PAGE
www.aieeepage.com
key AI - IITJEE - Model test - 03
PAGE
www.aieeepage.com
PAGE -
A Premier Institution for IIT - JEE & AIEEE at the National Level
1 of 26
Paper - I : Physics - 1 to 50 Solutions 1.a.
2.c.
The time for one orbit is simply the period of the motion. Recall the period = 1/frequncy. In the last sentence of the first paragraph wire told the suttle makes 1.85 x 10-4 revolutions per second. This is fgrequency in units of revolutions/second. Take one over this number to get period in seconds. The answer choices are not very close together, so we can approximate as needed to make the math easy. Approximate 1.85 x 10-3 by 2 x 10-4. Then have 1(2 x10-4)= 5000 seconds. 5000 seconds equals 5000/60 minutes which is close to choice A. Note that the next closest choice is 4 hrs which is over twice as large as choice A. We’re certainly not off by nearly as much as a factor of 2 in our approximations. Alternatively, since there are 60 x 60 = 36000 seconds in an hour, we know that 5000 seconds will be between 1 hour and 2 hours. The correct answer is C. Rule out choice A, since g = 0 implies no gravitational force, i.e. the shuttle is held in orbit by the gravitational force so without it, the shuttle would fly off into space. The passage itself states explicitly that gravity is not absent, things in free fall just behave as if gravity were absent. Choice B should also be eliminated by recalling that acceleration due to gravity decr3eases with height above the surface of the Earth, and the shuttle isn’t close enough to the surface to approximate g by this value. Choice D is just the same as choice B, that’s where the value of 9.8 comes from! Choice C makes sense because we know that the gravitional force is dependent on distance and the altitude of the shuttle is high so we can’t approximate the distance by the radius of the Earth. The acceleration due to gravity g is found by applying Newton’s 2nd law to the gravitional force of the Earth on an object. We write the force as F = mg where g is the acceleration of gravity, but we also know this same force is given by F = GMem/r2 where Me is the mass of the Earth, m the mass of the object, G the gravitational constant, and r the distance from the center of Earth to the position of the object. Equating the two expressions gives g = GMe/r2 which is choice C. Choice D states that g = GMeRe2 at the altitude of the shuttle. From above discussion we see that this is the value of g at the surface of the Earth, which is 9.8 m/s2. The orbit of the shuttle is sufficiently higher than the surface of the Earth so you can no longer approximate g by 9.8 m/s2. Unfortunately this is a case where we have to consider all the statements given to reach the correct answer. Statement II is certainly true. At an altitude of 1 km, the shuttle will experience considerable air resistance from the atmosphere. The air resistance acts as a frictional force which
will decrease the speed of the shuttle below the speed needed to maintain the 1 km orbit. The result will be a decay of the orbit with the shuttle finally crashing into the ground. The engines are needed to replenish the energy that is constantly being “bled off” by friction. (For orbits above the Earth’s atmosphere, the shuttle will maintain a constant speed with the engines turned of since there will be no frictional force to slow it down.) Statement III is incorrect because it is gravity, not the engines, that provides the centripetal force. We can eliminate choices A and D. Statement says that gravity is too strong at an altitude of 1 kilometer. A large gravitational force implies that the shuttle would have to have a larger (tangential) velocity in order to be in this orbit rather than fall to the ground. As long as the shuttle’s speed is great enough, it could orbit at 1 km above the surface (neglecting air resistance); i.e. if the shuttle can sustain unifor circular motion, it is dealing with gravity, no matter how strong, just fine. So it’s not the strength of the gravitational force that prevents an orbit at an altitude of 1 km. This statement is thus incorrect. Eliminate choice C. All objects in the shuttle including the shuttle itself are in free fall as the shuttle orbits the Earth. This means that things behave the same way they do for free fall in general. Consdier the workings of a pendulum clock on Earth. What makes the pendulum swing? The force of gravity that pulls it down as it swings up. Now consider a pendulum clock in free fall. As the passage indicates physical behavior is observed that makes it appear as if gravity is absent (weightlessness of astronauts, for example). In the apparent absence of gravity, the pendulum arm doesn’t swing, making the period of its motion infinite. The correct answer is B. At the end of the passage, we are told that a circular orbit is a special case where a which is half the length of the major axis, is the radius. (I.e., the major axis is the diameter.) In other words, all the relationships given apply to circular orbits as well. The comet in an elliptical orbit has energy.
PAGE
www.aieeepage.com 4.d.
5.b.
PAGE E1 = −
GM s mc . The comet in a circular orbit 2a
E2 = −
GM s mc 2r
www.aieeepage.com has energy:
3.b.
PAGE -
We are told in the question sterm that a = 2r:
E1 : E2 =
A Premier Institution for IIT - JEE & AIEEE at the National Level
E1 E2
=-
GM s mc GM s mc ÷ 2a 2r
2 of 26
=
6.d.
acceleration is in the opposite direction of displacement and initial velocity).
1 1 1 1 1 ÷ = ÷ = = 1: 2 2a 2 r 4 r 2 r 2
Note that we are told in the question stem that the comets have equal masses; that is why we cancel mc. ‘Which comet has the higher total energy?’ Surprise: the one in the elliptical orbit! this is so because the values of the energy are always negative. That E1 is half of E2 means that the energy of the comet in the elliptical orbit is ‘half as negative’. It is closer to zero and therefore more energetic!. In this question we are told that not only does a planet revolve around the Sun, but the orientation of the orbit itself also rotates. We are asked for a plausible explanation. Note that in questions like this, we are not expected to know precisely the science behind it, we just need to be able to eliminate the ones that are clearly irrelevant or even wrong. Choice A is incorrect be cause centrifugal forces do not cause things to rotate. Choice B is incorrect because while it is indeed true that the gravitational attraction of the Sun provides centripetal acceleration, this is what causes the planet to rotate in a circle or an ellipse. Choice C is incorrect because even though energy may be dissipated as a planet moves through space, this will only cause the motion to slow down and the planets to gradually spiral in towards the Sun. If the ellipse maintains its shape and merely changes in orientation, the energy is constant, and so this phenomenon is independent of the dissipation of energy. Choice D provides a feasible explanation: that the planets do not merely move in an ellipse means that something else other than the gravitational field of the Sun must be influencing their motion. The attraction of the planets among themselves is certainly a likely candidate for this additional influence. This is a projectile problem. You want to throw a projectile straight up from the 1st floor to the 6th floor, and you want it to be in the air during the time that the elevator is moving. First find the time the elevator is moving. To solve this problem, you must calculate the distance needed for the elevator to reach its maximum velocity (5 m/s as given in the paragraph above the diagram). This distance is the same as the distance necessary for the elevator to slow to a stop from maximum velocity. Use
8.b.
Conservation
of
energy
tells
us
that
1 mg 2 + mg (r + r cos θ) and therefore the 2 speed of the puck is determined by the equation mgh =
v 2 = 2 g[h − r (1 + cos θ)] . The track exerts a normal force N on the puck and thus the total force in the radial
PAGE
mg
Figure 2: The weight mg of the puck has an inward www.aieeepage.com radial component given by mg cos θ . There is also
7.d.
a tangential component given by – mg sin θ . direction is given by − N − mg cos θ . Since the radial acceleration is
−v 2 / r
we have
2
−mv / r = − N − mg cos θ . Thus the track exerts a
normal force given by N = mv 2 / r − mg cos θ . Using our expression for v2 we find that 2h − 2 − 3cos θ N = mg r
Note tht N ≥ 0 and one loses contact with the track at the moment N vanishes. This occurs at an angle θ0 given by cos θ0 =
2 h −1 3 r
PAGE
v 2 = vo2 + 2ax . This distance is the same as the distance necessary for the elevator to slow to a stop from maximum velocity. A trip from the first to the sixth floor is 25 m; exactly enough distance for the elevator to reach maximum velocity and then
If h = r then θ0 = 90º as expected. If you release it from the same height as the top of the loop-the-
9.d.
loop, h=2r, then θ0 = arccos (2/3) ≈ 48º. For the puck to reach the top of the loop-the-loop you must have h ≥ 5r / 2 . Rising hoop Let θ be the angle through which the bead has fallen, and let N be the normal force from the hoop on the bead, with inward taken to be positive. Then the radial F=ma equation for the bead is
www.aieeepage.com
slow to zero. (To find the time, use: x = ½ at 2 where x = 12.5 m, a = 1 m/s2 and t will be the time for half the trip.) Now be sure to take the time for the whole trip and plug it into x = vo t + ½ at 2 where x = 25 m and t =10s (Note: you must use a = –10m/s2 because
PAGE -
mv 2 (1) R The height the bead has fallen is R − R cos θ , so conservation of energy gives
A Premier Institution for IIT - JEE & AIEEE at the National Level
N + mg cos θ =
3 of 26
VW' ρ w g = Fb' = M Bg
mv 2 = mgR(1 − cos θ) ⇒ v 2 = 2 gR(1 − cos θ) (2) R Therefore, the radial F=ma equation becomes
and the volume of water displaced by the boat is VW = '
2
mv − mg cos θ R = 2mg (1 − cos θ) − mg cos θ = mg (2 − 3cos θ)
N=
(3)
MB ρW
(4)
The stone in the water also occupies volume thus
(3)
displaces water with volume VWn = VS and therefore the net volume of water displaced in this case (stone in water) by boat and stone is
By Newton’s third law, this is the force from the bead on the hoop, with outward taken to be positive. Note that this force is positive (that is, the bead pulls outward on the hoop) if
VW' + VWn =
MB + VS ρw
(5)
θ > cos −1 (2 / 3) ≈ 48.2º . Since there are two beads, the total upward force on the hoop from the beads is
To compare this with the previous case we take the difference
2 N cos θ = 2mg (2 − 3cos θ) cos θ
VW' + VWn − VW =
PAGE (4)
The θ that yields the maximum value of this upward force is obtained by taking the derivative, which gives
M MB M + VS − B = VS − S = ρW ρw ρW
VS −
Vs ρ s
ρw
(6)
ρs = VS 1 − ρw
where we worte mass of the stone, M www.aieeepage.com
d 0= (2 cos θ − 3cos 2 θ) dθ = −2sin θ + 6 sin θ cos θ
(6)
The hoop will rise up off the ground if this maximum upward force is larger than the weight of the hoop. That is, if
then water
11.b.
10.b.
ρ S > ρW the result in negative
meaning that when the stone is in the boat the displaced volume of water is larger then when the stone is in the water. Using work energy theorem 1
∫ F .ds = 2 MV
2 2
−
1 MV12 where v2 is the final 2
velocity of boat.
