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NARAYANA IIT ACADEMY INDIA
Sec: Sr. IIT-IZ-CO SPARK Jee-Advanced Date: 16-05-18 Time: 09:00 AM to 12:00 Noon 2015_P1 Max.Marks:264
KEY SHEET
PHYSICS 1 1 2 2 3 3 4 5 5 9
6 7 7 5 8 4 9 ABD 10 ABCD
11 ABD 12 BC 13 B 14 BCD 15 ABCD
16 ABD 17 ABCD 18 AB 19
A-RT; B-S; C-P; D-Q
20
A-PQRT; B-QS; C-PQRS; D-PRT
CHEMISTRY
21 2 22 2 23 8 24 4 25 3 26 0 27 2 28 1 29 AB 30 AB 31 B 32 AB 33 ABC 34 AC 35 ABD
36 ABCD 37 BCD 38 BD 39
A-PR; B-R; C-PT; D-QS
40
A-S; B-P; C-Q; D-R
MATHS 41 3 42 3 43 2 44 4 45 1
46 3 47 7 48 1 49 ABCD 50 BD
51 ABCD 52 D 53 ABCD 54 AC 55 AC
56 AD 57 BD 58 ABC 59
A-PQ B-PQ C-PR D-PQS
60
A-S B-R C-Q D-R
Narayana IIT Academy 06-05-18_Sr.IIT-IZ-CO SPARK_JEE-ADV (2015_P1)_GTA-18_Key & Sol’s
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SOLUTIONS PHYSICS
1. Complete the pyramids OBAD and 1OB AD for this ‘C’ is the geometric centre . The closed Figure enclose ‘6q’ charge electric flux through all 6 faces is
0
6q
. Through each
face 0
q
So n=1
2.
04mK22mK
12mK
2 12 sin 2 .............(1)mk mk
2 02 cos 4 ...........(2)mk mk 2 1 02 2 4mk mk mk ( squaring and add 1 and 2) 2 1 02 0.8 2 2.8k k k eV 2 1 02 0.8 2 2.8k k k eV Energy released = 1 2 0 3.6 1 2.6k k k eV
3 R2 2
0 00E 4 R 4 G 4 r p r dr
4
00 2
G p pEr
r2 2 2
1 0E 4 r 4 G 4 r dr
21 0E G S r
Potential inside the earth
i i S SV V V V V V
r R
i i 0RV E dr E dr
4R /2 R
0 0 2R
RG dr G drr
Narayana IIT Academy 06-05-18_Sr.IIT-IZ-CO SPARK_JEE-ADV (2015_P1)_GTA-18_Key & Sol’s
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3
031G R12
To escape from field of earth
3
2031G R 1m mv24 2
3031G RV
12
4. Ans : 3
Sol : sin 0 &x cm cmce F a V remains vertical
Constraint equation 22g
r lV
From L C E
rA
O
0 AV r
gV
2r
U K
2 21 12 2 22 g
l lmg MV I
2 2
22 2 1 12 2 2 62 2
l Mlmgl M
2 2 2 2 2 22 28 12 482 2
Ml ml Mlmgl
6 2 2 2
5
g
l
5. Conceptual
6. 1 71 3 8.023829H Li u
4 42 2 8.005206He He u
0.018623m u
0.018623 931.5 mevEu
Narayana IIT Academy 06-05-18_Sr.IIT-IZ-CO SPARK_JEE-ADV (2015_P1)_GTA-18_Key & Sol’s
Sec: Sr. IIT-IZ Page 4
=+17.35 Mev
7. Conceptual
8. Image formation due to convex lens
1 1 1
36 30v
30 36 1806
cm
This image will act like a vertical object for mirror and after reflection from mirror its
image (shown by 2I ) will be formed at 80cm below optical axis of convex lens.
For concave lens, this image will be object at a position of 15cm below the lens for
final image formed by concave lens.
1 1 120 15 f
1 5
300f
Also
1 1 11f R R
Or 5 3 21
300 2 R
60R
9. 1 0.1258
Smallest divisionon main scale cm cm
5 divisions of the vernier scale = 4 divisions of the main scale =4x0.125 = 0.5cm
I division of vernier scale 0.5 0.15
cm
Least count of vernrer = 1 main scale divisions -1 vernier scale divisions = 0.125 –
0.1=0.025cm
Least count of screw gauge = pitch/100
If pitch of the screw gauge is twice the least count of vernier calipers then the least
count of
Narayana IIT Academy 06-05-18_Sr.IIT-IZ-CO SPARK_JEE-ADV (2015_P1)_GTA-18_Key & Sol’s
Sec: Sr. IIT-IZ Page 5
Screw gauge 0.05 0.005100
mm
Also, least count of linear scale of screw gauge =2x0.025cm=0.05m 2 0.05 0.1 1pitch cm cm mm
1 0.01100mmLeast Count mm
Hence (C ) is correct .
