hints & solutions paper-1 part-i : physics€¦ · sol. mass of solution = 180 g since the salt...
TRANSCRIPT
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JEE PREPARATORY TEST-2 (JPT-2)(JEE ADVANCED PATTERN)
TARGET : JEE (MAIN+ADVANCED) 2018
DATE : 13-05-2018 COURSE : ALL INDIA TEST SERIES (VIKALP) | CLASS: XII/XIII
HINTS & SOLUTIONSPAPER-1
PART-I : PHYSICS1. In the circuit .......................Sol:
C1 = C C2 = C+( Q0 – q)
–( Q0 – q)
+q
–q
dqi
dt
R
0Q q
c
– iR –
q0
c
0Q 2q Rdq
c dt
q t
00 0
dq 1dt
Q 2q Rc
2t /Rc0Qq 1 e
2
2t /Rc0Qdqi e
dt Rc
2t /Rc00
QQ q 1 e
2
20Q
H4c
2. The electric .......................Sol. V = A(x2 – 2y2 + z2)
Plane parallel to x – z plane y = constant x2 + z2 = constant its a circlePlane parallel to x-y plane z = constantx2 – 2y2 = constant its a hyperbola
3. The figure .......................
Sol. E = 00
QQ AE
A
QABCD = 0AE1 QQ 1 1
k k k
Udielectric =
2 20
0
E At1 Ek At
2 k 2k
4. A neutron .......................Sol.
n v H
21mv k
2 (k)max =
21 m kv
2 2 2
E2 – E1 = 10.2 eV(k)max > 10.2 eV k > 20.4 eV
E4 – E1 = 12.75 eV
k12.75eV
2 K > 25.5 eV
5. Which of the .......................Sol. (A) [0E] = [Mº L–2 T1 A1]
(B) [] = [M–1 L-3 T3 A2](C) [force] = [M1 L1 T–2]
(D)E
B
= [Mº L1 T–1]
6. A solenoid of .......................Sol.
L R/20 = 2E/R
R
L R/20
R
0 =2E
R, U =
2 2 2
2 2
1 2E L 4E 2LEL
2 R 2 R R
sol
R
H 1
H 2 ,
2
S 2
1 2 LEH U
3 3 R
7. Force on a .......................Sol.
r
(x, y)
y
x
r ˆF A(r B) A(r B)rr
F
is conservative
fr
y
x
ir
dw = F.dr A(r B)dr
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w =f
c
r
r
A (r B)dr
2 2f i
f i
r rA B r r
2
(A) ri =B
2, rf =
3B
2
w =
2 29B B3B B4 4A B
2 2 2
= –A (B2 – B2) = 0(B) ri = rf = Bw = 0(C) ri = B rf = 2B
W = 2 24B B
A B 2B B2
=2 2
23B ABA B
2 2
8. Consider the .......................
Sol. (1) 10 kg100
(f12)max = 20
(2) 10 kg 100
(f23)max = 40
(3) 10 kg15
(f34)max = 60friction between (1) & (2) = 20 Nfriction between (2) & (3) = 40 Nfriction between (3) & ground = 25 N
9. A L shaped .......................Sol.
3V0
3V0
V0
V0B
C
3L0L0
AL0
2L0
y
x
VA - BB =0
0
3L
00 0
0L
B3V (x L )dx
L 0 0 018V B L
VC – VB =0
0
2L
00
0L
BV
L (3L0 + y) dy = 0 0 09V B L
2
VA – VC = (VA – VB) – (VC – VB)
= 18V0B0L0 – 0 0 00 0 0
9V B L 27V B L
2 2
10. Some amount .......................
Sol.
BA
FBD of A part (horizontal force balance)
dg (d )
2
= gd
d2
2d
g
FBD of A + B (Horizontal force balance)
(1 + sin) = 2gt
2
= 90 –
tg t
2
(1 + sin) = 2gt
2
2 (1 sin )t
g
11. The ends of .......................
Sol. x 0 x Hot end cool end
dTincreasing
dx
–dT
KAdx
= Constant A =Cons tant
dTK
dx
SincedT
dx = increasing
A decreasing
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12. A person riding .......................Sol.
Observer
S
10 m/s310 m/s
1240 m/s
S1
20 m/s
D
Frequency observed by S1 =310
310 10
(48) = 49.6 Hz
Frequency reflected & transmitted by S1= 49.6 HzFrequency of reflected sound received by observer =
310 10
310
310
310 10
48 = 51.2 Hz
Beat frequency observed by observer = (51.2 – 48) = 3.2 Hz
Frequency received by detector =1240 20
1240
31048 50.4
300
hz
wavelength observed =1260
25m50.4
13. There is a .......................Sol.
+Q +2 Q
+Q
A B C D
S1 S2
A D
0
B C
+q1 –q1 +q2 + Q – q1 + q2) 0
-q2
–( Q – q1 + q2)
2 1 2
0 0
q Q q qd d 0
A A
q2 + Q – q1 + q2 = 0 q1 – 2q2 = Q .......(1)
1 2
0 0
q qd d 0
A A
q1 + q2 = 0 ........(2)
from (1) & (2)
q1 =Q
3
q2 =Q
3
(qB)i = +Q (qB)f = – q1 + q2 =2Q
3
charge flown from S1 =5Q
3
(qD)i = +2Q (qD)f = – Q + q1 – q2
Q
3
change flown through S2 =7Q
3
14. A person A .......................Sol.
6 6
A1 m/s
6 3
BA
121
6
VAB = 1 m/s
VBA =
A
B1 m/s
16 3
6
= 2 m/s
dr
dt = – 1 m/s = m
m = 1, n = 4, P = 3
15. A wooden cubical .......................Sol.
N
x
mg
M = a2B0
for no toppling
mgx = a2B0 x =2
0a B a
mg 2
0a B 1
mg 2
0
mg
2aB
x =2
0a Ba
4 mg
=
0
mg
4aB
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16. In a radioactive .......................Sol. NA = N0 e
–t
NB = N0 (1 – e–t)
1t00
NN e
5 t1 = n5 1
n5t
600
NN e
8 6 = n = n8 =
n8
6
09N
10 = N0 (1 – e–t) t9
1 e10
2t 1e
10 2
n10t
t2 – t1 =n10 n5
n2n8
6
= 6n2
3 n2
= 2 sec
t1 =6 n5
n8
=6 n5
3 n2
= 2 log25
t2 = 2 2
n106 2log 10 log 100
3 n2
17. A non-uniform .......................
Sol.