2mg m 3 > mg ⇒ > (7) 3 M 2 Remark: Alternatively, we can solve for the minimum value of m/M by setting the upward force, 2 mg (2 − 3cos θ) cos θ , equal to the weight of the hoop, Mg, and then using the quadratic formula to solve for cos θ . A solution for cos θ exists only if the discriminant is positive, whichis the case only if m/M>3/2. From Archmedes principle the buoyant force is equal to the weight of water expelled by the abject floating in it. With the stone in the boat from the force balance we have (subscripts W, S, B refer to water, stone and boat, respectively),
and
M S = Vs ρ S . Since stone has higher mass density
(5)
Therefore, the maximum value is achieved when cos θ = 1/ 3 , in which case the upward force equals 1 1 2mg 2mg 2 − 3 = 3 3 3
S
boat → v = 5 m/s
M=200 kg F=500N
PAGE ⇒ 500 × 5 =
l1
1 × 200 v22 − 52 2
⇒ v2 = 50 m/s. = 7.07 m.s .
www.aieeepage.com 12.a. According to Gauss’ law.
VW ρW g Fb = WB + WS = M Bg + M sg (1)
so the volume of water displaced,
VW =
MB MS + ρW ρW
number of field lines is equal to enclosed charge q divided by ε 0 . for sphere of radius 2.00 cm ∴
VW is (2)
With the stone in the water the force balance for the boat gives
PAGE -
∴
A Premier Institution for IIT - JEE & AIEEE at the National Level
q = 0 ⇒ flux = 0 for sphere of radius 3.00 cm
4 of 26
q = 2µc ⇒ flux = (2 ×10−6 ) (0.11 ×1012 ) = 5
2
2.2 × 10 Nm C
∴
17.c.
−1
for sphere of radius 4.00 cm
K1πr 2 (T2 − T1 ) K 2 π[(2r )2 − r 2 ](T2 − T1 ) + l l
q = 2 + 1 = 3µc ⇒ flux = (3 × 10 −6 ) (0.11× 1012 ) = 3.3 × 105 Nm 2C −1
∴
∆Q ∆Q ∆Q + = t t ∆ ∆ inner outer ∆ t total
for sphere of radius 5.00 cm
=
q = 2 + 1 − 3 = 0 ⇒ flux = 0 .
13.c.
vn =
v1 ; n
v2 =
2.19 × 106 m/s 2
2.19 × 106 m/s n
=
14.a.
15.c.
v2 = 2πr 2
∴ K=
18.a.
2 × 2 π × 4 × 5.3 × 10−11
8 × 106
K1 + 3K 2 4
[gT 2 ] = MLT −2T 2 = ML ≠ [4πr 2 ] 2 2 4 2 2 −3 [Force] [distance] ⋅ L 3 πρ G r = [ρ G r ] = ML 2 [Mass]
When the springs at their rest length, just as the man touches the ground, the net force is mg. The spring force kx increases until it equals mg. At this moment the net force is zero, but the man has a maximum downward velocity. Now kx increases to greater than mg and the acceleration of the man increases from zero in the opposite direction until the springs are at maximum compression. Use conservation of energy and of momentum. There is no initial kinetic energy, just potential.
ML www.aieeepage.com =
Also, [Gm]=
19.a.
M v 1 1 M1 2 2 + M 2 v22 2 M 2 1
=
=
1 M 22 + M 2 v22 2 M1
v2 =
[Force] [distance]2 = [mass]
When the distance between the screen and object is greater than four times the focal length there are two positions of the lens when it forms real images of the object on the screen. There positions are conjugate and the product of the magnifications = 1 If d is the size of object and d1 and d2 the sizes of images, then
PAGE
∴ distance d (distance between virtual sources) d = 4.05 × 10 −3 × 2.9 × 10 −3 λ⋅D = d
www.aieeepage.com β=
2 gRM 1 2 gR and v1 = M 2 M1 + M 2 M1 (M1 + M 2 )
As v > EB, so force on electron due to electric field is greater than that due to magnetic field. Due to which the electron will not reach to the undeflected spot on screen but gets deviated in the direction opposite to that of electric field and meets the screen to a spot below the undeflected position.
PAGE -
⋅ L4 = L2 T –2 ≠ [g]
d = d1d 2
1 M 2 (M 2 + M1 ) 2 v2 2 M1
This then gives
M2
2 2 –2 4 [ρ G r ] L T πρ = = = LT −2 = [ g ] . G r 3 [r ] L
2
M 2 gR =
MLT -2
MLT −2 2 ⋅ L = L3T −2 = [ gr 2 ] M
1 1 M 1v12 + M 2 v22 2 2 Conservation of momentum involves only the motion in one dimension, so 0 = M1v1 + M 2 v2 These are two equations in two unknowns. Eliminate v1 between them to get
-3
[gr 2 ] = LT −2 L2 = L3T −2
M 2 gR =
16.b.
πr 2 (T2 − T1 ) l
K π 4r 2 (T2 − T1 ) l
PAGE
Number of revolutions made in 1s
=
or ( K1 + 3 K 2 ) =
2.19 × 106
K π (2r )2 (T2 − T1 ) l
20.c.
5893 × 10−3 × 1 4.05 ×10−3 × 2.9 ×10−3
= 1.72 ×10−4 m . Usually a medium has only one frequency for a given harmonic. The difference in this case is that by increasing the frequency, we are increasing the tension in the string and, in effect, actually altering the medium. From v = sqrt(T / µ ) we know that
A Premier Institution for IIT - JEE & AIEEE at the National Level
5 of 26
increasing the tension in the string increases the
23.c.
velocity of the wave. From v = f λ we know that frequency and velocity are proportional to each other. From question #705 ( L = nv / 4 f ) we see that velocity is proportional to string length, while frequency is inversely proportional. Thus, as velocity increases, frequency increases by the same proportion resulting in the same string length at any velocity and frequency. (Since the string is not perfectly elastic, there is a shortening of the wavelength that complicates calculations.) In this particularly odd situation, we can achieve resonance for any harmonic at any frequency on the same string. Note: It is a very good idea to try this experiment at home. A long string of beads or a telephone cord work particularly well. Try to establish different harmonics and then change the frequency. 21.a.
2
2
mv mv ke − = KE = 2 r 2 2r KE =
KA( 2T − TC ) KA(TC − T ) = BC CA
24.a.
22.a.
)
3
2
2 3
3T 2 +1
[charge]2
[ ε] =
2
[distance]
⋅
.
1 = Q 2 L−2 [Force]–1 [force]
1
1 1 µ −1 1 = (µ − 1) − = f − R 2 R −2 R
[force] [distance]2 2
= M −2 L2 [Force]
[force] ⋅ [distance] = Q −1L1 [Force] [charge] [power]
[current]2
=
[force] [velocity] [charge]2 [Time]−2
= Q −2 L1T 1 [Force]
⇒ [velocity] = ε −1 R −1 [V 2 ε] = [force] [εV] = Q1L1 ≠ [dipole moment] angular displacement is dimensionless
.
As f = (µ − 1) R − R 2 1
25.c.
∴ [angular displacement] = ε 0 R 0 G 0 V 0 Before collision, the mirror is at rest, let object is at a distance x from the mirror and image is at a distance y from mirror. As, y = x
PAGE
dy dx = dt dt r = − u (as x is decreasing) After collision object stops and mirror starts moving with u rightwards. Let from the object at any time mirror is at a distance x and image is at a distance y. From reflection at mirror, y = 2x
www.aieeepage.com
2R µ −1
C2
TC ( 2 + 1) = 3T
1 −1 1 Clearly, [εR] = L T = velocity
(n h )
In a concavo convex lens, centres of curvature of both the surfaces lie on the same side. Therefore, R1 and R2 have same sign (both + or both negative). Therefore, focal length shall not depend on direction from which light is incident.
f =
or
[ R] =
(
1
TC − T = 2T − 2 TC
[V ] =
1 ke2 1 ke 2 ⋅ 3 = ⋅ 6 r 6 4π2 RC 2 m
1
or
www.aieeepage.com[mass]
n2h2 ke 2 4π2 ke2 m = ; r = 4π2 ⋅ 2mr 2r 3 n2h2
m3
2T − TC TC − T = l 2l
[G ] =
1 2 ke mv = 3 ; 2 2r
6 ∴TE α n
∴
TC =
2
∴ TE =
AB = BC = l , CA = 2 l
If
PAGE
− do −∂ ke 2 − ke 2 F= = = 2 = dA dr 3r 3 r 2
Let Tc be the temperature of point C in the steady state. As TB > TA, therefore, heat flows from B to A directly as well as through C. In the steady state, Rate of heat flow in BC = Rate of heat flow in CA
(in magnitude)
C1
P1
P2
⇒
dy dx =2 dt dt r = 2 u (as x increasing)
PAGE -
A Premier Institution for IIT - JEE & AIEEE at the National Level
6 of 26
26.c.
Of course the hot air balloon also follows the buoyant force equation:
II)
( K .E.) max = hv − hv0 ⇒ ( K .E.) max = ∞ V
ρoutsideairVg = ρair inballoonVg . The volumes are
27.d.
equal at all altitudes, so the density will decide whether or not the balloon rises. Since the bottom of the balloon is open to the outside atmosphre, the pressures are equal at all altitudes. From PV = nRT , we know that increasing temperature at constant volume and pressure requires reducing the number of moles, and thus reducing the dentisty, raising the balloon. After one month the original material is still almost all there, but there is now some of the second element present, so it contributes to the number of decays per time also. The activity is now higher than at the start. One month is a lot longer that two days, so by now on average as much of the second element is decaying per time as is created per time. It has come to an equilibrium. On average for every decay of the first element there will be one of the second. Ther are then about 20,000 decays per second. The one month interval is a lot less than the five year half-life of the third element so not much of it has decayed and it doesn’t contribute much to the activity. 100 years however is a lot longer than five years. At this time there are about as many decays per time of element #3 as of element #2, which in turn has about as many as #1. The total activity is then about 30,000 decays per second. e = 0, when conductor moves along its length. In (c) and (d), conductor moves at right angle to its length and B is perpendicular to that, therefore, e = B λ v.
False
Thus maximum kinetic energy is proportional to frequency and not Intensity. 30.a.
x = angle at which ray of light leaves the plate n ∴ n0 sin α = n3 sin x ⇒ x = sin −1 0 sin α n3 Thus x depends on angle of incidence and refractive indices of the media on both sides of the plate. At any point in the plate, n sin θ = constant = n0 sin α . ∴ When n=n0 sin α , θ = 90º ⇒ T.I.R.inside the plate. Since hydrogen and oxygen are both diatomic the adiabatic exponent of the mixture formed by their masses in any ratio is also equal to that of a diatomic
PAGE 31.a.
gas i.e.,
7 . 5
R R www.aieeepage.com C = = γ −1 7
28.a.
