the wavelength at maximum intensity goes down. (Not that we aren’t overlooking the “maximum intensity” requirement the color of light we see from a thermally radiating object is generally from the most intense photons. Remember that photon intensity depends on the number of photons, not their energy.) So we need the color with the lowest wavelength. That’s blue and violet light, which radiates around 400 to 450 nm. The correct answer is choice (B).
2.9 ×10 −3 = 1× 10 -3 m 3
Since the question calls for extremes, we could probaly have eliminated choices (C) and (D) right away, since answer choices (A) and (B) are more extreme choices. 49d.
The quantization condition means that the energies of the system take the form: E = nhf So the energy difference between consecutive energy states is just hf. (For example, the energy difference between consecutive energy states n = 2 and n = 3 is 3hf-2hf = hf) For a mass oscillating on the end of a spring, the
PAGE
Space is homogeneous enough to be considered a blackbody. This is a straightforward calculation question - given the temperature, can you determine the wavelength of maximum intensity emitted by the cosmic background radiation?
λ max = 47.c.
2a K − k cos 60º
2.9 × 10 −3
www.aieeepage.com
αt ds = V0 e ; dt
αt
= 7250 K
PAGE
M1 kA + M2 M2g .
V
∫ ds = ∫ V0 e ⋅ dt; S = 44.d.
48.b.
Fmax ≤ µM 2 g ;
M 1 g + kA ≤ µM 2 g ;
400 ×10 −9
On the other had, an object that emits light at the high end of the visible spectrum has a temperature.
∴ v1 = 2 gh 3 . 42.b.
2.9 × 10 −3
frequency of oscillation f is : f
=
1 2π
k m
Consequently, we can write the energy gap between consecutive states as : ∆E =
h k 2π m
50.b. The specific heat of a substance is the energy www.aieeepage.com needed to raise the temperature of 1 kg of that
The real question here is, do you know what wavelengths correspond to visible light? Visible light ranges from about 400 to 700 nm in wavelength. If the light is emitted at the low end, 400 nm, the object’s temperature is:
substance by 1 degree K. In this case, we use 3840 J to heat a 3-kg block of lead up by 10 K. Apply the definition to calculate the specific heat: ∆Q = mc ∆T
c=
∆Q 3840 J J = = 128 m∆T (3 kg) (10 k) kg.K
PAGE - A Premier Institution for IIT - JEE & AIEEE at the National Level
7 of 21
Paper - I I : Mathematics - 51 to 100 Solutions 51.d.
The eqn. of chord OA is
b Tanθ1 y= a
or
x
− 2a +
O is centre (0,0)
these
are
55.d. diameters,
⇒
π a 2 sin 2 θ + b 2 cos 2 θ ∵θ 2 = 2 + θ1
∴ AO 2 + CO 2 = a 2 + b 2 AO = CO ⇒ AO 2 = CO 2 ⇒
56.b.
a cos θ + b sin θ = a sin θ + b cos θ 2
2
2
2
2
2
2
⇒ (a − b )v = (a − b ) cos θ 2
2
2
2
2
= 12
− 6m − 8) = 12 (m 2 + 1)
⇒
4 (3m + 4) 2 = 144 ( m 2 + 1)
⇒
(3m + 4) 2 = 36 (m 2 + 1)
⇒
27m 2 + 24m + 20 = 0
...(1)
Since the discriminant of (1) is (-24)2 - 4.27.20 = 1584 which is negative, there is no real value of m. Hence no such line is possible. Given circle is the intersection of the sphere
r 2 + r .( 2iˆ − 2 ˆj − 4kˆ ) − 19 = 0 or
⇒ Tanθ = ±1
x 2 + y 2 + z 2 + 2 x − 2 y − 4 z − 19 = 0
and the plane
⇒ θ = π or 3π 4 4 54.a.
m2 + 1
(6 m + 8) 2 = 144 ( m 2 + 1)
AO 2 = a 2 cos 2 θ + b 2 sin 2 θ, CO 2 =
2
m (−2) − 3 − 4m − 5
On squaring www.aieeepage.com
π ⇒ θ1 ∼ θ2 = . 2
53.c.
Suppose, iof possible, Equation of line through (4, -5) with slope m is y + 5 = m (x - 4) mx - y - 4m - 5 = 0 ⇒ Then
π ⇒ Tanθ1 = − cot θ2 = Tan + θ2 2
⇒ -a 2 + b 2 = 3ac
2b 2 = a 2 + 3ac
PAGE
conjugate
2 b Tanθ1 b Tanθ2 −b = 2 a a −a
52.a.
4b 2 = 6c a
⇒
A = (a cos θ1 , b sin θ1 ),
2b and = −2b a
Substituting the values of p and q in (3), we get
x
C = (a cos θ2 , b sin θ2 ) .
...(4)
from (2) & (3), q = −
b Tanθ2 The eqn. of chord OC is y = , where a
As
p = aq
r .(iˆ − 2 ˆj − 2kˆ ) + 8 = 0
Given equation is
PAGE
a( x4 + y 4 ) − 4bxy ( x2 − y 2 ) + 6cx2 y 2 = 0 ...(1)
Equation (1) is a homogeneous equation f fourth degree and since it represents two pairs at right
5
angles. i.e., sum of the coefficients of x 2 & y 2 should be zero.
P
Let a ( x 4 + y 4 ) − 4bxy ( x 2 − y 2 ) + 6cx 2 y 2 = 0
We have
c(-1,1,2)
M
Q
www.aieeepage.com CM = Length of the ⊥ from −iˆ + ˆj + 2kˆ to the plane
= ( ax 2 + pxy − ay 2 ) (x 2 + qxy − y 2 ) where p & q are constants On comparing similar powers, we get p + aq = - 4b .....(2) -2a + pa = 6c ...(3) Again, given two pairs coincide, then
(ii)
⇒ CM = p =q q
⇒ CM =
PAGE - A Premier Institution for IIT - JEE & AIEEE at the National Level
(−iˆ + ˆj + 2kˆ).(iˆ − 2 ˆj + 2kˆ) + 8 1+ 4 + 4 −1− 2 + 4 + 8 3
=3
8 of 21
f ' ( x) = −e − x − cos x and f ' ( x) = 0 ⇒ −(e− x + cos x) = 0
∴ PM = CP 2 − CM 2 = 25 − 9 = 4 57.a.
