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Paper - I : Physics - 1 to 50 Solutions 1.d.
The question is asking for the value of d that provides these experimental results - a sure clue that you’ll need to use the equation provided by the passage :
nλ = 2d sin θ
n = 5; f1 = 5 × f1 = 5 × 102.16 = 510.8Hz
n(3 cm)
n = 7; f1 = 7 × f1 = 7 × 102.16 = 715.00 Hz
d=
2.d.
n = 3; f1 = 3 × f1 = 3 × 102.16 = 316.48 Hz
= 2d sin 190
n = 9; f1 = 9 × f1 = 9 × 102.16 = 919.00 Hz
9 3n = n 1 2 2 3
for
The first maxima, at n = 1, yields d = 4.5 cm- answer choice (D) Noitce that other, higher interatomic separations are possible, but all of the other answer choices are smaller. As the layers are separated, the intensity of the reflected waves drops; the path lengths of the reflected wave are increased, and so they don’t meet in phase any longer. However, as the planes of atoms (or steel balls) are separated, eventually the difference in path length between microwaves will be another multiple of the wavelength, and constructive interference will occur again. So, the intensity will initially drop, but it will then rise againanswer choice (D) is correct. .Nothing about Bragg diffraction, as described in the passage, has anything to do with the intensity of the incoming microwaves. The microwaves will interfere with each other as they bounce off of consecutive layers, whether or not they are strong or weak. We expect, therefore, this to have no impact on the experiment, and choice (C) is correct. Destructive interference occurs when two waves meet out of phase, that is, that the peak of one meets the through of the other, and vice-versa. This will occur when the path length between the two interfering waves is a half-integer multiple of the wavelength. The difference in path length, according to the passage, is 2d sin θ , only answer choice (D) expresses this as one-half, or three-halves, or fivehalves the wavelength. We can calculate the frequency of microwaves if we knew their wavelength. The passage intimates that microwaves have wavelengths in the centimeter
n = 11; f11 = 11× f1 = 11×102.16
= which is > 1000 Hz ∴ Possible overtones will be for n = 1, 3, 5, 7, 9, i.e.,
PAGE 7.d.
5. Let f =frictional force between the bar and one of the disk.
1 2 mr α 2
Then Mg − 4 f = Ma; fr =
( a = r α) www.aieeepage.com
3.c
4d.
5.d.
region, so: f =
8
1 2 a 1 1 mr ⋅ = mra; f = ma 2 2 r 2
1 Mg = 4 ⋅ ma = Ma ; Mg = ( M + 2m)a 2 a= 8.b.
M 5 5g ⋅g= ⋅g = 9 9 9
Final angular velocity,
α=
ω = ω0 + αT
ω − ω0 O − ω0 ω = =− 0 T T T
Torque on the disc
1 ω (τ) = I α = ma 2 − 0 2 T
τ m a ω0 = . 2T a Given y = a sin (1000 πt − (3) x ) ,
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But τ = Fa ; F =
9.c.
∴k = 3 =
c 3 ×10 m/s = = 3 × 1010 Hz 0.01 m λ
2π 2π ;λ= λ 3
Phase diff. of
2π = Path diff. λ
V Resonating frequency for a closed pipe = n ⋅ 4L
Phase diff. of
= Path diff. =
340 × 100 = 102.16Hz 4 × 83.2 for closed pipe, next possible overtones will be for n = 3, 5, 7, 9.
λ×
Answer choice (D) is correct. 6.a.
∴ f ⋅r =
www.aieeepage.comπ
for n = 1; f1 = 1 ⋅
10.c.
3
π . 3
1 λ 1 2π π = = × = . 2π 6 6 3 9
W = weight of cone =
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1 A h2 h 1 2 ⋅ ⋅ d , g = Ah ⋅ d1 g 3 h q 3 81
∴ ∆ W = 500 R = 800 × 8.3 = 6640J
1 1 A h2 h W + Ahd 2 g = 2 ⋅ ⋅ ⋅ d1 g 3 3 h 4 2
3 × 6640 = 19920J ≈ 20KJ .
=
This is the work done per mole. ∴ Work done by three moles =
1 Ah ⋅ d1 g 24
57 d1 d 2 d1 d 2 + = : = 81 3 24 d1 648 . dv =
1 2 mv = change in energy 2
But, dv = Cv ⋅ d T =
∴
R ⋅d T r −1
PV =
m V ⋅ R ⋅ (K is constant) M K
∴ P = a constant B → C V=Constant. C → A : T=Constant. ∴ PV= Constant ∴ Plot suitable is (c) graph. ε A Q ; Q =V × 0 14.b. V = C d
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When the dielectric basis put capacitance increases to C1
2
ε A ; 3Q = VC C = ε⋅ www.aieeepage.com d
1 R mv ( r − 1) ⋅ d T = mv 2 ; T = 2 2R r −1
7 But r = for diatomic gas 5
15.a.
MV 2 7 mv 2 T= − 1 = 2R 5 5R 12.c.
1
0
3. V .
ε0 A =V d
circuit is ε −
ε0 A : ε=3 d
q C
Q
q Q = ∫ ε − dq C 0
∴W1 − 0 (2–3) : Pressure is constant :
PV = RT ; V α T
Q
1 q2 Q2 = ε − ( w) = q − Q 2 C O 2C
Since P is constant
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∆ w = P∆ V = ∆ PV ⋅ ∆ ( RT ) = R (T f − Ti )
∆w
ε ε0 A d
∴ Work done to develop total charge
∴V = Constant
V= Constant :
=V
q . Hence the voltage in the C
the PD across it is
P P α T : P = kT : PV = RT ; PV = R k
= R (2400 − 800)
ε.
1
When the charge developed on the capacitor is q,
(1–2) :
(3–4) : (4–1) :
m ⋅ RT M
A → B : V α T : PV =
To obtain P V relation we eliminate = T
1 1 1 Ah ⋅ d1 g + Ahd 2 g = ⋅ Ah ⋅ d1 g ; 81 3 24
11.d.
13c.
1 1 = ε 2C − ε 2C = ε 2C 2 2
(V) energy stored =
R(T f − Ti ) = R (400 − 1200) =
1 1 CV 2 = C ε 2 2 2
1 1 wV . = ε C : Cε www.aieeepage.com 2 2
∆ w2 = 1600 R
2
∆ w3 = 0
∆ w4 = 800 R −−−−−−− ∆ w = 500 R −−−−−−−
(∵ Q = εC )
16.b.
23 −
2
= 1:1 .
q q q +8− −5− = 0 2 3 4
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q
q
q
A
for circular coil of radius ‘R’, B =
B 2
8V
5V
Ratio of 'B' =
22.d.
∴ q = 24µC
P = i2 R =
q 24 = = 8Volt . 3 C
∴V = PDa / c 3µF =
∴
60 = 0.1A 600
0
Energy
=
E = L⋅
di ; dt
r 2
1 1 1 1 1 = + + + + .....16 lenses F f 2f 4f 8f
1 2 Li 2
=
1 1 2 ⋅ ≈ f 1− 1 f ; 2
For square coil of side ‘L’, B = 4 ⋅
∴F = f
PAGE =
24.c.
2.
Consider an element of thickness (dx) at a distance x from the axis of rotation force acting on the element.
– axis
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di E IR 2 ×10 = = = = 5 A Sec . 4 dt L L 21.a.
R
1 1 1 1 1 = 1 + + + + ..... F f 2 4 8
∫ di
1 × 0.25 ×0 (48) 2 = 288 J 2 20.d.
