Paper - I : Physics - 1 to 50 Solutions 1.a.
2.d.
The power dissipated by a resistor is equal to i2 R. Alternate forms of the equation which may be easier in certain cases are P = V2/R and P=IV. Here, since we know V and some information pertaining to R, it is easiest to use P=V2/R. P= V2/R=(120) (120)/R Substituting definition of resistivity, R=p1/A P=(120) (120) A/pI where A is the cross-sectional area of the wire, Πr 2 . Plugging in values of A and I, with all distances in meters: P= (120) 120) (3.14) (0.001( (0.001)/0.5p P = (12)(12) (3) x 10-4/p P=288(3) x 10-4/p P = 900 x 10-4/p P = 0.09/p The closest answer is 0.1/pp. The toast pops up when the magnetic force of the electromagnent pulling down on the tray becomes smaller in magnitude than the spring force pushing up. Since we want the toast to pop up sooner, we are looking for an answer that either decreases the magnetic force or increases the spring force. Answer (A) is incorect because decrease the spring force (F = kx). Answer (B) is incorrect because increasing the strength of the elecromagnent would increase its magnetic pull and keep the toast tray down longer. Answer (C) is incorrect because decreaseing the separation of the capacitor plates will increase the capacitance
voltage of 120 V. The rms value of voltage is equal to the peak voltage divided by the square root of 2.
Vrms =
Vpeak 2
;
Vpeak = Vrms 2
The correct answer is 120 2 V, answer choice (a) 5.c.
The current will be smallest when the resistance is largest. We therefore want to rank the resistor combinations in order of increasing resistance. The rule for combining resistors is that resistors in series add (R eq = R 1 + R 2 ),
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while resistors in parallel combine according to the equation
(1 / R eq = 1 / R 1 + 1 / R 2 ). The important implication of
this rule is that two identical resistors in series have a total resistance of 2R, while two identical resistors in parallel have a total resistance of only 1/2R. The correct order is is therefore #3 (total resistance R/3), # 4 (total resistance R), #2 (total resistance 3R/2), #1 (total resistance 3R).
www.aieeepage.com 6.a. Suppose that the frequency of the source is f.
(C = ε 0 A / d ). A larger capacitance means that the
3.c.
current through the electromagnet will not drop off as quickly (time constant = RC, so larger C means longer time to charge). Answer (D) is correct because increasing the resistance of the heating element actually decreases the power dissipated across the fixed voltage of 120 V. P = V2/R. Here, V remains fixed at 120V. so if R gets larger, the power gets smaller. With less power dissipated, the toast will be lighter. (Note : A real toaster uses a high resistance nichrome wire, and it is the resistor on the top branch that is adjusted to change the darkness of the toast by altering the duration of the toasting process). We know that the ratio of the energy inputs to compress the springs of the two toasters is 1 : 6. This is the same as the ratio of the energies stored in the springs :
1 k1x12 1 2 = 1 k 2 x 22 6 2
Then from v= f λ The wavelengths of the longitudinal waves generated, in the two sections, should be such that, their phase difference, starting from the two ends and reaching the middle one should be (2n + 1)π where n ∈1 . Now, λH =
and λ0 =
310ms −1 f
A path difference of diffrerence of
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λ
x 2 = 2x1 :
corresponds to a phase
76 2π . So, a length of path 2 cm
would effect a phase of
Substituting
1100ms −1 f
38 cm(2π ) radian λ
∴ The phase difference between the waves reaching the middle diaphragm
⇒
1 2π 1 2π 1 1 − = πl − λ0 2 λH 2 λ0 λH
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1 1 k 1x12 k 1x12 k1 1 2 = 2 = = 1 1 k 2 x 22 k 2 ( 2 x 1 ) 2 4k 2 6 2 2 k1 4 2 = = . k2 6 3 4.a.
The passage tells us that the wall outlet supplies an rms
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f f = π [76cm] − −1 −1 310 ms 1100 ms
For f to be minimum the phase difference should be a ‘least’ odd multiple of π 1 1 ∴π (76cm) − f =π 2 2 310 × 10 cm / s 1100 × 10 cm / s
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11.c.
∴ f = 568 Hz 7.b.
Q =2 q
⇒ q=
Q 2
i.e., the charge Q is divided equally into two parts and hence the force of repulsion between both of the parts is the maximum. 8.b.
Sincef =
1 T 2l µ
we have l2=kT (where k = constant) or 2 log l = log k + log T or log l =
1 1 log T + log k 2 2
which is of the form y = mx + C
∴ 9.b.
Slope m =
1 2
PAGE p
(2) (3) 2+3
6 = 1.5 A 4
⇒ I 2 = 0.9 A Potential drop across 2.8 Ω is V = (2.8) (1.5) V = 4.2 V Absolute temperatures of the black body corresponding to curve P and Q are in the inverse
i.e.,
TP 1987 = TQ 2980
x
x
x
x
x
E x x B
AP TP 16 = = AQ TQ 81
E B
12.a.
For a fixed mass of gas, if two of the three thermodynamic variables (P, V and T) coincide corresponding to any two states, the third automatically coincides, since they are bound by the equation of state. Since internal energy is a state function, so the return of the gas to the initial state implies no change in its internal energy.
13.a.
hv=K.E. (T) + Work function (W) hv= T + W ⇒ 4.25 eV = TA + WA (for Metal A) ⇒ 4.70 eV = TB + WB (for Metal B) ⇒ Since TB = (TA-1.5) eV also
λ=
⇒
λ =
⇒
A
h p
h 2mT
www.aieeepage.com λ T
4
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x
PAGE
λm (Wein’s displacement law).
Area under curves represent the total power radiated by a body and is proportional to the fourth power of absolute temperature (|Stefan’s law)
∴
x
figure. In such a situation, the velocity is v =
3 ⇒ I 2 = (1.5) 5
ratio of
x
m
1/ 2 Current in 2 Ω resistor = 12 = 1.5 1/ 2 +1/ 3
10.d.
x
Fe = eE (downwards) and if both are equal in magnitude then even the proton will suffer no change in its velocity and will continue to move along the dotted line as shown in
⇒ Rtotal = 4Ω ⇒ 1=
x
www.aieeepage.com F = evB (upwards) and
When steady state is reached no current will flow in the branch having the capacitor. So
Rtotal = 2.8 +
No change in velocity imnplies no acceleration i.e. no net force is acting on the proton, even under the joint influence of electric and magnetic field. This thing is possible under the following situations. Situation A: E = 0, B = 0, i.e., no field exists in the region. Situation B: E = 0, i.e. no electrostatic force. B ≠ 0 , but the charge particle enters parallel to the field, so that net force equals to zero. Situation C: E ≠ 0 , i.e., the charged particle proton must experience an electrostatic force eE and hence must accelerate. Situation D: E ≠ 0 , B ≠ 0 and both shown in figure Because in such a situation
λB
Since
=
∵
p2 = T = K .E . 2m
B
TA
1 λ A = λB 2
⇒ TA = 4TB
⇒ TB = TA − 1.50 gives
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⇒ TB = 4TB − 1.5 ⇒
V RT = P0 3 − V V0
or
TB = 0.5eV
⇒ TA = 2eV
P0 V2 T = 3V − R V0
⇒ WA = 2.25eV ⇒ WB = 4.20 eV 14.d.
For T to be maximum
Power developed = (heat capacity) rate of temperature rise 0 cal C = (180 g ) 0.1 0 0.5 g − C s
=9 cal/s P = τω
180 × 2π 37.8 = τ 60
∴ 15.c.
37.8 J/s
dT =o dV
3 dT P0 2V = 3 − = 0, or V = V0 2 dV R V0
∴
d 2T −2 P0 = = −ve, hence maximum. dV 2 RV0
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or τ =2 N-m
Consider light to be incident at near normal incidence. We wish to cause destructive interference between rays r1 and r2 so that maximum energy passes into the glass. A phase change of λ / 2 occurs in each ray for at both the upper and lower surfaces of the MgF2 film the light is reflected by a medium of greater index of refraction. When striking a medium of lower index of refraction, the light is reflected with no phase change. Since in this porblem both rays 1 and 2 experience the same phase shift, no net change of phase is introduced by these two reflections. Hence, the only way a phase is change can occur, is if the two rays travel through different optical path length is product of the geometric path difference a ray travels through different media and the refractive index of the medium in which it is travelling. For destructive interference the two rays must be out of phase by an odd number of half wavelengths. Hence, the optical path difference needed for destructive interference is:
∴
Tmax =
P0 R
3V0 3V0 2 1 9V0 P0 3 − × = 4R 2 2 V0
the spheres at a distance r from the centre www.aieeepage.com
2 µ d = (2n + 1)
λ , n = 0,1, 2.... 2
16.b.
λ / 2 λ 350 ×10−9 = = d= 2µ 4µ 4 ×1.38
= 100nm = 1× 10 −1 m The equation of curve is,
P − 2 P0 P0 − 2 P0 = V − V0 2V0 − V0
−P P = (V − V0 ) 0 + 2 P0 V0
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Electric field at some point P in the region between (according to Gauss’ las) is given by:
∫E
P
.ds =
q ε0
+q
or EP .4π r 2 =
or EP =
q ε0
O
+q +q a P P’
b
c
q 4πε 0 r 2
Similarly, electric field at point P outside the hollow sphere is
EP ' =
q 4πε 0 r 2 ;
Vinner =
1 4πε 0
q q q a − b + c
Vinner =
1 4πε 0
q q q c − c + c
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Note that 2 µ d is the total optical path length that the rays traverse when n=0.
∴
17.a.
∴
Vab = Vinner − Vouter ;
q 1 1 − 4πε 0 a b
www.aieeepage.com 18.b. ∵ wetting is complete, angle of contact θ=0 Rise of liquid
h=
2T Cos θ rpg
R
O
2T or h = ∞ cos θ ∵ = cons tan t rpg
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ie.,
A=amplitude]
nh cos θ = ⇒ cosθ = n h cos θ
or
⇒ θ = cos −1 (n) 19.a.
Velocity vector is perpendicular to magnetic field. Therefore, path of the particle is a circle of radius
r=
F block
∠CPQ = ∠CQP
or
90 − α = 90 − β
∴
α−β
P V β
= 2r cos (90 − α )
= 2r sin α 2mv0 PQ = sin α Bq
x
x
x
x
x
x
x
x
x
x
x
x
Now, the maximum velocity
∴ vmax =
Arc PSQ = (2π − 2α )r
2mv0 (π − α ) Bq
t PSQ =
PSQ 2m(π − α ) = v0 Bq
Volume of body =
g gT = 3.12 ms −1 or ω 2π
23.b. If there is no external electric field, then the charge www.aieeepage.com given to a conducting sphere gets uniformly
∴ ∠PCQ =2α
∴
5 =1×10-3 m 3 3 5×10
24.b.
