(1001CJA100220030)
Test Pattern
CLASSROOM CONTACT PROGRAMME
JEE(Advanced)
REVIEW TEST
(Academic Session : 2020 - 2021)
08-11-2020
JEE(Main+Advanced) : NURTURE COURSE (PHASE : TNPS)
ANSWER KEY
SECTION-I
SECTION-II
SECTION-I
SECTION-II
PART-1 : PHYSICS
Q.
A.
Q.
A.
1
A,B,D
1
4.00
2
B,C
2
1.73
3
4
A,C,D
A,B,C,D
3
4
0.70 to 0.71
8.00
Q.
A.
Q.
A.
1
A,C,D
1
8.00
2
B,C,D
2
4.00
3
A,B,C
3
5.00
Q.
A.
Q.
A.
1
A,B,C,D
1
5.00
2
C,D
2
1.00
3
A,D
3
0.50
5
A,C
5
0.50
PAPER-1
6
B
6
1.33 to 1.34
PART-2 : CHEMISTRY
4
A,C,D
4
5.00
5
B,C,D
5
2.00
6
A,C,D
6
73.00
7
A
7
3.00
7
A
7
9.00
8
C
8
4.00
9
A
10
C
8
D
8
6.00
9
B
10
C
8
C
8
5.00
9
C
10
A
PART-3 : MATHEMATICS
SECTION-I
SECTION-II
4
A,C,D
4
4.44
5
B,C,D
5
32.00
6
A,B
6
4.00
7
B
7
0.33
Test Pattern
(1001CJA100220031)
CLASSROOM CONTACT PROGRAMME
JEE(Advanced)
REVIEW TEST
(Academic Session : 2020 - 2021)
08-11-2020
JEE(Main+Advanced) : NURTURE COURSE (PHASE : TNPS)
ANSWER KEY
SECTION-I
SECTION-II
SECTION-I
SECTION-II
Q.
A.
Q.
A.
1
A,B,C
1
1.25
PAPER-2
2
A,B,C
2
0.16 to 0.19
PART-1 : PHYSICS
3
4
B,D
B,C,D
3
4
0.37 to 0.40 52.62 to 52.64
Q.
A.
Q.
A.
1
A,C
1
4.00
2
A,B,C,D
2
40.00
3
A,B,C
3
15.00
Q.
A.
Q.
A.
1
A,B,C
1
0.19
2
A,C
2
6.00
3
C,D
3
2.00
5
A,B,C
5
25.50
PART-2 : CHEMISTRY
4
A,B,D
4
3.00
5
A,C,D
5
135.00
6
A,C,D
6
17.50
6
C,D
6
1.00
7
A
7
1.50
8
C
8
0.36
9
D
10
C
7
D
7
3.00
8
C
8
25.00
9
B
10
C
7
C
7
0.00
8
B
8
0.86
9
B
10
B
PART-3 : MATHEMATICS
SECTION-I
SECTION-II
4
A,B,D
4
4.00
5
A,C
5
6.00
6
A,B
6
–1.40
Test Pattern
(1001CJA100220030)
CLASSROOM CONTACT PROGRAMME
JEE(Advanced)
REVIEW TEST
08-11-2020
(Academic Session : 2020 - 2021)
JEE(Main+Advanced) : NURTURE COURSE (PHASE : TNPS)
PAPER-1
PART-1 : PHYSICS
SOLUTION
SECTION-I
1.
2.
Ans. (A,B,D)
Sol. At maximum height, both blocks will be
moving with same speed in horizontal
direct ion & momentum w ill rema in
conserved in horizontal direction.
Ans. (B, C)
Sol. Let the normal force between the block and
the ball be N.
a
M
Þ m1v0 = (m1 + m2) v
For the block, from Newton's IInd law, we
have N = Ma = 2 ma
Applying work energy theorem
For ball (with respect to the block), from
1
1
1
m1 v2 + m2 v2 - m1 v 02
2
2
2
Newton's IInd Law, we have N + ma =
m2 v02
1
h
=
On solving
2 ( m1 + m2 ) g
3.
Ans. (A,C,D)
v=0
Sol.
v2
gl
m1v1 + m2v2 = m1v0
From energy conservation
1
1
1
m1 v 02 = m1 v12 + m2 v22
2
2
2
On solving
m1 + m 2
2m1 v 0
v2 =
m1 + m2
v=0
ml
w
v
l
1 ml2 2
mg cos 30° ´ 2 =
w ´2
2
2 3
m2
From momentum conservation
( m1 - m2 ) v 0
ml
w
v1
v1 =
mv 2
R
Solve The two equations.
When m1 again comes back to horizontal
position
m1
ma(pseudo force)
mg
m1 v 0
Þv=
m1 + m2
-m1gh =
N
N
4.
3 w2l2 v2
=
=
2
3
3
v=
3 3 (10 ) 2
3 3gl
=
2
2
v=
30 3
Ans. (A,B,C,D)
Sol. (A) Sf on A is in y-direction.
(B) Frame ® B
A will experience a pseudo force
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HS-1/14
Target : JEE (Main + Advanced) 2022/08-11-2020/Paper-1
ALLEN
5.
other possibility
Ans. (A, C)
R is reaultant of friction and normal
reaction
æmö 2
Sol. dT = - ç ÷ w xdx
èLø
R sin 7° = mg
x is the distance from O.