23.d.
i
∴ i = 48 ;
r/2
dP E2 2E 2 R = − =D; 2 2 dR r r R+ 2 R+ 2
R=
di = 0 ; E ⋅ dt = L ⋅ di ; dt
+ 2t )dt = 0.25
E
dR
the same emf will be developed a/c II coil as being wound round the (same) common core
0
R.
dP www.aieeepage.com When is =0, P is maximum dφ
0.3 di = L1 1 = 0.2 × = 66 Volts 0.001 dt
2
2
E r
dt
18.d. E = emf developed a/c first coil
∫ (3t
r R+ 2
R
Hence graph (a) in the choice.
2
π
E r
= − (0.2t − 6) = 6 − 0.2t ∴ as t increases, E decreases.
E−L
E
PAGE
−dφ 2 17.a. φ = 0.1(t − 6t ) ; E = dt
19.c.
R= 1
4 L = 2; L = 1 ; = 16: π2 2
1 1 1 26 = q + + 2 3 4
∴ 60 = I 2 × R2 ; I 2 =
4µoI 2 R 8R 8 1 1 ⋅ = = × × π L µoI π L π π 2
2 π R=2, π R =1 ;
25V
µoI 2R
h
µoI πL p
x
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dx
p+dp
L
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= (dm) ⋅ ω2 x
∴µ =
If a = cross sectional area of tube.
dm = ⋅ ∫ ⋅a ⋅ dx ( ∫ =density of liquid) Net pressure acting on element =
dp = dp =
2sin A cos A sin i sin A 2 2 = = A A sin r sin sin 2 2
µ . 2
A = 2 cos −1
dt dm ⋅ ω x = a a 2
27.a.
⋅ ∫ ⋅a ⋅ dx ⋅ ω2 x = ∫ ω2 x ⋅ dx a
for plane surface,
1 h −1 µ − = V − mR ∞
V = − µ mR
But net pressure acting on horizontal position of the
B
L
liquid position =
ρg h = ∫d p 0
PAGE A
C
mR
2
x ωL ρ g h = ∫ ρω2 x ⋅ dx =ρω2 = ρ 2 2 0 0 L
h= 25.c.
2
ω2 L2 . 2g
2 2
R
R
for curved surface =
1 h 1− µ − = R −(µmr + R) − R
1 h µ 1 2 + = ⋅ ;m= www.aieeepage.com R R(µm + 1) 2 R µ(µ − 2)
acceleration of system,
a=
mc .g unbalanced force = Total mass 2m + m + mc
28.a.
We have frindge width, β red = λ red ⋅
N
βblue = λ blue ⋅
”
µN
D d
ma
Let
mg
D d
x0 be the distance within which there are n red
fringes and (n+)th blue fringes. Then
or ,
mc.g
29.b. Ndecayed= N o ⋅ e
µ( ma ) = mg
and
PAGE
∴ N (left over ) = No − N decayed = No − N o ⋅ e −λt = No − No ⋅
mc .g g ∴ = ; 3m + mc µ
26.c.
−λt
∴n = 2
A = Ao e −λt = λN o ⋅ e −λt
g − (2) µ
mc =mass of block c =
n λ red = (n + 1) λ blue
nx 7800 = ( n + 1) 5200 ;
mc .g − (1) 3m + mc µr = mg a=
a=
x0 = n βred = (n + 1) βblue
A A = No − λN o λ
slope. www.aieeepage.com
This is the equation of a straight line in the –ve
3m . µ −1
30.b. E (total energy) = V + K ; K =
In minimum deisation position,
i + i = D + A = A + A; i = A and r =
A 2
(U is negative);
1 U 2
t = −2 K + K = − K
− K = −3.4 K = 3.4 eV .
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31.a.
32.c.
This is a bit of trivia that you should know for the Test: The gravitational force everywhere inside a uniformly dense sphere or ring, due to that sphere or ring, is zero. However, the uniformally dense ring or sphere does not block the gravitational field due to other sources, so the force due to the smaller ring still affects a particle at position D. Initially the two pistons are in equilibrium; hence, the pressure underneath them must be equal. Since there is vacuum above the pistons, the force of pressure acting on each piston must balance the weight of that piston. Thus, the piston having twice the mass must have twice the cross-sectional area. We can denote these two areas A and 2A. If we double the total load on the lighter piston, the extra load on the pistor would cause it to drop all the way to the bottom of its cylinder and all of the gas will be squeezed into the second cylinder. Since there is negligible variation inhydrostatic pressure across the height of a 40-cm column in a typical gas under reasonable conditions, one can assume the pressure is the same everywhere throughout the gas. But the pressure in this second cylinder must continue to hold up the second piston, whose weight has not changed. Thus, the gas pressure will be the same before and after adding the extra 1-kg load. Furthermore, the temperature and the number of moles are the same. Thus, the volume of the gas must be conserved. The initial volume is 3Ah. The final volume is 2AH. Equating these two expressions gives H=1.5h=0.60m. The inclined plane cuts out an area given by:
1 2 hc h mv = − 2 λ λ' Adding (1) and (2), we get 2 ⇒
hc 1 2 1 = mv + (mv 2 ) λ 2 α
hc mv 2 mv 2 = Eγ = + λ 4 2α
Since α << 1 ⇒
1 >> 1 ; α
⇒
Eγ =
⇒
1 2 mv = α Eγ 2
1 α
PAGE 36.a.
s s 2s = = cos θ cos 30º 3 Let the average intensity at the incline be x. Then:
Area =
2s ( x) = (1A) = (constant output power of bulb). 3 Finally,
34.c.
3 1A x= 2 s A ray through the center of the lens is not bent at all.
O 35.a.
mv 2 mv 2 << 4 2α
⇒
mv 2 2
Intensity α 1 d2 and Current α Intensity
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33.c.
... (2)
⇒
Current α
1
d2 i.e. when d becomes 3d 1 = 2mA 9 Where as stopping potential is independent of intensity.
1 becomes
37.d. 38.b. The pressure at a given point in the small pipes would double, but twice as much presure would be required to support the same volume of the heavier fluid. 39.d. v = sqrt(gh) and v = f λ , g is smaller on the moon, so velocity would decrease. The frequency is dictated by t, the period. Since t does not change, frequency remains constant and λ must decrease. 40.d. The net total force exerted on the chain (by both the surface and gravitation) at any time is equal to its mass times the acceleration of its center of mass.
PAGE X
Since V = α c (Where α << 1 ) the electrn is moving at non-relativistic speed. By Law of Conservation of Momentum
L
l=linear M mass density = L of the chain
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h h + = mv λ λ' hc h mv 2 + = mcv = λ λ' α By Law of Conservation of Energy ⇒
... (1)
X
To find the equation of motion of the center of mass, according to the figure, we can write (all the distances are evaluated with respect to the hanging point):
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second to the right). Applying Newton’s second law to the center of the board, we have
L−x ( xλ ) L + ( L − x ) λ x + Σmx 2 = = Lλ Σm
xcm
= x+ ⇒ x 'cm = x '− x ''cm = x ''−
1 x 1 x F1 + F2 = M x ⇒ µ k Mg + + µ k Mg − 2 d 2 d
L2 − x 2 2L
2µ k g 2µ g x⇒ x+ k x = 0. d d This is the equation for simple harmonic motion with angular frequency ω . =
xx ' L
xx ''+ x '2 xx ''+ x '2 ⇒ Mx ''cm = Mg − N = M x ''− L L
N; the normal force of the surface But x´´= g since the chain is falling freely and also we
x '2 = 2 gx (equation of motion with constant acceleation). So we have have:
41.c.