F
x = 0 x = L
dm = dx =
LL 2
0 0 0
0 0
x x 31 dx x L
L 2L 2
a0 =0
0
F 2F3 3 LL2
Mx =2
0
x x
2L
T =2
00
x 2Fx
2L 3 L
T =22F x
x3L 2L
LL 2 3 2 2
0 0
2F x x 2F L L 4FLTdx
3L 2 6L 3L 2 6 9
18. A triangular .......................Sol. t = 0
x=+L
x=0
x = –L
Vy
h
0 x –L
y(x,0) =hx
hL
–L × 0
hxh –
L0 ×+ L
0 x +Lat any time t .
x = Vt + Lx = Vt – L
y
h
x = Vt
0 x Vt – L ;y
0x
y
0t
y(x, t) =hx
hL
Vt – L × Vt ;y h
x L
y hV
t L
hx
h –L
Vt × Vt + L ;y h
x L
y hV
t L
0 x Vt + L ;y
0x
y
0t
Instantaneous power
P = –y y
fx t
0 x Vt – L
P =
2h
FVL
Vt – L × Vt
2h
FVL
Vt × Vt + L
0 x Vt + L
19. Choose the correct .......................
Sol.
01
Esin t
2R 3
02
Esin t
2R
12
=00E
2 cos30 sin t2R 6
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03Esin t
2R 6
= 0Esin t
Z 6
Z =2R
3, Phase difference =
6
, Power factor = cos =
3
2, tan =
Reac tance x 1
Resis tance R 3
2 2R x Z R = R, x =R
3
20. An electron .......................
Sol. L =nh
2
L = nh mh hn m
2 2 2
n – m = integer, so integer multiple ofh
2 are allowed.
PART-II : CHEMISTRY21. A 180 mL saturated ………….Sol. Mass of solution = 180 g
Since the salt is sparingly soluble in water, neglecting massof ions dissolved in water,Mole of water in solution = 180 / 18 = 10 mol(Pº – Ps) / P
º = (mol of ions / mol of water)Moles of ions dissolved in water = 0.0054 molMoles of MCl2 dissolved in water = 0.0018 mol[M2+] = 1.8 mmol / 180 mL = 0.01 M[Cl–] = 3.6 mmol / 180 mL = 0.02 MKsp = [M2+][Cl–]2 = 4 × 10–6
22. Which step(s) is/are ………….Sol. Bond breaking and ionisation energy are always
endothermic. Lattice making is always exothermic. Electrongain enthalpy of halogens is exothermic.
23. Which of the following ………….Sol. For exothermic reaction, equilibrium constant decreases
with increasing temperature.But, rate constant of any chemical reaction increases withincreasing temperature.
24. A solution of an unknown ………….Sol. White ppt with dilute HCl implies presence of group 1
cations. AgCl dissolves in ammonia solution. White ppt ofPbCl2 dissolves in hot water, PbI2 is yellow ppt. Possibility ofHg2Cl2 is ruled out because white ppt of Hg2Cl2 does notdissolve in ammonia solution..
25. Given for H2CO3 ………….Sol. [H+] for H2CO3
= M1021.0104CK 4701
[H+] for NaHCO3
= M104104104KK 911721
26. When solid NH4SCN………….Sol. Since gas is produced from solids entropy has increased. ∆ng >
0.In closed container, P-V work is zero. w = 0. Temperature dropimplies heat is absorbed by the system during the process. q >0. Thus, as per first law ∆E = q + w > 0.∆H = ∆E + RT∆ng.Thus, ∆H > 0.∆G < 0, because the process is spontaneous.
27. 1 mol red lead………….Sol. Pb3O4 + 8HCl 3PbCl2 + Cl2 + 4H2O
Pb3O4 + 4HNO3 2Pb(NO3)2 + PbO2 + 2H2O
28. Following is/are ………….Sol. Two oxygen atoms per tetrahedron are shared forming rings.
Hybridisation of each Si is sp3.
29. The following diagram ………….Sol. Low accuracy because measurements are far off from correctly
value. But, they are highly precise, because they are very closeto each other.
30. Two mole of electrons………….Sol. In first cell, at anode: 2Cl–(I) Cl2(g) + 2e–
at cathode, Na+ (l) + e⎯ Na(s)
In second cell,
at anode : 2 Cl⎯ (l) Cl2 (g) + 2 e⎯at cathode, 2H2O(I) + 2e– H2(g) + 2OH–
31. Which statement(s) ………….Sol. Soap and detergent both have hydrophillic as well as
hydrophobic groups, hence both can form micellles.Carboxylates are more basic than alkyl sulphates, becuasecarboxylic acids are weaker acids.In acidic pH, stearate gets protonated to form correspondingfatty acid, which is not much soluble in water due to longhydrophobic chain. Soap precipitates in hard water becuase itscalcium salt is insoluble.
32. Which diagram represents ………….Sol. In option C, Copper cannot reduce ferric ion to iron.
33. Which of the following ………….Sol. Conjugate base of Q is aromatic, hence it is the most acidic
compound among given hydrocarbons.