23°
2
æmö x
T = - ç ÷ w2
+ T0
2
èLø
T0 = mw2
L
2
TB = 3mw2
f
mg
L
1
, Elasting PE density =
×
8
2
stress × strain =
6.
N C
Rcos 7° = ma
1
Y(stress)2.
2
æ cos7° ö
Þ gç
÷ = a Þ a = g × cot 7° (This
è sin7° ø
option is not given)
Ans. (B)
Sol. R is resultant of friction and normal
reaction
Rsin53° = mg
Rcos53° = ma
æ cos53° ö
gç
÷=a
è sin 53° ø
7.
Ans. (A)
Sol.
é l - l 0 (1 + l 2 DT )ùû
Dl
F
=y
=yë
l
l 0 (1 + l 2 DT )
A
8.
Ans. (C)
Sol.
9.
f
23°
R C
N
30°
HS-2/19
2Ay éë l 0 (1 + l1DT - l )ùû
l 0 (1 + a1DT )
+
3Ay éë l 0 (1 + a2 DT - l ) ùû
l 0 (1 + a2 DT )
Ans. (A)
Sol. mv0r0 = m v 2 + 2gh r cos q
10.
Ans. (C)
Sol. N cos a =
mv 2
r
N sin a = mg
mg
a = g × cot 53° =
R
7°
3g
4
1001CJA100220030
Nurture Course/Phase-TNPS/08-11-2020/Paper-1
ALLEN
SECTION-II
1.
N1 =
Ans. 4.00
T
+ mg
2
T
= ma
2
T
= N2
2
R 2R
Sol.
mg -
H = -K,2 pl
R2
T=
T
2plk(T1 - T2 )
l
R2
ln
R1
T
= ma
2
\ mg -
dT
dr
2
Hdr
K
=
ò 2prl
ò dT
R1
T1
H=
....(i)
4.
....(ii)
T
T
=
Þ mg - T 2
2
2
mg (10 -1 )(10)
1
N
=
=
2
2
2
Ans. 8.00
Sol. Ball collide at highest point of projectile
Motion, after collision, motion is mirror
image of motion that would have been
without collision. Top view of motion is as
2.
H1 = Hf
shown in diagram OQ is required distance.
\ Ans n = 4
OQ = 2(OB) sin30° = 8m.
Ans. 1.73
Sol. Angular acceleration about edge cf table
Q
a=
mgx
mgx
=
ma 2
I0
+ mx 2 , for max a,
12
30°
8m
da
= 0 ® x = 3m
dx
3.
Q'
8m
O
B
8m
–
4Ö3m
Ans. 0.70 to 0.71
5.
N1
a
T
T
Sol. N2
a mg
1001CJA100220030
mg
Ans. 0.50
Sol. Dl1 + Dl2 = Dl
a1
a2
l1
l2
HS-3/19
Target : JEE (Main + Advanced) 2022/08-11-2020/Paper-1
ALLEN
l1a1DT + l2a2DT = (l1 + l2)aDT
mass removed =
æ l2 ö
a-ç
÷a
l1 + l 2 ø 2 ....(i)
l1
è
=
a1
l1 + l 2
Let
remaining mass = m –
l1
l2
= x , so
= 1 - x ....(ii)
l1 + l 2
l1 + l 2
a - (1 - x)a2
a1
æ a - a2 ö 17 ´ 10 -6 - 11 ´ 106
x=ç
÷ =
23 ´ 106 - 11 ´ 106
è a1 - a 2 ø
N
q
Þ N = 3mgsinq
mg
4mgsinqcosq = [Mg + 3mgsin2q]
M = 4msinqcosq –3msin2q
1
1
2
2
Sol. K.E. = mv cm + ICM w
2
2
7.
....(ii)
(3mgsinq) cosqq = µ [Mg + 3mgsin2q]
Ans. 1.33 to 1.34
v
R
1
mv 2 = mgR sin q
2
mv 2
....(i)
R
For wedge
6
= 0.50
=
12
w=
Ans. 4.00
Sol. For particle N – mgsinq =
a1x = a – a2 + a2x
6.
m
2m
=
3
3
æ 2m ö 2
ç
÷R
mR 2
I= è 3 ø
=
2
3
8.
x=
m
m
´2 =
6
3
v cm =
dM
1
= q Þ tan q =
dq
2
5
v
2
Ans. 3.00
ì
1
2
1 ü
´
- (3)
M - 4 í(4)
ý = 4kg
(5)
5
5
î
þ
M = 4kg
60°
Sol.
HS-4/19
M
1001CJA100220030
Nurture Course/Phase-TNPS/08-11-2020/Paper-1
ALLEN
PART-2 : CHEMISTRY
SOLUTION
Q
SECTION-I
1.
Ans. (A,C,D)
2.
Ans. (B,C,D)
3.
Ans. (A,B,C)
vapour density of gas is half of molar mass
vapour density of oxygen d O2 = 16
vapour density of gas mix = dmix
Grams law of effusion
4.
Ans. (A,C,D)
5.
Ans. (B,C,D)
6.
Ans. (A,C,D)
7.
Ans. (A)
8.
Ans. (D)
9.
Ans. (B)
10.
t2
=
t1
300
=
600
dmix =
Ans. (C)
SECTION-II
1.
2
1
´ 16 = 4
4
Ans. (5.00)
4.