2µ k g ) d Conclusion: The center of the board will oscillate with period. ( ω2 =
d 2µ kg
PAGE
M 3M N= ( xg + 2 gx) = gx L L Let k be the spring constant of the spring. In equilibrium, PA=kh. (P is presure, A is the area of the piston, and h is theheight of the piston above the
T = 2π
around the origin of our reference frame. The amplitude of the motion will depend on the “initial” situation, i.e., on the initial “asymmetry” of the board when it is placed on the cylinders. In the original figure, the center of the board seems to be (initially) slightly at the right of our origin, so the board will begin the oscillation moving to the left. 44.d. For a series RLC circuit
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bottom.) Therefore P1 / P2 = h1 / h2 . From the ideal gas law we have PV 1 1 / T1 = P2V2 / T2 , (T is the abosolute temperature.)
−L
V = hA, so P1h1 / T1 = P2 h2 / T2 .
With
T2 = 2T1 ,
we
have
P2 = 2 P1h1 / h2 , (h1 / h2 )2 = 1 . 2
The piston rises to a height 42.c.
43.a.
2 h when the gas
temperature is 2T. Notice that the voltage is a property of the field, and is independent of the charge or sign on q2. The formula for voltage is: V=kq/r. Voltage is inversely proportional to r. As we move against the lines of force (toward positive), voltage increases. Let us put the origin of our reference frame over the midpoint of the axes of the cylinders, at the level of the center of the board, and let us call x the horizontal position of the center of the board in this reference frame (x axis pointing to the right). The vertical equation for the forces acting on the board is
d d − N 2 − Mgx = 0 . 2 2
Q = CV ⇒ I = C − LC
d 2V dr
2
− RC
dV dt
dV −V = 0 dt
R 1 V ´+ V =0 L LC For the parallel RLC circuit shown V ´´+
or
I2
I1
C
Rp
PAGE
N1 + N 2 − Mg = 0 and the equation for rotational equilibrium is N1
dl − RI − V = 0 dt
Q = CV , I1 = −
L
dI dQ = −L 2 dt dt
V = R p ( I1 + I 2 ) = − L
dI 2 dt
dV
= R ( I´ + I´ ) dt www.aieeepage.com ( N is the normal at the right cylinder, and N at the 1
p
1
2
2
left cylinder.) From these two equations we obtain 1 x 1 x N1 = Mg + and N 2 = Mg − . 2 d 2 d
The kinetic friction forces are F1 = −µ k N1 and F2 = +µ k N 2 (the first points to the left, and the
= −CR pV ´´− R p V ´´+
V L
1 V =0 V ´+ RpC LC
Hence for the same L, C, Q we need
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48.a.
1 R L = ⇒ Rp = R pC L RC
45.c.
Use the right hand rule. If we imagine positive charge in the bar as it moves through the magnetic field, point your fingers of your right hand in the direction of the magnetic field, and your thumb in the direction of the moving charge (the direction of the moving bar). Now, with your hand in this position, push with your palm. This is the direction of the current, up through the bar. This creates a counterclockwise current in the loop. 46.d. Keeping in mind the relationship with poer and energy, let’s focus on how much energy is required to excite a proton from the lower to higher spin energy. The passage states that the RF pulse has to be applied at the specific Larmor frequency in order to excite a proton. By using E=hf to calculate the energy resulting from an RF wave at the Larmor frequency, we can then calculate the amount of power require to maintain this wave for 2 seconds. So, using f0= γ B0’ then E=h γ B0 so: E = (6.62 x10-34J/s) (1.5T) (42.56MHz/T) = 4.22 x 10-26 J Finally, divide by 2 sec to calculate the power P = 2.11 x10-26 W. 47.d. These types of questions ask you to examine an equation and are testing your understanding of the parametes of the equation. The question refers to Equation 1, which has temperature as part of the equation, so eliminate choice (B). Decreasing the magnetic field or using a different target atom would change B0 and γ respectively, which dictates the energy difference between spin states, so eliminate choices (A) and (C). changing the wavelength of the RF pulse would not affect the spin ratio, and choice (D) is correct.
You know that the energy gap between the spin states of the proton is directly proportional to the Larmor frequency, and that the Larmor frequency is proportional to the applied field. It folows that doubling the magnetic field strength will double the energy gap choice (A) is correct. 49.c. This is basically a discrete question - it has nothing to do with the passage. You are asked to consider how two forces act on the electron-magnetic and gravitational. Except that there is no magnetic force, since the electron moves antiparallel to the magnetic field. Moving charges have to have a component of their velocity be perpendicular to the magnetic field for the field to exert a force on it. So the only force on the electron is gravitatonal, and the correct answer is choice (C) 50.b. How many times greater” is IIT-JEE speak for” calculate the ratio” - in this case, the ratio of the Xray photon energy to the RF photon energy, used in the MRI described in the passage. Since the energy of a photon is proportional to its frequency, it will be sufficient to calculate the ratio of the frequencies. What is the frequency of the RF photons used in the MRI? At 1.5T.
PAGE
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f RF = γB0 = (42.56 ×106 Hz / T)
(1.5 T) = 6.38 ×107 Hz So the ratio of frequencies is:
f x − ray 1× 1019 = = 1.57 ×1011 7 f FR 6.38 × 10
PAGE
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Paper - I I : Mathematics - 51 to 100 Solutions 51.a.
(1 is subtracted since the particular match in which A and B play is counted twice) ∴ number of games played
n−1
(1 + x ) (1 + x 2 )...(1 + x 2 ) represented by n −1 1 (1 − x 2 ) (1 + x 2 ) (1 + x 4 ).....(1 + x 2 ) = 1− x n −1
n −1
= n C 2 − (2n − 3) + 6
n
(1 − x 2 ) (1 + x 2 ) 1 − x 2 = by taking log prod1− x 1− x
(Both play 3 games each)
uct & diff. on b.s. w.r to x
⇒ n 2 − 5n − 150 = 0 ⇒ (n + 10) (n − 15) = 0 ; ⇒ n = 15
= 84 (given);
2n−1
n −1
1 1 2 x −1 + + ... + = n−1 1 + x 1 + x2 1 + x2
PAGE
n−1
1 2n x 2 − 1 − . n 1− x 1 − x2 52.a.
If | x |< 1, Lt x n →∞
2n
57. c. In ∆OAP, A = 90o
Z 3 − Z1 t = i and 0 − Z1 r
∴
= Lt x 2
n −1
=0
n →∞
In ∆OBP, ∴
1 . 1− x
0-Z2 r = i Z3 -Z2 t
www.aieeepage.com A(Z )
So, the sum of the series to infinite terms = 53.d.
1
t
f (θ) = sin θ1 sin θ2 .....sin θn , taking log on b.s. & diff.
= log sin θ1 + ... + log sin θn ⇒
B(Z2) Z − Z1 − Z 2 Hence 3 = −1 − Z1 Z 3 − Z 2
n−1
(1 + x ) (1 + x 2 )...(1 + x 2 ) represented by n −1 1 (1 − x 2 ) (1 + x 2 ) (1 + x 4 ).....(1 + x 2 ) = 1− x n −1
n −1
by taking log
product & diff. on b.s. w.r to x n−1
1 1 2 n−1 x 2 − 1 + + ... + = n−1 1 + x 1 + x2 1 + x2 n−1
1 2n x 2 − 1 − . n 1− x 1 − x2 55.a.
or
n
(1 − x 2 ) (1 + x 2 ) 1 − x 2 = 1− x 1− x
P(Z3)
O
1 f 1 (θ) = cot θ1 + ... + cot θn . f (θ) 54.a.
1 = (n 2 − 5n + 18) 2
58. a.