29.a.
I).
uur r dl τ = uur dt
v
r τ=0
∴
ur L = constt M
γ = 4 / 3 m. Cv´ =
1 MR 2 w1 = w 2 1 1 M 2 4 +1 5 2 2 l2 = MR 2 + R = MR = MR 2 2 4 8 8 w2 = ? l1 =
∴
Cv Cv´
=
5R 5 = 2 × 3R 6
5 6 To conserve momentum in a system of colliding objects, all unbalanced forces must act only between objects in the system. No outside forces must affect the objects. In a system isolated in this manner, an object’s momentum will change only because of collisions with other objects, which will gain or lose momentum with respect to each other. The overall system momentum will stay constant, however. (I) True
PAGE 33.a.
l1 w1 = l2 w2
R = 3R 4 ; −1 3 ∴ n=
R
∴
5R 2
−1 5 When they combine chemically a triatomic gas (H2o) is formed.
32.a.
Since
=
When conducting rod AB moves parallel to x-axis in a uniform magnetic field pointing in the positive z-direction, then according to Fleming’s left hand rule, the electrons will suffer a force towards B. Hence the end A will become positive.
www.aieeepage.com (II) False
1 MR 2 × w l1 w1 2 8 4 w2 = = = w= w 5 2 5 5 l2 × 2 MR 8
When the angular displacement is 20º. the mass is at an extreme end mv 2 r At extreme end v = θ ∴ T − mg cos θ T − mg cos θ =
PAGE -
A Premier Institution for IIT - JEE & AIEEE at the National Level
7 of 26
θ
64 2 −8 E1 − E2 × 100 = × 100 = 75% E1 32
πT
37.a.
34.c.
(I)
π
π mg sin θ
mg
π
θ
Let r be the radius of ball. Since the block is large final temperature is 0ºC. If ρ is the density of ball and s its specific heat then the heat lost by iron ball
mg cos θ
=
True
2 3
3 Heat absorbed by ice = = π r ρ´× L
V2 = VX Vy 1
4 3 πr ρsθ 3
Where ρ´ is the density of ice
V1
PAGE
Vx
1
V
Y
V1 = Vx + Vy > Vx 2
2
2 3 4 π r ρ´L = π r 3ρ s θ 3 3
Vx
V3
θ=
V5 = Vx + Vy > Vx 2
2
38.d.
As shown in the figure, the velocity at 1 and 3 i.e., at any arbitrarty points before and after the topmost point is greater then vx.
ρ´L 2ρ s
The density of the plasma generated from the mercury vapor is very low, around 1015 charged particles per cubic meter. As a result, the extremely high temperature of the plasma is balanced by an extremely low heat capacity. Very little of the energy from the disassociated electrons and ions in the plasma is given off as heat. (I) True For the light to split, the material should have refractive index greater that 1 through which the light passes. Since the prism is hollow, we get no spectrum. (II) False Total energy of the ring = (K.E.)Rotation + (K.E.) Translational
www.aieeepage.com
(II) True Speed of sound waves in water is greater than in air. 35.d. (I) False If the sound reaches the observer after being reflected from a stationary surface and the medium is also stationary, the image of the source will become the source of reflected sound. Thus in both the cases, one sound coming directly from the source and the other coming after reflection will have the apparent frequency (Since velocity of source w.r.t. observer is same in both the cases). Therefore no beats will be heard. (II) False The electrons in a conductor are free and have thermal velocities. Thus electrons will be in motion even in the absence of potential difference. 36.d. R = radius of bigger sphere
8Q = (4.5 × 10 ) R 9
2
Q l = mr 2 VC = rw
Total K. Energy of the cylinder = (K.E.)Rotational + (K.E.) Translational 1 1 = l ' ω2 + MVC2 2 2 11 1 = Mr 2 w' 2 + M (rw ') 2 22 2 3 = Mr 2 w '2 4
www.aieeepage.com
E2 = Electrical energy of eight smaller spheres
Q2 (8) R Percentage of electrical energy that has been converted to the other forms of energy is
= (4.5 ×109 )
1 2 1 l ω + mVC2 2 2 1 1 = × mr 2 w2 + m(rw) 2 2 2 =
PAGE
4 4 π R 3 = 8 π R 3 ⇒ R = 2r 3 3 E 1 = Electrical energy of the bigger sphere ∴
39.a.
2 2 Equating (i) and (ii) mr w =
⇒
....(1)
3 Mr 2 w2 4
w' 2 4m 4 0.3 = = × =1 w2 3M 3 0.4
w ' = w. ⇒ ⇒ Both will reach at the same time. The statement is False.
PAGE -
A Premier Institution for IIT - JEE & AIEEE at the National Level
8 of 26
40.b.
(I) False. When water is heated at end x, the density decreases and the water moves up. This is comparated by the movement of water from y to x i.e., in clockwise direction.
X
∆S=
43.c.
Y
44.c.
41.a.
(II) True The statement is True. The metallic sphere which gets negatively charged gains electrons and hence its mass increases. The metallic sphere which gets positively charged loses electrons and hence its mass decreases. By Anpere’s law
When a wave pulse strikes the boundary between two strings, part of the pulse is transmitted along the heavy string at velocity v2. Since the second string is denser, however, most of the pulse is reflected back of velocity v1. The light string exerts an upward force at the boundary, causing an upright transmitted pulse, while the heavy string .... a downward force, causing an inverted reflected pulse. For simplicity, assume that the balls are separated by a very small distance, so that the relevant bounces happen a short time apart. This assumption isn’t necessary, but it makes for a slightly cleaner solution. Just before the basketball hits the ground, both balls are moving downward with speed (using mv2/ 2=mgh)
PAGE
φ B ⋅ dr = µ0 I = B(2π R2 ) = µ0 I
dφ dB = − AN 2 dt dt
45.c.
Magnetic moment
R2
E (2π R2 ) = − ( π R22 ) µ0 N 2 dI 4π dt
= 10 −7 × 5 × 0.2 42.d.
q 2 q m R 2 ω= R 2 ω . 2m 3 3
N 2µ0 1 dI ⋅ 2π R2 dt
46.d.
Reflected sound from the wall approaches the observer with velocity u. Both source and observer approach each ⇒ other with velocity u.
PAGE
v+u Apparent frequency = f . v−u The passage says that each wave form represents the same note, and that a note is the same set of harmonics. This means that A, B and D are not true. The wave form shown is a combination of all the harmonics for that note. The first harmonic would have the same wavelength, and thus the same frequency and perid. ∴
47.c.
= 10 −7 volt/metre 18gm of water arises from the melting of an equivalent amount of ice ( ∆ m) due to the friction between the blocks. Q = amount of heat required = ( ∆ m) L W = amount of work done =f ∆S = ( µ mg) ( ∆ S) [Q m=10.2kg>> ∆ m] ∴ Q=W
PAGE -
(1)
magnetic moment q = angular momentum = 2m
R1
E=
v = 2 gh
Just after the basketball bounces off the ground, it moves upward with speed v, while the tennis ball still moves downward with speed v. The relative speed is still 2v. (This is clear if you look at things in the frame of the basketball, which is essentially a brick wall.) Since the upward speed of the basketball essentially stays equal to v, the upward speed of the tennis ball is 2v+v=3v. By conservation of energy, it will therefore rise to a height of H=d + (3v)2/(2g). But v2 = 2gh, so we have h = d + 9h. Gyromagnetic ratio
www.aieeepage.com
dB µ0 1 dI = dt 2π R2 dt φE ⋅ dr =
(∆m) (L) (18 ×103 ) (3.4 ×105 ) = = 1km (µ) (m) (g) (0.06) (10.2) (10)
www.aieeepage.com 48.b. first harmonic
A Premier Institution for IIT - JEE & AIEEE at the National Level
second harmonic Flute
9 of 26
49.a.
50.b.
From the equation L = nλ / 2 where n= 1, 2, 3, .... for each harmonic, we see that the second harmonic has half the wavelength of the first, so the period is half as long. From question #967 we know that the period of the note is the same as the period of the first harmonic. It is the sum of these harmonics that gives the wave its irregular shape. For the oboe, the ratio of the amplitudes is different, so the oboe sound wave has a different shape. This is the difference that the ear perceives when we hear the same note on different instruments. From the equation L = nλ / 2 where n = 1, 2, 3, ... for each harmonic, we see that the third harmonic has one third the wavelength of the first, so the period is one third as long. From question #967 we know that the period of the note is the same as the period of the first harmonic.
PAGE
www.aieeepage.com
PAGE
www.aieeepage.com
PAGE -
A Premier Institution for IIT - JEE & AIEEE at the National Level
10 of 26
Paper - I I : Mathematics - 51 to 100 Solutions
51.b. 52.a.
∴− 1 ≤ x + [ x ] ≤ 1 clearly x ∈ [0,1)
⇒x=
f ( x ) = 2 cos 2 x + 3 sin 2 x + 1 = π π 2sin 2 x + + 2 or 2 cos 2 x + + 2 6 3
f = [−π 3, π 6]
Let X=domain of
∴ f −1 ( x ) = or
56.c.
∴ f : x → y is both one-one & onto.
⇒ z = −1, ω, ω2 are the roots of (1)
∴ f −1 ( x) exist & f −1 : y − x
Again when
PAGE
1 −1 x − 2 π 1 −1 x − 2 π sin − − or cos 2 2 2 3 2 6
x +1 1 = 1− 2 , also 2 x +2 x +2 2 ≤ x 2 + 2 < ∞∀x ∈ R
⇒
1−
When
www.aieeepage.com
1 ≤ 2
−
= (−1)1985 + (−1)100 + 1 = −1 + 1 + 1 = 1 ≠ 0
z = ω, z1985 + z100 + 1 = ω1985 + ω100 + 1
= ω2 + ω + 1
[Qω3 = 1]
=0
1 <0 x +2
z = ω2 , z1985 + z100 + 1
When
2
= (ω2 )1985 + (ω2 )100 + 1
1 1 1 x +1 ≤ 1− 2 <1; ≤ <1 2 x +2 2 x2 + 2 2
= ω3970 + ω200 + 1 = ω + ω2 + 1 = 0
−1 x + 1 ∴ Range of sin x 2 + 2 is [π 6, π 2) 2
x < 1, f ( x) = x identity mapping ∴ f −1 ( x) = x
1 ≤ x ≤ 4, f ( x) = x 2 , let y = x 2 ⇒ ∴ f −1 ( x ) = x
x2 ∴ f ( x) = 64 −1
y=x
Hence common roots are ω, ω2 ∴ (c) is correct
57.a.