11. Conceptual
12. from LCAM
20 0 0I I Mr 0 0
20
II mr
2
0 02
0
rr r
I dVa r VI mr dr
On integrating 2rV
And t fV R from LCAM 1f and 1tV
2 2 3g r tV V V
13. Electric field is discontinuous at the locations of charges Hence (B)
14.
1 1 1 1 1 125 20u f
1 1 1 1 10020 25 100
V cm
From O to I intensity increases and then decreases at
x = 90 cm and 110cm intensity is same (d) Radius at
x = 200cm is equal to radius of lens.
15. The no. of turns in the spiral coil. Per u nit radial width ' ' Nnb a
Consider a concentric
ring of radius ‘r’ and radial thickness dr. No. of turns in that thickness N drb a
.
Narayana IIT Academy 06-05-18_Sr.IIT-IZ-CO SPARK_JEE-ADV (2015_P1)_GTA-18_Key & Sol’s
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The flux linked with that ring 2r B
Emf induced 20
2.sinN d tdr r Bb a dt T
2
02 2. cosr N te dr B
b a T T
Total emf = b
a
e
2
20 2 2cosb
a
B N t r drb a T T
3 320 2 2cos
3
b aB N tb a T T
2 22
0 02 2 2cos cos
3
b ab a tE B N E tT T T
Current through the coil EiR
0 2cosE
i tR T
Current will be max when 30, , ,2 2T Tt T ……
16. At a given temperature the average Kinetic energy per molecule is given (f/2) kT
which is same for all Diatomic gases hence option (A) is correct. RMS velocity of gas
molecules at same temperature is inversely Proportional to the square root of molar
mass of gas hence option (B) is correct.
. From gas law pv=NKT we
Can see that option (D) is also correct.
17. Following from the graph in the question O<i<1 resistance of diode 5DR for 1i
there after
0. 0DV RI
(dynamic). Consider option (a).
When I = 1 A; 5DR ; and 5 10BV i = 15 volt
There after slope of 101
BV
( since 0DR )
30 15 15 102.5 1 1.5
So option (a) correct.
Option (b) slope 2 10Vi
from graph so it is correct
Narayana IIT Academy 06-05-18_Sr.IIT-IZ-CO SPARK_JEE-ADV (2015_P1)_GTA-18_Key & Sol’s
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Option c) when 215 , 5 , 10B DV V V V V V till this value of ,B DV V is linear there after 5DV V (
remains constant) option ( c) is correct
Option (D) when 215 ; 5 ; 10 .B DV V v V V V
There after for every 5V rise of ,BV 5 v rise in 2V appears so finally
when 230 , 25BV V V V
Option (d ) is correct.
18. Kmax = E – W
A AT 4.25 W
B A BT (T 1.50) 4.70 W
WB–WA = 1.95 eV
de–Broglic wavelength
h
2km
1k
B A
A B
KK
TA = 2 eV
WA = 2.25 eV WB = 4.2 eV
19. Conceptual
20.
Narayana IIT Academy 06-05-18_Sr.IIT-IZ-CO SPARK_JEE-ADV (2015_P1)_GTA-18_Key & Sol’s
Sec: Sr. IIT-IZ Page 8
CHEMISTRY
24. Key: 4
Solution : Form A: 3 4Cr NH Cl Br cl
Form B : 3 24Cr NH Cl Br
X= 1
Y = 0
Z = 3 Total 4
25. Key: 3
Solution : 4NO HSO
Isoelectronic with 2N B.O =3
26. ANS: 0
Sol: 'r
R is Identical to , , 12 2 2 3x y zp p p n
,Y 21 ,2 ,3 3S S S n
1 2 0n n .
27. ANS; 5 0 28.5 28500G kj j
28500 0.2996500cellE
0.059 0.29HcellE Xp 5Hp
28. ANS: 1
Sol: 2m , mole =0.2g Gmw=100
0.279=1.5x1.86x100
x x10 1.5i
00.5 c
.
30. Key: A,B,
34. a) Covalent chlorides do not give test b) Due to more solubility product some PbCl2 will remain in solution and black ppt
will be formed in second group also during qualitative analysis c) 3
4Cu CN
d) rosy red ppt
36. key: abcd
Narayana IIT Academy 06-05-18_Sr.IIT-IZ-CO SPARK_JEE-ADV (2015_P1)_GTA-18_Key & Sol’s
Sec: Sr. IIT-IZ Page 9
Solution : NCERT
37. Ans : BCD
SOL : CONCEPTUAL
39. Key: A-pr; b-r; c-pt; d-qs
Solution : Given processes are involved in their extraction processes.