Z3 Z + 1 = 3 −1 ; Z1 Z2
∴ Z3 =
2 Z1 Z 2 Z1 + Z 2
A cot 84o cot 48o = B tan 24o tan12o =
cot 84o cos 48o cos 24o cos 12o sin 84o sin 48o sin 24o sin12o
PAGE =
f(-1) = a - b + c < 1 .....(1) f (1) = a + b + c > -1 ....(2) ⇒ -a-b-c<1 F (3) = 9a + 3b + c < -1 From (1) and (2) - 2b < 2 ⇒ b+1>0 If n is the total number of players, there would have been nC2 games. Since the two players say A and B quit after playing 3 games each the loss of games. = 2 (n - 1) -1 =2n-3
(cos 108o + cos 60o ) ( cos 60o + cos 36o ) (cos 60o - cos 108o ) (cos 36o − cos 60o )
1 5 − 1 1 5 +1 − + 2 4 2 4 = 1 5 − 1 5 + 1 1 − + 4 4 2 2
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56.c.
=
(3 − 5 ) (3+ 5 ) (1 + 5 ) ( 5 − 1) ;
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=
4 =1 4
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59.c.
Solving, y = m1x and x + y = 1,
a + b , a − b and a + kb are collinear if and only
1 we get the x co-ordinate of A as 1 + m1 ∴ OL =
1 1 + m1
hence
A
O
B
C
63.a.
⇒
1 1 − 1 + m1 1 + m2 ;
Position vector of P, the mid-point of BC =
PAGE
1 and 1 + m2
Similarly, DQ is the perpendicular bisector of CA. ∴ D is the circumcentre of ∆ABC
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=
1 1 − 1 + m2 1 + m3
⇒ 1+m1 ,1 + m2 ,1 + m3 are in H.P. 60.b. C = (8 - 4, 24 - 6) or (4, 18)
∴ AC = (4 − 4) 2 + (18 − 6) 2 = 12 Since AC = AB = 12
Since A =
π 3
∴ the ∆ is equilateral and the radius of the
1 1 1 3 incircle = altitude = AC sin 60o = × 12 3 3 3 2
=2 3
Q c+a 2 d
P B
C
b+c 2
64.b. A normal vector n to the plane OQR is
i j k OQ × OR 1 3 4 2 1 -2
= -10 i +10 j -5k and nˆ =
-2 i + 2 j - k 3
∴ the perpendicular from P on OQR = projection of
PAGE OP on nˆ
x2 y2 + = 1; y = mx ± a 2 m 2 + b 2 is a 61. a. 576 49 tangent
A
D
1 1 1 , and are in A.P. 1 + m1 1 + m2 1 + m3
π ; 3
b +c 2
⇒ DP is perpendicular to BC. ∴ DP is the perpendicular bisector
1 1 + m3
B= C=
1+ k 1− k and m = 2 2
(b − c ) b +2 c − d = 0
Now, AB = BC ⇒ LM = MN. being projection on the x-axis
⇒
=
Hence, for these values of and m (correponding to any k) the given points are collinear.
N M L
Similarly, OM =
ON =
if a + kb = l (a + b ) + m(a − b ) for some real values of l and m . This implies that + m = 1 and − m = k and
i.e., y = x ± 25 is a tangent Solving with the equation of the ellipse, we get
= (3 i - 2 j + k . =
(-2 i + 2 j - k) 3
- 6 - 4 - 1 11 = 3 3
65.d. The probability of getting an odd number in a www.aieeepage.com throw = 1 / 2
625 x 2 ± 2( 25) (576 x ) + (576) 2 = 0 ( 25 x 2 ± 576) 2 = 0
∴x = + 62.c
576 49 and correspondingly y = ± 25 25
The probability of getting an odd number 2 times in 2
n trials
1 1 = C2 2 2
n-2
n
The points with position vectors
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10 of 19
similarly the probability of getting an even number 3 times in n trials 3
1 1 = C3 2 2
at both
n-3
n
1 2
n
1 2
5π f(x) is continuous in 0, 4
n
n n It is given C 2 = C3 or
n
68.a.
C 2 = n C3 ⇒ n = 5
n
n
1 1 1 C1 + n C3 + n C5 2 2 2
n
∑ n →∞ Lt
n →∞ 1
] [
⇒ x ∈ − 2π ,− π ∪ π , 2π
=
]
∫
(
y3
1
π y = 2 sin x − 4
70.a.
I=
π
π 4
π 2
-1
−π
= 5π 4
∫
0
2
1
(3
r/n + 4
∫ 4
=
∫
=
;
2π
+
2
π
2
2 1 . dt 3 t2
1 14 3π
+
)
∫
cot −1 cot x dx
2π
cot −1 cot(π + x) dx
π
+
2π
+
0≤ x≤
∫
cot −1 cot( x − π ) dx
π
3π
+
Plotting the curves on a graph
cot −1 cot x dx
0
π y2 = sin x − cos x = 2 sin x − 4 y3 = 1
∫
PAGE
π y = 2 cos x − 4
= sin x − cos x
z2 + 3
π
3π 4
f ( x) = sin x + cos x
∫ ∫ +
0
y1
)
21 1 = - ; 34 7 0
y2
dz
)
7
dx
x 3 x +4
0
π y1 = cos x + sin x = 2 cos x − 4
67.c.
(
r 3 r +4 n
r =1
⇒ x ≤ − π or x ≥ π and
[
n
1 1 www.aieeepage.com = Lt ∑ . 2 r/n
⇒ x 2 ≥ π and x 2 ≤ 2π
∫
= log tan 2 θ − 1 + tan 2 θ + 2 + C
69.a.
− 2π ≤ x ≤ 2π
tan 2 θ + 2
n
⇒ 0 ≤ x2 − π ≤ π
=
;
PAGE
1 5 1 C1 +5C3 +5C5 = 32 2
cos −1 y = x 2 − π
66.a.
dz
∫
= log z + z 2 + 3
where n = 5
=
Putting tan2θ −1 = xz I=
Then the probability of getting an odd number odd number of times n
5π π and 6 6
∫
cot −1 cot( x − 2π ) dx
www.aieeepage.com 2π
π 2
π 5π ≤x≤ 2 4
From the graph, it is clear that f(x) is differentiable
0
=
∫
( π + x) dx +
−π 2π
+
∫
π
∫
x dx
0
( x − π ) dx +
π
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3π
∫
( x − 2π ) dx
2π
11 of 19
= 71c.
(1 + 80)100 = 1 + 100 C1.80 + ... + 80100
π2 π2 π2 π2 + + + = 2π 2 2 2 2 2
Except the first term all other term have at least 3 zeros as the last three digits. ∴ the last two digits are 0 and1.
Distance from the origin, of the line y = mx = C is
C
±
1 + m2
= 1( given)
0 0 1 −1 1 1 76.c. The given determinant = −1 0 2
⇒ c 2 = (1 + m 2 ) ⇒ ( y − + mx 2 ) = (1 + m 2 ) Replace m by
( ∵[x + 1] = [x] + 1 etc and [x] = - 1, [y] = 0 [z] = 1) = 1 = [z]
dy (slope) dx
2
dy dy y − x =1+ dx dx
77.b.
Z − 2 = 2 is a circle with centre at (2, 0) and radius 2.
2
PAGE Y
72.b. We have 2 × 6 = 32 cos 2θ −1 + 34−2 cos 2θ
O
9 cos 2θ 81 ⇒ 12 = + cos 2θ 3 9
X
(2, 0)
Z (1 − i) + Z (1 + i) − 4 www.aieeepage.com ⇒ (Z + Z) - i (Z + Z) − 4 is a straight line.
2
⇒ x − 36x + 243 = 0 where x = 9cos2θ
⇒ ( x − 27) (x − 9) = 0 ⇒ 9 cos2θ = 27 or 9
Clearly Z = 2 satisfies this straight line ∴ It is a diameter of the circle. Hence the number of points of intersection = 2.