–HC
–C
34. Xconc. NaOH Y ………….
Sol.CHO|CHO
conc. NaOH
int ramolecularcannizaro reaction
2CH OH|COO
H 2CH OH|COOH
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35. The commonly used ………….Sol. Nitration of phenol with dilute HNO3, gives mixture of o-
nitrophenol and p-nitrophenol; of which o-nitrophenol (X) issteam-volatile. Y on reduction gives p-amino phenol.
36. Two ketones, P and Q………….Sol. Racemization of an optically active ketone in basic medium
is possible due to tautomerisation only if the alpha-acidic His attached to a chirality centre.In option A, both P and Q will undergo racemization. Inoption B, Q is optically inactive. In option C, Q will undergoracemisation.
37. C – NH2
||O
52OP
………….
Sol.
C – NH2
||O
52OP
+
(i) CH Mg3 I
(ii) H O3
C||O
+ 2CHI3
38. Which of the following ………….Sol. 2-bromopropane + Mg/ether followed by CO2; followed by
hydrolysis would give isobutyric acid.
39. Observed the following ………….Sol. Cyclobutanecarbonitrile is
CN
LAH/ether reduces it to primary aminecyclobutylmethylamine (A). A undergoes exhaustivemethylation by 3 eq. of CH3I to give quaternary ammoniumsalt (B). B is formed by substitution.
B is iodide salt ofN+
B further undergoes hoffman elimination on heating withAgOH.
C is CH2
40.O O O
AlCl3Benzene P Q
Zn(Hg)
SOCl2SOCl2
R
AlCl3
SNaBH4T
H2SO4U
NBS(1eq.)V
CCl4,hv
W
NaOEt/EtOH, heat
HCl
………….
Sol.
+
O
O
O
AlCl3
O
P OHO
Zn(Hg)HCl
Q OHO
SOCl2
R OCl
AlCl3
OS
NaBH4
OHT
H2SO4
U
NBS/CCl4 Br
VEtONa/EtOH
W
PART-III : MATHEMATICS
41. Let ‘m’ be .............................Sol. Suppose that both r and r + 1 are solutions to the equation
3 2x 3x 34x m
Then3 2r 3r 34r m , and also3 2(r 1) 3(r 1) 34(r 1) m
Subtracting the first of these equations from the second gives2(3r 1) 3(r 1) 34 0,
or23r 9r 30 0 or
2r 3r 10 0 r = –5, 2 and put in given equation
42. If a triangle .............................
Sol. AB BC AC
2u u v u vBC .
u u v u v
u v u v ˆ ˆ ˆ ˆAB . BC u v . u v 1 1 0.u v u v
B 90 1 cos2A cos2B cos2C 0
4cos A cosBcosC 0
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43. Let 1 1 1 2 2 2Z x iy , Z x iy ...................
Sol.1 2i i
1 2 1 2
1 2
Z e ,Z e ,Re(Z Z ) 0
/ 2,
1 1i i3 4 5 1Z e ,Z ie ,Z cos (1 i)
6 1Z sin ( 1 i)
44.n 2n n 2n 2
0 m 1 m.................C C C .C ... .Sol. The given series can be written as
nn 2n 2r r
r mr 0
S C C ( 1)
n
n rr
r 0
C ( 1)
× coefficient of xm in (1 + x)2n–2r
= coefficient of xm in
n n rn r 2r
r 0
C ( 1) (1 x)
= coefficient of xm in [(1 + x)2 – 1]n
= coefficient of xm in (x2 + 2x)n
= coefficient of xm in xn(x + 2)n
= coefficient of xm–n in (x + 2)n
n n (m n) 2n mm n
nC 2 2
m n
if m n and 0 if m < n.
45. If A is non –.............................
Sol.1 13ABA A 2A BA
1 13ABA A 2A 2A BA 2A 1 13A(BA ) 2(A B )A
1 13A(B A)A 2A (B A )A 1 13A(B A)A 2A (B A)A
Let B + A = X
1 13AXA 2A XA n 1 n 13 | A || X || A | 2 | A || X || A | n n3 | x | 2 | X | (as | A | 0) (i)
| X | 0 or | A B | 0
Let 1 1M ABA A BA 2 1 2 1AM A BA BA BA A BA AM
1 1
1 2 1
1
3ABA A 2A BA
2A (A BA AM)
2ABA 2M
1
1
1
ABA A 2M
A(BA ) 2M
A(A B)A 2M
46. 5 players of .............................
Sol.
5 6110
C 2 5P(A) ,
162
5 6110
C 2 5P(B) ,
162
5 32
10
C 2 2 5P(A B) ,
322
15P(A B)
32
47. The area bounded .............................
Sol. The desired area is shown as shaded region
x1O
1 /2 22 2
0 1
4A 4 (x x )dx 4 (x x)dx (2 3)
3 6
48. Suppose that .............................
Sol. f(1000)f f(1000) 1
f(1000)f(999) 1
999f(999) 1
1f(999)
999
The numbers 999 and1
999 are in the range of f.
Hence, by intermediate value property of continuous function,
function takes all values between 999 and1
999, then there
exists.
1,999
999
such that f( ) 500
Then 1f( ) f f( ) 1 f(500)
500
Similarly,1
199 ,999999
, thus 1f 199
199
But there is nothing to show that 1999 lies in the range of f. Thus
(d) is not correct and (c) is also incorrect.
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49. A function f : R R .............................