∴ Force of upthrust = wt. of water displaced = 1×10−3 m3 × 1000 kg / m3 = 1 kg ∴ Reading of balance A = 10 kg + upthrust
distributed over its surface. Therefore, statement (1) is correct. If an external electric field exists, then the charge gets distributed over the surface of the sphere in such a way that the electric field inside the sphere can become equal to zero. Hence, distribution of the charge on the surface of sphere will be nonuniform. Therefore, (2) is correct. Obvisously (3) is wrong. Since electric field inside the conducting sphere is equal to zero, therefore, potential difference between two points in the sphere is equal to zero. It means the potential is same at every point of the sphere. Therefore, (4) is correct. The gravitational field intensity at the centre of the solid sphere due to the ring is given by
E=
GM 1d (d + h 2 )3/ 2 2
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= 10 + 1 = 11 kg and reading of balance B = 5 kg - upthrust = 5 - 1 = 4 kg.
21.a.
22.b.
At
u = f ,v = ∞
At
u=0 v=0
R
(i.e., object is at pole)
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(image is also at pole) Satisfying these two conditions only option (a) is correct. The situation, when the block is just below the mean position is , the restoring forces acting on the piston, causes a normal reaction F to act on the block. For the block to separate.
=
F ≥ mg
i.e., mω 2 A ≥ mg [where ω = ang.
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vmax at that instant =
ωA
∠PCR =∠RCQ =α
=
mg
PAGE 90°
Q
PQ = 2 PR
Further
90°
R
Piston
∴
∴
g g gT 2 = or A ≥ ω 2 2π 2 4π 2 T
mv0 , v = v0, as the speed of the particle Bq
does not change in the magnetic field. Centre of the circle is C. x v0 x x CP = CQ
20.c.
A≥
and
25.a.
R
2
2R = d
θ
O
GM 1 (2 2 R) GM 1 (2 2 R) = (8R 2 + R 2 )3/ 2 27 R 3
The toral mass of the solid sphere can be assumed to be concentrated at its centre 0. ∴ Force acting between M1 and M2 is F=EM2
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26.a.
27.a.
To locate the instantaneous centre of rotation of a body, at least two points on it should be choosen, whose direction of velocity is known. Lines should be drawn perpendicular to the direction of velocity from those points. The point of intersection gives the instantaneous centre of rotation. In this case, points A and B move respectively along X and Y axes. So lines should be drawn from point A and B, perpendicular to X and Y axis. They intersect in the first quadrant. Chromatic aberration is a defect due to which the images of different colurs of a point object are formed at different places. Since refractive index of a matterial depends not only upon that material but also upon the colour of incident rays, therefore, when a light ray is refracted through a material, then the images of different colours may be formed at different place only when refraction takes place. Hence, the mirror cannot give rise to chromatic aberration. It means statement (a) and (2) are correct. Spherical aberration is a defect due to curvature of the refracting or reflecting surface. It can be produced by mirror as well as lens. Hence (3) and (4) are also correct.
dq1 dq2 ≠ dt dt
∴ 30.b.
∴
statement (1) is wrong If V1 and V2 be the velocities of A and B respectively after collision, then m(10) = mv1+mv2 v2-v1 = 10 (i) ⇒ If e be the minimum coefficient of restitution, then
v2 − v1 ≥ e(10) ⇒
v2 − v1 ≥ 10 e v2 ≥ 5(1 + e)
Adding (i) and (ii)
⇒
v22 = 64
Substituting for v2 from (iv) in (iii) ...(iv)
64 ≥ 25(1 + e)2
2
1 1 hR v = π r 2h = π h 3 3 H
∴
dV π R 2 dh = 2 ( h) 2 = K ( say ) dt H dt
dh H 2 K = dt π R 2 h 2
2 3 dh 3H K or = 2 2 4 dt 4π R h
or y=
C x2
31.c.
dV = k dt
R
q1 = ( EC )[1 − e − t / τ c ] and
⇒
e≤
8 5
3 5
When the ball is pressed down, its c.g. gets lowered hence, its P.E. decreases. However, the water displaced by the ball rises and hence its P.E. increases. The net increase in P.E. will be equal to the work done. 1 2
, so half its water
4 2 3 3 3R π R ρ Rg − π R ρ g 3 3 8
R
⇒ log y = log C C.G − 2 log x
4 2 3 26π R ρ g = π R4 ρ g 2 − = 3 3×8 8
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τ c1 = τ c2 = 2CR = τ c ( say )
29.a.
(1 + e) ≤
volume will remain immersed initially. Increase in P.E. of
H
Which is of the form Y = -2x +C Slope = -2
⇒
Since, the sp. gravity of the ball is
r
∴
1 2 mv2 2
m(10) (3.2) =
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∴
(iii)
By conservation of energy, for ball B
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28.b.
(ii)
Decrease in P.E. of ball =
2 4 3 ρ 4 πR Rg = π R ρ g 2 3 3 ∴ Net work done = net increase in P.E.
q2 = (2 EC )[1 − e − t / τ c ] q1 1 = ( at any time) q2 2
www.aieeepage.com 5 26 2 = − π R ρg = π R ρg 24
∴ The ratio of steady state charges is also 1:2
dq1 EC − t /τ c = e dt τc and
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dq2 2 EC − t / τ c = e dt τc
3
4
4
12
1 1 1 = ( µ − 1) + f R1 R2
32.a.
or
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1 1 2 = × or r = f f 2 R 5 of 24
1 1 1 1 1 2 = + = + = F f1 f 2 f f f
∴
F = ( f / 2) ∴ ∴ Focal length of the combination f R = = = half the radius of curvature 2 2 33.c.
35.a.
V
N
E
1 F = 2i − j 2 m
v be the vlocity at time t, then from v = u + at
1 ⇒ v = (5 + 2t )i + 2- t j 2 When motion is purely along X-axis, Y component of velocity should be zero
V
1 2− t = 0 ; 2
PAGE ∴
Train P
Train Q
)
2 − 1 = ax (2 − 2)
Given u = 5 i + 2 j ; a = If
ω0
ω0
(
ax 2 = ax 2 2 +1
∴ R=
s = ut +
Using
Evidently, the track of tain P as well as Q rotates with earth, due to its spin motion, (say with an
( r2 − r1 )
2
angular velocity ω 0 ). If R be the equatorial radius,
⇒ t=4s 1 a t2 2
1 = ut + at 2 2
www.aieeepage.com If ( x iˆ + yiˆ) be the position vector after time t,
then, the linear speeds of trains P and Q, relative to
their respective tracks will be (v+ ω 0 R) and (v-
ω 0 R) If T1 and T2 be their respective forces (centripetal) offered on the ground, Then
T1 =
m m (v + ω 0 R) 2 and T2 = (v − ω 0 R ) 2 R R
Evidently T1
then
1 ( x iˆ + y iˆ) − (2 iˆ + 3 iˆ) = (5 iˆ + 2 ˆj ) 4 + iˆ − 4
ˆj (16)
36 iˆ + 4 ˆj x iˆ + yiˆ = 38 iˆ + 7 ˆj
∴ 36.c.
> T2
34.a. 12
A
ax
12 1 1 ax D
B
A
B
D
C
ax a
2x
ax (i)
C
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(ii)
E= ∫
+π /2
dE , dq =
−π / 2
Current entering at A will divide into I1 along AC and I2 each in ABC and ADC since the resistances ABC and ADC are equal. Now, the resistance AB=resistance AD. Hence, the potentials of B and D are equal. Hence, there will be no current in BD. The circuit is equivalent to one shown in figure (ii). If R is the equilent resistance.
E= ∫
π /2
−π / 2
Q Q Qdθ dl = Rdθ = πR πR π
1 dq cos θ 4πε 0 R 2 1 Q cos θ dθ 1 2Q = 2 4πε 0 π R 2 πR 0
www.aieeepage.com 4πε
1 1 1 1 = + + R 2ax 2ax 2ax =
PAGE -
1 1 1 2 +1 1+ = ax 2 ax 2
=∫
π /2
−π / 2
37.b.
The component of velocity of the ball normal to the plane is -20 m/s. (before striking) Since the collision is elastic, immediately after the collision the component of velocity, of the ball normal to the plane bcomes +20 m/s.
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∴ Initial velocity of the ball immediately after it
or
u = 10iˆ + 30 ˆj + 2 − kˆ
∴ cos θ =
strikes the plane is
∴ Using v = u + at Final velocity after t= 2 sec will be
v = (10iˆ + 30 ˆj + 20kˆ ) − 10kˆ(2)
40.b.
(∵ a = − gkˆ )
Let 2a be the side of the traingle and b the length AE.
AH GH = AE EC AH ∴ GH= EC Ae (b − vt ) .a b
=
x
x
x
x
x
x
x
x A x
x
x
x
x
x
x
x
x
xF x
x Gx H
x vt
x
C
E
Let I1 and I2 be currents through B1 and B2.
∴
R 2 + 9.87
I 2 > I1
u2 = 20 (−iˆ + ˆj )
s1 = 20 ˆj + 10 (iˆ + ˆj )t − 5t 2 ˆj and
s2 = 20 iˆ + 20 (-iˆ + ˆj )t − 5t 2 ˆj
∴ Their relative position vector will be s '1 − s '2 = (−20 + 30t ) iˆ + (20 − 10t ) ˆj
Thus, I - t graph is a straight line with negative slope and positive intercept
∴ s '1 − s '2 = 10 (3t − 2)2 + (2 − t )2
ux and u y be the velocity of projection,
= 10 (9t 2 + 4 − 12t + 4 + t 2 − 4t )
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along the horizontal and vertical then
uy
i.e., s '1 − s '2
2
= 100 (10t 2 − 16t+8
For maxmum or minimum separation
(i)
ux
d 2 s '1 − s '2 = 0 dt
At a height equal to three - fourth that of maximum height, the vertical component of velocity ( v y ) will
= t = 4/5 s www.aieeepage.com
be
= 100 [20t - 16] = 0
d 2 s '1 − s '2 = 2000 = +ve; 2 dt
u y2 3 u v = u − 2g ⇒ vy = y 2g 4 2 2 y
tan (900 − θ ) =
From (i) and (ii)
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R 2 + 40
Their position vectors after time ‘t’ will be
I = k1 − k2t
Given
R 2 + (ω L) 2
s1 = 20 ˆj; s2 = 20 iˆ
e 2 Bv a I= = a − vt R R b
2 y
=
=
Bulb B2 will be brighter than B1. As frequency increases, XC decreases while XL increases, so I2 becomes less than I1 . Hence brightness of B1 will increases and that of B2 will decrease. Initial velocity of P and Q are :
∴ Induced current
tan θ =
R 2 + X L2
R 2 + (1/ ω C ) 2
Initial position vectors of P and Q are :
a = 2bv a − vt b
Let
R 2 + X C2
I2 = I1
So
and
Induced e.m.f. e = Bv(FG)
39.d.