3 ⇒ cos 2θ = or 1. But cos 2θ > 1. 2 Hence cos 2θ = 1 ∴ the numbers are 3, 6, 9 ⇒ c. d = 3 73.d. The different faces are 1, 2, 3, 4, 5, 6. For displaying the face containing 1, 3 dice can be chosen in
18
78.b.
x sec θ + y tan θ = a or (x sec θ − a 2 ) = y 2 tan 2 θ = y 2 (sec 2 θ − 1)
C3 .
sec 2 θ ( x 2 − y 2 ) − 2 ax sec θ + a 2 + y 2 = 0
After this for the face 2, the number of ways = 15 C 3
This is a quadratic in sec θ , whose roots are sec
and so on. Hence the required number is
= 74.b.
18
(x +
α and sec β
∠18 C3 . C3 . C3 . C3 . C3 . C3 ; = (∠3)6 15
12
9
6
) ( 6
)
6
6
3 z + 6C 2 ( x + 2 y ) 4 3 3 z 2
3
6
+ C3 ( x + 2 y ) .3 + C4 (x + 5 3
+ 6C5 (x + 2 y) 3 + 6C6 .32
4 2 3 2 y) 3
x2 − y2
79.b. Both the circles pass through the origin and hence they have common tangent at the origin. Both gx + fy = 0 and g’x + f’y = represent the same line.
∴
80.a.
g f = g' f'
or f' g = fg'
The circles pass through origin and the y-axis is the common tangent at the origin. A divides LM externally in the ratio 1 : 3
www.aieeepage.com
In RHS the first term contains 4 terms with rational coefficients (1st, 3rd, 5th, 7th), the 4th term 2 terms with rational coefficients and the last term 1 Hence the total number of terms with rational coefficients = 4 + 2 + 1 = 7 75.d.
a2 + y2
PAGE 2
3
sec α . sec β =
3
2 y+3 3 z = x+ 2 y
+ 6 C1 ( x + 2 y ) 5
α and β satisfy the equation
P A
3400 = (34 )100 = (1 + 80)100 The expansion of
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(-1, 0) L
R
Q
C
M(3, 0)
O B
S
∴ C = (- 3, 0) 12 of 19
1
Equation of APC is y = Equation of ARS is y =
3 −1 3
84. a.
( x + 3)
of Lagranges mean value theorem in [p, q]. ∴ there exists a, ‘e’ in (p, q) such that at (c, f(c)) the tangent is parallel to the chord joining the given points (p, f(p)) and (q, f(q)). Hence
( x + 3)
For the ∆ABC , AO is a median L is the centroid. ∴ centroid = (-1, 0) The other choices can be verified to be wrong. 81.c.
for
x2 a2
+
y2 b2
y = f ( x ) = ax 2 + bx + c satisfies the conditions
= 1, slope of the tangent at the
point of intersection (x1 , y1 ) is m1 = -
b2x1 a2y1
a(q 2 − p 2 ) + b(q − p) q−p
2ac + b =
⇒ 2ac + b = a (q + p) + b
⇒ c=
p+q 2
85. a. Since the result does not involve these particular equation, the parabola can be taken as y2 = 4ax
PAGE
S = (a, 0); T = (at1t 2 , a (t1 + t 2 ) ),
For x 2 − y 2 − c 2 , slope of the tangent at
x1 ( x1 , y1 ) is m2 = y 1
(
)
(
P = at12 , 2at1 , Q = at 22 , 2at 2
ST 2 = a 2 (t1t 2 − 1) 2 + a 2 (t1 + t 2 ) 2
(
)(
= a 2 1 + t12 1 + t 22
-b2 x x Since m1m2 = −1, 2 1 1 = −1 a y1 y1
)
www.aieeepage.com SP = (at − a ) + 4a t
⇒ b 2 x12 = a 2 y12 ⇒
x12 a2
=
2
SQ 2 = a 2 1 + t 22
b2
2
2
b a
Case II : Let x ∈ (0, ]
b a
Case III : Let x ∈ ( , ∞) .
f ( x) = f ( x − 2 + 2) = − f ( x − 2 + 6) = − f ( x + 4)
Then f(x) = ax - b + cx
= − f ( x + 2 + 2)
b In (−∞,0) f(x) is decreasing and in , ∞ f(x) a is increasing If f(x) is to be minimum at x = 0, then f(x) must be
PAGE
= f ( x + 8) ∀ ∈ R ∴ the period of f is 8.
{x} = 0. Then a
π f ( x) = 2 sin θ + 4
⇒ {x} =
)
)
b bc Then f(x) = b - ax + cx and f = a a
(i.e., ) a 2 − b 2 = 2c 2
maximum at θ =
(
= a 2 1 + t12
86.d. Case I : Let x ∈ (−∞, 0] Then f (x) = b - ax - cx and f (0) = b
a2 b2 − = c2 2 2
Let
2 2 1
Hence ST 2 = SP × SQ
Substituting in the equation of the hyperbola,
83.d.
2
2 1
(
y12
x12 y12 + 2 2 b =1 a ⇒ 2 2
82.b.
)
increasing in 0,
which
π ; 4
∴
its
Hence f(x) = c - a should be positive
∴ c
aπ . But 0 < {x} < 1 4
aπ 4 ∴ 0< <1 ⇒ 0 < a < π 4
attains
a
=
4
b a
x/2
87. c. 2
x/2
+3
Which
( )
2 ⇒ 13
is
of
= 13
x/ 2
x/ 2
3 + 13 the
x/ 2
=1 form
x cos x / 2 α + sin x / 2 α = 1. ∴ = 2 . 2
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88.d. We have f(0) = 2, therefore we cannot say
g =0 ⇒
f ( x) rel="nofollow"> 2, ∀ x ∈ R.
b = 10 -3 (0) = 10
Also, y f(x) does not represent a hyperbola and neither y = f(x)2 +4 represent a parabola. 2a
89.
Area of the region =
2a − x x
∫
a
0
g =1 ⇒
2
g=2 ⇒
dx
Put x = 2a sin θ ∴ dx = 4a sin θ cos θ dθ
∫ 0
π/2
∫
= (8) (4)
b = 10 - 3 (3) = 1
∴ total number of ways = 4
93. d. Equations of x, y and z axes are
x y z x y z x y z = = , = = and, = = 1 0 0 0 1 0 0 0 1
PAGE
x - 3 y - 2 z -1 = = 3 1 0 We obsereve that 3 × 0 + 1 × 0 + 0 × 1 = 0 The given line is
5.3.1.1 π 5π 2 . = a 8.6.4.2 2 8
Also area of circle = π a 2
∴
Hence, the line is perpendicular to z - axis. 94. a. Here, the centre of the sphere is in OX Y’Z octant. So, let its coordinates be (r, -r, r) Since the spehre touches 2x - y + 2z - 4 = 0. Therefore,
www.aieeepage.com 2r + r + 2r − 4
5π 2 5 reqd. ratio = a :π a2 = 8 8 =5:8
4 +1+ 4
a a 1 sin 2 t sin 2 t Lt e dt e dt − 90. b. x→0 x y x+ y
∫
∫
e
sin 2 t
= Lt
x
x →0
95.a.
2
= a ( x 2 + x + 1) ( x − a ) where α is a real root of the given equation. Equating constant term on both sides,
d =-a α ∴ α =96.a.
= f (1) e' f ' (0) + f ' (1) . 1 e
y = f 1 ( x) is symmetrical about y-
97.b.
98a.
f 11 ( x) > 0 as x < 0 & f 11 ( x) < 0 as x > 0
∴ x = 0 is the point of inflexion. As f ( x ) is odd, so a . ∫ f ( x) dx = 0
www.aieeepage.com
⇒ g + 2 g + b = 10 ⇒
f ( 0)
The graph of
axis, so f ( x ) is an even function.