Sol.2 nx
nxn
ax bx c ef(x) lim
1 c.e
n2 x
nn x
n2 x
nn x
2
nx
n
nx
ax bx c elim ; x 0
1 c. e
ax bx c elim ; x 0
1 c. e
ax bx c1
elim ; x 0
1c
e
2ax bx c; x 0
1; x 0
1; x 0
c
since f(x) is continuous function x R
x 0 x 0lim f(x) lim f(x) f(0)
2
x 0 x 0
1lim lim (ax bx c) 1
c
11 c 1
c
c 1, a, b R
50. Let f(x) be .............................
Sol. 34
1 f(x)f '(x)f '(x) f (x) 1
f(x) 1 f (x)
Integrating on the interval (a, b), we getb b
1 2 ba4
a a
f(x)f '(x) 1dx dx tan f (x) b a
1 f (x) 2
1 2 1 2
x b x a
1b a lim tan f (x) lim tan f (x)
2 24
51. A straight line.............................Sol. Let equation of line passes through A(–2, –3) is
x 2 y 3r
cos sin
Point on it (rcos –2, rsin –3)If this point is B lies on x + 3y = 9
AB =20
cos 3sin If this point is C lies on x + y + 1 = 0
AC =4
sin cos as AB.AC = 20
3sin2 + cos2 + 4sin cos= 4 3cos2 – 4sin cos+ sin2= 0 (cos – sin)(3cossin) = 0 tan = 1,3Equation of line y + 3 = x + 2 or y + 3 = 3(x + 2) y = x – 1 and y = 3x + 3
52. If graph of .............................
Sol. Let image of point A( , ) about y 2x is B(a, b).
Mid point of AB
i.e.a b
M ,2 2
lie on y 2x 0
b a
2 b 2 2a2 2
(i)
and slope of AB1
2
b 1 1 a
ba 2 2 2
(ii)
Subtracting (i) and (ii),
5 3 4b 3a
2b a2 2 5
8b 6a 3b 4a
2a b5 5
( , ) lies on xy = 1
4b 3a 3b 4a
15 5
2 212b 7ab 12a 25
2 212x 7xy 12y 25 0
53. A circle touches.............................Sol. Family of circles touching x + y – 2 = 0 at (1, 1) is
2 2(x 1) (y 1) (x y 2) 0 2 2x y ( 2)x ( 2)y 2 2 0 (1)
Given circle is:2 2x y 4x 5y 6 0 (2)
Equation of common chord PQ is S – S’ = 0.
( 6)x ( 7)y 8 2 0 (3)
(a) PQ || line: x + y – 2 = 0
61 6 7;
7
which is impossible
(b) PQ line : x y 2 0
6 131 ;
7 2
which is impossible
But when13
2 , we can see that the circles (1) and (2) are
not intersecting each other and their radical axis is perpendicular
to the given line x y 2 0 Eq. (3) can be written as
6x 7y 8 (x y 2) 0
which is in the form 1 2L L 0
Solving 1L and 2L , we get the point of intersection (6, –4)
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54. Let A, B and C .............................
Sol. Let 2 2 21 1 2 2 3 3A (2t ,4t ),B (2t ,4t ) and C (2t ,4t )
Slope of AB = 2 t1 + t2 = 1 and t1 + t2 + t3 = 0So, t3 = – 1
Also,
2 2 22 21 2 3
1 2
2(t t t ) 4t t 1
3 3
1 2t 1, t 0 A (2,4),B (0,0) andC (2, 4) HenceP (6,0)
55. The volume .............................
Sol. Volume = | 2b c 3c a 4a b | 18
224 a b c 18
3| a b c |
2
Now ,
(1 sin ) cos sin 2
2 2 4a b c sin cos sin 2
3 3 3
2 2 4sin cos sin 2
3 3 3
Applying 1 1 2 3R R R R and expanding
3a b c 3 | cos3 |
2
1 2cos 3 3 ,
2 3 3
2,
9 9
56. Identify the correct .............................
Sol. (A)
–2 – O 2
No of solutions are 5
(B)
8
38
8
5
2
8
7
No of solutions are 5
(C) 2
No of solutions are 4
(D) |12–x| 2
2 = |x|2 |x2–12| = |x|
No of solutions are 4
57. Which of the .............................Sol. If f(x) is not constant on [–1, 1]
f '(x) 0 at some point in –1 < x < 1
f is increasing at that point and we can’t have f(–1) = f(1)(b) Not trueIf f is bounded curve changes its concavityIf f is unbounded then curve crosses x-axis(c) Not truee.g. f(x) = x4
(d) f '(x) =2x2 + 2x – 12 = 2(x + 3)(x – 2) < 0 for x ( 3, 2) where f(x) decreases.
Also f "(x) 4x 2 0 for1
x2
58. If A(x + y) .............................Sol. A(x + y) = A(x)A(y)
A(0 + 0) = A(0)A(0) A(0) = 1Put y = –x, we getA(0) = A(x)A(–x) …..(i)
2
A( x)B( x)
1 (A( x))
2
1
A(x)1
1(A(x))
2
A(x)
1 (A(x))
= B(x)
Thus, B(x) is even.2011 2010 2011
2010 2010 2010
B(x)dx B(x)dx B(x)dx
2010 2011
0 2010
2 B(x)dx B(x)dx 2010 2011
0 0
B(x)dx B(x)dx
59. Letdy
y f(x)dx .............................
Sol.dy
y f(x)dx is linear differential equation.
I.F. = ex
Solution is yex =xe f(x)dx C
now if 0 x 2 thenx x xye e e dx C
yex = x + Cy(0) = 1, C = 1 yex = x + 1
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x
x 1y ;
e
2
y(1)e
Alsox x
2x
e (x 1)ey '
e
2 2
e 2e e 1y '(1)
ee e
If x > 2
x x 2ye e dx yex = ex – 2 + C y = e–2 + Ce–x
As y is continuous
2 xxx 2 x 2
x 1lim lim(e Ce )
e
2 2 23e e Ce C 2 for x > 2y = e–2 + 2e–x
y(3) = 2e–3 + e–2 = e–2 (2e–1 + 1)
xy ' 2e
3y '(3) 2e
60. If
n
0k 3k
1Ck .............................