58.c.
The equation of xy - plane is z = 0 i.e., 0x+0y+z=0. Clearly, the given plane is perpendicular to this plane.
sin x − cos x)
f ( x) =
=
sin (π + x) + cos(π + x sin x − cos x
=
sin x + cos x
63.c. 64.d.
....(1)
sin x + cos x
sin (π + x − cos(π + x)
f (π + x) =
⇒ 1 + e x cos x = 0 ⇒ e x cos x = −1 .
sin x) − cos x)
1
- sin x − cosx)
b
59. d.
f ( x) = e x
−3 x + 2
= f ( x)
∴
f : (−∞, −1] → (0, e ], e
x3 −3 x+2
=e
slope of normal of f(x) = 0 as x = 0 is −
=
61.c.
e12 =
66. b. 2
2
2 xf ' ( x ) − 40 xf ' (4 x ) + 56 xf ' (7 x ) 2
16 + 9 9 =1+ 16 16
a b , and Radius is OC = 2 2
i.e,
2
f ' ( x ) − 20 f ' (4 x ) + 28 f ' (7 x )
PAGE
1 1 1 = =− f ' (0) − 20 f ' (0) + 28 f ' (0) 9 f ' (0) 3
x = 0 is a point of (-1, 2). At x = 0, log t and
sin t t
a 2 b2 + 4 4
Y
(0,b)B
a b C , 2 2
M
www.aieeepage.com is continuous and differentiable
∀t ∈ (−1,2).
1 2 (x ) = 1 k2
Let the co-ordinates of A be (a, 0) and of B be (0, b), then AOB being a right angled triangle, the centre of the circumscribed circle is the mid-point of AB
1
1+ t + t2
( x2 − y 2 ) +
∴ e12 + e22 ≤ 3
2x
1- t + t 2
=1
16 − 9 9 =1− 16 16 2
both are not defined. ∴ (a) & (b) are false.
62. a.
b
2
and e2 =
2
2
2
b2
(± k , ± k ) satisfying the above equation.
f ( x ) − 5 f ( 4 x 2 ) + 4 f (7 x 2 )
x →0
k2
Which is independent of a and has 4 points
65.c.
2
1
1 y2 1 x2 2 + 2 − 2 = 1 k b b
∴
1 f ' (0) ..(2)
1 By (1) and (2), f ' (0) − 3
= lim
b
+
2
www.aieeepage.com 1 ....(1)
x →0
1
y2
−
⇒
if x = -1
=3
= lim
=
2
PAGE 4
3 x − y + 3 = 0 ⇒ y = 3 x + 3 ⇒ slope of normal
x
a
a2
Range of f is (0, e 4 ] ⊂ (0, e 5 ] ∴ f is into map.
lim
1
x2
≥ 0∀x
5
x →0
a2
Hence, the equation of the family of hyperbolas is
Evidently f is increasing function and so f is one one map.
60.a.
1
<
2
Let they differ by a constant k 2
f (π + x) = f ( x) ⇒ f ( x) has period π 2
Since, it is given the the reciprocal of the square of the length of semi-conjugate axis is less than the reciprocal of the square of the length of semitransverse axis i.e.
O
A(a,0)
X
L
∴ c is true
If we take f ( x ) = e − x − sin x
[For e x sin x = 1 ⇒ sin x = e− x ⇒ e− x − sin x = 0]
Equation of the circle is
( x − a) ( x − 0) + ( y − 0) ( y − b) = 0 ⇒ x 2 + y 2 − ax − by = 0
then
PAGE - A Premier Institution for IIT - JEE & AIEEE at the National Level
9 of 21
a b x (0) + y (0) − ( x + 0) − ( y + 0) = 0 2 2
which is independent of A, B, C and D. 69. a.
⇒ ax + by = 0
Lety = [sin 2 α + sin 4 α + sin 6 α + ...to ∞ ] log e 2
Let AL and BM be the perpendiculars from A and B on equation (1)
a2
thern, AL =
a2 + b2 b2
and BM =
a2 + b2
sin 2α log e 2 = tan 2 α log e 2 ⇒ y= 2 1 − sin α
=m
2
⇒ y = log e (2 tan α ) ∴ e y = 2 tan
=n
2
α
2 Given 2 tan α = 8 = 2 3
∴ AL + BM = m + n = a 2 + b 2 = diameter
⇒ tan 2α = 3
of circle
⇒ tan 2α = 3
Y
67. d
.
C’
B
α 3
−1
www.aieeepage.com
8
C
8
A
X
70.d.
Coordinates of Co − ordinates of the centre = point C of the smallest circle Coordinate s of Co − ordinates of the centre = point C of the smallest square ∴ (Coordinates of point C) = (4, 4) PQ = 8 - 2 - 2 = 4 = QR {∴ radius of each large circle = 2 cm} 2
2
∴ PR = 4 + 4 = 4 2 and hence, PC =
Putting x = 0 and both sides of the equation we have
1 2 0 g = 0 1 2 =9 2 0 1
71. a.
Differentiating both sides and then putting x = 0, we get f = -5. The system has non - zero solution if
sin 3θ ∆ = cos 2θ
−1 1 4 3 =0
2
4 2 −2 2 2
2π 2π π = =α =π − 3 3 3 π 2π 3π ≤ ∵ ≤ 3 2 2
Q
P O
PAGE ⇒ α=
Noew, sin sin
R
S
7
7
Applying R 2 → R2 + 4 R1 and R3 → R3 + 7 R1
PAGE
∴ Radius of smallest circle = 2 2 − 2 Hence, the required circle is
( x − 4)2 + ( y − 4)2 = (2 2 − 2)2
68. d.