Ans. (5.00)
(b) , (c) , (d) , (h) , (i)
5.
1 48 ´ 10 -3
1
murms2 = ´
2 6 ´ 1023
2
d mix
16
3.
Ans. (8.00)
Av. K.E. =
d mix
d O2
æ 184 ö
´ç
÷
è 0.92 ø
Ans. (2.00)
2
eq. of MnO4– = eq. of AX+
= 1.6 × 10–21
1 × 5 = 1.67X(5 – x)
Ans. (4.00)
x=2
Time taken for effusion by O2 (t1) = 10 × 60
= 600 sec
Time taken for effusion by gaseous mixture
6.
Ans. (73.00)
7.
Ans. (9.00)
8.
Ans. (6.00)
(t2) = 5 × 60 = 300 sec
1001CJA100220030
HS-5/19
Target : JEE (Main + Advanced) 2022/08-11-2020/Paper-1
ALLEN
PART–3 : MATHEMATICS
SECTION–I
1.
SOLUTION
3.
y = ax2 + bx + c
Ans. (A,B,C,D)
(
)
b2 - 4ac
b
= 4; =2
2a
4a
B
A
Sol. P
Ans. (A,D)
\ b = –8a; b2 – 4ac = –8a
T
PA.PB = (PT)2
So,
2.
PA + PB
³ PA.PB = PT = S1
2
Þ\ c=
b 2 + 8a
b2
Þc =2+
4a
4a
\ c =2+
64a 2
Þ c = 2 + 16a
4a
Þ PA + PB ³ 6
\ D = abc = a(–8a) (16a + 2)
Ans. (C,D)
Þ D = –16a2(8a + 1)
Þ D = –16(8a3 + a2)
\ Dmax = –16(8(1)3 + 12) = –144
P
A
Sol.
O
l
Dmin = –16(8(3)3 + 32) = –3600
B
4l
4.
C
(3/2, 2)
Sol. am2 + 2m + 1 = 0
m2 + 2m + a = 0
Q
Solving we get m = ±1.
If m = 1 is common root then a = –3.
OA.OB = OP.OQ
Þ 4l2 = (r – OC)(r + OC)
Þ 4l2 = r2 – (OC)2 =
Hence other lines are y = -
41 25
=4
4
4
Þ
3x2 + 10xy + 3y2 = 0
a = 3, b = 10, g = 3.
Þ AB = 5
Þ distance of AB from center
41 25
=2
4
4
uuur
Let AB be y – mx = 0
So,
x
and y = –3x.
3
Hence (x + 3y)(3x + y) = 0
Þl=1
=
Ans. (A,C,D)
3m
- 2 = 2 1 + m2
2
5.
Ans. (B,C,D)
sin q cos q cos q
Sol.
D = cos q sin q cos q
cos q cos q sin q
1
1
1
= ( sin q + 2 cos q ) cos q sin q cos q
cos q cos q sin q
Þ m = 0,
HS-6/19
-24
7
[R1 ® R1 + R2 + R3]
1001CJA100220030
Nurture Course/Phase-TNPS/08-11-2020/Paper-1
ALLEN
=
1
0
0
0
( sin q + 2cos q ) cos q sin q - cos q
cos q
0
sin q - cos q
éC2 ® C2 - C1 ù
ê
ú
ëC3 ® C3 - C1 û
Paragraph for Question 7 and 8
7.
Ans. (B)
Sol. C1 : x2 + y2 = 4
C2 : y = x + 1
while region R is the disc centered at
(
3, 1) and radius
P
= (sinq – cosq)2 (sinq + 2cosq)
A
Now, if tanq = –2 then D = 0
But, h 2 – ab = cos2q – sin 2q = cos2q =
B
-
3
<0
5
Let P(x1,y1)
Clearly AB : y – x = 1
is chord of contant
\ AB : xx1 + yy1 = 4
(1) & (2) are identical
i.e. imaginary pair of lines
Hence, for real straight lines, q = np +
p
4
......(1)
......(2)
x
y
= 1 = 4 Þ (x1y1) = (–4, 4)
-1 1
Also, in this case h2 – ab = cos2q = 0
i.e. lines are parallel.
further, one can observe that by taking
q=
p
, S = 0 becomes x2 + y2 + 1 = 0
2
(Ö3,1)
i.e. no real locus.
6.
Ans. (A,B)
Sol. a + b > g, b + g > a, g + a > b
Þ
Þ
8.
Ans. (C)
Sol. Smallest possible value of 'a' for which
A Ç C = A is a = 4.
g
a
b
< 1,
< 1,
<1
b+a
b+g
g+a
\
a
b
g
+
+
<3
b+ g g+a a+b
P=
CD : y = x + 1
using A.M. > H.M.
3 +1 -1
3
=
2
2
MD = 16 -
[(a + b) + (b + g) + (g + a)]
1
1 ù
é 1
ê ( a + b ) + (b + g ) + ( g + a ) ú ³ 9
ë
û
R : (x - 3)2 + (y - 1)2 £ 4 2
\
9.
(Ö3,1)
3
2
3
29
=
2
2
M
D
CD = 58
Paragraph for Question 9 & 10
Ans. (C)
a+b+ g a+b+ g a+b+ g 9
+
+
³
( a + b)
(b + g )
(g + a) 2
a
b
g
3
+
+
³
b+ g g+a a+b 2
1001CJA100220030
d
r
90–q
Þ
Sol.