Sol.
n
0k 3k
1Ck
=
n
0k )3n)(2n)(1n(
)2k)(1k( n+3Ck+3
=)3n)(2n)(1n(
1
n
0k
2 )2)6k2(–6k5k( n+3Ck+3
=)3n)(2n)(1n(
1
n
0k
n
0k3k
3n3k
3n3k
3nn
0k
2 C2C)3k(2–C)6k5k(
=)3n)(2n)(1n(
1
n
0k
n
0k3k
3n2k
2nn
0k1k
1n C2C)3n(2–C)2n)(3n(
=)3n)(2n)(1n(
1
2
)3n)(2n(–)4n(–22)]3n(–2)[3n(2–)1–2)(3n)(2n( 3n2n1n
on solving we get
=)3n)(2n)(1n(
2–2)2nn( 1n2
a = 1 p = 1b = 1 q = 2c = 2 r = 3d = – 2
PAPER-2PART-I : PHYSICS
1. In the figure ...........................Sol.
k2
massless rod
(1) k
/2 /2
m
/2
Let the block is displaced x.Elongation in spring (2) = x –
m
k(x–)
/2k/2
k(x–)2
k k(x )2
22x
4
x =5
4
Restoring force = k (x – )
= k24 kx
x x m x5 5
=k
5m, T =
5m2
k
2. A particle A ...........................Sol.
y
xA B
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BAˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆV 2i j k 2i 2 j k 4i 3 j
37ºA
4
3
B
dmin = 5 sin 37 = 3m
3. Two small ...........................
Sol.
m m v0
v0
0v
2
0v
2
Vcm
In COM frame
0v
2m
m0v
2
0v
2
0v
2
0V
2
m
k1 45º
0V
2
m45º
K1 +x
v
v
m
COME
20mV
2 = mV2 +
21kx
2COAM
0V x2 m 2 mv
2 2 2
0mV2
mV ( + x) v = 0V
2( x)
20mV
2 =
2 220
2
mV 1kx
24( x)
after substituting the values we get v0 = 1.3 m/s
4. A non-conducting ...........................
Sol.
45º53º 37º
53º
37º
E
= 45º
Electric field will be along angle bisector.
5. Two uniform ...........................
Sol.
22
zero
mg2 2
2 21 m 1 mcos mg cos2 (2 )
2 2 2 3 2 3
=
constant
2 22 21 m
mg 1 1 2 (1 4)2 2 2 3
= constant
2 2 25 3m mg
6 4 = constant
22m 5 3g
2 3 2
= constant
2 25 3g
3 2 = constant
5 d 32 g2 0
3 d 2
103g 0
3
2 9g(angularfrequency)
10
T =10
29g
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6. There are ...........................Sol. Suppose we have connected battery across points A and B.
All points other than A and B will be at same potential soequivalent circuit will look like as given below
r
r
18 Points
So equivalent resistance will be 6
7. Moment of ...........................
Sol. = 2 2Ma b
2
= M (aa + bb) + m2 2a b
2
A0 = 74B0 = 222.
8. A glass sphere ...........................Sol. A = 2(1 – cos2C)R2
2C C
C
= 2 (2sin2C) R2
= 4R229 36 R
25 25
9. A uniform ...........................Sol.
A
Acosg
mg
N
fs
2A cos
C
A cos sin
C
N =2A cos
mgC
s
A cos sinf
C
=A
mg2C
s
Af
2C
fs N
A Amg
2C 2C
A
A 2mgC
10. Four identical ...........................
Sol.
q
q
R
Rq
4
3
q2
R
R
2R
k1
0
12 3
3q R. 2Rsin
4 ˆB k4 2R
= 0q k8 2 R
0
13 3
q. R.2Rsin2 ˆB k
4 2R
= 0q k16 R
0
14 3
q. R. 2Rsin4 ˆB k
4 2R
= 0q k8 2 R
01
q 1 ˆB 2 k8 R 2
1 1 1F q v B
F1 =
2 20q 1
28 2
12. Find the minimum ...........................Sol.
x
y2, 2y1, 1
1 1 2 2
1 2
y A( T x) y A( T x)
(1 T) (1 T)
y1 (1T – x)(1–1T) = y2 (2T + x)(1–2T)
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x = 1 1 2 2
2 2 1 1
(y y ) T0
y (1 T) y (1 T)
Tension = 1 1
1
y A T
(1 T)
= y1A1T (1 – 1T)
= y1A1T = (109)(10–5) (6 × 10–4) (20)
= 120 Newton
Fundamental frequency of 1st wire alone
=1 120 20
10Hz2(1) 0.3 2
Fundamental frequency of 2nd wire alone
=1 120 40
20Hz2(1) 0.075 2
Frequency for which both the wires will vibrate (joint on
node)
= LCM of individual frequency = 20 Hz
14. Which of the ...........................
Sol. E =KQq 1 KqQ
– ma 2 ma
=KQq KQq KQq
– –a 2a 2a
If length of semi major axis is A
E =KQq
–2a
=KQq
–2A
A = a
+Q
a
u0 m
a
60°
3
2
Length of semi minor axis =3
2a
2 2
KQq GMm
a a
M =KQq
Gm
T2 =2
0
41 Qq
G4 Gm
a3 =
3 3016 ma
T =
3 3016 ma
16. Which of the ...........................Sol.
T
5T0
2T0
A C
B
5P02P0
A – B T = constant T
V = constant T = KV
PV = nRT P = constant V TB – C = constant V = constantC – A = constantP
C
B AV
V
CB
A
TP
AB
C
T
A – B P = constant , U0 = 0
3nRT
2WAB = nRT = nR (–3T0) = –3nRT0 = –2U0
BC V = constantWBC = 0
C – A WCA = nR(5T0) n i
f
= 5nRT0 n5
2
= 0
10 5U n
3 2
Process Q U W
AB –5U0 –3U0 –2U0
BC +3U0 +3U0 0
CA0
10 5U n
3 2
0 10
3v 0
10 5U n
3 2
0
n(2.5)
Net heat = 0
10 5U n
3 2
– 2U0
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18. Radius of the ...........................