3 cos x − sin x ≥ 2
⇒
3 1 cos x − sin x ≥ 2 2 2
PAGE
π π ⇒ cos + x ≥ cos ; 4 6
x > 4, f ( x ) = 8 x , let y = 8 x ⇒
y2 =x 64
⇒
π π π ≤ +x≤ 4 6 4
and in general
π π π www.aieeepage.com 2nπ − ≤ + x ≤ 2nπ +
here Y = [log e 5, log e 19] (range) Let y=log e ( x 2 + 3 x + 1) ⇒ e y = x 2 + 3 x + 1
⇒ x 2 + 3 x + (1 − e y ) = 0
⇒
π ∴ x ∈ 0, π 5π ≤x≤ ; 12 12 12
4
Given g :[1,3] → Y ,
PAGE -
z = −1, z1985 + z100 + 1
= (ω3 )661 ω2 + (ω3 )33 . ω + 1
1 1 ≥ 2 >0 2 x +2
⇒−
55.b.
z3 + 2z 2 + 2z + 1 = 0
⇒ ( z + 1) ( z 2 + z + 1) = 0
2
54.a.
−3 + 5 + 4e x 2
[π 6, 2π 3] & Y = Range of f = [0, 4]
is
53.c.
−3 ± 9 − 4(1 − e y ) 2
⇒ 2nπ −
6
π 5π ≤ x ≤ 2 nπ + and for n = 1 12 12
19π 25π ≤x≤ ; 12 12
A Premier Institution for IIT - JEE & AIEEE at the National Level
4
⇒
19π ≤ x ≤ 2π 12
11 of 26
19π ∴x∈ , 2π 12
=
π 19π ∴ x ∈ 0, ∪ , 2π 12 12 1 – (probability of not selecting the right number in the two trials)
58.b.
9 8 = 1− × . 10 9 (choose all 9 incorrect out of the total ten available
=
60.c.
61.c.
BD Area of ∆ABD Then DC = Area of ∆ADC A
= S B
C
D
=
2 2 Now ( x + y ) = ( x − y ) + 4 xy
PAGE
Area of ∆ABS Area of ∆ASC
y = m( x + 1) +
1 using the formula m
a ...(1) m and the equation of any tangent to y = mx +
(by ratio and proportion rule)
y 2 = 8( x + 2) is
1 2 R sin ∠ ASB sin 2C = 2 = 1 2 R sin ∠ ASC sin 2 B 2
2 ...(2) m' Since (1) and (2) are at right angles
(sin 2C ) c + (sin 2 B ) b sin 2C + sin 2 B
...(1)
Area of ∆ ASC Area of ∆ CSD Area of ∆ BSA
=
+ area of ∆ ASC + area of ∆ CSD Area of ∆ BDS
sin 2C + sin 2 B = sin 2 A
...(2)
−1 m The equation of the second tangent, now is mm´= −1
∴ m´=
−1 ( x + 2) − 2 m ...(3) m The equation to the locus of the point of intersection is found by eliminating m, between (1) and (3). Subtracting (3) from (1). y=
mx +
x 2 1 + m + + + 2m = 0 m m m
1 1 x m + + 3 m + = 0 m m
www.aieeepage.com 63.b. Since t = 200
Hence, by section formula P.V. of S
(sin 2C ) c + (sin 2 B ) b (sin 2C + sin 2 B) sin 2C + sin 2 B +(sin 2 A) a (sin 2C + sin 2 B ) + sin 2 A
y = m '( x + 2) +
PAGE
=
(by ratio and proportion rule)
PAGE -
[Q x > 0, y > 0]
The equation of any tangent to y 2 = 4( x + 1) is
www.aieeepage.com
Area of ∆SBD Area of ∆SDC
AS Area of ∆ BSA Similarly, SD = Area of ∆ BDS
=−
≥ ( x − y )2 + 400 ≥ 400 ∴ the smallest possible value of x+y=20.
62.a.
Hence, the position vector of point
D=
log10 x + log10 y ≥ 2 ⇒ log10 xy ≥ 2 ⇒ xy ≥ 100
8 . 9 If S is the circumcentre of the ∆ , let AS meet BC at D.
59.a.
Every element of A can have image in B in 3 ways. So, the total number of ways in which 3 elements of A can have images in B = maximum number of definitions of f = 3 × 3 × 3 . The number of ways of arranging 1, 2, 3 in places of a, b, c is 3!= the maximum number of definitions of one-one function g.
9 and choose all 8 incorrect out of 9 available 10
in the second trial =
Σ sin 2 A a Σ sin 2 A
∴ x+3= 0
(given)
4
1 6 log x +1 ∴ C3 x
⇒ 20 x
A Premier Institution for IIT - JEE & AIEEE at the National Level
3 2(1+ log x )
3
3
1 x 12 = 200
1
x 4 = 200
12 of 26
⇒
3 1 + x 2(1+ log x ) 4
=10
taking log to base 10 on both sides, we have 3 1 + log10 x = log10 10 2(1 + log x) 4 ⇒
66.b.
T2 n +1 = a + 2nd = a + 4n
6 + (1 + log x) log10 x = 1 4(1 + log x)
Last (2n+1) terms are in G.P. with common ratio =
⇒ 6log10 x + log10 x + (log10 x) 2 = 4 + 4 log10 x
its first term = a + 4n = A (say) Middle term of A.P. = Tn +1 = a + nd = a + 2n n Middle term of G.P. = Ar
⇒ (log10 x) + 3(log10 x) − 4 = 0 ⇒ (log10 x) 2 + 4 (log10 x) − (log10 x) − 4 = 0 (log10 x − 1) (log10 x + 4) = 0
64.a.
PAGE
x = 10−4
2
2
a + 2 n = ( a + 4n ) ⋅
∴
Focus of x2 = 4by is at (0, b) and the equation of the directrix is y = –b Hence, the equation of the circle is 2
2
1 2n
a + 4n 2n = n a + 4n − a − 2n 2 − 1
i.e., ( y + 3b) ( y − b) = 0 y = b is the only solution as y is positive.
When y = b, x 2 = 4b2 or x = ±2b ∴ the point of intersection is (±2b, b) .
4 2 x 9 represents a parabola with vertex (0, 0) and focus (0, 5). Hence R ∩ R ' is the set of points indicated in the figure by shaded portion, such that
⇒
a + 4n 2n = n 2n 2 −1
⇒
a + 4n =
2n − 1
= a + 4n =
centre (0, 0) and radius 5 and the equation y =
67.b.
n ⋅ 2n +1 2n − 1
( x − 3) 2
+
( y − 4) 2
=1 42 32 Shifting the origin to (3, 4), equation becomes x2
and range R ∩ R ' = [0,5] ∪ [0, 4] = [0, 5]
.
9( x − 3)2 + 16( y − 4)2 = 144
PAGE
R ∩ R ' = {( x, y ): − 3 ≤ x ≤ 3, 0 ≤ y ≤ 5}
(0,5)
n ⋅ 2n +1
∴ middle term of the sequence = T2 n +1
The equation x 2 + y 2 = 25 represents a circle with
Thus, dom R ∩ R ' = [−3,3]
n
2
y 2 + 2by − 3b 2 = 0
42
+
y2
32
=1
... (1)
If ACX = θ , then A = (CAcosθ,CAsinθ) and
B = (CB cos (90+θ), CBsin(90+θ)) www.aieeepage.com = (-CB sinθ, CBcosθ)
(3,4)
(-3,4)
A
B (3,0)
Test Line
(-3,0)
PAGE -
n
a + 4n 2 www.aieeepage.com = or x + y − 2by − 3b = 0 1 a + 2n 2
x + ( y − b) = (2b) This meets the parabola x2 = 4by at the point given by
65.a.
1 = (a + an) 2
log10 x + 4 = 0
x = 101
1 2
∴
2
log10 x − 1 = 0
The given relation is not a function as according to VPL Test, if we draw a Vertical Parallel Line it intersects the curve at two real and distinct points i.e., for one value of x we have two different values of x, which is against the definition of function. Let a be the first term. ∴ first (2n+1) terms are in A.P. with common difference =2
A Premier Institution for IIT - JEE & AIEEE at the National Level
C
X
13 of 26
∴ RHL = lim f (1 + h) = lim(1 + h) = 1
Since (CA cos θ, CA sin θ) satisfies the equation (1) of the ellipse
h →0
CA 2 cos 2θ CA 2 sin 2 θ + =1 16 9
∴ the ,imit of f ( x ) as x → 1 exists.
cos 2 θ sin 2 θ ∴ = + 16 9 CA 2
...(2)
sin 2 θ cos 2 θ = + Similarly 16 9 CB2
...(3)
1
1
1
69.b.
⇒ cos −1 x − cos −1 y =
1
cos −1 x + cos −1 y =
x 4 x 2 < 1 f ( x) = 2 x x ≥ 1 x 4 ⇒ f ( x) = x
x
<1
x
≥ 1
PAGE
−1 Hence, cos x =
70.a.