P
cos q =
d
r
Q
HS-7/19
Target : JEE (Main + Advanced) 2022/08-11-2020/Paper-1
ALLEN
10. Ans. (A)
Sol. Required locus is perpendicular bisector
of PQ
SECTION–II
1.
Ans. 5.00
æ log10 x ö
Sol. ç
÷
è 3 ø
(1)
(2)
(3)
2.
2
log10
x - 3log10 x + 3
1
ælog x ö
= ç
÷
è 3 ø
Þ
2
log10
x - 3log10 x + 2 = 0
Þ
log10x = 1, 2 Þ x = 10, 100
log10 x
= 1 Þ log10 x = 3 Þ x = 1000
3
log10 x
= 0 Þ log10 x = 0 Þ x = 1
3
5.
Ans. 32.00
Sol. a1, a2, a3, a4,..... are in G.P.
Let a2 = x Þ a 3 =
1
1
Sol. cos a cos( a + b) cos( a + b + g )
sin a sin(a + b) sin( a + b + g )
= sing – sin(b + g) + sinb = 0
æb+ g ö æb+ g ö
Þ sin b + sin g - 2sin ç
÷ cos ç
÷=0
è 2 ø è 2 ø
æb+ g ö æb-g ö
æb+ g ö æb+ g ö
Þ 2sin ç
÷ cos ç 2 ÷ - 2sin ç 2 ÷ cos ç 2 ÷ = 0
2
è
ø è
ø
è
ø è
ø
3.
Ans. 0.50
Sol. either use cot q – tanq = 2cot2q
or differentiate the series
cosq. cos2q.cos4q... =
sin 2 n q
by taking log
2 4 sin q
Ans. 4.44
P(k,k)
O
common ratio
Given
y=x
Q(5,4)
x2
2
x 2 x3 x 4
,
,
with common ratio
2 4 8
x
2
x4
= 162 Þ x £ 6
8
x4
are only in x, must be even
8
and then only
x4
will be an integer
8
Þ x = 4 and 6
6.
Ans. 4.00
Sol. minimum value occur at x = 2
7.
Ans. 0.33
Sol. Tr =
1 æ (3r + 1)2 - (3r - 2)2 ö
ç
÷
3 è (3r + 1)2 (3r - 2)2 ø
Tr =
1æ
1
1
ö
ç
2
2 ÷
3 è (3r - 2) (3r + 1) ø
S¥ =
1
3
8.
Ans. 5.00
Sol. tan2q = tan6q
sin 2q sin 6q
=
cos 2q cos 6q
R(4,5)
Sol.
Þ 2, x,
and x and
(4) log10x = –3
Ans. 1.00
HS-8/19
40
æ 4 - k öæ 5 ö
ç
÷ ç ÷ = -1 Þ k =
è 5 - k øè 4 ø
9
2
log10
x - 3log10 x + 3 = 1
1
4.
mPQ.mOR = –1
q=
Þ
sin4q = 0
np
4
so possible solution are q = p , p, 3p , 2p, 5p
2
Reject solution are
2
2
p 3p 5p 7p 9p 11p
, , , , ,
4 4 4 4 4 4
1001CJA100220030
Test Pattern
(1001CJA101620000)
CLASSROOM CONTACT PROGRAMME
JEE(Advanced)
REVIEW TEST
08-11-2020
(Academic Session : 2020 - 2021)
JEE(Main+Advanced) : NURTURE COURSE (PHASE : TNPS)
PAPER-2
PART-1 : PHYSICS
SOLUTION
N = Fspcosq
SECTION-I
1.
Ans. (A,B,C)
Sol. l =
1000
= 200m
5
(A) W = 600 × 200 = 120000 J
æ 2R
ö
Fsp = k ç
- 2R ÷
è cos q
ø
xi =
2R
- 2R
4/5
N = 2kR (1 – cosq)
xi =
R
2
120000 = 1 × 400 DT
xf = 0
1200
= 300°C
(B) DT =
4
(C) m ' = 8.5 ´
2.
5
42
=
= 0.042 kg
1000 1000
Ans. (A,B,C)
TT
4kg
R
2T 2T
a1
0
a
fr
a2
fr
a1
F = 40N
a2 + Ra = a1
....(i)
2a1 = Ra – a2
....(ii)
40 – 2T – fr = 8a1
....(iii)
fr – T = 4a2
....(iv)
= m4kR2 éëtan q - m ( sec q + tan q ) ùû37°
0°
dy
= 2R sec2 q
dq
dy = 2R sec2q dq
é
3
æ 5 3 öù
= m4kR 2 ê0 - - 0 + m ç + ÷ ú
4
è 4 4 øû
ë
4R 2
a
frR + TR =
2
fr + T = 2Ra
3.
y
2R
y = 2R tanq
T
Þ fr =
tan q =
Wfr = m2kR ò (1 - cos q ) 2R sec 2 q dq
T
T
0
Sol.