I1 × R 2 + X C2 = 220
u1 = 10 (iˆ + ˆj )
a FG=2GH=2 a- vt b
or
2
1 3
www.aieeepage.com 41.c.
a = a − vt b ∴
=
( 2 ) +1 2
PAGE
x
D
1
Then
= 10iˆ + 30 ˆj 38.b.
tan θ = 2
vy ux
=
uy 2u x
Hence the separation is minimum (ii)
Putting t =
4 s in equation (i ) 5
tan θ = 2cot θ
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16 4 s '1 − s '2 = 10 10 − 16 + 8 25 5 = 20 42.b.
2 m 5
Any transition causing a photon to be emitted in the Balmner series must end at n=2. This must be followed by the transition from n=2 to n = 1, emitting a photon of energy 10.2 eV, which corresponds to a wavelength of about 122 nm. This belongs to the Lyman series.
49.b.
which both forces act is the same, then the absolute value of the work done by the normal force is greater than the absolute value of the work done by gravity. Answer choice (B) is not true because in this case the normal force is NOT perpendicular to the direction of motion. Answer choices (C) and (D) follow from the misconception that just because an object is moving down, the net force is down. This is not ture. The direction of the net force is the sae as the direction of the change in velocity, not of the velocity itself. A pulley system with one moveable pulley has a mechanical advantage of 2. (Note that fixed pulleys do not increase the mechanical advantage.) Therefore, if the weight of the load to be lifted were W, an applied force of only W/ 2 would be necessary in an ideal, friction-free system:
43.c. 44.b.
Using RHSR v × B is along east
MA =
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But F = −e (v × B ) [ ∴ electrons carry -ve charge]
2=
∴ Direction of F is opposite to that of (v × B ) . 45.b 46.c.
47.b.
W W →x= x 2
Remember that if the load is to move at a constant speed, all the forces on the laod must balance. Because any real system is NOT friction-free, some energy will be converted to heat. When the load is being lowered, this energy loss would subtract from the gain in kinetic energy experienced by the load. In other owrds, friction is doing negative work working in conjunction with the applied force to hold the load back against gravity. The applied force would therefore be less than W/2. When the load is being raised, the energy lost to friction is again subtracting from the gain in kinetic energy of the load, holding the load back. But this time, this effect is working in conjunction with gravity and in the opposite direction as the applied force. To counterbalance the effect of friction and raise the load at a constant speed, the applied force would therefore need to be greater than W/2. Comparing these two situations, wer see that the applied force necessary when the load is being lowered is less than the applied force necessary when the load is being raised. The pulley on the flagpole is a FIXED pulley. According to the passage, fixed pulleys do not contribute to mechanical advantage, but merely redirect the rope. Therefore, this system does not change the amount of force needed to lift the object. The applied force is the same as the force on the load, and the mechanical advantage is 1.
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The upper limit on mechanical advantage will be the mechanical advantage of a friction free system. In this case, a free body diagram of the boxed portion of the system reveals that six rope segments actually support the load. Therefore, each segment need only support 1/6 the weight of the load. The tension in the rope created by the applied force is one sixth the weight of the load and the upper limit on mechanical advantage is therefore 6. Answer choice (A) is simply a count of the moveable pulleys, but it must be remembered that each moveable pulley gives a factor of TWO in mechanical advantage. Choice (D) arises from a rope segment count that includes the segment being pulled. However, this segment does not support the load, as do the other six. It is therefore not included in calculations of mechanical advantage : If 20% of the energy is lost, then 80% of it is retained as useful work in lifting the load. The efficiency can then be calculated: for each unit of work input, 0.8 units of work are done on the load:
efficiency
work output 0.8 = = 0.8 work input 1
force applied =
50.b.
PAGE
The efficiency equation can be re-arranged as: (force applied) (distance of applied force) (efficience) - (force on load) (distance load travels)
48.a.
force on load applied force
(1000 N) 1 6 (0.8)
In elevator problems, it is usually critical to determine the direction of the acceleration and the direction of the net force and then relate this information to the relative sizes of the forces in a free-body diagram. If the elevator isw moving down and slowing down, then the direction of the net acceleration is up. The direction of acceleration is the same as the direction of the net force. Therefore, the net forece also points upward. The only two forces that act on the person are gravity and the normal force. If the net force is up, then the normal force must be greater than the gravity force. Since work is force times distance, and the distance over
PAGE -
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Paper - I I : Mathematics - 51 to 100 Solutions 51.a.
52.a.
He will be left with no money at the of end of the tenth round I and only I he gets a tail on each of them. The probability of this = q10 (q = 1 − p) [Note that the player cannot drained out of his money before 10th round] Again he cannot be drained at the 13th round I A gets at least 2 heads in the firist 10 rounds. To finish at the 14th round, A must get exactly 2 heads in the first 10 rounds and a tail on all the rounds from 11th to 14th. This has a probablity 10
C 2 p 2 q12 . Therefore required probability =
= q10 (1 + 10pq + 45p2 q 2 ) Here, P be (x, y, z) then,
Thus, we have
dy = (− cot x + cosec x )dx dx
⇒ log y = − log sin x + log tan
x c c 2 = ⇒y= = x sin x 1 + cos x 2 cos 2 2 c tan
PAGE we get y =
e 1− cos x
www.aieeepage.com ⇒ x = 2 sin
x = r sin θ. cos φ, y = r sin θ sin φ, z = r cos θa
−1
⇒ 1 = r sin θ. cos φ, 2 = r sin φ, z = r cos θ
⇒ 12 + 22 + 32
= r 2 sin 2 θcos 2φ + r 2 sin 2 θ sin 2 φ + r 2 cos 2 θ
55.c.
∵ f ' ( x) =
C 2y
1 1 1 x = = sec 2 1 + cos x 2 cos 2 ( x / 2) 2 2
Integrating both sides with respect to x, we have x f ( x) = tan + c 2
P(1, 2, 3)
θ O
∴ f (0) = 0 + c = 3 then
φ ∴
M
= r 2 sin 2 θ (cos 2 φ + sin 2 φ) + r 2 cos 2 θ
π 22 11 32 = 3+ = 3+ = ≈ 4.57 2 14 7 7 ∵ 3.78 < 4 < 4.57
∴ from (i), we have 1
, sin θ sin φ =
PAGE
2
, cos θ =
3
14 14 14 (neglecting -ve sign assuming acute angles) sin θ sin φ 2 = sin θ cos φ 1 and
⇒ tan φ = 2 54.c.
Solving for
π π π < f < 3+ 4 2 2 It cal also be checked that other options do not satisfy the conditions.
Hence 3 +
= 56.a. The given curve is y = be www.aieeepage.com cos θ 3
tan θ =
and ⇒ tan θ =
sin θ
5 3
dy , we obtain dx
5
Let us consider a point Then
x/a
( x1 , y1 ) on the curve.
y1 = be x1 / a
dIferentiating the curve
y = be x / a with respect
to we get
dy −2 y cot x ± 4 y cot x + 4 y = 2 dx 2
PAGE -
π 22 11 53 = 3+ = 3+ = ≈ 3.78 7× 4 14 14 4
and 3 +
⇒ r = ± 14
∴
x f ( x) = tan + 3 2
π π f = tan + 3 = 4 2 4
Now, 3 +
= r 2 sin 2 θ + r 2 cos 2 θ = r 2
sin θ cos φ = +
x + log c 2
dy Solving y = −(cot x + cosec x)dx,
q10 +10 C1 pq11 +10 C 2 p 2 q12 53.a.
= y ( − cot x ± cosec x)
2
2
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I conic be an ellipse, then SO + S’O = 2a and SS’ = 2ae
dy 1 = be x / a . dx a ∴
b dy = e x1 / a dx ( x1 , y1 ) a
I conic be a hyperbola, then S' O - SO = 2a and SS’ = 2ae
dx = y 1 Thus, the length of subtangent dy ( x1 , y1 = y1
59.a.
PAGE
57.a.
tan
g2 + f 2 −c
θ CA = = 2 PA
S1
g2 + f 2 −c
⇒ θ = 2 tan −1
S1
www.aieeepage.com Also,
1 1 (be x1 / a ) 2 = y 2 [using (i)] a a
1 - tan 2 θ/2
cos θ =
Therefore, subnormal varies as the square of ordinate. Case I : I n is even, say n = 2k. Then limit is
1 + tan 2 θ/2
cot θ =
−6k + (−1) −6 k + 1 lim = lim 2 k k →∞ 8k − ( −1) k →∞ 8k − 1 2k
=
(x1 , y1 )
1 → 0, where k → ∞} k
S1 + ( g 2 + f 2 − c )
n = 2k + 1. Then limit is
2 S1 g + f 2 −c
60.b.
Let the sides of a triangle be a, ar, ar 2
∵ ar 2 is the greated side (r > 1) ∴ a + ar > ar 2 ∵ r 2 − r −1 < 0 ⇒
1− 5 1+ 5
1 1 www.aieeepage.com Also r < (6 + 2 5 ) = (3 + 4 2
I two foci be S (5, 12) and S’(24, 7) and it passes through origin O.
PAGE -
2
C(-g, - f)
2
24 + 144 = 13; S ' O = 576 + 49 = 25
and SS’ = 386
S1 − c + g 2 + f 2
1+ 5 2 ∴ (1) is correct
−3n + (−1) −3 = , n even or odd ∴ lim n n →∞ 4n − ( −1) 4
Then So =
S1 + c − g 2 − f 2
⇒1< r <
n
58.b.
=
B
PAGE
2 k = −3 = lim k →∞ 5 4 8+ k
g2 + f 2 −c
S1 − ( g 2 + f 2 − c )
θ/2 θ/2
P
−3(2k + 1) − (−1) 2 k +1 −6k − 3 + 1 = lim lim + 2 k 1 k →∞ 4(2k + 1) − ( −1) k →∞ 8k + 4 + 1 −6 −
S1
θ = 2
A
1 k = −6 = −3 = lim k →∞ 1 8 4 8− k
Case II : n is odd, say
and cot
1 1 − tan 2 θ/2 = ≠ tan θ 2 tan θ/2
−6k +
{as
θ 2
From right angled triangle PAC.
dy again, length of subnormal = y1 dx x1 y1
=
let PA and PB be the tangent from P( x1 , y1 ) to the given circle with centre C(-g, -f) such that ∠APB = θ then ∠APC = ∠APC = ∠CPB =
⇒ Subtangent is of constant length a
be x1 / a a
SS' 386 = S' O + SO 12
∴e =
a a = be x1 / a . x1 / a = a (constant) [using (i) and (ii)] be x1 / a be
= be x1 / a
SS' 386 = S'O + SO 38
∴e =
r4 <
1 1 (14 + 6 5 ) = (7 + 3 5 ) 4 2
2 4 ∴ 1+ r − r < 1+
cos C =
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5 ) and
1 1 (3 + 5 ) − (7 + 3 5 ) = −1 − 5 < r 2 2
a2 + a2 r 2 − a2r 4 2a 2 r
=
1+ r2 − r 4 1 < 2r 2 10 of 24
π π ⇒C > 3 3 ∴ (4) is correct Also cos C < cos
cos B =
=
a 2 + a 2r 4 − a 2r 2 2a 2 r 2
sin 2 A + sin 2 B = 1 + cos C Note that
62.b.