Sol. +e
–e
m
Pr
2
2
ke( e) mv
r r
22ke
mvr
(m) vr = n
nr
mv
22ke
mvn
mv
2
2kemv mv
n
2kev
n
2 2
2 2
nr
ke m
2ke( e) 1E mv
r 2
=2
2 2ke 1 1mv mv
r 2 2
2 4 3
2 2
k e mE
2 n
For hydrogen atom = 1
V0 =2Ke
r0 =
2
2Ke m
E0 =2 4
2
K e m–
2For this atom = 3
V =n
V3 0 , for n = 1 V = 3V0
r = 0
2
r9
n for n = 2 r = 0r9
4
E = 02E
n
27
PC = E3 – E1 =9
27– E0 –
0E
1
27–
= –3E0 + 27E0 = 24E0P=C
E24 0
20. The velocity ...........................
Sol.
J
a
COM
VCM
3
a2
J = 6aVCM VCM =a6
J
CM = 3(2a) 12
a2 2
+ (2a )
2
3
a2
+ 2 (2a)
9
aa
22
= 2a3 + 3a9
8 + 3a
9
40
= 3a9
66 = 3a
3
22
AMAIT w.r.t com
Ja = 3a3
22
2a22
J3
Velocity of P
J/6a
2
3J 2a
322 a
= ia11
JJ
a6
J
=
6
J
11
i
a
J
PART-II : CHEMISTRY
21. If P = total number of ………….
Sol. 6 GI with following trans pairs are possible: (aa)(bb)(cd),
(ab)(ab)(cd), (ac)(bd)(ab), (ad)(bc)(ab), (aa)(bc)(bd),
(bb)(ac)(ad). Out of which 3rd and 4th are chiral. So there are 2
pairs of enantiomers and total 4 chiral stereoisomers.
22. The following wave………….
Sol. The wave function is independant of angle which implies it
belongs to spherically symmetric s orbital. Further, it has only
one radial node at r = Z/2 which indicates 2s orbital. 2s orbital is
first excited state.
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23. A first order reaction………….
Sol. K =2.303
2 log
100
80 K =
2.303 5log
2 4min–1
K =2.303
2 ×0.1 min–1
t½ =K
693.0 = 6 min
24. 5.6 L of an unknown………….
Sol. Mol of gas = n = 5.6 / 22.4 =1
4E = nCvT
Cv = 3 cal.mol–1.K–1
Cp = 5 cal.mol–1.K–1
=5
3
ideal gas is monoatomic.
25. A crystal is made up ………….
Sol. Effective no. of atoms of X = 4 – 4×(1/8) = 3.5
Effective no. of atoms of Y = 4 – 1 = 3
Effective no. of atoms of Z = 8 – 4 = 4
Hence, formula is X7Y6Z8
26. When the following ………….
Sol. Bromine disproportionates in basic solution of carbonate.
3Br2 + 3Na2CO3 5NaBr + NaBrO3 + 3CO2
27 Consider the ………….
Sol. x = 6, y = 3.
Final product is benzene.
H+/vkf/kD;leku iqujko`fr
28. How many of the following………….
Sol. Reactivity order towards Grignard reagent is as follows
C
O
R H > C
O
R R > C
O
R OR > C
O
R NR2
Only (i), (iv), (vi) and (vii) undergoes nucleophilic addition
reaction.
29. How many hyperconjugable………….
Sol.
CH–CH(CH3)2
+
CH2CH3
CH3
1
2
3
Total hyperconjugable H-atoms = 6
30. How many of the following………….
Sol. (S1, S3, S5 & S7 are correctly matched)
31. Metal M ………….
Sol. K2Cr2O7 is a very common lab reagent whose aqueous solution
is orange coloured but turns yellow in alkaline medium due to
formation of chromate.
K2Cr2O7 is obtained from Chromite ore FeO.Cr2O3 which can be
written as Fe(CrO2)2. Thus, M is Cr.
32. The gas C ………….
Sol. 2NH4Cl + K2Cr2O7 N2 + Cr2O3 + 4H2O + 2KCl
33. Suppose two hypothetical………….
Sol.
*2Px *2Py *2Px *2Py
One state has both electrons in different π* orbitals but with
opposite spins (violating Hund's rule).
Other state has 2 outermost electrons paired in a single *
orbital (again violating Hund's rule).Total spin S in both the
cases is zero, hence both are singlet states.
34. Which of the following………….
Sol. Acid catalysed dehydration of alcohol (E1 elimination) proceeds
through carbocation intermediate, which are singlet species as
they contain no upaired electron. Thus, total spin is zero.
Reimer-Tiemann formylation and Carbylamine formation involve
formation of singlet dichlorocarbene by alpha elimination of
chloroform by an alkali.
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35. Which of the following………….
Sol. Y is aromatic cyclopentadienyl anion. Z is ferrocene, in
which ferrous ion is central atom and two cyclopentadienyl
anion (Y) act as ligand each donating 6 pi electrons to
ferrous ion.
EAN = 26 – 2 + 2 × (6) = 36.