Wfr = ò m2kR (1 - cos q ) dy
....(v)
40
80
40
40
, a1 =
, a2 = , a=
27
27
27
9
Ans. (B,D)
é3
ù
-m4kR2 -m4mgR
= -m4kR2 ê - 0.7 ú =
=
5
5
ë4
û
Wg = mg
3R
2
W = Dk
N
q
Fsp
Sol.
q
2R
Y
3R
4R 1 4mg æ 2 R 2 ö
mg
- mmg
ç0 ÷
2 1
5
2 R è
4 ø
1
424
3
424
3
144
42444
3
Wg
Wfr
W
sp
=
1
m3gR - 0
2
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+91-744-2757575
info@allen.ac.in
www.allen.ac.in
HS-9/14
Target : JEE (Main + Advanced) 2022/08-11-2020/Paper-2
ALLEN
dx Kq0
=
dt rLx
mg3R
4R 2mg R2 m3gR
- µmg
+
=
2
5
R 4
2
mmg
4R mgR
=
5
2
dx
dx 1
µ q0 &
µ
dt
dt x
7.
5
µ=
8
Ans. (A)
mg3R mgR
1
= mv 2 Þ v = 2 gR
+
2
2
2
4.
B
A
L/3
Sol. (P)
v0
Ans. (B,C,D)
m
w
L
ML2
Mv CM =
w
6
12
Sol.
wL
2
v CM =
VA = 0
4U
4Ml2
w Þ w=
(C) MUl =
4l
3
(D) 2Mgh =
Impulse from Hinge 'A' = 0
(1, 2, 3, 4)
1 4Ml 9U
2 3 16l2
2
2
(Q)3
(R) 3,4
(S) 1,3
3U 2
h=
16g
5.
8.
Ans. (C)
Sol. 1st collision
Ans. (A,B,C)
Sol. (A) F = 10 × 5 = 50N
m
(B) 10 × 15 +25 V = 35 × 5
V = 7 – 6 = 1 m/s
6.
VCM
w
For (C & D)
Ans. (C,D)
ice
x
dx
Sol.
water
rAdxL KA
=
q0
dt
x
HS-10/19
v
2m
v=4
m
v1
m(v = v1 + 2v2)
.....(i)
1 v 2 - v1
=
2
v
.....(ii)
2m
v2
Þ v2 = v/2, v1 = 0
2nd collision
m
2m
v=0
v/4
v3
m
v4
2m
æv
ö
m ç = v 3 + 2v 4 ÷
è2
ø
1001CJA100220031
Nurture Course/Phase-TNPS/08-11-2020/Paper-2
ALLEN
1 v 3- v 4
=
Þ v/8 = v3 – v4
2 v/4
2T sin
3v
= 3v 3 Þ v3 = v/4, v4 = v/8
4
T=
Mw2R 2
L
e=
Mw2R 2 Mw2L2
= 2
AY
4 p AY
e=
ML2 p2F
FL
=
4 p2 AY ML 4AY
IA =
9.
5v
v
, IB =
4
4
Ans. (D)
T
Sol. (P)
F
3F
x
e=
T
T =F+
Fx
e
L
Fxdx
de Þ ò de = ò
AL
ALY
0
0
dx
FL
2AY
x
T=
ò de = ò
e=
=
10.
Mw2 (L2 - x 2 )
2L
Mw2 (L2 - x 2 )dx
2L
AY
Mw2 æ 3 L3 ö Mw2L2
çL - ÷ =
2ALY è
3ø
3AY
FL
3AY
dq
2F
x
L
2FL
AY
Ans. (C)
6 m/s
Sol.
1 m/s
2kg
3kg
VCM = 3 m/s
in COM frame
6 m/s
2 m/s
2kg
3kg
-3 £ V2/CM £ 3
-2 £ V3/CM £ 2
0 £ V2 £ 6
1 £ V3 £ 5
PEmax =
(R)
x
2F ö
æ
ç F + L x ÷ dx
è
ø
ò de = ò
AY
e=
(Q)
F
(S)
M F Fx
T=
x
=
L M
L
Y=
dq M
=
Rdqw2R
2
L
1
1
1
´ 2 ´ 36 + ´ 3 + 1 - ´ 5 ´ 9
2
2
2
PEmax = 15
0 £ PE £ 15
1001CJA100220031
HS-11/19
Target : JEE (Main + Advanced) 2022/08-11-2020/Paper-2
ALLEN
1.
SECTION-II
fr = MaCMcosq = MR
Ans. 1.25
Sol. F = mgsinq
mgsinqR =
a=
2.
Mg – N = MaCMsinq
mR 2
a
2
Þ N = Mg -
2g sin q 10 5
=
= = 1.25
R
8 4
Sol.
T
q
y
T
Þ m=
x
q
2Tsinq = lyg
4.
P = m60000
y/x = 2(µsinqcosq – sin2q)
P´
= sin2q + cos2q –1
P´
tan2q = 1
=
4
= mS20 + m × 2100 × 20
3
80000 = S20 + 42000
Þ s = 1900
5.
Ans. 25.50
Sol. Dl = l a DT
Q = rAlSDT
= 3-2 2
= rAS
= 0.17
3.
5
= mL
3
10000 = L
df
= 2cos2q – 2sin2q = 0
dq
= ( 2 - 1)2 = 2 + 1 – 2 2
2 +1
Ans. 52.62 to 52.64
P = m1800 + m 42000
(µlxgcosq – lxgsinq) 2sinq = lyg
2 -1
Ans. 0.37 to 0.40
3.14
p
=
= 0.39
p -2
8
2
Sol. P × 1 = m900 × 20 + m2100 × 20
T + lxgsinq = µlxgcosq
y
y/x
2 -1
=
=
Fraction =
2x + y 2 + y / x 2 + 2 - 1
M2g g
2Mg
= Mg - 2
p pR
p
2 ö
æ
Mg = mMg ç 1 - 2 ÷
p ø
è
Ans. 0.16 to 0.19
x
g
Mg
=
pR
p
6.