Roots of equation x 2 − 9 x + 8 = 0 are 1 and 8
⇒
sin θ ∆ = cos 2θ + 4 sin 3θ 2 + 7 sin 3θ
−1 0
1 7 =0
0
14
cos A sin A cos (A + D) Since, ∆ = cos B sin B cos (B + D) cos C sin C cos (C + D)
⇒ ⇒ ⇒
14 cos 2θ + 56 sin 3θ - 14 - 49 sin 3θ = 0 2 cos 2θ + sin 3θ - 2 = 0 2 (cos 2θ − 1) + sin 3θ = 0
Applying C 2 → C3 − (C1 cos D − C 2 sin D )
⇒
- 4 sin 2θ + 3 sinθ − 4 sin 3 θ = 0
⇒
− sin θ + ( 4 sin 2 θ + 4 sin θ − 3) = 0
⇒
sinθ ( 2 sin θ − 1) (2 sin θ + 3) = 0
www.aieeepage.com
cos A sin A 0 ⇒ ∆ = cos B sin B 0 = 0 cos C sin C 0
⇒
PAGE - A Premier Institution for IIT - JEE & AIEEE at the National Level
1 −3 sinθ = 0, sin θ = , sin θ = 2 2 10 of 21
72.a.
π 6
θ = nπ ,θ = nπ + (−1)n
⇒
tan 2 x =
∴
Clearly, 7 λ + 7 µ will be divisible by 5 if either : 7 λ has 3 or 7 in the unit place and 7 µ has 7 or 3 in the unit place
tan 2 x
⇒
2
tan y
=
a−d b−c , tan 2 y = c−a d −b
(b − c) (d - b) (c − a ) (a - d)
or :
7 λ has 1 or 9 in the unit place and
But a tan x = b tan y
∴
7 µ has 9 or 1 in the unit place. For any chice λ , µ the digits in the unit
⇒
λ
µ
b2
place of 7 + 7 is 0, 4, 6, 8. It is divisible by 5 only when this digit is 0.
∴ The required probability = 73.d.
(a - b) = ( d − b ) (1 + tan 2 y )
⇒
th
th
5 ⇒ P(E) = 6
k −1
probabilty that that 6 is tossed on the k toss
=
⇒
75.a.
.....(2)
(b − c) (d − b) (c − a ) ( a − d )
{ using (1) & (2)}
a 2 (c − a ) ( a − d ) = b 2 (b − c ) (d − b)
From the given equations we have
u = 1 − v, 6v =
w 3 , w = (1 − u ) 1− w 4
www.aieeepage.com Which are satisfied by the pair th
≡ 10 0
u v w 76.c.
Let E = (sin −1 x ) 3 + (cos −1 x ) 3 ⇒
1 6
E = (sin -1 x + cos −1 x ) 3 -3sin -1 x cos −1 x (sin −1 x + cos −1 x )
The probability that Anthony will toss the first 6 is the sum of the probabilities that he will toss the first 6 on his first turn (i.e., 3rd toss of game), second turn (i.e. 6th toss of game). etc. ∴ Required Probability i.e. P(EP is given by 2
tan x b = tan y a
PAGE
1 4
Let P (E) be the probability that first 6 is tossed on the kth toss is
probabilty that 6 was never ∴ P(E) = tossed before k toss in (k-1) tosses
a
2
.....(1)
5
5 1 5 1 5 P( E ) = + + .... + 6 6 6 6 6
3n −1
1 + ...∞ 6
⇒
E=
π3 π - 3 (sin -1 x cos −1 x 8 2
⇒
E=
π 3 3π π − sin -1 x − sin −1 x 8 2 2
⇒
E=
2 3π π 3 3π 2 sin -1 x + sin −1 x − 8 4 2
⇒
E=
π 3 3π + 8 2
⇒
E=
π 3 3π -1 π 3π 3 + sin x − − 8 2 4 32
⇒
π 3 3π -1 π + E= sin x − 32 2 4
(
)
2
5 1 . 6 6 ⇒ P(E) = 3 5 1- 6 ∴ 74. b.
Required Probability = 2
2
a sin x + b cos x = c (b - a ) cos 2 x = c − a
⇒
(b - a ) = (c - a) (1 + tan 2 x )
⇒
(a - b ) cos 2 y = d - b
2
2
www.aieeepage.com
⇒
Now, b sin 2 y + a cos 2 y = d
PAGE
25 91
−1 2 π −1 (sin x) − 2 sin x
π3 So, the least value is 32 2
π 3π since, sin -1 x − ≤ 4 4
PAGE - A Premier Institution for IIT - JEE & AIEEE at the National Level
2
11 of 21
∴ The greatest value is 77.c.
π 3 9π 2 3π 7π 3 + × = 32 16 2 8
Let x = sin 2 A + sin 2 B
f 2 − c ⇒ (y + f) 2 = −2 g x − 2 g
{given}
⇒ x = 1 − cos 2 A + sin 2 B
80.b.
⇒ x =1+ sin B − cos A 2
2
⇒ x = 1 + cos ( A + B ) cos ( A − B )
π π a cos + θ , b sin + θ respectively 2 2
⇒ x =1− cos C cos( A − B) {∵ A + B + C = π } ⇒
1 − cos ( A − B) 2
∴ CP 2 + CD 2 = a 2 cos 2 θ + b 2 sin 2 θ
Now, −
1 ≤ cos ( A − B) ≤ 1 2
⇒ 1−
1 1 1 ≤1+ cos ( A − B ) ≤ 1 + 2 2 2
PAGE ∴
CP 2 + CD 2 = a 2 + b 2
D
81. d.
∴
1+4
⇒
−
1+4
82.a.
This shifted line cuts x-axis at (k, 0).
1 2
∫
Sinc, I =
∴ I=
∫
6t 5 (t 6 + t 4 + t ) t 6 (1 + t 2 )
Thus, the equation of the line in the new position is
⇒ I=
y − 0 = tan (θ − 30 ) (x - k) o
79. d.