P( Z ≤ m) = P{X ≤ m, Y ≤ m} = P{X ≤ m} P{Y ≤ m} But
=
1+ r 4 − r 2 2r 2
=
1 2 1 r + 2 − 1 2 r
P{Y ≤ m} = P{X ≤ m} = P(X = 0) + P(X = 1) + ........ + P(X = m)
= p + pq + pq 2 + ..... + pq m −1 =
2 1 1 1 r − + 1 > 2 2 2
∴ P(Z ≤ m) = (1 − q m ) 2 Now, {(Z = m) P ( Z ≤ m) − P ( Z ≤ m − 1)
π π ∴ cos B > cos ⇒ B < 3 3
= (1 − q m ) 2 − (1 − q m −1 ) 2
= 1 − 2q m + q 2m − (1 − 2q m −1 + q 2m − 2 )
π
PAGE
2 Also, a < ar < ar ⇒ A < B < C ⇒ A < B <
61. a.
sin 2 A + sin 2 B + sin 2 C
= 2pq m-1 − p(1 + q)q 2m − 2 Clearly, ∑ [2pq m −1 − p(1 + q)q 2m − 2 ] =
1 1 = (1 − cos 2 A) + (1 − cos 2 B ) + sin 2 C 2 2
m ≥1
63.b.
n
1 = 1 − {2 cos( A + B ). cos( A − B)} + sin 2 C 2
Now,
= 1 + cos C. cos(A - B) + 1 - cos 2C
But
n n +1
=
1 1 1+ n n
n
∴ sin 2 A + sin 2 B + sin 2 C ≤ 2 − cos 2 C + cos C
n
2
= 1+1+
{∵ the greatest value of cos (A - B) = 1}
= 2 − (cos2 C − cos C)
3
1 1 1 1 2 1 − + 1 − 1 − + ..... 2! n 3! n n n
1 1 9 1 = 2 − cos 2 C − cos + + = − cos C − 4 4 4 2 ∴ sin 2 A + sin 2 B + sin 2 C ≤
1 Clearly, 1 + > 2 n
2
n
1 1 1 1 Also 1 + < 2 + + 2 + 3 + ....∞ 2 n 2 2
9 4
Now, for positive quantities, AM ≥ GM
1 = 2 + 2 = 3, provided n > 3 1 1− 2
PAGE
sin 2 A + sin 2 B + sin 2 C 3 ≥ sin 2 A sin 2 B sin 2 C 3 ∴ from (2),
n
1 ∴ 2 < 1 + < 3 n
9 2 2 2 4 ≥ sin A + sin B + sin C ≥ (sin A sin B sin C ) 2 / 3 3 3 3 ≥ (sin A sin B sin C ) 2 / 3 4
2 1 1 www.aieeepage.com ⇒ < 1 + n n n
3 3 or sin A sin B sin C ≤ , i.e., 4 8
Also in ∆ ABC, sinA > 0, sinB > 0, sin C > 0 and from (1)
n
<
3 < 1, for n > 3 n
(n + 1) n
< 1 ⇒ (n + 1) n < n n +1 for n > 3, n n +1 which is satisfied only by the option (2) and (3) ∴
3 3 2
PAGE -
(n + 1) n
n +1
1 n(n + 1) 1 n(n + 1)(n + 2) 1 1 + .... 1 + n = 1 + n. n + 2! n + 3! n
= 2 − cos 2 C + cos C. cos (A - B)
∴
2p p(1 + q) − = 2 −1 = 1 1− q 1− q2
We note that eacch option is of the form
www.aieeepage.com (n + 1) > or < n
1 = 1 − (cos 2 A + cos 2 B) + sin 2 C 2
∴
p(1 − q m ) = 1− qm 1− q
64.a.
(i) xy > 0, 3x 2 − xy − 2 y 2 = −7 or (3 x + 2 y ) ( x − y ) = −7
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x and y being integers, we can trake 3x + 2y = 7 and x - y = - 1 which given x = 1, y = 2. I x and y are charged to -x, -y the equation remains the same so that x = -1, y = -2 is also a solution pair. It is easily seen that 3x + 2y = - 1 and x = y = 7 do not given integral solutions. (ii) xy = 0 will not lead to integral solutions. (iii) xy < 0. The equation is
67.c.
= ( m C 2 ) ( n C1 ) + ( m C1 ) ( n C 2 )
3x 2 + xy − 2 y 2 = −7 or (3x - 2y) (x + y) = - 7 3x - 2y = -7 and x + y = 1 leads to x = - 1, y = 2. As before x = 1, y = - 2 is also a solution set. The required complex numbers are 1 + 2i, 1 - 2i, -1 + 2i, -1-2i which form two conjugate pairs. 65.b.
Discriminant D = 4p 2 − 4(p 2 − 1) = 4 > 0
-2p 2
⇒ 16 − 8p + p 2 − 1 > 0 and 4 > p ⇒ (p − 3) (p − 5) > 0 and p < 4
68.b.
=
1 mn (m − 1 + n − 1) 2
=
1 mn (m + n − 2) 2
∆ = ( 13 + 3 ) [25 − 5 2 ]
+ 5 ( 15
5 + 5 5 − 15 − 5 65 )
− 50 3 − 50 5 + 30 2 + 10 5 www.aieeepage.com
= −25 3 − 35 2 − 65 5 − 5 6 + 10 5
-2p 2
26
+ 5 15 + 25 69.c.
= A real irrational number. Let f(x) = x tan x
∴ f ' ( x) = x sec2 x + tan x
⇒ p < −3 or p > −1 and p > -3 ⇒ p ∈ (-1, ∞) Further exactly one root lies in the interval (-2, 4) I D > 0 and f(-2) f(4) < 0
=
x + sin x cos x cos 2 x
>0
[∵ 0 < x < π/2]
∴ f (x) is increasing function.
⇒ ( p + 3)( p + 1)( p − 3)( p − 5) < 0
Now
⇒ p ∈ (−3,−1) ∪ (3, 5) Finally, 1 lies between the roots I D > 0 and f(1) < 0
x1 < x2
⇒ f ( x1 ) < f ( x2 )
⇒ 1− 2 p + p2 −1 < 0
∴ x1 tan x1 < x2 tan x 2
⇒ p ( p − 2) < 0 ⇒ 0 < p < 2 ⇒ p ∈ (0, 2)
⇒
x1 tan x2 tan x2 x1 < ⇒ > ; x2 tan x1 tan x1 x2
PAGE 70.d.
2
26
+ 5 15 + 25 − 15 5 − 25 13
⇒ (p + 3)( p + 1) > 0 and p > -3
| α -β | > 3p
10 )
= 25 13 + 25 3 − 65 2 − 5 6
⇒ (4 + 4 p + p 2 + 1) > 0 and 3 < p
66.b.
m(m − 1) n(n − 1) .n + m. 2 2
− 2 5 (5 15 + 25 − 3 10 − 65
⇒ p < 3 or p < 5 and p < 4 ⇒ p ∈ (−∞, 3) Again both the roots are greater than -2 I D ≥ 0, f( −2) > 0 and - 2 < -
=
PAGE
∵ Roots of the equation are real and distinct Now both the roots are less than 4 if D ≥ 0, f(4) > 0 and 4 > -
⇒ p>4 To get a triangle , we either take two points on AB and ne pt. on AC or one pt. on AB and two pts. on AC ∴ no. of triangles formed
Since f(x) is defined for +ve as well as -ve value
I α, β are the roots of x + px + 1 = 0
of x and f(x) ≥ 0 ∀ x ∈ D f
so that α + β = -p, α β = 1
Hence the graph of f(x) will lie on right as well as on left y-axis and above the x-axis. ∴ graph of f(x) lies in Ist and IInd quadrant.
⇒ (α - β) 2 > 3 p ⇒ (α + β) 2 − 4αβ > 3p 2
⇒ (− p) − 4.1 > 3p ⇒ ( p − 4) ( p + 1) > 0
xe www.aieeepage.com Let I = ∫ 71.a. 2
⇒ p − 3p - 4 > 0
⇒ p > 4, p > −1 or p < 4, p < −1 ⇒ p > 4 or p < -1 But p is not -ve [∵ If p is - ve, then 3p is not real]
x
1+ ex
dx ;
∴e x dx = 2t dt Also x = log(t 2 − 1) ;∴ I =
∫
= 2 log(t 2 − 1)dt ;
PAGE -
Put 1 + e x = t 2
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∫
log(t 2 − 1).2t dt t
∫
= 2 log(t 2 − 1) . 1 dt I
II
12 of 24
2t .t dt = 2 log(t 2 − 1).t − 2 t −1
∫
= 2t log(t 2 − 1) − 4 = 2t log(t 2 − 1) − 4
∫t
∫
2t 2
−1
⇒ cot A cot B cot C ≤
t 2 −1 +1 t 2 −1
74.d.
dt
72.b.
Area of the region =
1+ ex +1
7π π ∴ θ = nπ + (−1) n − = nπ + (-1) n 6 6
75.b.
a2
www.aieeepage.com (c iˆ + c ˆj + c kˆ). (iˆ + ˆj ) Also,
5/ 2
dx
=
Put x = 2a sin 2 θ ∴ dx = 4a sin θ cosθ dθ π/2
=
∫
2
2a cos θ. (2a sin θ) a
0
5/2
4a sin θ cos θ dθ
2
∫
5.3.1.1 π 5π 2 . = a 8.6.4.2 2 8
76.a.
Also area of circle = πa 2
c1iˆ + c2 ˆj + c3 kˆ). (iˆ + kˆ) 1 = 2 2
∵ In = =
5π 2 5 a : π a2 = = 5 : 8. 8 8 We know tht I A + B + C = π, then tan A + tan B + tan C = tan A tan B tan C ...(1) Since A.M. ≥ G.M. ∴
3
2
1 c = − (iˆ − 4 ˆj + kˆ). When c 3 = 1, c = iˆ + kˆ 3
0
73.a.
2
1 1 3c32 − 2c3 − 1 = 0 ⇒ c3 = − or 1 when c 3 = − 3 3
π/2
∴ reqd. ratio =
1
∴ c12 + c22 + c32 =| iˆ + ˆj |2 = 2
⇒ c1 + c2 = c2 + c3 = 1 ⇒ c1 = c3 and c 2 = 1 − c3 Hence from (1),
= (8) (4) sin 6 θ cos 2 θ dθ
= 32a 2 .