38. Choose the correct ………….
Sol. (37) & (38)
H3OOO
COOH(X)
HOH
HOH
CH3–CHOH
OH+
OH OHCOOH
–H2O
CH3–CH=O
(Z)
(–H2O)
(Y)
O
OH
O
40. Compound P and R ………….
Sol.
O O
H /OH
OH
(P)
KOHCHCl3
OH
(Q)CHO
Ac2O / AcO,
(T)
OO
42SOH
OH
(S)COOH
Ni/H2
OH
(R)CH=CH–COOH
PART-III : MATHEMATICS
41. If a, b, c ..............................
Sol. c a b, b c a, a b c are all positive
Using A.M. G.M.
1/3a b c abc
3 (i)c a b a b c b c a (c a b)(a b c)(b c a)
Also,2 2 2a a (b c)
2a (a b c)(a b c)
Similarly2b (b c a)(b c a)
2c (c a b)(c a b)
2 2 2 2 2 2a b c (a b c )(b c a) (c a b)
Thus abc
(a b c)(b c a)(c a b)
abc1
(c a b)(a b c)(b c a)
hence from (i)abc
1(c a b)(a b c)(b c a)
42. If a, b, c, d > 0..............................
Sol. We have
2 2 2 2 2 2 2(a b c )x 2(ab bx cd)x b c d 0
2 2 2(ax b) (bx c) (cx d) 0
2 2 2(ax b) (bx c) (cx d) 0
b c dx
a b c
2b ac or 2 log b log a log c,
Now 1 1 3 2
33 14 log a
65 27 log b [Apply R R R 2R ]
97 40 log c
0 0 0
65 27 log b
97 40 log c
43. Suppose that ..............................
Sol. h'(x) f(x)
h''(x) f'(x)
h(1) 0,f(1) f'(1) h'(1) h''(1) 1 g(1)
f g(x) x
'g' is inverse of 'f'
f ' g(x) .g'(x) 1
f ' g(1) .g'(1) 1
f '(1).g'(1) 1
g'(1) 1
44. A be a square..............................
Sol. Letm n q n
A , adj(a)p q p m
Let | A | = d = mq – np
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m qd n(1 d)A d adjA 0
p(1 d) q md
2 2 2 2
2 2 2 2
3 2 2 2
2 2
mq m d q d mqd np 2npd npd 0
(mq np) (mq np)d m d q d 2mqd 2d 0
(d d 2d ) d(m q 2mq) 0
d[(d 1) (m q) ] 0 d 1,m q 0
2| A d adj | A | (m q) 4(mq np) 4d 4
45. Two persons ..............................
Sol.4 2 2 4 3
2 3C p (1 p) C p (1 p)
6(1 p) 4p
3p 0.6
5
46. If the radius..............................
Sol.
22 h ab 4 2tan
a b 7
1tan
2 2 2
1 2(8 )sin
2 3 2
6
Hence, equation of circle is2 2(x 6) (y 6) 8
47. Possible value..............................
Sol. x2 – 5|x| + 1 = (3 – k)x
draw the graph of x2 – 5|x| + 1 and (3 – k)x
y = x2 – 5|x| + 1y = (3–k)x
There are two ways for line (3 – k)x to have 3 distinct
solutions with x2 – 5|x| + 1 as shown in figure
Case-I
so y = (3 – k)x, line touches y = x2 – 5|x| + 1 at some x > 0
and cuts at two points for x < 0
x2 – 5x + 1 = (3 – k)x
x2 + (k – 8)x + 1 = 0 D = 0 (k – 8)2
k = 8 ± 2, k = 6 is only possible value for which x > 0
k = 6 or 10
Similarly for x < 0
x2 + 5x + 1 = (3 – k)x
x2 + (k + 2)x + 1 = 0
(k + 2)2 = 4 or k = – 4
x > 0 so rejected
k + 2 = ±2
k = – 4 or 0 k = 0
x < 0 is possible value
so total possible values of k are 0 and 6
48. If the eccentric..............................
Sol.
If P is the point (acos , b sin ), the coordinates of Q are
a cos , bsin2 2
i.e., ( asin , b cos ) where2 2a 28, b 7
The area of 1CPQ
2
2 2| ab cos ab sin |
1 2 2 21
area (x y x y )2
1ab
2
1( 28)( 7 ) 7 sq. units
2
49. The straight line..............................
Sol.
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Let the perpendicular distance of P from the line be h,
1h 5 6( 2 1)
2
12h ( 2 1)
5
Also tangent parallel to the given line is 4y 3x 12 2
Its distance from the line 4y 3x 12 IS
12H ( 2 1)
5
Hence there are three points as shown in the figure.
50. Number of the ..............................
Sol. 2 sin 11x cos 3x 3 sin 3x 0
1 3cos 3x sin 3x sin 11x 0
2 2
osin(30 3x) sin 11x 0
2sin 7 x cos 4 x 012 12
sin 7 x 0 or cos 4 x 012 12
7 x n or 4 x n ,n Z12 12 2
n n 7x or x ,n Z
7 84 4 48
51. [ a b b c c a
]..............................
Sol. [ a b b c c a
] = [ a b c
]2
52. If xa yb zc md
..............................