Dl
= 25KJ
a
Ans. 1.00
x1
y
2x1
F
2R/p
fr
2R
= 2MR 2a
p
= a=
HS-12/19
2x1
q
Sol. aCM
Mg
Sol.
g
pR
x1
x
FSP = kx1
F = 3FSP = 3kx1
....(i)
x – y = 2x1
....(ii)
x – 4x1 + 2y = 2x1
....(iii)
Þ x=
10F
9k
1001CJA100220031
Nurture Course/Phase-TNPS/08-11-2020/Paper-2
ALLEN
7.
Ans. 1.50
mv1 ö
æ
= M 5gl
Sol. ç mv1 3 ÷ø
è
v1 =
8.
3
100 ´ 10 = 1.5 km/s
2
Ans. 0.36
CM
h
Sol.
(0,0)
q
(hcotq, 0)
a
x CM = a =
a + h cot q
3
Þ cot q =
a2
A
PART-2 : CHEMISTRY
SECTION-I
SOLUTION
3.
Ans. (15.00)
1.
Ans. (A,C)
4.
Ans. (3.00)
2.
Ans. (A,B,C,D)
5.
Ans. (135.00)
3.
Ans. (A,B,C)
let mmoles of each is = x
4.
Ans. (A,B,D)
n-factor of FeO = 1
5.
Ans. (A,C,D)
n-factor of Fe0.80O = 0.4
6.
Ans. (A,C,D)
meq of FeO + meq of Fe0.80O = eq of KMnO4
7.
Ans. (D)
x × 1 + x × 0.4 = 70 × 0.3 × 5
8.
Ans. (C)
x = 75 mmoles
9.
Ans. (B)
mmoles of Fe3+ produced = 75 + 75 × 0.8
10.
Ans. (C)
= 135 mmoles
SECTION-II
6.
Ans. (17.50)
1.
Ans. (4.00)
7.
Ans. (3.00)
2.
Ans. (40.00)
8.
Ans. (25.00)
22
Sol. 1.5 ´10 =
mole ´ N A
N
= A
gm
M
M =40
1001CJA100220031
HS-13/19
Target : JEE (Main + Advanced) 2022/08-11-2020/Paper-2
ALLEN
PART–3 : MATHEMATICS
SOLUTION
SECTION–I
1.
now 24 =
Ans. (A,B,C)
ìa
ï
ïb
4
3
2
Sol. 24x + l1x + l2x + l3x + 1 = 0 í
ïg
ïî d
or 8 3 =
2bc 3
.
b+c 2
bc
Þ 8 3 ( 24 + c ) = 24c
b+c
A
abgd =
1
; a + 2b + 3g + 4d = 4
24
AM ³ GM Þ
30º30º b = 24
24
75°
C
45°
bk
T
ck
c
1
a + 2b + 3g + 4d
³ ( a.2b.3g.4d ) 4
4
B
Þ a = 2b = 3g = 4d = 1
24 + c = 3 c Þ c =
1
1
1
\ a = 1, b = , g = , d =
2
3
4
24
3 -1
Q a =1 is a root Þ 24 + l1 + l2 + l3 + 1 = 0
D=
Þ l1 + l2 + l3 = -25 Þ l2 = 35
0
Sol.
1
1
bc sin A = .24.12
2
2
2
(
)
3 +1
(
)
3 +1 .
(
3
= 72 3 + 3
2
)
a
b
2a
=
= 24 2 Þ a = 12 6
or
sin 60° sin 45°
3
P
B
Ans. (A), (B) are false
S1 is director circle of S2
b
24 2
= 2R Þ R =
= 12 2
sin 45°
2
Þ (C) is correct ]
POAB is square of side 2 unit
Þ AB = 2 2 Þ Area DPAB = 2 sq. unit
4.
Ans. (A,B,D)
B
Ans. (C,D)
8
6
Sol. Obviously b = 24
D=
) = 12
3 +1
again from the DABC
and
3.
(
Ans. Þ (D) is correct
Ans. (A,C)
A
24
hence
Þ l1 = –50; l3= –10
2.
=
(
)
1
24
3
bc sin A =
.c.sin 60° = 12.
c= 6 3 c
2
2
2
...(1)
Sol.
A
a g
b
C
10
HS-14/19
1001CJA100220031
Nurture Course/Phase-TNPS/08-11-2020/Paper-2
ALLEN
Þ equation of 's' is x 2 + y 2 - x - y = 0
12
6 + 8 + 10 = 2pR Þ R =
p
Þ now the length of the tangent from (2,
g p
b 5p
a p
ÐA = = , B = =
, C= =
2 3
2 12
2 4
Þ
=
5.
1 2
1
1
R sin a + R 2 sin b + R 2 sin g
2
2
2
2 x2 + 2 y 2 - 2 x - 2 y - 1 = 0
7.