Let
c = c1iˆ + c 2 ˆj + c3 kˆ
1+ ex +1
0
7π 6
n Thus, θ = nπ + (-1)
+c
1 + e −1
2a − x x
∫
[∵ sin θ - 2 = 0 is not possible]
PAGE
1+ ex −1
x
1 2
7π π = sin − = sin 6 6
1 + e x −1 1 +c − 4. log x 2 + + e 1 1
2a
2 sin 2 θ - 3 sin θ - 2 = 0
⇒ sin θ = -
= 2 1 + e x .x − 4 1 + e x
∴ f ( x) = 1 + e x , g ( x) =
3 3
⇒ (2 sin θ + 1) (sinθ - 2) = 0
t −1 = 2t log (t 2 − 1) − 4t − 4 log +c t +1
= 2( x − 2) 1 + e x − 2 log
3 3
1
⇒ K≤
.t dt
1
∫
π/4
0
∫
π/4
0
tan n x dx
tan n - 2 x(sec 2 x − 1)dx
PAGE
tan A + tan B + tan C ≥ (tan A tan B tan C)1/ 3 3
=
∫
π/4
0
tan n - 2 x sec 2 x dx −
∫
π/4
n
tan n - 2 x dx
π/4
tan n −1 x In = − I n −2 (n − 1) 0
1 ⇒ (1) is correct n −1 Replace n by (n + 1)
www.aieeepage.com 1/3
⇒ tanA + tanB + tanC ≥ 3(tanA tanB tanC)
⇒ tanA tanB tanC ≥ 3(tanA tanB tanC)1/3 (By(1)) ⇒ ( tanA tanB tanC)3 ≥ 27(tanA tanB tanC)
⇒ ( tanA tanB tanC)2 ≥ 27 ⇒ tanA tanB tanC ≥ 3 3
PAGE -
⇒ I n + I n−2 =
Then I n +1 + I n −1 =
1 n
Now, in interval 0,
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⇒ (2) is correct
π tan n x < tan n − 2 x 4
13 of 24
∫
∴
π/4
0
tan n x dx <
∫
0
⇒ I n < I n−2 ⇒ I n < ⇒ 2I n <
π/4
tan n - 2 x dx
or
1 − I n (from (i)) n −1
1 (n − 1)
Again given f '( x) g '( x ) = c DIferentating both sides w.r.t., we get f '( x) g "( x) + g '( x) − f '''( x). g ( x) + g ''( x). f '( x ) + f ( x). g '''( x ) + 0
and I n + 2 < I n ⇒
= f '''( x). g ( x) + g '''( x). f ( x) + 0 [from (3)]
1 − In < In n +1
Now dividing both sides by F(x) = f(x) g(x) Then
1 < 2I n n +1 from (1) and (2) we get
F '''( x ) f '''( x) g '''( x) F ''' f ''' g ''' = + or = + F ( x) f ( x) g ( x) F f g
⇒
PAGE
1 1 < 2I n < ⇒ (3) is correct (n + 1) (n − 1)
77.c.
F " f " g " 2c = + + ⇒ (2) is correct F f g fg
R ∩ R'
79.a.
x 2 + y 2 = 25 represents a circle
The equation
with centre (0, 0) and radius 5 and the equation
axeax + e ax ; x ≤ 0 ⇒ f '( x) = [ f '( x)is continuous at x = 0] 2 1 + 2ax − 3 x ; x > 0
and
y=
2
4 2 x represents a parabola with vertex in 9
the Figure = {( x, y ) : −3 ≤ x ≤ 3, 0 ≤ y ≤ 5} . www.aieeepage.com Thus dom R ∩ R ' = [−3, 3] and range
2ae + a xe ; x ≤ 0 f "( x ) = [ f "( x ) is continuous at x = 0] x>0 2a − 6 x ; ax
ax
Now f "( x ) > 0 ⇒ 2a − 6 x > 0 I x > 0 or
2aeax + a 2 xeax > 0 if x ≤ 0
⇒ x<
2 a I x > 0 or x > − if x ≤ 0 3 a
R ∩ R ' = [0, 5] ⊃ [0, 4] (0, 5) (3, 4)
(−3, 4)
(3, 0)
(−3, 0)
2 a ⇒ f "( x ) > 0if - < x < a 3
Since (0, 0) ∈ R ∩ R ' and (0, 5) ∈ R ∩ R ' ∴ 0 is related to 0 as will as 5 Hence R ∩ R ' does not define a function. Also, any line parallel to y-axis contains infinite points in the region -3 < x < 3
2 a ∴ f '( x ) increases at x ∈ − , a 3 78.a.
⇒ (3) is correct
Given F (x) = f(x). g(x) DIferentiating both sides w.r.t x we get
F '( x ) = f '( x ) + g '( x). f ( x )
f ( x) g ( x) ⇒ F '( x) = f '( x) g '( x) + f '( x) g '( x)
So, there are infinitely many images for each value
PAGE
f g ⇒ F ' = c + ⇒ (1) is correct f ' g '
80.b.
of x ∈ ( −3, 3), thus R ∩ R ' cannot be a function. The equation of the line can be written in the slope form as
y=−
Again dIferentiating both sides sides w.r.t x we get
a1 / 3 1/ 3
b
x+
(− a
a 1/ 3
/ b1/ 3
)
a a i.e.y = mx + where m = www.aieeepage.com m b F "( x) = f ''( x).g ( x) + g ''( x). f ( x) + 2 f '( x). g '( x)
⇒ F "( x ) = f ''( x).g ( x) + g ''( x). f ( x) + 2c(2..) Dividing both sides by F ( x ) = f ( x ).g ( x )
{∵ f '( x ).g '( x ) = c}
So it touches the parabola y 2 = 4ax The equation of the line can also be written in the form
x=−
F '( x) f "( x ) g "( x) 2c = + + then F ( x) f ( x) g ( x) f ( x) g ( x)
PAGE -
1/ 3
1/ 3
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b1/ 3 a
1/ 3
y+
(− b
b
1/ 3
/ a1 / 3
) 14 of 24
i.e. x = my +
- b1/3 b where m = 1 / 3 m a
⇒ y = tan 4θ =
So it touches the parabola x2 = 4by also. 81.b.
2
2
2
=
2
= x tan α - 2xy tan α sin θ + y sin θ
2 × 2 x(1 − x 2 ) 2 2
(1 − x ) − 4 x
2
=
=
2 tan θ 2 1 − tan 2 θ 2 tan θ 1− 1 − tan 2 θ
2
4 x(1 − x 2 ) 1 − 6x2 + x4
or = (cos 2 θ sin 2 α + sin 2 θ − tan 2 α)
so that y is finite I x 4 − 6 x 2 + 1 ≠ 0
x 2 + 2(tan α sin θ)xy + cos 2 θ sin 2 αy 2 = 0 Since the slope of these lines are given as
6 ± 36 − 4 ≠ 3± 2 2 2 Put x = 0 in the given determinant, we get a = 0. DIferntiating ∆ column by column
Sum of the slopes =
x2 ≠
84.b.
tan θ1 and θ 2 .
PAGE
−2 tan α sinθ cos 2 θ sin 2 α
1 − 2 tan α × 2 ⇒ tan θ1 + tan θ 2 = 3 2 × sin α 4
π ∵ θ = 6
ex cos x ex sin x 1 2x 2 ∆' = − sin x log e (1 + x) 1 + cos x 1 x2 + 1 x2 1 x 2x
1 0 1 85.a.
26 4 7 1 8 ⇒ P − , and Q , and PQ = 3 3 3 3 3
...(i)
8 7 − 3 3 = −1 1 8 ⇒ y − = −5 x − ..(ii) 1 4 3 3 + 3 3
2
10 10 10 p = E 2 (10!) = + 2 + 3 = 5 + 2 + 1 = 8 2 2 2
10 s = E 7 (10!) = = 1 7
86.a.
A.M. of x, z = y, G.M. of x, z =
Also, AM ≥ HM ⇒ y ≥
PAGE
x+ y y+z + 2y − x 2y − z ≥ ∴ 2
www.aieeepage.com
13 2 0, or , 1. 3 3 I we put
x = tan θ, the given equality becomes tan -1 y = 4θ
2xz x+z
x+ y x+ y x+ y y+ z y+z = = = , 2y − x x + z − x z 2y − z x
13 From (ii) when x = 0, y = and when 3 2 x = , y =1 3 So the required coordinates of R are
xz and
A.M ≥ G.M. So, y 2 ≥ xz.
1 26 1 1 2 26 x − = , x − = ± ⇒ x = 0 or x = . 3 9 3 3 3
PAGE -
1 1 1 0 1 = −1 0 0 1
10 r = E 5 (10!) = = 2 5
and QR is perpendicular to PQ.
From (i) and (ii), we get
1
10 10 q = E 3 (10!) = + 2 = 3 + 1 = 4 3 3
2 × 2 + 1(−3) 2 × 3 + 1× 2 , 3 3
8 3× ⇒ 1 x− 3
1 + 0 = b + 2cx
Now, putting x = 0
1 0 1
8 = − cosec 2αα 3 P, Q divide AB in the ratio of 1 : 2 and 2 : 1 respectively and hence the coordinates of P and Q
y−
1
www.aieeepage.com b= 0 0 1+ 1
1× 2 + 2(−3) 1× 3 + 2 × 2 , are and 3 3
83.c.
1 - tan 2 2θ
The given equation can be written as
( x 2 + y 2 ) (cos2 θ sin 2 α + sin 2θ)
82.a.
2tan 2θ
= 1+
∴
87.a.
x+ y y+z . z x
3y2 y( x + y + z ) = 1+ xz xz
[∵ x + z = 2 y]
x+ y y+z y2 + ≥ 2 1 + 3. ≥4 2y − x 2y − z xz
[∵ y 2 ≥ xz ]
Sinc LHS > 0 ⇒ k > 1 Now 2 log, k - 4 log, 3 = 7
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Let log k = t > 0 as k > 1. Then, 2t - 4/t = 7
90.c.
1 ⇒ 2t 2 − 7 k − 4 = 0 ⇒ t = − , 4 2
(5 x / 4 + 35 / 4 ) n = 64
But t > 0, so log 3 k = 4 ⇒ k = 34 = 81
∴ (1 + n) n = 64 ⇒ 2 n = 64 = 26 ∴ n = 6 Term with the greatest binomial co-efficient = (n - 1) + Third term
π/4 < tan -1k < π/2 because k > 1.
88.b.