π ; 3
...(2) ⇒ x = cos
π 1 = 3 2
1 ⇒ ( x, y ) = ,1 2 Hence, only one pair. Since distance covered by a particle at any time t is
t4 − 2t 3 + 4t 2 + 7 4 Case 1: Extreme Distance dS dS = t 3 − 6t 2 + 8t ; ⇒ = t (t − 2) (t − 4) dt dt Distance covered will be maximum or minimum if
f ( x) LHL = xlim →−1 Let x = –1–h (h>0)
dS =0 dt
As x → −1 ⇒ h → 0
∴ LHL = lim f (−1 − h) = lim(−1 − h) = −1 h →0
f ( x) and RHL = xlim →−1
⇒ t (t − 2) (t − 4) = 0 i.e., t = 0, 2, 4
Now,
Let x = −1 + h (h > 0) As x → −1+ ⇒ h → 0
Since LHL ≠ RHL
dt
= 3t 2 − 12t + 8
2
PAGE
∴ RHL = lim f (−1 + h) = lim(−1 + h) = 1 h →0
d 2S d 2S
4
dt 2
> 0 for t = 0, 4, and
d 2S
∴ the limit of f ( x ) as x → −1 does not exist. Case 2: at x = 1
LHL = lim f ( x) x →−1
> 0 for t = 2 dt 2 ∴ distance is minimum at t = 0, 4 Option (III)
Case II : Extreme Velocity Now, velocity of the particle is given by
www.aieeepage.comdS
Let x = −1 + h (h > 0)
v=
+ As x → −1 ⇒ h → 0
∴ LHL = lim f (1 − h) = lim (1 − h) 4 = 1
f ( x) Also, RHL = hlim →0
π 3
S=
Case1: at x = –1
h→0
...(1)
www.aieeepage.com
x x ≤ −1 ⇒ f ( x) = x 4 − 1 < x < 1 x x ≥1
h →0
π 3
and cos −1 y = 0 ; x − y = cos 0 = 1
x 4 −1 < x < 1 ⇒ f ( x) = x x ≤ −1 or x ≥ 1
h →0
π 3
sin −1 y − sin −1 x =
π π π ⇒ − cos −1 y − − cos −1 x = 2 2 3
1 1 25 + = + = Hence CA 2 CB2 16 9 144
68.d.
h →0
Since LHL = RHL
h→0
dt
dv = 3t 2 − 12t + 8 dt Velocity of the particle is maximum / minimum if ⇒
Let x = 1 + h (h > 0) As x → 1+ ⇒ h → 0
PAGE -
A Premier Institution for IIT - JEE & AIEEE at the National Level
14 of 26
dv =0 dt
Y Y´
⇒
3t 2 − 12t + 8 = 0
⇒
t=
12 ± 144 − 96 6 ± 2 3 = 6 3
2
Now,
d v dt
2
> 0 at t = d 2v
and
dt
2
and OY´ =
dv d 2 S = 2 dt dt
(6 − 2 3) 3 Option (I)
r and the component of a in the direction of OY´
PAGE
r r r r r j +i y− x = ( x i + y j + z k ) ⋅ = 2 2
Given
da =0 dt
Now,
73.b.
t=2 d 2a dt 2
Clearly a = ( x + 2)! ;
b = x P11
x! ⋅ ( x − 11)! ( x − 11)!
2
x + 3 x + 2 = 182 2
x + 3 x − 180 = 0
⇒ ( x + 15) ( x − 12) = 0 72.d.
dy dy Since, dx × dx = − 1 C1 C2
PAGE 74.b.
Hence, the two curves are orthogonal. x>2 1
⇒ x = 12.
∫ 0
x www.aieeepage.com = 1 + 4x +
π r oz through an angle , the z component of a is 4 not altered. r r r r Let a = x i + y j + z k in the first system. r r πr πr i+ j OX´ = cos i + sin j = 4 4 2
x
∫
⇒ f ( x) = (6 − †) dt + (4 + †) dt
[Q x cannot be –ve] Since the coordinate system oxyz is rotated about
PAGE -
=3 2
y2 + y +1 dy dx = 2 C2 x + x + 1
⇒ ( x + 2) ( x + 1) x ! = 182 x ! ⇒
2
For the family of curves represented by the second differential equation, the slope of the tangent at any point is given by
Option (II)
Since a = 182 bc
⇒
2
y−x
x2 + x + 1 dy = dx C1 y 2 + y + 1
c = ( x − 11)! ∴ ( x + 2)! = 182.
= 2 2 and
x = −1 and y = 5 y − x = 6 r Hence a in the first system = (–1, 5, 4) For the family of curves represented by the first differential equation, the slope of the tangent at any point (x, y) is given by
> 0 throughout
∴ Acceleration is minimum if t = 2
71.b.
x+ y
www.aieeepage.com y + x = 4
da = 6t − 12 dt Acceleration of the particle is minimum or maximum if ⇒
⇒
2
r r r r r i+ j x+ y = ( x i + y j + z k ) ⋅ = 2 2
CaseIII: Extreme Acceleration Let a is the acceleration of the particle, such that a=
r r i+ j
r Then the component of a in the direction of OX´
(6 − 2 3) 3
∴ Velocity is maximum if t =
X
O
(6 + 2 3) 3
< 0 at t =
X´
1
2
2
x ≤ 2 ⇒ f ( x) = 5 x + 1 − + At x = 2. f (2) = f (2 ) = f (2 ) = 11
⇒ f ( x) is continuous of x = 2 LHD = Lt
A Premier Institution for IIT - JEE & AIEEE at the National Level
h →0
f (2 − h) − f (2) −2 15 of 26
=
5(2 − h) + 1 − 11 =5 −h
RHD = Lt
h →0
a/b a/b x x = exp. log 1/ 2 = y y1/ 2
f (2 + h) − f (2) h
a y and exp. log 1/ 2 c x
(2 + h)2 + 4(2 + h) + 1 − 11 2 = Lt h →0 h = Lt
h →0
a/b a/c x y ∴ B = 1/ c + 1/ b y x
h (2 + 4) =6. h
x a / b a Tr +1 = Cr 1/ c y
LHD ≠ RHD 75.d.
y = x1/ b
i.e., f ( x ) is not differentiable at x = 2. Let the source of light be situated at A(a,0,0), where a ≠ 0. Let OA be the incident ray and OB the reflected ray. ON is the normal to the mirror at O.
a
a −r
y a / c 1/ b x
PAGE =
N 1,-1,1 A(a,0,0)
a
Cr x
B
y
a/c
r
a a ( a − r ) r b bc
a a ( a − r ) r bc c
lm
n
>
If powers of x and y are same, then www.aieeepage.com a a a a (a − r ) − r = − (a − r ) + r b bc bc c
<
— — 2 2 0(0,0,0)
∴ θ 2 D.R.´s of OA are (a, 0, 0) and so D.C.´s are (1, 0, 0) ∴ ∠AON = ∠NOB =
1 1 1 ,− , Similarly, D.C.´s of ON are 3 3 3
θ 1 cos = 2 3 Let l, m, n be the D.C.´s of the reflected ray OB. Then, ∴
l +1 m+0 1 = = , 2 cos θ / 2 3 2 cos θ / 2 n+0 = 2 cos θ / 2 ⇒
1 3 1
77.c.
r=
a (c − 1) . b+c+2
1 1 f ( x) = 1 + n 1 + sin x cos n x
1
= 1+
n
sin x
∴ f ´( x) = −
−
2 −2 2 l = − 1, m = ,n= 3 3 3
1 2 2 ⇒ l=− ,m=− ,n= 3 3 3 Hence, D.C.´s of the reflected ray are
= +
cos x 1
sin
1 sin
2n
2n
x cos 2 n x
sin
x cos
n +1
sin
1 n
sin x cos n x
⋅ n sin n −1 x cos x ⋅ n cos n −1 x sin x
(sin x
n
n +1
+
n(sin n −1 x cos n +1 x − sin n +1 x cos n + 2 x )
n
n +1
x
cos 2 n x
PAGE
3
n
1
+
and
1
+
x cos n +1 x
n+2
x − cos n + 2 x
)
(sin 2 x − cos 2 x) ...(1)
π www.aieeepage.com ∴ f ( x) = 0 ⇒ sin x = cos x ⇒ x = . 4
1 2 2 − 3 ,− 3 , 3
76.c.
a x exp. log 1/ 2 b y
PAGE -
π , f ´( x ) changes sign from negative to 4 positive
At x =
(Q For x <
A Premier Institution for IIT - JEE & AIEEE at the National Level
π , sin x < cos x and 4
16 of 26
for x >
π , sin x > cos x ) 4
k +2 ⇒ ( k + 1)! < 2
π f ( x ) is minimum. 4
Hence at x =
∴ Statement is also true for n = k + 1. Hence by Principle of mathematical Induction given is true, ∀n > 1 .
2
78.b.
n 2 + 1 2 The minimum value is . I) False
79.c.
dy = cot x cot y dx ⇒ tan y dy = cot x dx
x * y = x − y + 2 is rational.
Given that,
Consider x = 2 2 , y = 2 then x * y = 2 2 − 2 + 2 = 3 2 (irrational)
⇒ log (C sec y ) = log sin x where C is a suitable constant. ⇒ C s e c y = s in x π π y = w hen x = 4 4 1 ⇒ C 2 = 2
PAGE
y * x = 2 − 2 2 + 2 = 0 (rational)
x *y ≠ y*x ∴ Hence * is not symm ⇒ * is not equivalence relation. II) True The given inequality can be written as
1 2 ∴ 2sin x = sec y i.e. C =
www.aieeepage.com 80.c. I) True
n
n +1 ( n !) < for n > 1. 2
Let us use mathematical induction to check the validity of given inequality.
A , B, C are three unit vectors s.t. A. B = A. C = 0
2
9 3 For n = 2, we have 2! < = which is true 2 4 ∴ Inequality is valid for n = 2. k
k +1 Let it be valid for n = k then k ! < ....(1) 2
k +1 2
Consider (k + 1)! = (k + 1)k ! < (k + 1) Using (1) Now we will try to check k
k +1 k + 2 (k + 1) < 2 2
k +1
..........(2)
Which is equivalent ot write k +2 2< k +1
Now,
k +2 k +1
k +1
k +1
Now eq. (1) shows that A is perpendicular to both B and C
B × C A ⇒ B × C = λ A where λ is any scalar.
⇒
B× C = λ A
⇒
cos π / 3 = ±λ
(as π / 3 is the ∠ between B & C ) ⇒ λ = ±1 / 2
PAGE
1 B × C = ± A ⇒ A = ±2 ( B × C ) 2 Given statement is true. ∴ II) True
⇒
k +1
...........(3)
1 = 1 + k +1
k
and angle between B and C is π / 6 .
k +1
z1 = x1 + iy1 then give that z 2 = x2 + iy 2
Let z = x + iy www.aieeepage.com
z1 ∩ z 2 ⇒ x1 ≤ x2 & y1 ≤ y 2
2
= 1 + (k + 1)
1 (k + 1)k 1 + + ... 2 ! k +1 k +1
(Using Binomial expansion) ∴
k +2 k +1
k +1
>2
⇒ (3) holds and hence (2) holds
PAGE -
then 1 ∩ z ⇒ 1 < x & 0 < y (by def.) Consider
1 − z 1 − ( x + iy ) = = 1 + z 1 + ( x + iy ) (1 − x) − iy (1 + x) − iy × (1 + x) + iy (1 + x) − iy
A Premier Institution for IIT - JEE & AIEEE at the National Level
17 of 26
2a
A(Z1 )
∫
l=
Let
0
O
2a
B(Z2 )
= =
=
C(Z3 )
1− x2 (1 + x 2 ) + y 2 1− x2 2
(1 + x) + y
2
− −
a
(1 + x) 2 + y 2
[Using
2iy 2
(1 + x) + y
2a
l=
1− x2 1− z ≤ 0 and ∩0 ⇒ 1+ z (1 + x) 2 + y 2 (1 + x) 2 + y 2
0
2l =
⇒ x ≥ 1 and y ≥ 0 which is true as x > 1 & y > 0 ∴ The given statement is true V z. Since the given systems has a non-trivial solution. ∴ 2b − c b a
b
f ( 2a − x ) f ( 2a − x ) + f ( x )
∫ 0
a
f ( x ) + ( 2a − x ) dx = = 1. dx f ( x ) + f ( 2a − x )
∫ 0
⇒ l=a ∴ The given statement is true.