∫
t2 +1
PAGE ∫
P(x,y)
∫
dt
dt
6 2
t +1
dt
O
X
Let the given line be the y-axis and the circle to 2
have the equation x + y + 2 gx + 2 fy + c = 0 , then according to given condition, we have 2
3 2/3 x + 6 tan −1 ( x1/ 6 ) + c 2
www.aieeepage.com ∴
2
dx
t4 ⇒ I = 6 + 6 tan −1t + c 4
C(-g,-f)
2
6 (t 5 + t 3 + 1)
⇒ I = 6 t 3dt +
Y
2
x(1 + x1 / 3 )
⇒ dx = 6t 5 dt , we have
{slope f x+2y=4}
where, k = 4 + 3 5
x + x 2/3 + x1 / 6
Put x = t6
After rotation the slope of the line is tan (θ - 30 o ) , where
tan θ = −
M
By symmetry area (∆ ADE) = Area (∆ABM) So, integral is minimum when m = 0
=3
k =4+3 5
C
O
=k The distance between these two lines is 3
4
B
E
www.aieeepage.com The equation of a line parallel to x + 2y = 4 is x + 2y k
+ a 2 sin 2 θ + b 2 cos 2 θ
A
1 2 +1 ≤x≤ 2 2
∴ 78. a.
x = 1−
which represents a parabola with its axis perpendicular to y-axis. Since, P and D are the ends of semi-conjugate diameters of ellipse respectively, therefore, the coordinates of P and D are (a cos θ be sin θ ) and
2
2
x = ( x + g ) + ( y + f ) − ( g + f − c)
83. d.
I=
1 2e 4e 6e Since, = + + + ...∞ x 3! 5! 7 ! ⇒
PAGE - A Premier Institution for IIT - JEE & AIEEE at the National Level
1 1 2 3 = 2e + + + ...∞ x 3! 5! 7 !
12 of 21
The Area bounded is shown in the figure such
1 = 2e(S ) x
⇒
Such that S =
π 1 and B is , 6 3
that, A is
∞
∑
n =1
n (2n + 1) !
Hence, Required Area π /6
A=
1 ∞ 2n + 1 − 1 = S Now, 2 n=1 (2n + 1) !
∑
⇒
S=
⇒
S=
∫
π /3
⇒
1 1 1 − 2 n =1 (2n) ! 2n + 1
∑
( )
1 −1 e 2
A = (log sec x)π0 /6 +
∫
f ( y ) log y 1 dy
84.c.
Here, y = f [ g ( x )] = f (cos x ) = cos 2 x Again,
86. c.
A=
∫
π 6
π π , 3 6
π π + log sin - log sin 2 3
6
and 3y = 2 y - x
∫
6
Both give the differential equation
O
2xdy + ydx = 0
whose solution is xy2 = c
87.a.
Since,
1 1 1 i- j + k = (a × b) × (c × d) 6 3 3
⇒
1 ˆ 1ˆ 1 ˆ i- j + k = [a b d ] c -[a b c] d 6 3 3
y = tan x
⇒
C 3
dy dx
www.aieeepage.com 1 1 1
y = cot x
1
dx dy
PAGE
1 π Since, f ( x ) = MIN tan x, cot x, ∀x ∈ 0, 3 2
y=
dy (X − x) dx
⇒ 3x = x - y
1 cos 2 xdx = (1 + cos 2 x) dx 2π
Y
Equation of the tangent at (x, y)
π 3
π
85. a.
6 3
4 3
+ log
dx dy P1 = x − y ,0 and P2 = 0, y − x dy dx
1 sin 2 x 3 π A= x+ = 2 2 π 12
⇒
π
Y−y=
Hence, the required area is π 3
1 π /3 (x)π /6 + (log sin x)ππ /3/2 3
1 1 π A = log 2 - log 3 + 2 36
∴A =
18x 2 = 9 πx + π 2 = (3x − π)(3x − π)(6x − π) = 0
x=
π /3
www.aieeepage.com+ log 1 - 12 log 3 + log 2
I =0
⇒
A
B
π 6
π 3
cot x dx
{where, y > 1}
0
⇒
∫
π/6
⇒
1
3
dx +
1 π π π ⇒ A = logsec - logsec0 + − 6 3 3 6 0
PAGE
1 1 = 2e = 1 x 2e
π /2
1
π /6
0
∞
∫
tan xdx +
Substituting respective value of S in (1), we have
∴ I=
π 1 , 3 3
6
ˆi - ˆj + kˆ = [a b d] c 3 3
....(1)
{∵ a, b, c are coplanar} π 2
Now., X
⇒
PAGE - A Premier Institution for IIT - JEE & AIEEE at the National Level
[a b d] = a × b . d
[a b d] = a × b d cos θ
13 of 21
⇒ E = 2 {log100 a + log g 100}
{∵ d ⊥ a, d ⊥ b ∴ d || a × b} ⇒
[a b d] = ab sin 30 o.1.) ± 1)
Since A.M. ≥ G.M.
⇒
1 1 [a b d] = 1.1. .1.(±1) = ± 2 2
⇒
Substituting respective value in (1), we have
⇒ log100 a + log a 100 ≥ 2
2 i - 2 j + 2k 1 2 c = ± i - j + k = ± 3 3 3 3
∴ 91.a.
88. c. A( a )
min (log100 a + log a 100) = 2
We have
α
i 2π r =e 5 ,
where r = 0, 1, 2, 3, 4
Clearly, α 5 = 1 and 1 + α + α 2 + α 3 + α 4 = 0 Now,
PAGE
O (b)B
V1 =
log100 a + loga +100 ≥ log100a × loga 100 =1 2
C( c )
1 1 (a × b) , V2 = (b × c), 2 2
α 5 =1 ⇒α 4 =
1 α
∴ 1+α +α 2 +α 3 −
Now, V1 + V2 + V3 + V4 = 0
= - α 4 − α 4 = 2 α 4 = 2 α = 2 [∵ α = 1]
∴
∴ log 2 1 + α + α 2 + α 3 −
www.aieeepage.com
89. a.
V1 + V2 + V3 + V4 = 0
Let OA = 2 unit then OB = 4 unit and OC = 6 unit. Now,
1 1 OD = OA + AC' + AB' = 2ˆi + 2ˆj + 3kˆ 2 2
92.d. 93.c.
D’
n − factors
There are 3n factors in this product. To get a term of degreer, we choose r factors out of these 3n factors and then multiply second terms in
C
O
each factor. There are
A
3n
C r such terms each having
coefficient 1.