∴ y' (0) = f(e 0 )e f ( 0) f ' (0) + f ' (e 0 )e 0 e f ( 0)
=2.1=2 92. d. Let no. of green balls = g No. of red balls = r and no. of blue balls = b r = 2g and g + r + b = 10 ∴
d a
PAGE
y ' ( x ) = f ( e x ) e f ( x ) . f ' ( x ) + f ' (e x ) . e x . e f ( x )
= 0 ( e) f ' ( 0) + ( 2) e
2 Since x + x + 1 is a factor of
∴ ( ax 3 + bx 2 + cx + d )
y
0
1 2
ax 3 + bx 2 + cx + d
0 form 0
2 d ( x + y )e sin ( x + y ) − 0 = Lt dx x →0 1
= e sin
1 2
(x − r) 2 + (y + r) 2 + (z − r) 2 = r 2 , where r = 2,
dt
y
= r ⇒ 5r − 4 = 3r ⇒ r = 2,
Thus, the equation of the sphere is
∫
x+ y
91.c.
r = 2 (3) = 6
respectively.
sin 6 θ cos 2 θ dθ
0
= 32 a 2 .
r = 2 (2) = 4 b = 10 - 3 (2) = 4
g =3 ⇒
2a cosθ (2asin2θ)5/2 4a sin θ cos θ dθ a2
r = 2 (1) = 2 b = 10 - 3 (1) = 7
5/2
2
π/2
r = 2 (0) = 0
b = 10 − 3 g
99.a.
−a
f 1 ( x) ≤ 0 ∀x , so f ( x) − a is always decreasing.
100.b. f (0) = 0 , so f ( x ) is an odd function (Derivative of odd function is even)
The possible values of g are 0, 1, 2, 3
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Paper - I I I : Chemistry - 101 to 150 Solutions 101.b. The iron oxide surface acts as a catalyst to speed up Reaction 1. Even though we are not given any psecific information about how it does its job, we can see how choice B is a reasonable description of its role: In order to form ammonia, the nitrogen-nitrogen bond must be broken at some point. Facilitating the dissociation of moleculear nitrogen would then certainly allow the reaction to go faster. We can examine the other choices and see why they could not be an accurate description of the role of the iron oxide surface: Choice A is incorrect because if gaseous ammonia is on the surface, the reaction has already taken place. Facilitating its desorption from the surface would not make the reaction itself go any faster. Choice C is incorrect since N-H bonds are found only in the product of the reaction. Destabilizing the product will not make a reaction proceed more rapidly. Choice D can be eliminated because a catalyst cannot change the thermodynamics of a reaction. 102.b. The correct answer is option (b). The important part of this question is that it is asking about the decomposition of ammonia, i.e., the reverse of Reaction 1. More specifically, the reaction we are interested in is:
is a false statement and the answer we are looking for It is stated in the passage that raising the temperature shifts the equilirbrium in favor of the reactants. We know also from equation 2 that the value of the equilibrium constant decreases at a higher temperature. Either of these two pieces of information should tell us that the reaction is exothermic. The decomposition reaction is therefore endothermic. 103c.. This question examines howle Chatelier’s principle can be used to drive the reaction towards the formation of ammonia. We are looking for the choice that will NOT shift the reaction to the right. Low temperature and high pressure (from small volume) favors the formation of the product. This means that the scheme in choice C will work.
PAGE
104.d. K 800 = K 298 × 7 ×10−11 ≈ 6 ×105 × 7 × 10 −11
= 42 × 10 −6 ≈ 4 × 10 −5
105.c. Of the 4 answer choices provided an examination of www.aieeepage.com the nucleophilicity of the nitrogen shows that
1 3 N2 + H2 2 2
NH3
We can eliminate choice D since it contains a true statement there is 1 mole of gas on the left and 2 moles on the right, the decomposition thus leads to an increase in entropy. How is the equilibrium constant for this reaction related to the equilibrium constants given in the passage for Reaction 1? Since the two are reverse reactions, the larger the K for one, the smaller the K of the other. If K of reaction 1 is gtreater than 1, then the K for the decomposition reaction is smaller than 1, and vice versa. At 298 K, the K for reaction 1 is given in the second paragraph of the passage as 6 x 105. The K for the decomposition reaction is therefore less than 1. choice A is a correct statement and hence not the correct choice. What about at 800 K? a little bit of work is needed to first find out the equilibrium constant for reaction 1 at this temperature. This is done with equation 2.
diethylamine is the most basic of the structures and will therefore react with the most affinity. The electron withdrawing carbonyl in the amide pulls charge away from the nitrogen electrons making it less negative B. While the thylamine is stabilizing because of its inductive electron donating properties, it does not provide as much electron density as the more electron donating diethyl side groups. While a benzene ring offers great stability it is not more electron donating when compared to 2 ethyl groups 106.d. All the other factors except (E) (thermal) affect the energy content of the system of N molecules. 107.a. The functional groups of eugenol are circled below. aromatic ring hydroxyl
PAGE
K 800 = K 298 × 7 ×10
−11
alkene
OH OCH3 ether
108.a. The final answer has the same number of significant digits as the measurement with the least number of significant digits. 109.c. The Joule-Thompson experiment is carried out for real gases undergoing a throttled adiabatic expansion. 110.d. The delocalized electrons of benzene stabilize the molecule by creating shorter stronger bonds between the carbons. The bonds are not as strong as double bonds, but they are stronger than single bonds.
www.aieeepage.com
= 6 × 105 × 7 × 10 −11 = 42 × 10 −6 < 1 for the forward reaction at 800 K. The equilibrium cosntant for the decomposition reaction at 800 K is therefore greater than 1. Choice B
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111.a. D-Glucose and D-Altrose are stereoisomers that are not mirror images and they are different at more than one carbon. 112.c. All are diastereomers because none of them are mirror images and all are stereoisomers. Epimers are diastereomers. 113.c. All three statements are accurate, note that ∆G , 114.b.
115.d. 116.d.
117.c.
not ∆Gº , is dependent on pressure. The first step of an E1 mechanism involves the formation of a carbocation after protonation of the alcohol. The alcohol in answer B will form the most stable carbocation. Electronegativity measures the attraction of an element for electrons within a chemical bond. Using the ideal gas law, PV/RT equals 1. In the real methane we can see that PV/RT equals less than one at 200 atm and more than one at 600 atm. So the calculated V must be greater than the real V at 200 atm and less than real V at 600 atm. Applying Kohlransch’s law, we have: λ
λ
λ
correct choice. Ammonia, along with primary and secondary amines are especially powerful catalysts for the Michael addition. Not only do they abstract a proton from the reagent, generating a carbanion, but they also react with the carbonyl group to form an intermediate which is particularly reactive towards nucleophilic addition. 126.c. The hydrogen bonding in cyclohexanol raises its melting point causing it to be a solid at room temperature. 127.b. When using proportional reasoning, be careful to make sure that all other factors are constant. Although D is a true statement, it’s technically only true when temperature and number of moles are both constant. 128.c. Supercritical fluid is the fluid state temperatures and pressures above the critical temperature and pressure.