Ans. (C)
(x + 1) (2x2 – (a + 1)) (x – (3a – 1)) = 0
Ans. (A,C)
–1=
1
-c
-b
Þ c
b
-1
a = 0 Þ 1 – 2abc – a2 – b2 – c2 = 0
a
-1
Þ a + b + c + 2abc = 1 – b – c + b c Þ (a
+ bc)2 = (1 – b2).(1 – c2)
2
2
2
a +1
, 3a – 1
2
for exactly two equal roots
Sol. For non trivial solution
2
2 2
Similarly ; (b + ac)2 = (1 – a2) (1 – b2)
(c + ab)2 = (1 – a2)(1 – b2)
So 1 – a2, 1 – b2, 1 – c2 are of same sign and
atleast one in proper fraction, so all are
positive.
Þ 1 – a2 > 0, 1 – b2 > 0 ; 1 – c2 > 0 Þ a2 + b2 +
c2 < 3 and 1 – 2abc < 3 Þ abc > –1
6.
and the equation of the director circle is
x = –1,
72 æ
1
3 ö 36(3 + 3)
1+ +
÷=
2 ç
2 2 ø
p è
p2
2
s = 2.
Sol. 2x3 + x2(3 – 7a) + x(3a2 – 5a) + (3a2 + 2a – 1)
=0
Area of DABC
=
2) is
6 p
10 5p
8 2p
= ,b=
=
, g= =
R 2
R
6
R
3
& a=
Ans. (A,B)
a +1
Þ a = –3
2
–1 = 3a – 1 Þ a = 0 and
a +1
3
= 3a – 1 Þ a =
2
5
for all negative roots a + 1 < 0 and a <
\ a < –1
–1 lies between the roots
a +1
< –1, 3a – 1 > –1 Þ a < –3 and a > 0
2
\ No value
Or
a +1
> –1, 3a – 1 < – 1 Þ – 3 < a < 0
2
Let the equation of circle 's' is given by
All the roots lie in [–1, 2] if –1 £
( x - 1)
–2 £ a + 1 £ 4
2
+ ( y - 1) + l ( x + y - 2 ) = 0 ... (1)
2
and circle 's' is orthogonal to the circle
1
3
a +1
£2Þ
2
–3 £ a £ 3 and – 1 £ 3a – 1 £ 2 Þ 0 £ a £ 1
a Î [0, 1]
x 2 + y 2 + 2 x + 2 y - 2 = 0 then l = 1
1001CJA100220031
HS-15/19
Target : JEE (Main + Advanced) 2022/08-11-2020/Paper-2
ALLEN
8.
Ans. (B)
1 ö
(sin 3q + sin 2q)
æ 1
2ç
+
sin q
÷ sin q = 2
è sin 2q sin3q ø
sin 3q sin 2q
Sol. S : x2 + y2 – 2x – 4y + 1 = 0,
centre (1, 2), radius = 2
2(sin 4q + sin 2q) 2(2sin 3q cos q)
=
= 2cosecq
sin 3q sin 2q
sin 3q sin 2q
S' : x2 + y2 – 4x – 2y + 1 = 0,
centre (2, 1), radius = 2
(3) cos((2x)2) = 2cos2(2x) – 1
(P) (1, 3) lies inside S and outside S'.
= 2(2cos2x – 1)2 – 1
Þ Sum of tangents is 2.
= 2(4cos4x – 4cos2x + 1) – 1
(Q) Required circle is
2
2
2
2
= 8cos4x – 8cos2x + 1
x + y – 2x – 4y + 1 + l(2x – 2y) = 0
a0 = 1, a2 = 8, a1 = –8
x + y – 2(1 – l)x – 2(2 + l)y + 1 = 0
centre is (1 – l, 2 + l)
(4) Let
it lies on 2x + y – 6 = 0
y = tanq + tan5q + tan9q + tan13q
Þ 2 – 2l + 2 + l – 6 = 0
y = (tanq – cotq) + (tan5q – cot5q)
–l – 2 = 0 Þ l = –2
radius =
y = 2(cot2q + cot10q)
9 + 0 -1 = 8
= –2(cot2q – tan2q)
= –4cot4q = 4
r2
=2
4
10.
(R) m = –1
Sol. (P)
Ans. (B)
(S) Circles are intersecting
9.
y -3 =
cos2 q
sin2 q
= x3 ;
= y3
sin q
cos q
sinqcosq = x3y3
3
3
cos q x
y
= 3 Þ tan q =
3
sin q y
x
x2y2(x2 + y2) = 1
(2) Let
HS-16/19
180°
=q
7
1
3
Equation of line is
(C1C2 )2 - (r1 - r2 )2 = 2
Ans. (B)
(1)
Slope of line in new position is
tan 30° =
Þ length of tangent
=
p
p
= q & 8q =
16
2
——–
Öx2+y2
q
x
1
(x - 5)
3
x - 3y + 3 3 - 5 = 0
y
(Q)
(1,1)
P
(3,2)
Q
Image of P about x-axis is P'(1, –1)
1001CJA100220031
ALLEN
Nurture Course/Phase-TNPS/08-11-2020/Paper-2
Line joining P' and Q is 3x – 2y 5 = 0
æ5 ö
Point on x-axis is ç ,0 ÷
è3 ø
SECTION–II
1.