Also,
PAGE
1 1 OD' = OC + CB' + B' O' = ˆi + 2ˆj + 6kˆ 2 2 Now, cos DOD' =
⇒ cos DOD ' =
∴
n − factors
{(1 + z )(1 + z )(1 + z )....(1 + z )}
D B
1 = log 2 = 1. α
The given expansion can be written as n − factors
B’
C’
4
{(1 + x)(1 + x)(1 + x)...(1+ x)}{1 + y )(1 + y )(1 + y )...(1 + y )
O’
A’
90. a.
1 = 1+ α + α 2 + α 3 −α 4 α
1 1 V3 = (c × a) and V4 = {( c − a) × (b − a)} 3 2
OD . OD' OD
OD'
94.a.
www.aieeepage.com
24 ∠DOD ' = cos −1 697 2
Cr .
S∞ = 2 for G.P.
or 2 log10 x = y
4 + 4 + 9 1 + 4 + 36
3n
First equation is log10 x 2
∴
(2) (1) + (2) (2) + (3) (6)
1 Let E = 2log100a − log a 100
Hence, the sum of the coefficients =
...(1)
From second equation S n =
n (a + l ) 2
2 y2 20 40 ∴ We have y (3 y + 5) = 7 log 10 x = 7 y [By (1)]
⇒ 7 y 2 − 60 y − 100 = 0
PAGE - A Premier Institution for IIT - JEE & AIEEE at the National Level
14 of 21
⇒ ( y − 10) (7 y + 10) = 0 ⇒ y = 10, y ≠ −
⇒ cos θ + sin θ =
10 7
100.d.
Convert the eqn. in normal form. Then
⇒ 2 log10 x = 10 ⇒ log10 x = 5 ⇒ x = 105 95.a.
l
The given inequality
a +b 2
| x| 1 1 1 1− ≥ ⇒ ≥ 1+ | x | 2 1+ | x | 2
P=
⇒ 1+ | x |≤ 2 ⇒ | x | ≤ 1
⇒ 3r cos θ + 4r sin θ = 5 ⇒ 3cos θ + 4sin θ = 97.a.
5 r
The eqn. 11+0 cos θ + sin θ =
= r cos(θ − α),
α = Tan −1
b & a
l a 2 + b2 l b , Tan −1 2 2 a a +b
Put x = r cos θ, y = r sin θ , then 3 x + 4 y = 5
2
∴ foot of ⊥ r is (ρ, α ) i.e.,
⇒ − 1 ≤ x ≤ 1 ⇒ x ∈ [−1,1] 96.c.
3 2 . r
PAGE
2 is r
www.aieeepage.com
K cos θ + sin θ = & passing thro` r
(1, π 2 ) ⇒ 0 + 1 = K1 ⇒ K = 1 ∴ cos θ + sin θ = 98.b.
The
1 is required eqn. r
⊥ + 0 a cos θ + b sin θ =
l r
is
π π K a cos θ + θ + b sin θ + θ = 2 2 r passing thro` pole ⇒ (0, 0) ⇒
O+b =
K ⇒ O (O + b) = K O
PAGE
⇒K =O ∴ required eqn. is −a sin θ + b cos θ = 0 ⇒ Tanθ = 99.d.
b b ⇒ θ = Tan -1 a a
www.aieeepage.com
P = r cos(θ − α) is eqn. of line in normal form. Here
P = 3, α = π . 4
(
∴ r cos θ − π
4
)=3
1 1 ⇒ r cos θ⋅ _ sin θ ⋅ =3 2 2
PAGE - A Premier Institution for IIT - JEE & AIEEE at the National Level
15 of 21
Paper - I I I : Chemistry - 101 to 150 Solutions 101.b.
The bismuth endpoint is the point at which all the bismuth in the solution has reacted with EDTA, and the additional EDTA that is added then begins to react with the coper ions. While the EDTA reacts with bismuth, no increase in absorbance takes place. This produces the straight line betwen x=0 and x= 3 on the graph. When the EDTA reacts with copper, a complex is formed that absorbs light. As this reaction progresses, absorpotion increases; this is indicated by the linear portion of the graph with positive slope. The bismuth endpoint is the point at which absorption just begins; this is clearly at the point where 3 mL of EDTA has been added, so choice B is correct. Chice A is wrong because when 2.5 mL EDTA has been added there is no absorption occurring. This means that there is still some bismuth in the solution that is reacting with the EDTA. Choice C, 4.0 mL, is wrong because at this point the copper complex is forming, meaning that all of the bismuth has alreadyreacted. The other horizontal line, on the upper right side of the graph, indicates that, once all the copper has reacted, no further complex is formed and the basorption remains constnats as more EDTa is added 5 mL is therefore the copper endpoint, but we are looking for the bismuth endpoint, so choice D is wrong. Choice B is contradicted in the passage. We looked at the K’s and determined that the copper complex is less stable. Choice C is half right and half wrong, the copper complex absorbs, but the bismuth complex does not, and thus has a zero slope. The statement in choice D is true, but not an answer to this question. For one thing, it is not necessary that the absorbance occur in the visible region. It is true that the copper EDTA complex absorbs light, and so produces color, but there is nothing in the passage to indicate that an especially intense color would be needed, so choice D is wrong. This question is a good candidate for last to be done. Whenever we do answer it howerver, it should be done with a minimum of actual calculation.