Also, 0 < cot −1 k < π as k > 3 6 Since z satisfies | z - 8 - 6i | + | z -14 - 6i | = 10 ....(1) ∴ locus of z is an ellipse with foci (8, 60 and (14, 6) and length of its major axis = 10 (1) can be written as
⇒ 6C3 (3− x / 4 ) 3 (35 x / 4 ) 3 = 6 − 1 + 6 C 2 (3− x / 4 ) 4 (35 x / 4 ) 4
PAGE
⇒ (4 y + 1) ( y − 1) = 0 ⇒ y = 1
14 + 8 6 + 6 , = (11, 6) 2 2
2ab = (14 − 8) 2 + (0) 2 = 6 3 ∴10e = 6 ∴ e = 5
[∵ y > 0 ∴ y = −
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9 = 251 − = 16] 25
91.c.
b 2 − 4ac < 0 For b = 1 any a and c which can be chosen in 4 ways For b = 2 either a = 1, c = 2 or a = 2, c = 1 or a = 2, c = 2 ⇒ Required probability = 7/8
92.a.
ab = Σ
∞
⇒ 16 x 2 + 25 y 2 − 352 x − 300 y + 2436 = 0 ...(2) Pair of tangents to (2) from the origin can be found as ...(3) 10 x 2 − 66 xy + 48 y 2 = 0 Max, and Min. arguments of z are slopes of the lines given by (3)
89.b.
33 + 604 48
Let the three numbers in G.P. be a, ar, ar 2
∴ a(1 + r + r 2 ) = 14 ∴ a (r 2 + 1) = 2 (ar + 1) r 2 + r +1 2
r − 2r + 1
=
14 =7 2
...(2)
⇒ (r − 2) (2r − 1) = 0 ⇒ r = 2 or
1 2
1 ,a=8 2 ∴ terms are 2, 4, and 8, 4, 2. Hence lowest of the original terms = 2
PAGE -
2n − 1 + 1 ∞ 2n + 1 − 1 Σ n =1 ( 2n − 1)! n =1 ( 2n + 1)! ∞
= Σ
∞ 1 1 = Σ + n =1 (2n − 2)! (2n − 1)! ∞ 1 1 = Σ − n=1 (2n)! (2n + 1)! 1 1 1 1 1 = 1 + + + + + + .. 1! 2! 3! 4! 5! 1 1 1 1 = − + − + .. 2! 3! 4! 5!
1 1 1 = 1 + + + + .... 1! 2! 3!
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⇒ 6r 2 − 15r + 16 = 9 ⇒ 2r 2 − 5r + 2 = 0
I r = 2, a = 2 and I r =
∞ 2n 2n Σ n =1 ( 2n − 1)! n =1 ( 2n + 1)!
PAGE ...(1)
2 Again a + 1, ar + 1, ar − 1 are in A.P.
⇒
1 is not possible] 4
∴ 3 x / 2 = 1 = 30 ⇒ x = 0
b 2 = a 2 (1 − e 2 )
∴ Max. Arg. z = tan -1
C3 . 33 x = 5 + 15.33 x / 2
Let 3 x / 2 = y ∴ 20 y 2 = 5 + 15 y
∴a = 5
2a = 10
6
⇒
⇒ 4 y 2 − 3y −1 = 0
( x − 1) 2 ( y − 6) 2 + =1 25 16 [∵ centre of the ellipse
Since sum of binomial co-efficients in the expansion of
1 1 1 1 = (1 − 1) + − + − + .... 2! 3! 4! 5! 1 = (e) (e −1 ) = e. = 1 e Hence ab = 1
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93.d.
Let f ( x) = px 2 + qx + r
2 2 1 1 ⇒ . | AB × AC | p = 3 3 2
2
∴ f ( a + h) = p ( a + h) + q ( a + h ) + r f (a) = pa 2 + qa + r
⇒
f (a + h) − f (a ) p (h 2 + 2ah) + qh = a+h−a h
∴
⇒ p=2 ∴ Position vector of E, I E divides median AF in the ratio λ : 1 .
h = p ( h + 2a ) + q = 2 p a + + q 2
Now
h = f ' a + 2
Hence c = a + a
94.b.
Let I =
a
a
∫ 1+ e
f ( x)
a
dx f (a − x)
=
95.d.
a
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1 + e f ( x)
∫ 1+ e
f ( x)
y z x logb . . i.e. logb 1 i.e. 0 z x y
dx = a
0
a 2
z x y logc . . i.e. logc 1 i.e. 0 x y z
dy dz = dx dx
98.b.
dz − 1 = sin z + cos z dx
⇒
⇒
2α 2β c =0 x− y+ 1+ λ 1+ λ 1+ λ Its centre is x2 + y2 −
dz = dx 1 + sin z + cos z
α 2 + β 2 − c(1 + λ) β α , amd radius is |1+ λ | 1+ λ 1+ λ
dz = dx z z z 2 cos 2 + 2 sin cos 2 2 2
∫
1 z sec 2 dz 2 2 = dx + c z 1 + tan 2
⇒ log 1 + tan
∫
z = x+c 2
PAGE
Volume of tetrahedron =
PAGE -
The radius vanishes I 1 + λ =
α2 + β2 c
cα cβ , 2 So, the other limiting point is 2 2 2 α +β α +β
99.a.
For any value of λ, the given equation represents circles passing through points of intersection of
www.aieeepage.com the circle x + y 2
x+ y ⇒ log 1 + tan = x+c 2
96.c.
∴ given determinant = 0. The equation of the given coaxal system is
x2 + y 2 − 2αx-2 βy + c + λ(x 2 + y2 ) = 0 or
dz ⇒ = 1 + sin z + cos z dx ⇒
Operate C1 + C 2 + C 3 the given determinant has first column as
x y z loga . . i.e. loga 1 i.e. 0 y z x
−− f ( x)
0
dx ; ∴ 2I =
Put x + y = z ∴1 + ∴
dx
∫ 1+ e
0
∴ I=
λ (i − j − k ) λ +1
PAGE 97.d.
f ( x)
∫ 1+ e
e f ( x)
=
∴ (2) 2 + AE 2 = (4) 2 ⇒ AE 2 = 16 − 4 = 12
0
=
λ AF ; λ +1
λ ⇒ AE 2 = − 3 But p 2 + AE 2 = AD2 ; λ +1
h which is independent of p, q, r. 2
0
Also I =
AE =
2
dx
∫ 1+ e
1 | (i × k ) × ( j + k − 2 i ) p = 2 2 2
2 2 1 (∆ ABC) p = 3 3
100.c.
2
+ 2ax + 2by + c = 0 and the line ax - by + 1 = 0. So, ax - by + 1 = 0 is the common chord for each circle of the system, henc it is the radical axis. The line of centres will be perpendicular to radical axis, so its equation is bx + ay + k = 0 It must pass through the centre (-a, -b) of the memeber of the system, so, k = 2ab Hence the desired equationis bx + ay + 2ab = 0
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Paper - I : Chemistry - 1 to 50 Solutions 101.d. 102.d. 103.a.
All of the choices are correct Pb
++
+ 2HCl →
PbCl 2
comes
↓ → PbS ↓ H 2S
N 2 H 4 → N2 0
Equivalent weight of N 2 H 6SO 4 =
130 = 32.5 4
This is acetic acid.
S N 2 reaction and is thus favored by a polar
107.b. Step 1 is an
5 1 = 50 × 1000 500 and I weight of hydraize sulphate be x gm then
aprotic solvent like DMF. N-methylformamide is a polar protic solvent. (Note that there is a hydrogen bonded to a nitrogen atom. The molecule can thus participate in hydrogen bonding). It will solvate the nucleophile and reduce the rate of the reaction. 108.b. Tertiary alkyl halides will not be able to undergo the SN2 reaction outlined in the passage because of steric hindrance (eliminate choice A). However, they can undergo an SN1 reaction with water (Choic B) to make a stable tertiary carbocation, which can accept water as a nucleophile. Choice C makes a tertiary Griganard reagent, a BASIC carbon reagent, which picks up a PROTON form water, not a hydroxyl group! Choice D, as alluded to in the passage, gives mainly E2 (alkene) products, especially in this case since the halide is hindered. 109.a. The saponIication mechanism (outlined below) is required to understand this question. The bottom line is that the nucleophilic hydroxide ion attacks the ester carbonly (not the secondary carbon in the alcohol product). So the isotopic lable ends up still attached to the alcohol product, but not to the carboxylic acid. Choice D would be the correct answer I the isotopic lable were in the attacking hydroxide and the ester was unlabelled.
PAGE
equivalents of hydrazine sulphate = ∴
x 32.5
1 x 32.5 = or x = = 0.065 g 500 32.5 500
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Hence wt of N 2 H 6SO 4 in 10 ml solution = 0.065 g
105. a.
CH 3COOH.
Number of equivalents of KMnO 4 = 20 ×
104.b.
to the carbonyl carbon ar (1) the oxygen atom with which it forms a double bond, (2) an OH group, and (3) the R group. In this case, the R group is a methyl group, and so the molecule left behind after the alcohol departs is
Black ppt
White ppt dissolves on boiling
sp 2 hybridized again. The three groups attached
∵ wt of N 2 H 6SO 4 in 1000 ml solution = 65 g Only cis isomer can easily form a chelate ring with oxalate group ; the trans isomer can not form a chelate ring . Both are square planar. In the IUPAC system, esters are named as alkyl alkanoates, The “alkyl” group is the one that is attached to the oxygen atom, while the “alkan-” part contains the the carbonyl carbon :
O O
C
CH 3
ethanoate butyl
O
Choice b is incorrect because it would have an ehtyl group attached to the oxygen atom :
ethyl
H3C
O
OH
H3C
O18
O O
18
O
PAGE NaOH HzO
O O
O O 18
C
110.a
butanoate
Choice c is incorrect because acetate is the common name for ethanoate. Ethyl acetate is ethyl ethanoate which is not the ester formed in Ste 1. (it would not have been the IUPAC name anyway) Choice d is incorrect because carboxylate is the salt of a carboxylic aicd. Acid-catalysed hydrolysis of the ester ROOCR’ proceeds via the formation of a tetrahedral intermediate. The OR’ group is then protonated and departs as the alcohol R’OH as the carbonyl bodn reforms and the carbon atom be-
proton
+ HO
CH3 exchange
products (Choice A)
Hydrated ferric chloride (FeCl3 .6H 2 O) upon heating gets hydrolysed by its own molecules of water of crystallisation to give Fe(OH) 3 , which
www.aieeepage.com Fe Cl .6 H O → Fe(OH) + 3HCl + 3H O
106.c.
PAGE -
changes to Fe 2 O3 of on further heating 2
3
2
3
2
2Fe(OH) 3 → Fe 2 O 3 → Fe 2 O 3 +3H 2 O
111.a
E n of H = ∴
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-21.76 × 10 -19 J n2
E n of He+ =
-21.76 × 10-19 × Z2 J 2 n
18 of1 of 24 7
distance travelled by the gas .Hence
-21.76 × 10-19 × 4 E 3of He = J 9
∴
+
r1 (of HCl gas ) at pressure
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 E =
112.b.