\a+b+c=8
Ans. 0.19
Sn =
r ( r + 2)
å ( r + 1)( r + 2 )( r + 3)( r + 4)
(3,4)
(R)
=
5
3
(0,0)
4
(r
-å
é 3(3) + 5(3) + 4(0) 3(4) + 5(0) + 4(0) ù
I=ê
,
úû
ë
12
12
+ 3r + 2 ) - ( r + 1) - 1
å ( r + 1)( r + 2 )( r + 3)( r + 4 )
=å
(3,0)
2
1
( r + 3)( r + 4 )
-å
1
( r + 2 )( r + 3)( r + 4 )
1
( r + 1)( r + 2 )( r + 3)( r + 4 )
ö
1 ö 1 æ
1
1
æ 1
= -å ç
+ åç
÷
÷
ç
è r + 4 r + 3 ø 2 è ( r + 3)( r + 4 ) ( r + 2 )( r + 3 ) ÷ø
I = [2, 1]
\ a = 2, b = 3
(S)
x4 – 2x3 – x2 – 2x + 1 = 0
x2 +
x+
+
1
1ù
é
- 2 êx + ú - 1 = 0
ë
xû
x2
ö
1 æ
1
1
çç
÷
å
3 è ( r + 2 )( r + 3 )( r + 4 ) ( r + 1)( r + 2 )( r + 3 ) ÷ø
1ö 1æ
1
1ö
æ 1
= -ç
- ÷ + çç
- ÷÷
è n + 4 4 ø 2 è ( n + 3)( n + 4 ) 12 ø
1
=y
x
1æ
1
1 ö
+ çç
- ÷÷
3 è ( n + 2 )( n + 3)( n + 4 ) 24 ø
y2 – 2 – 2y – 1 = 0
y2 – 2y – 3 = 0
(y – 1)2 = 4
y – 1 = 2, –2
=
1 1
1
- 4 24 72
=
7
36
y = 3, –1
x+
\
1
=3
x
x2 – 3x + 1 = 0
3± 5
x=
2
1001CJA100220031
2.
Ans. 6.00
Here
p ( x ) = x6 - x5 - x3 - x 2 - x + x 4 - x 4
= x2 ( x4 - x3 - x2 ) + ( x 4 - x3 - x2 ) - x
HS-17/19
Target : JEE (Main + Advanced) 2022/08-11-2020/Paper-2
ALLEN
4.
= x2 - x +1
Ans. 4.00
Sol. Locus of (µ – 1, µ + 3) is y – x = 4
\ p ( a ) + p (b) + p ( g ) + p ( d)
= å a -å a + 4 = ( a + b + g + d )
2
|l |
l l
centre is always æç , ö÷ & r =
2
è2 2ø
2
If points are concyclic, then line y – x = 4
-2 ( ab + bg + gd + ad + ag + bd ) - å a + 4
must be chord/tangent of circle
=6
3.
Ans. 2.00
Sol.
From (i)
4
Þ
2cos2x – 1 + 1 –
2cos2x – (2 –
3 = (2 –
3 ) cosx –
2cos x (cos x – 1) +
(cosx – 1) (2cosx +
cos x = 1 or
From (ii)
Sol.
4.2 sin 2 qt 2 + 2 cos2 q - 4 sin q cos qt
4.2sin2 qt2 + 2 cos2 q + 4 cos q sin qt
4 sin2 qt2 2sin q
t +1
cos q
cos2 q
4 sin 2 qt 2 2sin qt
+
+1
cos2 q
cos q
- 3
2
sin 3 x = 2 sin x
3 sin x – 4
Þ |l | ³ 4
( 4 - 4 cos 2q ) t2 + (1 + cos2q ) - 2t sin 2q
( 4 - 4 cos2q ) t 2 + (1 + cos 2q ) + 2t sin 2q
3) = 0
sin3
2
Ans. 6.00
3 (cos x – 1) = 0
cos x =
|l |
5.
3 ) cosx
3 =0
2
£
put 2tanqt = x
x = 2 sin x
sinx = 0 or 3 – 4
sin2x
x2 - x + 1
x2 + x + 1
=2
4 sin2x = 1
sin x = ±
common solution in [0, 5p] is
0, 2p, 4p,
If tan x ³
é1 ù
it's range ê ,3ú
ë3 û
1
2
5p 7 p 17p 19p 29p
,
,
,
,
6 6
6
6
6
6.
Ans. –1.40
Sol. - cos 2q =
1
3
P
5q
0
p
ép
ö
x Î ê + n p, + n p ÷
2
ë6
ø
7p 19p
then common solution: x Î
,
6
6
HS-18/19
52 + 52 - 82 50 - 64 -14
=
=
2.5.5
2.5.5
2.5.5
A
D
cos 2q =
7
25
2q 5
8
B
1001CJA100220031
Nurture Course/Phase-TNPS/08-11-2020/Paper-2
ALLEN
OD = 5 cos 2q =
7.
7
5
\ y=
-7
5
Ans. 0.00
8.
Ans. 0.86
Sol. B(–1,1)
A(1,1)
45º
Sol.
(1,1)
C(–1,–1)
D(1,–1)
x+y=0
Requried area = 4 × area in quadrant 1
(h2 + k2 – 4)2 = (h2 + k2 – 4h) (h2+k2–4k)+16
2
2
Þ (h + k) (h + k – 2h – 2k) = 0
pö
æ
= 4 ´ ç1 - ÷ = 4 - p
4ø
è
circle C1 : x2 + y2 – 2x – 2y = 0 and
C2 : x + y = 0
circumcentre will be mid point of
hypotenuse in right angle triangle
(a,b) = (0,0)
1001CJA100220031
HS-19/19