104.b.
choice D or, better still, 2 x 3 with a decimal somewhere is 6 with a power of ten after it, also choice D only. This is a typical “qhat if” question if you understood what was happening in figure 1 you should’ve been able to apply your understanding to this question. Q: What will happen at the begining of this new reaction? A: At the beginning of the reaction, only the reactnat, which absorbs light, is present/ The graph should thus begin with a high positive value Q: Now, s the titrant is added one drop at a time, it reacts with the reactant to form the product. What happens to the curve at this time? A: The conentration of the reactant decreases. As the reaction proceeeds, the amount of light absorbed by the reactant therefore also decrease, and the vurve should slope wownward. Q: When the endpoint is reached, what happens? A: A minimum is reached. When the endpoint is reached, only the product will be present. At that point the basorbance will be zeor since the product doesn’t absorb light? Q. Then what? A: As additional titrant is added, the titrant, which no longer has anything to react with, will accumulate in the solution. The titrant does absorb light, so the absorbance will increase after the end point. Therefore choice B is the correct answer. (this question illustrates again, the importance of understanding versus memory. This question can be answered via Beer’s law. The stem says that the container is twice as long, and the answer choices al have something to do with absorbance; absorbance and pathlenght through the container are compared in Equation 1. Absorbance is directly proportional to the pathlength. So if we double b, we double the absorbance at every point in the titration. The absorbance would therefore increase twice as fast, i.e. the slope is doubled. The width of the container is the variable b; the path length that the beam must travel through . ε is a constant and doesn’t change, c is the molar concentration,and b is the path length. You an see that b and c are directly related to the absorbance, so any change in b or c will result in a similar change in A. If, for example, the solution contains the copper(II)- EDTA complex, this solution would have a certain constant absorptivity, ε . However, the amount of light actually absorbed by a particular sample will depend on the concentration of the solution and the length of the path through which light travels. If the concentration is increasede, which means that the value of cc increases, then the absorbance will also increase. Likewise, if the distance traveled by the
PAGE
www.aieeepage.com
102.a.
103.d.
mL ×
105.b.
PAGE
L mol EDTA mol Bi g × × × =g mL L mol EDTA mol
For question 1, we found that the volume of EDTA (Y) needed was 3 mL, the question gives us the concentration of EDTA as 0.1 M, the passage, or its equations, tell us that the reaction is one to one, and the periodic table says that bismuth weights 209 g/mol. We can thus plug in and calculate:
www.aieeepage.com
3×
1 0.1 1 209 × × × = 0.0627 1000 1 1 1
This could have been simplified to (0.3) (0.2)=0.06,
PAGE - A Premier Institution for IIT - JEE & AIEEE at the National Level
16 of 21
light moving through the solution increases, this will also increase the absorbance. So, if the diameter of the flask containing the solution is increased from 10 centimeters to 20 centimeters, twice as mich ight will be absorbed. Therefore the value of A, for any given concentration, will be twice as great, even though the absorptivity per centimeter will, of course, remain the same. The absorbance is a measure of the actual amount of light absorbed, not of the solution’s characteristic capacity to absorb light. Now, if the absorbance during a titration had originally increased form zero to x, and the length of the light path were doubled, the absorbance would then increase from zero to 2x when the same amount of titrant is added. This would double the slope of the graph, so the correct answer is choice B. 106.d. Since Z is the number of protons and A is the number of protons plus neutrons, Z cannot be larger than A. 107.c. The atomic number defines the element. 108.b. The functional groups are indicated below. ester
and ∆ S by the equation ∆ G = ∆ H –T ∆ S , at equilibrium, however ∆ G =0, and T= Tequib hence
∆H ∆ H = Teq ∆ S , Tequib = ∆ S . 113.c.
114.a.
O OH
−21.76 × 10−19 × Z2 J n2
∴
E3 of He+ =
−21.76 × 10−19 × 4 J 9
λ=
130 = 32.5 4
hc λ
116.c.
and if weight of hydraize sulphate be x gm then equivalents of hydrazine sulphate =
x 32.5
hc 6.625 ×10 −34 × 3 × 108 × 9 = E 21.76 × 10−19 × 4 = 0.255 × 10 −10 m=2055Å
115.d.
5 1 KMnO4 = 20 × = 50 × 1000 500
117.d.
β in this equation represents the compressibility factor of the system. Along with a collision, the molecules must have enough energy and must strike each other with the proper spatial orientation in order for a collision to occur. Note that only I and II are correct statements. At T1 ∆Gº is on the positive side of the ordinate.
PAGE
1 32.5 x or x = ∴ = = 0.065g 500 32.5 500
Note that when the concentration of Y is held constant at 2.0M the concentration of X is tripled, the rate of tripled. Also, note that when [X] is held constant at 2.0M and [Y] is tripled, the rate of unchanged. Hence the rate is directly related to the concentration of [X] but independent on the concentration of [Y]. A sneaky loophole, and probably not one you would have though of before seeing the answer choices. Always read all answer choices before
∂ ∆ Gº
Note also that the slope ∆ T , at the minima P
N 2 H 6 SO4 in 10 ml solution = 0.065g
∴ Wt. of N 2 H 6 SO4 in 1000ml solution = 6.5g.
111.b.
En of He+ =
or
Number of equivalents of
110.c.
−21.76 × 10−19 J n2
∴
E=
Equivalent weight of
Hence wt. of
En of H =
carboxylic acid
0
N 2 H 6 SO4 =
The longest chain is seven carbons, a heptene. Answer C gives the lowest numbered carbon attached to the double bond.
Hence energy equivalent to E3 must be supplied to remove the electron from 3rd orbit of He + . Wavelength corresponding to this energy can be determined by applying the relation.
N 2 H 4 → N2 2
The change in Gibbs´ free energy is related to ∆ H
www.aieeepage.com
phenyl ring
109.c.
112.a.
PAGE
O O
making your selection!
118.c.
(T3), is zero. The carbon atom of a carboxylic acid is at a higher oxidation state than the carbon atom in either an alcohol or an aldehyde. Potassium chromate is a strong oxidant. Although ozone produces an oxidized product, it only works on alkenes. Volume of balloon =
www.aieeepage.com 119.a.
4 3 4 22 21 21 21 πr = × × × × = 4851ml 3 3 7 2 2 2 Calculation of total volume of hydrogen in the cylinder at N.T.P.
PAGE - A Premier Institution for IIT - JEE & AIEEE at the National Level
17 of 21
change in energy (a state function) should be zero. But, from the first law of thermodynamics,
P1V1 P2V2 = T1 T2
∆ Esystem = Qin − Wout , or, in this case,
P1 =1 atm
P2 =20 atm
Qin = Wout = 60, 000kJ . Some of you who have
V1 =?
V2 =2.82l
studied thermodynamics may be bothered by this answer, because you think the second law of
T1 =273K T2 =273+27=300K
thermodynamics demands
20 × 2.82 × 273 ∴ V1 = = 51.324l = 51324ml 300 × 1
problem here is that confusion can sometimes arise between two definition of Qin . In this problem, we
Actual volume to be transferred into balloons = 51324–2820ml=48504ml. No.of balloons that can be filled up =
48504 = 9.999 = 10 4851 120.a. 121.b.