2b = 0
Consider ( n + 1)( n + 2)...( n + r )
a
r! =
1.2.3...(n − 1) n(n + 1)( n + 2)...(n + r ) (n + r )! = 1.2.3...n.r! n!r!
a b a+c ⇒ 2b − c b 2b = 0 0 0 −c
=
⇒ − c(ab − 2b2 + bc) = 0
Thus given statement is true.
n+ r
Cr
= some integral value
⇒ (n + 1)(n + 2)...( n + r ) is divisible by r!
⇒ − bc(a − 2b + c) = 0 ⇒ 2b = a + c
84.c.
(Q b ≠ 0, c ≠ 0)
f ( x + 1) − f ( x ) = b( x 2 + 2 x + 1) + c( x + 1) + d
⇒ a, b, c are in A.P.
− (bx 2 + cx + d )
x2 In −8 < x < 8, y = 64 + 2 = 2 .
= 2bx + b + c = 8x + 3
PAGE
∴ the required area = the shaded area =
1 (1 + 3) 2 = 4 . 2
⇒ b = 4 and c = −1 ⇒ The roots of the equation x 2 + bx + c = 0 are
Y Y=2
−4 ± 20 = −2 ± 5 2
www.aieeepage.com Hence, roots are irrational.
(3,2)
Y=X–1
O
......(2)
= [ x]02a = 2a
Operate R3–R1
82.b.
0
www.aieeepage.com II) True
b a+c
a
∫ f (a − x)dx ]
Adding (1) and (2), we get
PAGE
≤0
2
81.a.
∫
2a
− 2y
∫
a
f ( x )dx =
0
2
....(1)
f ( 2a − x ) dx f (2a − x) + f [2a − (2a − x)]
∫ 0
iy (1 − x + 1 + x)
f ( x) dx f ( x ) + f ( 2a − x )
(1, 0)
85.b. X
I) False If ( x − r ) is a factor of f (x ) repeated m times then f '( x ) is a polynomial with ( x − r ) as factor repeated
83.c.
I) True
PAGE -
at least (m − 1) times. ∴ Statement is False.
A Premier Institution for IIT - JEE & AIEEE at the National Level
18 of 26
II) True tan A =
then 86.c.
π ⇒ x 1 + cos x + 1 = 1 ...(3) 2
1 − cos B 2 sin 2 B / 2 = = tan B / 2 sin B 2 sin B / 2 cos B / 2 tan 2 A =
2 tan A 2 tan B / 2 = = tan B 1 − tan 2 A 1 − tan 2 B / 2
∴ I=
∴ Statement is true.
The box contains 2n shoes. We can choose 2r shoes out 2n shoes in 2n C2r ways. We can
88.c.
choose one complete pair out of n pairs in n C1 ways. Now we have to avoid a complete pair. While choosing (2r–2) shoes out of remaining (n–1) pairs of shoes, we first choose (r–1) pairs out of (n–1) n −1
pairs. This can be done in
n(
Cr −1 ) ⋅ 2 2n
87.c.
.
C2 r
πx ⇒ x 1 + cos + 1 = 1 2
0
f ' ( x) + f ' (− x) 2 f ' (− x) + f ' ( x) ψ (− x) = = −ψ ( x ) 2
ψ ( x) = φ' ( x) =
www.aieeepage.com 5 4
0 1 2 − 10 = 0 2 1 5
89.c.
LHS on expanding gives 5 (10+10)–4(5+20) =100 – 100 = 0 ∴ Statement is True. Any line PC through the intersection of x − 7 y + 5 = 0 and x + 3 y − 2 = 0 is of the form x − 7 y + 5 + λ ( x + 3 y − 2) = 0
...(1)
or x(1 + λ) − y (7 − 3λ) + 5 − 2λ = 0
πx π ∈ − ,0 x ∈ [−1, 0) ⇒ 2 2
B
πx ⇒ 1 < 1 + cos <2 2 πx ⇒ 1 + cos >1 2 πx ⇒ 1 + cos = 1 ; 2
f ( x) + f (− x) which is an even 2
PAGE
πx ⇒ 1 + cos = 0 2
πx <1; 2
−1
∫ dx + ∫ 0 dx + ∫ dx
∴ ψ is odd. ∴ Statement is true. II) True Statement can be restated as– ‘three lines are concurrent’ for which
x −π x ∈ [−2, − 1) ⇒ π ∈ −π, 2 2 ⇒ 0 ≤ 1 + cos
−2
=1+1=2 I) True
Now
Hence the prob. of the reqd. evert r −1
1
function.
each of these (r–1) pairs choose (r–1) single (Unmatching) shoes from each pair. This can be done in 2r–1 ways. Thus the no. of favourable ways is
n −1
0
Consider φ ( x) =
Cr −1 ways. From
(n C1 ) (n −1 Cr −1 )2 r −1
−1
PAGE
P
Equation of PA: x − 7 y + 5 = 0 with a slope
πx ≤1 2
π ⇒ 1 + cos x = 1 2 π ⇒ x 1 + cos 2
PAGE -
x = 0
1 7
Equation of PC: x + 3 y − 2 = 0 with a slope −
www.aieeepage.com APC = BPC
(Q x is negative) x ∈ [0,1] ⇒ 0 ≤ cos
A
x x
πx x 1 + cos = x 2
πx x 1 + cos + 1 = [1 + x ] = 0 2
C
...(1)
1 3
⇒ tan APC = tan BPC ...(2)
1 1 − − 7 3 1 1 1+ − 7 3
A Premier Institution for IIT - JEE & AIEEE at the National Level
19 of 26
∴ − x9 − x > 0 ∴ E > 0 in this case also.
1+ λ 1 − − 7 - 3λ 3 = 1 + λ 1 1+ − 7 - 3λ 3 1 1 = giving 2−λ 2
This becomes
90.b.
92.d.
λ =0 or λ =4 λ =0 gives PA itself and λ =4 gives PB Equation of PB is 5x + 5y – 3 = 0 I) False 1x − a ! x−a Consider f ( x) = x − a , g ( x) = 1x − a !
is continuous both x=
lim g ( x) does not exist.
x→ a
π π f '( x) = 2sin x cos x + 2sin x + cos x + 3 3
π π = − sin x cos x + − cos x.sin x + 3 3 2π π = sin 2 x + sin 2 x + − sin x + x + 3 3
the points www.aieeepage.com π π π = 2 sin 2 x + cos − sin 2 x +
through
A(1, 3 ), B (1,− 3 ) and C (3, 3 ) . Here line AB is parallel to y-axis and BC is parallel to x-axis, there ∠ ABC = 0 AC is diameter of circle. ∴ Eq. of circle is ∴
But f (0) = 0 + sin 2
3
= ( x + x) ( x − 1) + 1
E = x( x3 − 1) ( x8 + 1) + 1 ≥ 1 > 0 for
x ≥1
∴ for x ≥ 1, E > 0
95.c.
96.d.
5 ( gof ) ( x) = g ( f ( x)) = g = 1 4 Check options, by taking the values of x, y and z accordingly with the given conditions. Check options, by taking the values of x, y and z accordingly with the given conditions.
Q limiting points are ( C , O ), ( − C , O ) real
⇒C ≥O.
97.a. In coaxal system of Oles line of centres always ⊥ www.aieeepage.com to radical axis & the Ole eqn is of the form
Again E = x12 + x 4 (1 − x5 ) + (1 − x)
When 0 < x < 1 , the terms 1 − x5 and 1 − x are both +ve. ∴ E>0 When x < 0 , the terms with minus signs i.e. x9 and x are both less than zero.
PAGE -
5 5 ∴ f ( x) = ∀ x ∈ R 4 4
PAGE
= x ( x − 1) + x( x − 1) + 1 9
∴K =
94.c.
3
π π + cos 0.cos 3 3
2
25 + 1 − 10 < 0 4 ∴ Pt lies inside the circle. ∴ No tangent can be drawn to the given circle from pt (5/2, 1). ∴ Given statement is True. 3
3
3 1 3 1 5 = + = + = 2 2 4 2 4
S1 =
9
3
∴ f ( x ) = constant = K(say)
⇒ x 2 + y 2 − 4x = 0 Let us check the position of pt (5/2,1) with respect to the circle (1), we get
Let E = x12 = x 9 + x 4 − x + 1
3
π π = sin 2 x + − sin 2 x + = 0 3 3
( x − 1)( x − 3) + ( y − 3 )( y + 3 ) = 0
91.d.
1 1 and– 2 . 2
PAGE 93.a.
lim ( f ( x) g ( x) ) exists but lim f ( x) and then x→ a x→ a
∴ Statement is false. II) True The circle passes
Hence E > 0 for all x ∈ R Hence reqd. largest interval is −∞ < x < ∞ . The graph of y = |x| is shown in the Figure, when x < 0, y = –x and when x >0, y = x. At x = 0, f(0) = f(0–) = f(0+) ⇒ f(x) is continuous at x = 0 LHD = –1 and RHD = +1 at x = 0 ⇒ f(x) is not differentiable at x = 0 Note: Choices A and B are not correct because f(x)
S + λL = O , where S=O is member of system of Oles & L=O is R.A. 98.c.
Q R.A. is x + y − 1 = 0 line of centre is
y − x + k = 0 (R.A. ⊥ line of centres) α passing thro centre
A Premier Institution for IIT - JEE & AIEEE at the National Level
( −1,3 2 )
.
20 of 26
∴ it is (c) 99.c.
From definition of R.A.,
S11 = S22 (Powers same)
⇒ S11 = S22 ⇒ lengths of tangent same ⇒ α =β 100.a.
Radius of Ole is
g 2 − C , it is real only when
g 2 − C ≥ 0 ⇒ (g − C ) (g + C ) ≥ 0 ⇒ g ≤ − C
centre of this Ole is ( − g , 0) ∴ there is no such of in b/w − C & C
Q( − C , 0), ( C , 0) are limiting points.
PAGE
www.aieeepage.com
PAGE
www.aieeepage.com
PAGE -
A Premier Institution for IIT - JEE & AIEEE at the National Level
21 of 26
Paper - I I I : Chemistry - 101 to 150 Solutions 101.c
102.a.