P = 40 =
hc λ
= 2055 × 10-10 m = 2500Å CO & N2can be separated by bubbbbbling through ammonical cuprous chloride in which CO is absorbed but not N2 .N2 O2can be seperated by bubbling through alkaline solution of pyrogallol in which oxygen is absorbed but not
P=
Malachite
Dissolution CuO.H 2 SO4 → CuSO 4 +H 2 O;
www.aieeepage.com of NaOH Fe O Smelting :
Dipole moment µ = e × d coulomb metre
2Cu 2 S + 3O2 → 2 Cu 2 O + 2 SO2 ;
119.b.
( electronic charge ) (e) = 1.602 × 10 −19 C
µ = 1.602 × 10 −19 × 2.6 × 10 −10
= 4.1652 ×10−29 µ KCl = 3.336 × 10 −29 m
115.a.
116.c.
117.a.
3.336 × 10-29 = 80.09 % 4.1652 × 10−29 × 100 Both lattice energy and hydrolysis energy decrease s moving down the group due to gradual increase in size of M ++ ion .So former tends to increase the solubility while latter tends to decrease it . But lattice energy has dominating role here .And therfore , solubility increase down the group . EA of representative elements increases from left to right in a period ( barring few exceptions ) and decreases down the group . However ,the second period group members ( e . g N , O , F ) ,being small in incoming electron and the the other electorns in the atom . Since the pressures of gases are dIferent ,and the temperature is constant , the rate at which molecules of the two gases dIfuse is directly proportional to the pressure . This rate of dIfusion is also directly proportional to the
PAGE -
+ 3C → 2Fe + 3CO
2 3 Reduction with C
Cu 2 S + 2Cu2 O → 6Cu + SO2
For KCl d = 2.6 ×10−10 m For complete separation of unit charge
∴ % ionic character of KCl =
Soluble
Self reduction :
through water in which NH 3 dissolved but not
114.a.
ppt
More electropositive metal replaces lesser one
excess
113.d.
(Soluble)
CuSO 4 + Fe → Cu + FeSO 4
PH 3 & NH 3 can be seperated by bubbling
Hence
Hydrometallurgy : Roasting → 2CuO+H 2 O+CO CuCO 3 .Cu(OH) 2
N 2 .Ammonia is recovered by heating the solution
PH 3 . All the choices are correct
17
60 36.5 × = 2.197atm 40 17
PAGE 118.a.
thropugh H 2SO 4 when NH3 is absorbed but not (NH 4 )2 SO4 with
k ×1
r1 60 kP 17 = = × r2 40 36.5 k × 1
N 2 .NH3 & N 2 can be seperated by bubbbling
of
36.5
and r2 (of NH 3 ) at 1 atm. pressure
hc 6.625 × 10−34 × 3 × 108 × 9 = λ= E 21.76 × 10−19 × 4
or
kP
P = 60=
120.a.
Electrolytic Reduction : Electrolysis of Al2 O3 in molten cryolite A positive Beilstein’s test for halogen does not always indicate the presence of halogen since some halogen free compounds viz . urea , thiourea, amides etc. also respond this test . The reason being the fact that these halogen free compounds form cuprous cyanide which is volatile and decomposes to copper which burns with green flame . Initial concentration of each gas = 1 mole Let the No .of moles of NO2 reacted at equilibrium = x , Then
PAGE SO2 ( g )
+ NO2 ( g )
( 1- x )
( 1- x )
SO3 ( g ) ( 1+ x )
+ NO2 ( g ) (1+x)
Now we know that [ SO3 ][ NO ] , [ SO ] [ NO ] = K c
or
(1 + x)(1 + x) = 16 (1 − x)(1 − x)
www.aieeepage.com (1 + x) 1+ x = 16 or =4 2
2
2
(1 − x) 2
1− x
3 = 0.6 5 Thus the concentration of No at equilibrium = 1 + x = 1 -0.6 = 0.4 moles
1 + x = 4 - 4x or
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5x = 3
x=
19 of 24
Heat
O Br2 / NaOH a) R - C - NH 2 → R − NH 2
(Hofmann reaction)
126.c.
at 20 C =
||
|| H 2 SO4 → R − NH 2 c) R - C - OH+HN 3
∴
( Schmidt reaction )
87 ×1 × 2750 mL 0.877
PBo 78 × 2750 × = 1 × 0.0821 × 293 760 0.877 × 1000
PAGE
HO d) R - C - NHOH → R − NH 2
PTo 92 × 7720 × = 1 × 0.821 × 293 760 0.867 × 1000
( Lossen reaction )
At pH < isoelectric pH, -NH 2 group is protonated
or PTo = 22.37 mm;
www.aieeepage.com 127d. Rate = k[A] [ B]
+
to - N H3 and the sol is positively charged. At
pH > isoelectric pH, -COOH is deprotonated to -COO- and the sol is negatively charged. At isoelectric pH both groups are equally ionised, the sol particles carry no net charge. 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 x -119 = - 119 = - 357 kj mol-1.
x
or
∆H (Reaction) = ∆H of (Product) − ∆H of (Reactants)
= -156 - (49+0) = 205 kj mol-1 ∴ Resonance energy =
(rate) 2 8 ×10−3 k[0.024] x [0.070] y = = = 2x (rate)1 8 ×10−3 k[0.012] x [0.070] y
Q=S+
1 D + I − E −U 2
128.d.
d [ A] = k [ B ]3 dt
2AB2 → 2AB + B2 1− α
α
PAGE PAB2 =
(1 − α)p 2(1 − α)p = (2 + α)/2 2+α
PAB =
αp 2αp = (2 + α)/2 2 + α
α 2+α = 2 2
α www.aieeepage.com p
200 × 0.1 = 0.02 Moles of the acid neutralized = 1000 Moles of acid remaining undissociated =
PB2 =
25 0.021 − = 0.015 100
HENCE,
Heat evolved during neutralization for 100% ionization = 13700 × 0.02 = 274cal Heat used up for the dissociation of 0.015 mole = 274 - 244 = 30 cal
PAGE -
α/2
Totalmoles = 1 - α + α +
∆H Exp − ∆ cal = −357 − (−205) = −152kj mol −1 All of the choices, since
2 y = 8 23 , y = 3
or, −
Actual enthalpy of the reaction can be evaluated as below:
y
(rate)2 [0.024] x [0.070] y 8 ×10−3 =k = =8 (rate)3 [0.024] x [0.035] y 1×10−3
+ 3 H2
125.d.
or PBo = 74.74cm
Similarly,
-
124.d.
=
92 × 1 × 7720 mL 0.867
O
123.a.
mole
1 mole (or 92g ) toluene has volume at 20°C =
heat b) R - C - N 3 → R − NH 2 (Curtius reaction)
122.a.
per
In vapour phase 1 mole (or 78 g) benzene has volume 0
O
O ||
dissociation
30 = 2000 cal = 2.0k cal 0.015
||
121.c.
of
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αp 2 = (2 + α) 2 + α 2
20 of 24
2
kP =
129.c.
p 2AB pB2 2 p AB 2
=
αp 2αp × 3 2+α 2+α α p = = (α << 1) 2 2 p 2 ( 1 ) − α 2 + x
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
As for emission of one α - particle atomic mass decreases by 4 and number by 2. Further for the emission of one β - particle, the atomic mass do not change but the atomic number increases by 1. So we first find the α - particles; Decreases in atomic mass = 234 - 206 = 28 No. of α - particles emitted =
90 − (7 × 2) = 76
Now, atomic number of pb = 82, which is more by (82 - 76) = 6
Enthalpy
B+ A+
C
132.d.
I. [ Ag + ]. = K sp ( AgCl ) = 1.0 × 10 − 10 = 1.0 × 10 − 5 M Ag + = K sp ( AgI ) = 1.0 × 10 −16 = 1.0 × 10 −8 M
II.
III.
In the above reaction, heat is gained, thus the reaction is endothermic (i) The product may be meso or achiral (ii) In case an optically active chiraL reagent, solvent, or catalyst such as enzyme is used, the productwill likely to be optically acitve. (iii) The configuration can be changed only by breaking a bond. (iv) Changing the configuration at one of the chiral carbons convetrs one diastereomer to the other. (v) The terms D and L do not refer to the sign of rotation. They refer to the configuration of as stereoisomer relative to that of D-glyceraldehyde.
134.a.
[ Ag + ][CN − ]2 [ Ag + ] × 0.12 = = 1.0 × 10 − 19 M [ Ag ( CN ) 2− ] 1.0
Density of 39% H 2 SO4 before discharging = 1.294 g/ml ∴ Amount of H 2 SO4 in the solution before
discharging =
3500 × 1.294 × 39 = 1766.3 g 100
Density of 20% H 2 SO4 after discharging = 1.139 g/ml ∴ Amount of H 2 SO4 in the solution after
discharging
3500 ×1.139 × 20 = 797.3 g 100
PAGE
Amount of H 2 SO4 consumed by the battery 969 = 9.887 moles 98 The charging and discharging reactions are
=1766.3-797.3=969.0 g =
Pb + SO42 − → PbSO4 + 2e − (charging)
www.aieeepage.com → PbSO + 2 H O (discharging) PbO + 4 H + SO + 2e = 3.012 × 10 21
+
2
Density of Mercury (Hg) = 13.6 g/c.c ∴ Volume of 1 atom of mercury (Hg) =
1 10 ×10 c.c. = c.c 3.012 ×1021 ×13.6 3012 ×1021 ×136 3
PAGE -
2
Ag (CN ) −2 → Ag + + 2CN − ,
6.023 × 1013 6.023 × 1013 = 200 2 × 10 2
21
Ag + + 2 NH 3 ;
Ag + [ NH 3 ] [ Ag + ] × 0.12 K d = 1.0 × 10−8 = = ⇒ [ Ag + ] = 1.0 ×10−6 M 1.0 Ag ( NH 3 )2+
Avogadro’s number = 6.023× 1013 At. wt. of mercury (Hg) = 200 ∵ In 1 g of Hg, th total number of atom =
= 3.0115 × 10
Ag ( NH 3 ) 2+
K d = 1.0 × 10 − 21 =
Reaction coordinate
131.b.
133.c.
iv.
D+E+F
1 − 24 3 = (24.4 × 10 ) cm.
= 2.905 × 10 −8 cm ≅ 2.91Å cm.
www.aieeepage.com
Therefore, β - particles emitted = 6 Activated complex
=
1 − 23 3 (2.44 × 10 ) cm
PAGE
28 =7 4
Hence, atomic number decreases to
130.b.
10-17 10 -17 c.c. = c.c 3.012 × 136 409632
2− 4
−
4
2
Adding the above two reactions, we get disch arg ing Pb + PbO2 + 2 H 2 SO4 → 2 PbSO4 + 2 H 2 O ch arg ing
Thus w see that 2 moles of H 2 SO4 are consumed to give 2 moles of electrons, i.e., 2 Faraday of current.