Wout < Qin . The
are using it as the net flow of heat into the system. In some situations, this is broken down into a
Qin and a Qout , and in that sense, it is true that
PAGE
Wout < Qin . But that’s not the way this problem is
If matter can leave our universe, it must be open. Incidentally, Stephen Hawking has theorized that black holes also radiate mass into our universe. The only correct choice here is II, although the two halves of the cell are connected by the salt bridge which permits a current to flow, the salt bridge does not allow mixing of the two solutions. Note also that only the oxidation of zinc, to zinc ions (Zn2+) occurs at the zinc electrode. Tautomers are formed when a hydrogen is moved within a structure. The shifted hydrogen is circled below.
125.b.
set up. When two ions come together to form an ionic bond, energy is released. The more energy released, the stronger the ionic bond. The amount of energy released is determined by the equation :
www.aieeepage.com E = 1.44 q q
122.c.
O
126.c. 127.a.
OH
H H
123.b.
Structure A Structure B Initial concentration of each gas = 1 mole Let the No. of moles of NO2 reacted at equilibrium = x Then, SO2 ( g ) + NO2 ( g ) (1− x )
(1− x )
Now we know that,
(1 + x) (1 + x) = 16 (1 − x) (1 − x) (1 + x) =4 (1 − x )
where E=energy in eV, q1 and q2 are the charges on each ion, and r is the distance between the two nuclei. While changes in ionic charge and ionic radius will effect the energy of the bond, changes in temperature will not. Fatty acids are stored as triglycerides. Standard enthalpy of hydrogenation of cyclohexene (–199kJ mol–1) means the enthalpy of hydrogenation of one double bond. Now benzene has three double bonds, the enthalpy of the reaction would be
= 3 × −119 = −357 kJ mol –1 + 3 H2
SO3 ( g ) + NO3 ( g ) (1+ x )
(1+ x )
[SO3 ][ NO ] = K [SO2 ][ NO2 ] c
Actual enthalpy of the reaction can be evaluated as below.
PAGE
(1 + x)2 ; (1 − x ) 2 = 16
or
∆ H (Reaction) =∆ H fº (Product) –∆ H fº (Reactants)
= −156 − (49 + 0)
= −205 kJ mol –1
or
∴ Resonance energy = ∆ H Exp − ∆ cal
www.aieeepage.com 3
1 + x = 4 − 4 x or 5 x = 3 ; x = 5 = 0.6
Thus the concentration of NO at equilibrium = 1 + x = 1 + 0.6 = 1.6 moles Concentration of NO2 at equilibrium 124.b.
1 2
r2
=1 − x = 1 − 0.6 = 0.4 moles If each cycle returns to the same state, then the
128.c. 129.d.
= −357 − ( −205)
= −152 kJ mol –1
Vapor pressure always increases with increasing temperature. In vapour phase 1 mole (or 78g) benzene has volume at 20ºC =
78 × 1 × 2750 ml 0.877
1 mole (or 92g) toluene has volume at
PAGE - A Premier Institution for IIT - JEE & AIEEE at the National Level
18 of 21
20ºC =
92 × 1 × 7720 ml 0.867
a = 0.31 a−x
PBº 78 × 2750 ∴ × = 1× 0.0821× 293 760 0.877 × 1000
a − 23814 × 10 −19 = 0.31 a
or
PBº = 74.74mm
−18
131.b.
Similarly,
∴
Prº 92 × 7720 × = 1× 0.0821× 293 760 0.867 × 1000
or Pr = 22.37mm ; º
Pmixture = 46.0 mm (Given) ∴ PM = PBº ⋅ X B + Prº ⋅ X r ;
PAGE 132.c.
∴ 46 = 74.74 ⋅ X B + 22.37(1 − X B ) (∴ X B + X r = 1)
mol On usual calculations, a = 3.451× 10 Cyclopropane suffers from a great deal of ring strain, making it far less stable than propene, Cyclohexanone is far more stable than the ring structure of the oxirane. The meta – and para–xylenes are approximately equal in stability. Benzene is far more stable than a multiply unsaturated compound and cyclopentane is significantly more stable than 1-ethyle-cyclobutene due to the ring strain in the latter. A high heat capacity means that water gain or lose energy with smaller changes in temperature. This corresponds to I and III only. The chemical reactions taking place at the two electrodes are
133.c.
∴ on calculation X B (liquid phase) = 0.45;
At cathode :
www.aieeepage.com
Cu 2 + + 2e − → Cu
X r (liquid phase) = 0.55
PB = PBº ⋅ X B (1)
Also, or
130.c.
However, note that only Cu 2 + ions will be discharged so as these are present in solution and
74.74 × 0.45 = 46.0
⇒
H − ions will be discharged only when all the
X B ( g ) = 0.73
Minimum number of β -particles required =346 min–1
Cu 2 + ions have been deposited.
2OH − → H 2O + O + 2e −
At anode :
O + O → O2
No. of β -particles required for carrying out the
6.909 × 60 minutes = 346 × 6.909 × 60 = 143431 ∴ Amount of β -particles required experiment for
=
143431 = 2.3814 × 10−19 mol 23 6.023 × 10
Thus in first case, Cu 2 + ion will be discharged at the cathode and
electrolysis. According to Faraday’s second law
31.75 g Cu = 8g of oxygen
Further we know that,
λ= where
PAGE
0.693 0.693 = = 0.0104 hr −1 T1 66.6 2
2.303 a log t a−x
= 5.6 litres of O 2 at NTP
0.4 g Cu =
5.6 × 0.4 litres of O2 at NTP 31.75
=0.07055 litres=70.55 ml
As mentioned earlier, when all the Cu 2 + ion will be
deposited at cathode, H ions will start going to www.aieeepage.com
a=Initial concentration of β -particles
a λT 0.0104 × 6.909 = =− = −0.0312 a − x 2.303 2.303
+
cathode liberating hydrogen ( H 2 ) gas, i.e.
x=Consumed concentration of β -particles
log
O2 gas at the anode. Let us cal-
culate the volume of gas ( O2 ) discharged during
Now we know that,
λ=
H + + OH −
H 2O
H + + e−
H
H + H → H2
However, the anode reaction remains same as previous. Thus in the second (latter) case, amount of
H 2 collected at cathode should be calculated.