Sol. (a b) ( c d) (b c) (a d) (c a) (b d)
= [a b d] c [b c d]a [c a d]b – 6d
Now d a b rc
, where , , r are scalars
Take dot product with a b
(a b) d r [a b c]
= 2r
r =(a b) d
2
Similarly find and
So 2 d
= [a b d]c [b c d]a [c a d]b
m = 2
53. The value of ..............................
54. The minimum ..............................
Sol.x yf(x y) 2 f(y) 4 f(x) .........(i)
Interchanging x and y, we get
y xf(x y) 2 f(x) 4 f(y) .........(ii)
x y y x
x x y y
x x
2 f(y) 4 f(x) 2 f(x) 4 f(y)
f(x) f(y)k
4 2 4 2
f(x) k(4 2 )
f(1) k(4 2) 2
k 1
x x
4 4
x 2 x x 2
f(x) 4 2
f(4) 4 2 240
f(x) (2 ) 2 (2 1/ 2) 1/ 4
Thus, f(x) has least value as 1/ 4
55. The value of ..............................
56. The value ..............................
Sol.2 2n 2
2nn
(x ax 1) x (2x x b)f(x) lim
1 x
2
2
x ax 1, | x | 1
2x x b, | x | 1
a b 3, x 1
2a b 5
, x 12
x 1lim
f(x) exists if
x 1lim
f(x) =
x 1lim
f(x)
2 2
x 1 x 1lim (2x x b) lim (x ax 1)
2 1 b 1 a 1
a b 1 ........ (i)
x 1lim
f(x) exists if
x 1lim
f(x) =
x 1lim
f(x)
2 2
x 1 x 1lim (x x 1) lim (2x x b)
1 a 1 2 1 b
a b 1 ...... (ii)
Solving Eqs. (i) and (ii), we get a = 1 and b = 0.
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57. Let f be differentiable....................
Sol. f is differentiable at x = 1 and g(x) is continuous at x = 3.
i.e. a = 2b and 2a = b
Also 3a + 3 – c = 10 + b
a b 0 and c 7
2x 7 x 49k (1 ) 0 has real and distinct roots
for R
2
2
2
49 4(49k)(1 ) 0 R
4k 4k 0 R
16k 16k 0
k(k 1) 0
k ( 1, 0)
58.x 2
f(x)lim
| g(x) | 1 exists ..............................
Sol. h 0 h 0
f(2 h) b(4 h)LHL lim lim
| g(2 h) | 1 | c(2 h) d | 1
4b
| 2c d | 1
h 0 h 0
f(2 h) b(4 h)RHL lim lim
| g(2 h) | 1 | a(2 h) 3 c | 1
4b
| 2a 3 c | 1
f is differentiable at x = 1 i.e., a = b = 0
So, LHL =RHL = 0 (it is given that limit exists)
59. ABC ..............................
60. The inradius ..............................
Sol. LetA B C
tan x, tan y, tan z2 2 2
CP
B
A
R
Q
r
r
r
I
r r rx , y , z
AR BP CQ
put the values of AR, BP and CQ in given relation, we get
2x 5y 5z 6
r r r r
2x 5y 5z 6 (1)
Also in the triangle, xy + yz + zx = 1 (2)
If we interchange y and z, then both equations (1) and (2)
remain unchanged
ABC is isosceles with B C y z
So from (1) and (2), we get
x = 3 – 5y (3)
and 2xy + y2 = 1 (4)
Solving we get,4 1
x , y z3 3
3rAR , BP 3r, CQ 3r
4
Now perimeter 2s = 2 . AR + 2 . BP + 2 .27r
CQ2
Given perimeter is smallest integer r 2
27s
2
27rs 2 27 sq. unit
2
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JEE PREPARATORY TEST-2 (JPT-2)(JEE ADVANCED PATTERN)
TARGET : JEE (MAIN+ADVANCED) 2018
DATE : 13-05-2018 COURSE : ALL INDIA TEST SERIES (VIKALP) | CLASS: XII/XIII
ANSWER KEYCODE-0
PAPER-1PART-I : PHYSICS
1. (ABD) 2. (BD) 3. (BC) 4. (AC) 5. (ACD) 6. (BC) 7. (ABCD)
8. (AB) 9. (AD) 10. (AC) 11. (AD) 12. (BCD) 13. (C) 14. (ABCD)
15. (BD) 16. (BCD) 17. (ABCD) 18. (ABC) 19. (ABCD) 20. (BC)
PART-II : CHEMISTRY
21. (BD) 22. (AD) 23. (AC) 24. (AC) 25. (AD) 26. (ACD) 27. (AD)
28. (ABCD) 29. (BC) 30. (ABD) 31. (ABD) 32. (ABD) 33. (ABC) 34. (BD)
35. (ABD) 36. (D) 37. (BD) 38. (ACD) 39. (ABC) 40. (AD)
PART-III : MATHEMATICS
41. (AC) 42. (AC) 43. (ABCD) 44. (AB) 45. (BC) 46. (ACD) 47. (AD)
48. (AB) 49. (BCD) 50. (AC) 51. (AC) 52. (BCD) 53. (ABC) 54. (ABC)
55. (AB) 56. (BCD) 57. (AD) 58. (BD) 59. (ABD) 60. (ABC)
PAPER-2PART-I : PHYSICS
1. (5) 2. (3) 3. (2) 4. (9) 5. (9) 6. (6) 7. (3)
8. (1) 9. (4) 10. (2) 11. (A) 12. (C) 13. (B) 14. (A)
15. (B) 16. (D) 17. (D) 18. (A) 19. (A) 20. (D)
PART-II : CHEMISTRY
21. (6, 4) 22. (1) 23. (1, 2, 3, 4) 24. (1) 25. (7, 6, 8) 26. (5, 1, 3)
27 (9) 28. (4) 29. (6) 30. (4) 31. (D) 32. (B)
33. (C) 34. (C) 35. (C) 36. (B) 37. (A) 38. (D)
39. (A) 40. (C)
PART-III : MATHEMATICS
41. 0,1,2,3 42. 1,7 43. 2,3 44. 1,2,4 45. 2,3,5 46. 1,2,4,8 47. 0, 6
48. 0,1,2,3,4,5,6 49. 3 50. 2,5,8 51. (A) 52. (A) 53. (B)
54. (C) 55. (B) 56. (C) 57. (A) 58. (C) 59. (B) 60. (C)