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21 of 24
1 mole of
(H 2 C = CH 2 ) will be the major product (Hofmann product or Hofmann elimination)
H 2 SO4 = 1 F of H 2 SO4 = 96500 coulombs ∴ No. of ampere-hours for which battery is used =
β2
96500 × 9.887 = 265.02 ampere hours 60 × 60
135.b.
PH 2 × [ Zn 2+ ] [ H + ]2
=
1× 0.01
= 1, log Q = 0
0.12
→ Cu(2aq+ ) + 2 Ag ( s ) 2) Cell reaction : 2Ag +(aq) + Cu( s )
Q=
PH 2 [Cd 2+ ]
3) Cell reaction : 2H
+ 2
[H ] + (aq)
=
1 × 0.01 0.12
+ Cd( s ) → H 2 + Cd(2aq+ ) ; Q =
=1
136.c.
PH 2 × [ Zn 2+ ] [H + ]2
=
1 × 01 0.12
= 10
139.c.
→ Mn 2+ + CO 2 + H 2 O MnO −4 + C 2 O 24− + H + 5
(‘n’ factor = 7-2 = 5) Equivalent mass of
(from β1 )
(from β 2 )
(from β 3 )
In Hofmann elimination H atom (in the form of proton) is eliminated from that β carbon atom which has maximum number of H atoms(s). This is due to the fact that Hofmann elimination (an example of E2 reaction) requires anti-coplanar arrangement of the β H atom and ther leaving group. Further, higher the changes for this arrangement to remain in staggered conformation more will be the case of elimination. Thus greater the number of H atom at β position, higher will be the chances for Me3N+_ and H atom to remain in this particular conformation (anti - coplanar arrangement of Me3N+- and H atom in staggered conformation). In case of exothermic dissolution, the solubility of the solid increases on lowering the temperature. On cooling, the solution becomes unsaturated and solid solute does not separate. At 00C, water in the solution does not freeze.
140.b. rCs + / rCI − = 0.90Å ⇒ rCs + = 0.90 × rCI − = 0.90 × 1.80 = 1.62 Å
rCs +
For precise fitting of the cubic voids, r
Formula mass 88 = = 44 2 2 [‘n’ factor = 2(4-3)=2] C 2 O 24− =
Cl −
MnO −4
= 30 × 0.05 × 5 ×10
−3
= 7.5 ×10
Hence, cubic packing of Cl− ions is loose to some extent.
−3
The Cs + ion at the centre of the body is in touch
Mass of C 2 O 24 − = 7.5 × 10 −3 × 44 = 0.33 g;
% C 2 O 24 − =
with 2Cl− ions at the opposite corners of cube.
0.33 × 100 = 66 0 .5
Compound (A) on treatment with AgNO3 give white precipitate of AgCi, which is readily soluble in dil.aq. NH3. Therefore it has at least one CI_ ion in the ionization sphere furthermore chromium has coordination number equal to 6.So its formula is
Therefore it has Br − in the ionization sphere.So its formula is [Cr ( NH 3 ) 4 CI 2 ]Br x x 3d
x x
x x 4s
distance
between
Cs + and Cl −
= rCs + rCl − = 1.62Å + 1.80Å = 3.42Å 141.a.
The four compounds dIfer in two respects : Compounds II has sp3 hybridised N, compounds I and III have sp 2 while compound IV has sp hybridised N. Now we know that greater the s character of an orbital, more tightly its electronsare held and hence lesser will be their availability for protonation causing weak basic character. Thus the basic character of the three N’s is
sp N > sp > N > spN . However, in pyrrole www.aieeepage.com electrons on N are part of aromatic sextet, i.e., these x x
x x
x x
4p
State of hybridization of chromium in both (a) and (B) is d2sp3. Spin magnetic moment of (a) or (B)
µ spin = n (n _ 2) = 3(3 + 2) = 15 = 3.87 BM The given amine has three β − hydrogens, hence it can form three alkenes of which least substituted
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Hence
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[Cr ( NH 3 ) 4 BrCI]CI Compound (B) on treatment with AgNO3 gives pale yellow precipitate of AgBr soluble in conc. NH3.
Cr 3+ ( Z = 24)
must
be 0.732 Which is smaller than the given value of 0.90.
Equivalents of C 2 O 24 − = Equivalents of
138.c.
+ H 2C = CH 2
+
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Equivalent mass of KMnO 4 =
137.a.
(ii) Ag 2O (iii) heat
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PH2 ×[Cd 2+ ] 1× 0.01 = =1 [ H + ]2 0.12
+ + Zn (s ) → H 2 + Zn 2(aq ) ; Q = 4) Cell reaction : 2H (aq)
(i) CH 31
β2
NHCH 2 CH 3
+ + + Zn (s) → H 2(g ) + Zn (2aq 1) Cell reaction : 2H (aq) )
Re action quotient, Q =
β1
3
2
are delocalised and hence lesser available for protonation and thus pyrrole is a weaker base than pyridine because in pyridine in nonbonding electrons present in sp2 orbital do not form a part of aromatic sextet. Thus the basic character should be in the following order.
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<
H 3C − C ≡ N IV (sp hybridised N)
O N H
N H
III 2 (sp Hybridised N but electron pair forms part of sextet)
O
<
<
N
N H
I 2 (sp Hybridised N; electron pair not a part of sextet)
142.d.
or
II 3 (sp Hybridised N)
The product, 2-bromo-4 methylhexane, has two chirality centers
CH 3
Br
CH 3CH 2 * CHCH 2 * CHCH 3
144.a.
filled five hybrid orbitals form σ - bonds with 4F atoms and O atom. The pure 5d orbital forms π - bond with 2p orbital of O-atom. Although ordinarily chlorine present on benzene nucleus is dIficult to be replaced, however I it has electron-withdrawing group in the o- and ρ position it is easily replaced. In the present case CI (electronegative group) present on C2 and C4 are ortho to CI present on C1, hence the latter can be replaced by OH − . Note that other two chlorines (at C2 and C4) are meta to one chlorine, so these are not replaced.
CI 1
CI
PAGE 2
CI
145.d. Energy for photon E = hν =
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a) The S-atom in the excited state (3s 2 3p 3 3d1 ) has 2
sp3 hybridization - (sp 3 ) (sp 3 )1 (sp 3 )1 3d1. The three hybrid orbitals form three (S-O) σ − bonds and the 3d pure orbital gives (S-O) π − bond .
CI
CI hc 6.63 × 10 −34 × 3 × 108 J = 66.3 × 10 −19 J = λ 300 × 10 −10
K.E. acquired by the electron
= 6.63 × 10−19 − 1× 1.6 × 10−19 = 64.7 × 10−19 J 1 1 mv 2 = × 9.1× 10 −31 × v 2 ⇒ v = 3.8 × 10 6 ms −1 2 2
146.a. Equation 1 shows the exchange of a sulfide ion for two acetate ions. Since sulfide is in a-2 oxidation state on both sides of the equation, we can eliminate choice (C), which suggests a change in oxidation state. Since exchange reaction is not one of the choices, we must determine which of the remaining choices is correct. Choice (A) works because sulfide is a lewis base that attacks the positively charged lead ion choice (B) is incorrect, an Arrhenius acid base reaction would involve hydrogen ions. Choice (D) is incorrect because no neutralization is taking place. 147.b. The solubility of calcium sulfide and magnesium sulfide is very low, so these solides will form readily from solution. When hard metal ions are present in the water used to rinse the hair before dye application, they react with the sulfide in hair keratin before lead and bismuth ions have a chance to attack. Therefore, choice (B) is correct. Choice (A) is incorrect because citrate and acetate ions will not precipitate with hard metal ions - these will remain in solution. Choice (C) is incorrect because in solution, additional ions will not decrease the effectiveness of other ions through collision reduction. I this wee the case, the hair dye could be left on the hair longer and it would have time to react fully. Choice (D) is incorrect because calcium and magnesium do not precipitate in elemental form. They are much more stable as positively charged ionsremember that they are alkaline earth metals, which prefer to be in a + 2 oxidation state.
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b) In PO34− , P-atom in its excited state (3s1sp 3 3d1 )
is sp3 hybridized. Four hybrid orbitals forms (P-O) σ - bonds whereas 3d1 gives (P-O) π - bond .
c) In NO 3- , N-atom is sp 2 hybridized as
(sp 2 ) 2 (sp 2 )1 (sp 2 )1 2p1z .
aq.OH
4
Since C4 of the product has R configuration in the reactant, and since this chirality center is unaffected by the reaction, its configuration us unaffected (assuming that the relative priorities of the four group are not changed by the reaction); i.e., the configuration at C4 in the product is also R. Addition of HBr involves the formation of carbocation, it is attacked by Br − form top as well as bottom face leading to configuration at C2 as R as well as S. However, the carbocation does not have a plane of symmetry, it is chiral because of the chirality center at C4, hence it will not be attacked equally from the two faces, leading to a mixture of 2R and 2S products in a ratio other than 50 : 50. Thus the net result is the formation of two diastereomeric products [2R, 4R] - and [2S, 4R]- in unequal amounts. 143.c.
OH _
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Completely filled sp 2 hybrid orbital on N-atom donates an electron pair to the vacant 2p orbital of O atom to form dative bond. Other two hybrid orbitals form (N-O) σ - bonds . The pure 2p1z orbital
forms (N-O) π - bond . d) In XeOF 4 , X e -atom in its excited state
(5s 2 5 p3 5d 3 ) is sp 3d 2 hybridized. The singly
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148.c. This is a thinking question - you can’t just search for the answer in the passage prose or equations. The TEST loves to throw a few of these at you - after all, they are trying to test your critical thinking! Let’s examine how we can attack this question effectively. Bismuth is in the p-block right next to lead. Before it loses any electrons, it has a half-full p subshell. Since it is so far down on the periodic table, it is considered a metal, not a metalloid. Theefore, you can eliminate choice (B). The question is basically asking you why bismuty loses p electrons but keeps s electrons. Choice (A) doesn’t give an answer to that-it suggests that the opposite is true by mentioning the noble gas configuration bismuthy could adopt by losing all of its valence electrons. Choice (D) anasers a dIferent question about gaining electrons, so it must be incorect. This leaves only choice (C), which correctly explains that the s orbital for bismuth is much lower in energy, so the element is more stable withou losing those electrons.
150.a.
There are three crucial things to remember when attacking an equilibrium expression question. * Products appear in the numerator reactants in the denominator * A coefficient in the reaction equation corresponds to a superscript in the equilibrium expression * Pure solids and liquids do not appear in the equilibrium expression.
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149.d. This is another le chatelier’s principle question. The simplest thing to do is to consider how the equilibrium will be affected, then attack the first part of the statements regarding the effectiveness of the hair dye. Adding more acetate ions will shIt the reaction to the left since there will be more products in solution. Therefore, we can eliminate choices (A) and (B). Now, how will this affect the effectiveness of the hair dye? It will decrease, since less lead sulfide will be produced. Therefore, choice (D) is correct.
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