PAGE 129.c. q = It
∴ q = (15.0 A) (1.25hrs) (3600 s hr −1 )
λ
O ( NH 4 OH ) = O ( NH 4 Cl ) + O ( NaOH ) − O ( NaCl )
=150+248–127=271. 118.a. The ketyl group substitutes for a hydrogen on the aromatic ring. It is an aromatic substitution reaction. 119.c. The methoxy group is ring activating. The other substituents are deactivating. Memorize the table of activating and deactivating groups. 120.c. First, balance the reaction :
note that in order to convert units to number of Coulombs (C) it’s necessary to convert hours to seconds. Since
www.aieeepage.com 1C = 1 A × 1Sec.;
C12 H 22O11 (l ) + 12O2 ( g ) → 12CO2 ( g ) + 11H 2 O ( g ) . From there, the stoichiometry is easy. 121.d. A primary alcohol loses one α -hydrogen during oxidation (note that α -hydrogen is the one attached to the carbon bearing the –OH group) – OH | C5 H 5 NH + CrO3Cl – R — C — OH | H α–Hydrogen a 1º alcohol
H | R—C=O an aldehyde
Also, a primary alcohol is oxidized readily to a carboxylic acid in the presence of potassium permanganate. Under milder conditions - such as in
answer (c) is the correct one. 130.c. Since 1-bromobutane proceeds by a SN2 mechanism the rate depends on the concentration of the nucleophile. However the 2-chloro-2methylpropane proceeds by SN1 so the rate only depends on the concentration of the electrophile and not the nucleophile (1–). 131.a. Calculation of % of nitrogen. 50 ml. of 0.05 MH2SO4=50ml. of 0.1NH2SO4 (∵ Normally of H2SO4=2 × Molarity) Excess of acid requires 25ml. of 0.1M or 0.1N NaoH (∵ Normally of NaOH – Molarity of NaOH) 25ml. of 0.1N NaOH=25ml. of 0.1N H2SO4 ∴ Vol. of 0.1N H2SO4 used for the neutralisation of NH3=50–25=25ml. Now we know that, % of nitrogen =
1.4 × Normality of acid × Vol. of acid Wt.of compound
C2 H5 NHCrO3Cl and CH 2Cl 2 - the primary
PAGE
alcohol is oxidized to an aldehyde. 122.b. The only step of the SN2 reaction is a substitution occuring with the new bond and old bond forming and breaking simulataneously. 123.b. Nearly all reaction rates increase with increased temperature regardless of their enthalpy change. The problem is that heat is a product, so the equilibrium would shift to the left according to LeChatelier’s principle. 124.d. We want to maximize Pn in the second equation. Although delta T remains constant, from the second equation we see that at higher temperatures, Pn is greater. Black objects have a higher emissivity than white objects, which also increases Pn. 125.d. All the choices except (d) are true. Hence (d) is the
=
1.4 × 0.1 × 25 = 11.55% 0.303
Hence
% of oxygen = 100–(69.4+5.8+11.55)=13.25 Calculation of empirical formula of the compound.
www.aieeepage.com Percentage
Relative No. Of atoms
Simplest whole ratio
C
69.4
69.4 = 5.8 12
=7
H
5.8
5.8 = 5.8 1
=7
N
11.55
O
13.25
Element
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11.55 = 0.825 14
13.25 = 0.825 16
=1 =1
16 of 19
Thus the empirical formula of the aromatic compound = C7H7NO Calculation of molecular formula. Empirical formula wt. of the compound = 84+7+14+16=121
n=
Hence
Mol. wt. 121 = =1 Emp. formula wt. 121
Hence molecular formula of the compound= (C7H7NO)1 = C7H7NO Structure of the aromatic compound: Since the compound C7H7NO is aromatic, it may be written as C6H5CH2NO or C6H5CH=NOH (benxaldoxime). Benzadoxime can exist in the following two isomeric structures.
syn-Benzaldoxime
1
1
1 1 23,032 = 109,678 2 − 2 n 1 n2
MHCO3
44g of
CO 2 is given by 16.8g of MHCO3
44g of
CO 2 is given by =
Since two molecules of
16.8 × 44 = 168g 4.4
MHCO3 are taking part in
the reaction, the molecular weight of
MHCO3 (X) =
168 = 84 2
Calculation of atomic weight of metal M
MHCO3 = 84
heat 2NaHCO3 → CO2 + H 2 O + Na 2 CO3 (A)
(X)
(B)
(Y)
Na 2 CO3 + BaCl2 → BaCO3 + 2NaCl (white)
heat (X) (g) → A(g) + B(g) + Y(s) 1.8g
The above equation leads to the following facts : (i) Since the gas A turned lime water milky, it
Thus A is
CO 2 , B is H 2 O , Y is Na 2 CO3 .
136.c. Compound (A) on treatment with AgNO3 gives white precipitate of AgCl, which is readily soluble in dil.aq.NH3. Therefore it has at least one Cl– ion in the ionization sphere furthermore chromium has coordination number equal to 6. So its formula is
[Cr (NH3 )4 Br Cl] Cl . Compound (B) on treatment with AgNO3 gives pale yellow precipitate of AgBr soluble in conc. NH3. Therefore it has Br– in the ionization sphere. So its formula is
CO 2 .
(ii)
Calculation of molecular weight of
www.aieeepage.com
133.a. Studies reveal these patterns for a black-body. Classical physics could not account for these patterns. Planck used a quantum hypothesis to explain the black-body radiation. 134.d. The structure in answer D has the lowest pKa therefore it has the most acidic proton. 135.c. Representing the given facts in the form of equation.
must be
1.8g
M + 1 + 12 + 48 = 84 M + 61 = 84 M = 84 – 61 = 23 Thus the metal must be Na and hence the given salt X is NaHCO3. The above facts coincide with the given thermal decomposition.
23,032 1 1 = 2 − 2 109,678 n1 n 2
4.4g
4.4g
16.8g
syn-Benzaldoxime
132.a. v = RH × 2 2 n n 1 2
16.8g
heat 2MHCO3 → CO 2 + H 2O + M 2 CO3
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C6 H5– C – H || HO – N
C6 H5– C – H || N – OH
(iv) The above facts point out that B may be water. Thus the above reaction can be written as below:
[Cr (NH3 )4 Cl2 ] Br .
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The compound Y gives alkaline solution
in water which when treated with
BaCl 2 forms a
Cr 3+ (Z = 24)
white precipitate of Z. Since the compound Z when treated with acid gives effervescenes of
CO 2 , Z
may be written as metal carbonate
MCO3 or
M 2CO3 . (iii) When X is heated, it yields a carbonate (Y) along with the evolution of
4s
4p
State of hybridization of chromium in both (A) and
www.aieeepage.com Spin magnetic
and hence Y must be carbonate, CO 32 − . Hence Y
3d
(B) is
d 2sp3 .
moment of (A) or (B),
µspin = n(n+2) = 3(3 + 2) = 15 = 3.87BM . 137.d. The compound is a dicarboxylic acid and when heated it yields an anhydride.
CO 2 (A) and another gas (B),
it must be a bicarbonate.
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O || C H2C | H2C
increase, the reaction rate. Oxygen would compete with chlorine, reacting with the methyl radical; step (3). This would slow down the overall reaction since the rate of the propagation step is decreased.
O || C OH
H2C | H2C
heat
OH C || O
O + H2O
142.a.