Exciting an electron requires increasing its kinetic energy. This would allow it to move to a higher energy state. I, II, and III all lead to an increase in the kinetic energy of the electron. Thus, I, II, and III must all be true. The New Zealand scientist noted that most alphsparticles were propelled straight through or deflected at very small angles. He also noticed that very few electrons were backscattered at 180 degrees. The “Plum-Pudding” Model proposes that the electrons and protons are uniformly distributed throughout the atom. If this were true, then the alpha particles would never be backscattered at 180 degrees because the electrons and protons have considerably less mass than the alphaparticles (He). The New Zealand scientist observed that the alpha particles went back with almost all their original energy. Thus, they must have collided elastically with a much much larger mass due to the fact that they maintained almost all their kinetic energy AND after the collision they were not attached to the structure with which they collided. Choice b and d claim ineleastic collisions. Furthermore, Choices c and d claim that the collision occurred with otehr electrons of the gold atom. This could not be the case since they possess considerably less mass than the alpha-particles. If the collision were with one of the protons or electrons of the gold atoms, the electrons and protons would have escaped from the gld atoms, ionizing the atom. You should always think of differences and similarities between theories when asked to compare. The “Planetary Model” differs from the Bohr Model on the point that the Bohr Model places the electrons in discrete orbits around the nucleus, where the “Planetary Model” simply has electrons revolving around the nucleus without any fixed orits. Thi is even stated in the passage. Thus, in both models the electrons are in circlar motion about the nucleus. As a result, centripetal force of the nucleus would pull the electrons into the nucleus, unless they were in fixed orbits from the nucleus. As the electrons are pulled into the nucleus, their electrical potential energywould decrease. Choice a is wrng because it says there is no electrical potential energy. There will always be electrical potential energy when two charges separated over a distance. Choice B is incorrect because it claims a gain in electrical potential energy. Choice d is wrong because it refers to the Bohr Model. The passage provides an equation to estimate the number of particles at a particular angle of deflection. The number of particles is inversely related to the sin of the angle of deflection. Thus, you want an angle with the smallest sin value. As an angle increases from 0 to 90 the sin increases Although IIT-JEE students are notrequired to know specific
105.d
106.b.
trigonometry function, they must know the relationship between angles and sin and cos. Thus, you want the smallest angle of the four answer choices. The equation that tells you the number of electrons in any given principal quantum numebr (including all the subshells) is 2n 2 . Replacing n in this equation with 4, will given you 32. LeChatelier’s principle states: If a system at equilibrium is disturbed by a change in temperature, pressure or the concentration of one of the components, the system will shift its equilibrium position so as to counteract the effect of the disturbance. If the addition of heat caused the reaction to shift to the right, (i.e. away from the reactants), this, implies that heat is required for the reaction to ‘go’. This is indicative of an endothermic reaction.
PAGE 107.c.
In [Ni(CO)4 ] , 4s electrons are shifted to 3d making
it3d 4s and in [Ni(CN) ] www.aieeepage.com
103.c.
10
108.c.
109.b.
110.d.
0
2–
CO(CH2)4 CH3
H2 NNH2
112.d.
, all the 8 electrons
get paired up in the crystal field. You may have been looking for hydrogen bonding, which would probably be the best answer, but it’s not there. Polarity is a prerequisite for hydrogen bonding, so choice C is the best alternative. Use process of elimination: C is false; you can even ionize an atom without affecting its nucleus. A, although true as far as elements in their ground state are concerned, says nothing about excited states. It is difficult to see what would make choice D correct. But choice B agrees with something that you should know about orbitals (higher n means greater size), and gives a plausible explanation for the observations. An electrolyte is a substance that conducts electricity in aqueous solution. The fact that it’s an electrolyte does not make it a voltage source itself, so no current would flow if it were merely attached to a resistor. The reaction proceeds as follows:
PAGE 111.d.
4
CO(CH2)5 CH3 KOH
According to the ideal gas law, both pressure and volume are directly proportional to (absolute) temperature. The experiment describes the mixture along the dotted line in the diagram. Don’t think about this until you look at your choices! You may come up with a perfectly reasonable explanation, but it may not be the one the question-writer had in mind. Instead, use process of elimination. You can eliminate choices either because they are factually incorrect or because they have nothing to do with the phenomenon described.
www.aieeepage.com 114.c.
104.a.
PAGE -
113.a.
A Premier Institution for IIT - JEE & AIEEE at the National Level
22 of 26
115.d.
116.d.
117.c.
118.c.
119.a.
Taking the choices one at a time: choice A would be an explanation if it were true, although a hydrophobic residue would not activate an acid. Choice B is factually untrue (many proteins are found in nearly neutral environments), and it also wouldn’t explain the phenomenon (an acidic environment makes weak acids less likely to dissociate, by Le Chatelier’s principle). Choice C would be an explanation if it were true; plus, since basic amino acids are stronger bases than water, this looks like a good answer. In considering choice D, realize that at high concentrations acids tend tobe less dissociated, at least on a percentage basis. At any rate, weak acids do not dissociate to a very great extent, even at low concentrations. The Nernst equation is really about non-standard conditions, so A seems doubtful. Standard potential is just the potential at standard conditions. This definition does not directly involve equilibrium, so B seems doubtful as well. C is kind of silly if you thing about it; standard potentials are measured under standard conditons. But D is a true statement: you need two half-reactions to make an actual reaction. The rate of a reaction is correctly described by choices A, B, and C. Choice D however is incorrect. There is no such equation as the Eurim equation. They Eyring equation is used at times to calculate the rate of a reaction. Whether or not it is applicable is determined by the partition function of the species involved. A first order reaction follows the same trend as a half life reaction. The rate is directly proportional to the concentration. On the graph, the slope is the reaction rate. Since the concentration is moving toward zero, the slope should also be moving toward zero. This is curve C. Since electrons are flowing into the sample, the probe must be the site of reduction, i.e., the cathode. Furthermore, since the sample is receiving electrons, H+ in the sample is being converted into H2, raising the pH. Concentration cells are trying to equalize the concentration in the two cells, so if the pH in the sample is increasing, it must have started at a pH below 2.0. The resonance structure below shows the double bond on the nitrogen.
121.c.
[C ] [ D ] 3 × 3 Reaction quotient, Q = [ A] [ B ] = 1× 1 = 9
Since Q > Ke, the reaction will proceed in the backward direction. Let ‘x’ mole of C be converted into ‘A’ till the attainment of new eqm. state. 1 mole 1 mole 3 mole 3 mole ...initially A B C D (1+x) (1+x) (3–x) (3–x) ...at eqm.
PAGE K e = 2.25 =
(3 − x ) (3 − x ) (1 − x )2 = (1 + x ) (1 + x ) (1 + x )2
www.aieeepage.com 3 − x
O | H–C=NH2 120.d.
= k [A] [B] it is possible to derive an expression for the variation in concentrations of A and B with time. 122.d. Gases can be written either as concentrations or as partial pressures in equilibrium expressions. The two can be related by the ideal gas law, and thus the numerical values will differ. 123.d. The larger alkane chain the stronger the London dispersion forces leading to a solid. Answer D is the longest alkane chain in the answers. The alkanes in answers A, B and C are all liquids. C+D; Ke= 2.25 (given) 124.b. A+B
125.a.
1+ x whence x = 0.6 Now, [A] at eqm. = 1+x = 1+0.6 = 1.6 [C] at eqm. = 3–x = 3–0.6 = 2.4 or [A] : [C] = 1.6 : 2.4 or 2 : 3. If MgCO3 can decompose to yield CO2 at 1.0 atm (to push back the air), it will do so.
K p = PCO2 = 1atm log K p = log1 = 0.00 = 7.310 − T=
126.b.
8500 T
8500 = 1163 K = 890ºC . 7.310
1 liter 0.3 moles AgNO3 (26.7 ml AgNO3 solution) 1000 ml 1 liter solution
= 8.0 × 10 –3 moles AgNO3
PAGE
As we boil the mixture the escaping vapor contains more of the more volatile component, benzene. The remaining liquid becomes richer in toluene. According to the graph, a liquid mixture 25 mole % in benzene will boil at ~100ºC. As we boil the liquid its composition and boiling point move to the left and up on the “boiling point line” until we reach the desired point. If a reaction is second order by virtue of a first order rate dependence on each of two reagents where rate
PAGE -
1.5 =
or
1 mole Cl – 35.5 g Cl – 8.0 × 10 –3 moles AgNO3 1 mole AgNO3 1 mole Cl
= 2.84 ×10 –1 g Cl –1
2.8 × 10 –1 g Cl –1 × 100 = 2.0% Cl –1 in H 2O 14.0 g sample of H 2 O
127.d. This is very basic thermodynamics: breaking up www.aieeepage.com water to produce hydrogen, and then using the hydrogen to make water, cancel each other out, which means that there is no net gain of energy (in keeping with the First Law of Thermodynamics). You may wonder, then, what the big deal is about fuel cells, both in the general media, among scientists, and on Wall Street. The answer is that a fuel cell can provide a convenient way to store energy. For example, a gas, oil or nuclear
A Premier Institution for IIT - JEE & AIEEE at the National Level
23 of 26
128.a.
129.a.
plant produces a lot less pollution per Joule than a car engine. So by running an electric plant and using the energy to split water, and then using the products in a fuel cell in a car, the total amount of pollution can be reduced. Likewise, equipping your house with solar panels normally only gives you power when it’s sunny! But a fuel cell could be a convenient way to store the energy for later use. The addition of four hydrogen atoms indicates two double bonds, so answer A is the only correct answer. The fact that the saturated alkane is still 4 hydrogens short (would be 32 hydrogens if there were no rings) indicates two rings in the structure. Volume of Ag to be deposited = area × thickness = 80 cm2 × 5 × 10–3 cm = 0.4 cm3 Mass of Ag to be deposited = Vol. × densitty=0.4 cm3 × 10.5 gm. cm–3 = 4.2 gm. Number of gm. eq. of Ag to be deposited =
132.c.
133.a.
134.c.
a cm. p(200–a)cm.