40. Conceptual
MATHS
41. 1( ) ( )f x f x x
42. (i) 111 2 ,
11x y y k
2x y is divisible by 11
(ii) 213 2
13x y y k
x – 2y is divisible by 13
6 12 111
x y 6 1
11x y
6 12 1
13x y
6 13 1
13x y y
6 113x y
.
6 143x y k
6 6 143 5 138 5 5x y k y k y k
k y is divisible by 6
6 6 138 5 6x y k 28x y
43. 2 2211 10 11r r
42 100
5021
r
r
44.
2017
1 40350 1
xI dxx
=
1 2017 2017
4035 40350 11 1
x xdx dxx x
Put 1xt
Narayana IIT Academy 06-05-18_Sr.IIT-IZ-CO SPARK_JEE-ADV (2015_P1)_GTA-18_Key & Sol’s
Sec: Sr. IIT-IZ Page 10
2017 2017
1 4035 40350 1 1
x tI dx dxx t
1 2017 2016
1 240350
( )1
x xI dx Ix
1
2
1II
45 Here for x = 1, f(x) is zero which lies in [-1, 1) so if f(x) does not contain any value in
[-1, 1), (x-1) should be cancelled 1x k has a root 1 1 1 0k .
k = 0
For k = 0, 1 1 [1, )1
xf x xx
46 1 2 21 2 2 1 0( ) ....n n n
n n nlet f x a x a x a x a x a x a
11 1'( ) 1 ....n n
n nxf x na x n a x a x
22 1 21 1'( ) '' 1 ...n n
n nxf x f x n a x n a x a
'' ' 0af f
0, ' 0, ''( ) 0f f f
min 3value of k is 47 Let P denotes the chances of single bacteria to die, then
1 1 1. . . .4 2 4
P P P P P P
3 2 22 4 1 0 1 3 1 0P P P P P P
3 132
13 3 5 131 12 2
P
P
48. MULTI CORRECT
49. conceptual
50. 2 3 101 ..... 0a
10 2 9 3 3 4 7 5 6, , , ,
2 3 4 5Reso 6 7 2 8Re
Narayana IIT Academy 06-05-18_Sr.IIT-IZ-CO SPARK_JEE-ADV (2015_P1)_GTA-18_Key & Sol’s
Sec: Sr. IIT-IZ Page 11
6 5 10 7 4 2 9 8 31 1 1Re 12 2 2
(b) 11
2 101 ......1
x x x xx
2 4 6 8 10 12 14 15 18 20x x x x x x x x x x
= 2 4 6 8 10 3 5 7 9x x x x x x x x x x
10
1
k
k
x
Put x = 7
77 10
71
1 01
k
k
(c) put x=1 11 310
1
1 1 11 1
k
k
i iii i
(d) put x= 1110
1
1 01
k
k
51. conceptual
52. : 2 3 2 2 32x ydy x x y dx xy dx y dx
2 2 3 2 2 3x ydy y xdx x dx x ydx xy dx y dx
xy 2 2xdy ydx x x y dx y x y dx
2
2 1 1
y yddxx xxy y
x x
Let 1yx 2 1 1
tdt dxxt t
2
1 1 1 log4 1 4 1 2 1
dt dt dt x ct t t
log 41 2 log
1 1t x kt t
53.
12
2sin1
xx
1
1
1
( 2 tan ), 12 tan , 1 1
2 tan , 1
x xx x
x x
1 12 tan 5 tan ,x f x g x x x
Narayana IIT Academy 06-05-18_Sr.IIT-IZ-CO SPARK_JEE-ADV (2015_P1)_GTA-18_Key & Sol’s
Sec: Sr. IIT-IZ Page 12
1for x
1. . 3tan 1i e x x for x
54. Let A a bc d
, then a bc d
a cb d
= 1 00 1
2 2 1a b
ac+bd=0 2 2 1c d
det 1
11a b
A Ic d
ad a d bc
55. Conceptual
56.
consider the function f(x)= 3 21 3 2 5xe x x xx
57. The parabolas are 2 2 2 2 2 24sin ( sin ) 4cos ( cos )y x and y x , hence the locus is
2 2cos sin 0 1 0x x .
58.
59. Key: (A)-p,q (B)-p,q (C)-p,r (D)-p,q,s
60.
Key: (A)-s (B)-r (C)-q (D)- r
(A) tan tan 3A C sin sin 3cos cosA C A C
(B)
Similarly,
Narayana IIT Academy 06-05-18_Sr.IIT-IZ-CO SPARK_JEE-ADV (2015_P1)_GTA-18_Key & Sol’s
Sec: Sr. IIT-IZ Page 13
So,
But (using )
Also, in an acute angled
(Using )
But
So,
Therefore,
So,
The minimum value of is 8.
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