C || O
All the other compounds, especially the ester and the xanthate, when heated yields alkenes. Note how similar the structure of (C) to the xanthate and ester structure (respectively). This in itself should give us a hint that these compounds would react similarly under similar conditions. 138. b. Nylon 6.6 has nitrogen atoms next to carbonyls, indicating an amide. The brackets with the subscript n denote a repeating group. 139.a. In such a case
6.023 ×1013 6.023 × 1013 = 200 2 × 10 2 = 3.0015 × 1021 = 3.012 × 1021 Density of Mercury (Hg) = 13.6 g/c.c. ∴ Volume of 1 atom of mercury (Hg) = 1 10−17 103 × 10 c.c. = c.c. = c.c. 3.012 ×136 409632 3012 ×10 21 × 136
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A = Ca(OH ) 2 , B = NH 4 HCO3 ,
=
C = Na2 CO3 , D = NH 4 Cl and E = CaCl2 CaO + H 2 O → Ca (OH )2 ( A)
Avagadro’s number = 6.023 × 1013 At wt. of mercury (Hg) = 200 ∵ In 1 g of Hg, the total number of atom =
10−17 10 −17 c.c. = c.c. = 3012 × 136 409632
1000000 × 10−23 c.c. = 2.44 × 10−23 c.c. 409632 Since each mercury atom occupies a cube of edge length equal to its diameter, therefore, diameter of one Hg atom =
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NH 3 + CO2 + H 2 O → NH 4 HCO3
Sod. Bicarbonate (B)
1
NH 4 HCO3 + NaCl → NaHCO3 +
NH 4 Cl Amm. chloride (D)
∆
2 NaHCO3 → CaCl2 + 2 NH 3 + 2 H 2O (E)
140.a.
can be used again
143.c.
The required reaction can be obtained the following way. → Cu + Cu 2 + + e−
∆G º = −0.15 F
→ In + , In 2+ + e−
∆G º = +0.40 F
6000 + 300 x 12 + x According to Raoult’s law
(∆G º = −nFE º )
On adding, Cu2+ + In2+ → Cu + , E º = −0.59F Now we know that − nFE º = −59 F
6000 + 300 x 12 + x − 400 0.525 = 6000 + 300 x x + 12 + 0.525 12 + x
0 0 or − Ecell = −0.59V or Ecell = 0.59V
0.0591 log K c ; 1
0 0 = 0, than Ecell = Ecell
0.59 =
0.0591 log K c 1
141.c.
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On usual calculations, x = 9.9 Now according to first order kinetics,
0.0591 log K c n
K=
144.c.
Element
2.303 10 = 1.005 × 10−4 min −1 log 100 9.9
%
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0.59 = 10; K c = 1010 0.0591 For the mechanism shown, step (1) represents initiation, while steps (2), (3) AND (4), (5) represent propagation and termination respectively. Step(2) is the most difficult step to occur, this step thus limits the overall reaction rate, and we would expect this tep to be the rate determining step. Addition of oxygen to the reaction, would inhibit, rather than log K c =
=2.905 ×10−8 cm = 2.91Å cm Let the number of moles of A left after 100 min = x Total number of moles after 100 min =x+12+0.525 x 12 × 300 + × 500 = Pmix = p A + pB = 12 + x 12 + x
→ In3+ + 2e − , ∆G º = −0.84 F In +
0 Ecell = Ecell −
1
(2.44 × 10−23 ) 3 cm = (2.44 × 10−24 ) 3 cm
Relative no. Of atoms
Simplest Ratio
C
42.86
42.86 = 3.57 12
3.57 =3 1.19
H
2.40
2.40 = 2.40 1
2.40 =2 1.19
N
16.67
16.67 = 1.19 14
1.19 =1 1.19
O
38.07
38.07 = 2.38 16
2.38 =2 1.19
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∴ Empirical formula of the minor product is C3 H 2 NO2 Molar empirical formula mass of the minor pdt =
3 × 12 + 2 × 1 + 1× 14 + 2 × 16 = 84g mol−2 Let M be the molar mass of the minor pdt. For 5.5g of the minor product dissolved in 45g. 5.5g/M benzene, the molality (m) of the solution = 0.045kg
Substituting this in the expression of elevation of boiling point, 5.5g/M ∆ Tb = K b m ⇒ 1.84 K = (2.53K kg mol−1 ) 0.045kg
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or M = 168 g mol−1 No. of unit of empirical formula in molecular formula =
168 g mol−1 84 g mol−1
=2
Hence the molecular formula of the minor product
145.c.
nitrate. Figure 1 indicated the enthalpy of solution of ammonium nitrate is 25.69 mJ/mole. Therefore, multiply. 75 moles by 25.69 kJ/mole to get the answer. Indeed, at this point we can determine only c is a reasonable answer. 148.c. The solubility of silver chloride in water is .000089 grams per 100 mL. Therefore the amount of water needed to dissolve a practical amount of silver chloride and produce neasurable enthalpy changes would be far too great. Perhaps this answer could be reached through elimination. Choice A says silver chloride is not an ionic compound and it is not soluble in water. Silver chloride is an ionic compound and, though only slightly, it is soluble in water. The enthalpy of solution of silver chloride is only 61 kHJ/ mole, which is just over twice as much as that of ammonium nitrate, so choice B is not the answr. Finally, choice D can be ruled out because as table 1 indicates, the dissolution of silver chloride is endothermic 149.b. In order to calculate the change in the freezing point, we use the formula ∆T = K/m, where ∆T is the change in the freezing point, Kb is the freezing point depression constant for water and m is the molality of the solution. A saturated aqueous solution of sodium floride will have 0.3 grams of LiF per 100 mL of water, which translates to 3 grams per liter. Since the definition of m is moles of solute per kilogram of solvent, we need to convert 3 grams of LiF to moles. The molar mass of Lif is 25.941. This can be rounded to 27 grams. In order to calculate moles of LiF, we take 3 grams / 27 grams per mole which is equivalent to 1/9 moles. We can round 1/9 moles of LiF to 0.1 moles ofs LiF. In order to calculate molality, we can assume the mass of water is 1 kg/L. Therefore, the molality is 0.1. Therefore, ∆T =0.1 x - 1.86 K/m = 0.186 K. Finally, since the freezing point of pure water is 0 degrees Celsius and the magnitude of one degree Celsius and one degree Kelvin are equivalent, the presence of LiF in solution will lower the freezing point to •0.186 degrees Celsius 150.a. According to the Gibbs free energy equation,
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is 2(C3 H 2 NO2 ) , i.e. C6 H 4 ( NO2 ) 2 . The product is m – dinitrobenzene. 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 emthalpy of the reaction would be
3 × −119 = −357 kJ mol−1
+ 3H 2 Actual enthalpy of the reaction can be evaluated as below. ∆ H (Reaction) = ∆ H ºf (Product) − ∆ H ºf (Reactants)
= − 156 − (49 + 0)
= −205 kJ mol−1 ∴ Resonance energy = ∆ H Exp − ∆ Cal = −357 − (−205) = 152kJ mol−1
146.c. Table 1 indicated the solubility of sodium chloride as 35.7 grams per 100 mL. (.1L). The molar mass of sodium chloride is 58.4527 grams per mole. To speed up the math, we can estimate 35.7 grams/58.425 grams per mole to be roughly 35/60 which is the same as 7/12, or, in decimal form, approximately. 6. Therefore, 35.7 grams is approximately. 6. moles of sodium chlordie. Since. 6 moles of sodium chloride is soluble in 100 mL of water at 0 degrees Celsius, 6 moles of sodium chloride must be soluble in 1L of water at the same temperature. It should be noted that, given the degree of difference between the answer choices, our estimation was appropriate 147.c The molar mass of ammonium nitrate is approximately 80 grams/mole. We can round 61.11 grams of ammonium nitrate to 60 grams of ammonium nitrate. Therefore, we have 60/80 or 75 moles of ammonium
∆G = ∆H − T ∆S, the two factors which
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determine the spontaneity of a chemical reaction are entropy changes (∆S) and enthalpy changes
(∆H ) . Since this reaction includes a solid reactant
and a gaseous reactant being cnverted into a single, solid product, entropy has decreased and ∆S is,
therefore, negative. Since ∆H is negative www.aieeepage.com (spontaneous), temperature becomes important. At
T∆S will be low and will not significantly affect ∆G . However, at high temperatures, T∆S will be high and may make ∆G low temperatures,
positive and, therefore nonsportaneous.
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