Since the tube is of uniform cross section, the rate of diffusion is directly proportional to the length of the tube. Let the distance of point “P” from the HCl end be ‘a’ cm. Then distance of “P” from the NH3end = (200–a)cm.
r ∝a Thus, www.aieeepage.com
= 0.0388
108 gm. eq.–1
HCl
rNH3 ∝ (200 − a)
Now, according to Graham’s law of diffusion,
a 0.68 or (200 − a) = 1
Cl — CH 2 COO – +H +
or a = 200 × 0.68 − 0.68a or 1.68 a = 200 × 0.68
van’t Hoff factor using Ostwald’s dilution law of weak electrolyte
∴
= 0.37
135.d.
uA =
200 × 0.68 = 80.95 cm 1.68
h λ A.m
PAGE
= 0.51× 0.01× 1.37 = 0.007º
uB =
b.p. of solution T = T0 + (∆ T )b = 100 + 0.007º = 100.007ºC. The conversion proceeds as follows: hence,
∴
uA λB = =2 uB λ A
(Qλ B = 2λ A )
hence,elevation in b.p (∆ T )b = K b m (molarity) i
131.d.
a=
or
i = (l + x) = 1.37
M HCl
a 17 = (200 − a) 36.5
0.0388 × 96500 = 2496.13seconds 1.5 = 2500 seconds.
−3
M NH 3
rHCl = rNH 3
=
Ka 1.36 × 10 = 0.01 c
NH3(g)
HCl(g)
Charge in coulombs Current in amp.
x=
y
PAGE
4.2 gm.
Cl — CH2 COOH
200 cm.
x
Number of Faradays needed to deposit 0.0388 eq. of Ag = 0.0388 Number of coulombs needed = 0.0388 × 96500 Time to coat the metal surface =
130.a.
Hence, the correct sequence is magnesium, dry ether, carbon dioxide, HBr, excess methanol and sulfuric acid (twice). In the reaction in choice II, internal energy merely changes forms, from potential (chemical) to kinetic (thermal). Or perhaps some of the chemical energy is used to do work, in which case the internal energy of the system decreases. In any case, II is incorrect, which leaves C as the only answer. The fact that the addition is anti-Markovnikov indicates that the hydroxyl group will add to the least substituted carbon.
h λ B. m
TA mu A2 4 Now T = mu 2 = 1 ; Also TA − TB = 1.50 B B
www.aieeepage.com
Br
+
Mg
dry ether
MgBr
O | |– CO
CO 2
++
Mg Br
Bromobenzene
O || CO–OH +
H Br
O | CO–OH | OCH3
PAGE -
H2SO4
–
∴
TB = 0.50
TA = 0.50 + 1.50 = 2.00 eV MgBr 2
+
H 2SO 4
CH3OH excess O || C–OCH3
Also 4.25 = hv0 A + TA ,
∴ +
H2 O
A Premier Institution for IIT - JEE & AIEEE at the National Level
hv0 A = 4.25 − 2.00 = 2.25eV
24 of 26
H 2 → 2 H ; ∆ H = 436 kJ mol –1 Energy required to bring 0.0409 moles of hydrogen gas to atomic state
4.20 = hv0B + TB , hv0B = 4.20 − 0.50 = 3.70eV 136.a.
137.b.
138.b. 139.a.
The primary functional group of this compound is nitrogen. Thus, this compound is a member of the aniline family. The ring substituents are named according to their relative position to the functional group. The methyl group is directly opposite the amine group and this is indicated by the p (for para). Both the isopropyl and methyl groups are attached to the functional group of the compound and this is indicated by the ‘N, N’ prefix. Choices A and C misrepresent the role of a catalyst: catalysts do not change the stability of a certain set of products, but they are actively involved with reactions (because they change the transition state). As far as choice D goes, catalysts certainly do affect the reverse reaction, but that isn’t ruled out by the observations; also, what is a “kinetic promoter”? It’s made-up phrase, but it sure sounds like a description of a catalyst! Step I is the rate determining step, so it must be slower than Step II. Surface area of spherical drop = (a) When radius is 3.0mm
= 436 × 0.0409 = 17.83 kJ Calculation of total number of hydrogen atoms in 0.0409 mole of H2 gas
1 mole of H2 gas has 6.02 × 1023 molecules
6.02 ×1022 × 0.0409 1 Since 1 molecule of H2 gas has 2 hydrogen atoms 0.0409 mole of H2 gas =
6.02 ×1023 × 0.0409 molecules
PAGE
4π r
When
= 13.6 ×
2
3 = 10.2eV = 1.632 × 10−21 kJ 4
www.aieeepage.com Therefore energy required to excite 4.92 × 10
(= 3.0 ×10−4 cm) surface area A2 = 4 π (9 × 10 −8 ) cm 2
141.d.
number of drops formed =
volume of a drop of radius 3.0mm volume of drops of radius 3 × 10−3 mm 4 π r13 33 3 = = = 109 −3 3 4 π r23 (3 × 10 ) 3
142.d.
= 8.03 × 10 = 80.3 kJ Therefore total energy required = 17.83 + 80.3 = 98.17 kJ. The Second Law of Thermodynamics states that in an isolated sytem, spontaneous processes occur in the direction of increasing entropy. The substances that react in the first reaction both show an increase in entropy: from solid to liquid and from liquid to gas. However, in the second reaction the net entropy remains the same since one substance changes from a solid to a liquid and the other changes from a liquid to a solid. Hence, the total “randomness” of the sytem remains the same. This time, the experiment was explicitly performed under standard conditions: 1 atmosphere of oxygen gas. Yet the graph shows a negative free energy change at that temperature, which implies that the reaction is spontaneous. Many spontaneous reactions, are so slow at certain temperatures that they effectively do not occur. The reaction between oxygen and gasoline is a typical example (you need a spark to get it started). The leaving group would be CH 3– before the substitutions of the 1atoms for hydrogens. After the substitutions the leaving group isCl3–, Cl3– is a better leaving group beacuse the 1 atoms are electron withdrawing and stabilize the negative chage. Both carbon oxygen bonds are involved in the resonance of the negative charge. The bond length is between the single and double bond. An electrochemical cell consists of two electrodes dipped into an electrolyte. If both of the electrodes are dipped into the same electrolyte, the solution
PAGE
Hence, surface area of 109 drops =
4 π (9 × 10−8 ) ×109 =4 π× 90 cm 2 (4 π× 90) − (4 π× 0.09) =
22
electrons = 1.632 ×10 −21 × 4.92 × 10 22 kJ
3.0 × 10−3 mm
is
increase in surface area =
gas
1 1 1 1 = 13.6 2 − 2 eV = 13.6 × − n 1 4 1 n2
(= 3.0 ×10−1 cm)
radius
H2
= 2 × 6.02 × 1023 × 0.0409 = 4.92 × 10 23 atoms. Energy required to excite an electron from the ground state to the next excited state
surface area A1 = 4 π (0.09) cm 2 (b)
of
4 π 89.91cm 2 =1129.842 cm 2 ∴ work done in dispersion = γ × increase in area =
143.a.
www.aieeepage.com
72.8 × 1129.842=8.2252 × 104 erg = 8.2252 × 10–3 J. 140.a.
144.d.
Determination of number of moles of hydrogen gas, PV 1×1 = = 0.0409 RT 0.082 × 298 The concerned reaction is n=
PAGE -
145.c.
A Premier Institution for IIT - JEE & AIEEE at the National Level
25 of 26
potential φ (s) is common to both electrodes and the arrangement is called a cell without a liquid juction, see the figure below. Pt wire H2
Pt wire
149.a. Pt
146.b.
AgCl/Ag
A simple electrochemical cell without a liquid juction. Long wavelengths correspond to low frequencies and short wavelengths correspond to high frequencies. The wavelength and frequency of electromagnetic radiation are inversely related according to the equation This is also ilustrated in the diagram below. Since infrared radiation has the longest wavelength of the choices listed, choice (b) is correct.
triplet state is released the energy transferred is emitted as phosphorescence. Internal conversion is also a transfer of energy, however it is attained by other nonradiative means such as the release of heat or transfer to another molecule via collision or internal vibrational relaxation. Quenching is another nonradiative process that releases the energy from the excited state back down to the ground state. Choice (a) is correct because the absorption of energy is quantitized from the ground state to the excited state. Each molecule exhibits its own inherent quantitized energy levels, which give rise to each element displaying a unique line spectrum. Choice (b) is incorrect because fluorescence results
PAGE
Wavelength
1010
from the emission of energy from the excited S1 state to the ground So state, not from internal conversion. Choice (c) is incorrect because intersystem crossing, not internal conversion, is involved with phosphorescence. Finally, choice (d) is incorrect because conversion from the T1 to
the S state is called phosphorescence, no www.aieeepage.com vibrational relaxation. (meters) 10 −10
o
150.d.
In order to anwer this question, we must consult the diagram; we want to focus on the energy levels fo the excited S1 state and the excited T1 state.
Frequency
147.c.
148
10 20
(sec onds −1 ) 10
0
In order for the equation using energy transferred to be dependent upon the color of light, it must somehow be dependent on the wavelength or frequency of the incident light. Color is dependent upon the wavelength of visible light. The different colors of light (visible to teh naked eye) are located in the visible region of the electromagnetic radiation spectrum. The equation relating wavelength to frequency is c = λv where c is the speed of light, λ is the wavelength of light, and v is the frequency of the light. Because of this relationship ( v = c / λ ), both equations (II and III) relate wavelength to energy. Choice (c) is correct. Equation I only relates energy to mass and velocity, so it does not apply and choices (a) and (d) are incorrect. Each one of the process shown in Figure 1 results in a change in energy. The energy transferred durign absorption an fluorescence is quantitized and of equal magnitude involving the absorption or emission of an electron or photon. Electromagnetic radiation (light) is composed of photons, which exhibit both particle and wavelike properties. Electrons can only be promoted to levels of a certain energy from the lower ground state to the higher excited state (and vice versa) according to thier quantitized energy levels. Intersystem crossing to the triplet state is also an energy transfer that does not emit energy as a radiative process. When the
emission from the excited S1 state (which is of a higher energy level than the T1 state) to the So state is called fluorescence. Emission from the T1 state to the So state is called phosphorescence. There is no emission of light from internal conversion (which is nonradiative) or from the absorption process (the molecule absorbs light to become excited). How do we determine whether fluorescence or phosphorescence corresponds to the longer wavelength ? The key to this equestion
s that the T1 state is of lower energy. This is clear in Figure 1 and it is mentioned in the last paragraph of the passage. Using the relationship between
PAGE
energy and wavelength (E = hc / λ ), we see that they are inversely proportional. That means that as the waelength increases the energy of the light decreases or vice versa, so choice (d) is correct Emission of light from the lower energy T1 state to
www.aieeepage.com a smaller transfer of energy, hence, a longer
PAGE -
the So ground state(phosphorescence) results in wavelength.
A Premier Institution for IIT - JEE & AIEEE at the National Level
26 of 26