PAGE - A Premier Institution for IIT - JEE & AIEEE at the National Level
19 of 21
8 g of O 2 = 1 g of H 2 5.6 litres of O2 at NTP = 11.2 litres of hydrogen Quantity of electricity passed after 1st electrolysis, i.e. Q=Ct
1.2 × 7 × 60 = 504 couloumbs 504 coulombs will liberate
= Similarly,
=11.2 ×
5.6 × 504 = 29.24 ml 96500
H 2 liberated by 504 coulombs
504 = 58.48 ml 96500
Total volume of
O2 liberated =
140.d. 141.d.
free energy is negative. A negative ∆ G can only 142.c.
H 2 liberated = 58.48 ml
143.a.
We can rearrange ∆ G = ∆ H − T∆ S to:
134.d.
∆H−∆G T= ∆S
135.d.
HC ≡ CH
6028 − 22 = 273º K 22
+
OH + HOMgCl
O
136.a.
137.d.
138.c.
139.d.
NH3
CH3 | CH3CHCH2CH2C ≡ H
On an oxyacid, when the central atom is different and each central atom has the same number of oxygens, the central atom with the greatest Electronegativity produces the strongest oxyacid. The Nearnst equation is really about non-standard conditions, so A seems doubtful. Standard potential is just the potential at standard conditons. This definition does not directly involve equilibrium, so B seems doubtful as well. C is kind of silly if you think about it: standard potentials are measured under standard conditions. But D is a true statemnt: you need two half-reactions to make an actual reaction. The rate d[XY2] / dt is unaffected by changes in [X], therefore it is zero order in X. Tripling [Y] causes the rate to increase nine times, that is, by a factor of 32. Since the rate increases with the square of the increase in [Y], the rate is second order in Y, the rate law is: d [XY2] / dt = k[Y]2 As we boil the mixture the escaping vapor contains more of the more volatile component, benzene. The
O
Br
In (2) it is O O
145.b.
H , H2O
NaNH2
O
MgCl
O MgCl
CH3 | CH3CHCH2CH2Br
has no protection group. In (1) the protected form of the reactant is
+ Mg
MgCl
c 3 × 1010 cm. sec −1 = = 1× 1013 sec −1 −3 λ 3 × 10 cm
www.aieeepage.com
The first step is the formation of a Grignard reagent (cyclohexylmagnesium bromide). This reacts with ketone as shown below: Cl
v=
144.b.
substituting given values :
T=
be possible if ∆ H is negative and ∆ S is positive. The Arhenius Law states that K=A exp (–Ea/RT) This can be rearranged to : In K= In A –Ea/RT If we plot in K vs 1/T, the slope of the graph is –Ea/ R.
PAGE
70.55+29.24 = 99.79 ml. Vol. of
remaining liquid becomes richer in toluene. According to the graph, a liquid mixture 25 mole% in benzene will boil at –100ºC. As we boil the liquid its composition and boiling point move to the left and up on the “boiling point line” until we reach the desired point. Geletinous precipitate e.g. Al(OH)3 are among the most difficult to filter. A spontaneous reaction is one in which the Gibbs
Si(CH3)3
Crystal field spiltting energy for tetrahedral complexes ∆t ( where only four ligands approach the central metal ion producing weak field around it ) is lesser ( about half ) than the crystal field splitting energy of octahedral complexes
∆0 (
where six ligands approach the central metal ion producing strong field around it )
PAGE
related to each other as ∆t =
∆t and ∆ 0 are
4 ∆0 9
146.d ‘1000 tunch’ silver is 100% pure silver. 147.d. The ideal gas law is PV = nRT. 148.d. As described in the question stem, a semilog plot is generated when one of the coordinates (usually the y-coordinate) is plotted as the logarithm. Equation 5 is an exponential equation. You may know that the semilog plot of such a function would result in a straight line. This can also be seen directly by actually taking the logarithm of both sides. The equation of a generic straight line has the form y = mx+b, where m is the slope and b the y-intercept. In this case, if we take y=In k, and x as (1/T), then the equation above conforms to that for a straight line,
www.aieeepage.com
PAGE - A Premier Institution for IIT - JEE & AIEEE at the National Level
20 of 21
with m=-Ea/R and b = In A. The y-intercept is b, which in this case is In A. In k In A
1/T
149.a.
Note that the y-intercept corresponds to a point where T is infinity, and so obviously it is not an actual data point we can measure. This is obtained by extrapolating a straight line that can be constructed with values from physically resonable temperatures. As the pressure increases, we can see from equation 2 that the numerator and denominator wuld become very close in value, i.e., the fraction would approach 1. Choice B is incorrect because it depicts the initial region as linear, but the form of the equation is not that of a straight line. Also, it has the y-value become one abruptly, rather than approaching it asymptotically. Choices C and D are incrrect because they show the value of θ increasing without bound. You probably know already that increasing the temperature would increase the rate of a eaction. Specifically, a general rule of thumb is that for every 100C or 10 K increase, the rate doubles. So we know that statement II is correct. In order to determine if statements I and III are correct, we need to know more about HOW a temperature increase leads to a higher reaction rate. The temperature is an indication of the energy of the molecules. The higher the temperature, the more energetic these molecules are, and thus the more likely it is that they have enough energy to overcome the activation barrier for a reaction Note that a high temperature does not actually lower the activation barrier, it just makes it easier for the molecules to overcome it. So statement III is incorrect. An increase in temperature does lead to a higher rate constant. This can be seen from equation 5. As T increases, -Ea/RT becomes less negative: it thus increases in value, and so does exp (-Ea/RT). k. therefore also increases. Statement I is correct and so B is the correct choice.
PAGE
www.aieeepage.com
150.b.
PAGE
www.aieeepage.com
PAGE - A Premier Institution for IIT - JEE & AIEEE at the National Level
21 of 21