Post on 23-May-2020
Vidyamandir Classes
VMC/Final Step [Part-A] 1 HPT/Class XIth/JEE-2015
SOLUTIONS
Home Practice Tests/JEE-2015 Final Step [Part-A]
HPT-1 Chemistry
1.(C)
2.(B) 2
2
anP V nb nRT
V
For 2
1 a b 1n , P V RT
2 2 24V
2
aP 2V b RT
4V
3.(B) 1
3RTv
M
2 2T 2T,M M / 2
2 1
3R 2Tv 2v
M / 2
4.(D) Both H2Se and H2S are polar but the former has higher dispersion forces due to its higher molecular weight and hence
the higher boiling point.
5.(B) 1
22(He )
EE
n Z
22
2
2 413.6 13.6 13.6 eV
4n
6.(A) Initially BF3(sp2, trigonal planar) and NH3(sp
3, pyramidal) will convert into BF3(sp
3, tetrahedral) and NH3 (sp
3,
tetrahedral)
7.(A) Basic strength of the oxides of alkali metals increases down the group.
8.(B) 0G G RTln Q ; where Q is reaction quotient; at equilibrium cG 0andQ K
0
cG RTln K
9.(A) Fe(OH)3 + NaOH no reaction. Other hydroxides make soluble mettalate in cases of Zn(OH)2 and Al(OH)3 or
mettalite in case of Sn(OH)2.
Vidyamandir Classes
VMC/Final Step [Part-A] 2 HPT/Class XIth/JEE-2015
10.(D) Meq. of HCl = 100 0.1 = 10
Meq.ofNaOH = 30 0.2 = 6 meq.
Meq.of HCl left = 4
4 = 0.25 V
4
V 4 4 16 ml0.25
11.(C) 3 3 aqaq
AgIO Ag IO S Solubility
2
spK S
or, 2 8 4S 1.0 10 or S=1.0 10 mol / lit
41.0 10 283 g / lit
41.0 10 283 100g /100ml
1000
428.3 10 g /100ml 32.83 10 g /100ml.
12.(D) 2H2SO4 4 2 3HSO H H O SO
13.(D)
H
H H
H
CH3
60º
CH3
14.(B)
CH3–CH–CH2–CH3 2BrCH3–CH–CH–CH3 + CH2–CH–CH2–CH3
CH3 CH3 CH3
Br Br
*
*
15.(A) (CH3)2CHBrCuI)ii(
Li)i( Li[(CH3)2CH]2Cu
BrCHCH)CH( 223(CH3)2CHCH2CH(CH3)2
(A) (B)
16.(B) Dipole moment of 3BF and 4CCl is zero. In 3NF , the dipole moment is diminished due to lone pair and it is
reinforced in the case of 3NH
17.(A) Let the mass of C-14 isotope be ‘x’ gm
Hence, the mass of C-12 isotope would be (12-x) gm
Number of C-14 atoms A
xN
14
and Number of C-12 atoms A
12 xN
12
A
A A
xN
14 100 2x 12 x
N N14 12
12x
100 212x 168 14x
1200x 336 4x
Vidyamandir Classes
VMC/Final Step [Part-A] 3 HPT/Class XIth/JEE-2015
336x gm
1204
Hence number of C-14 isotope 23336 16.023 10
1204 14
221.2 10
18.(C)
b
BpH 14 pK log
BOH
V = volume of acid required for the equivalence point.
(i) b1
10pH 9 14 pK log
V 10
(ii) b2
25pH 8 14 pK log
V 25
(i) – (ii)
25 10 25 V 101 log log log
V 25 V 10 10 V 25
V 30ml
19.(C) Spherical node = n – l – 1 = 3 – 1 – 1 = 1 Non-spherical node = l = 1
20.(D) Density Mass
Volume
For a sealed rigid container mass as well as volume remain a constant hence the density remains same even if the
temperature changes.
21.(D) 2
H 2 21 2
1 1 1R Z
n n
For Balmer Series 2H 2 2
1 1 1R 1
2 3
H 1R 5cm
36
22.(A) (g) (g) (g) (g)A B C D
Mole 1 1 1 1
Conc. 0.5 0.5 0.5 0.5
At 0.5 + x 0.5 + x 0.5–x 0.5 – x
2
C
[C] [D] (0.5 x) 1K
[A] [B] (0.5 x) 4
0.5 x 1
0.5 x 2
1 2x 0.5 x
1
x6
[C] at equilibrium 1 1
0.5 x2 6
3 1 1
M6 3
23.(A) S + O2 SO2; H 298.2kJ
2 2 3
1SO O SO
2 H 98.7kJ
3 2 2 4SO H O H SO H 130.2kJ
2 2 2
1H O H O
2 H 227.3kJ
Adding all the above equations we get :
S + 2O2 + H2 H2SO4 H = 754.4kJ
Vidyamandir Classes
VMC/Final Step [Part-A] 4 HPT/Class XIth/JEE-2015
24.(A) 4% of NaOHimplies 100 gm solution 4 gm NaOH
So 1000 gm solution 40 gm NaOH
1 mole NaOH
But 1000 ml solution weights 1200 gm (d = 1.2 gm/ml)
So 1000 ml solution 1.2 mole NaOH
Molarity = 1.2
25.(B) 2 3 2 2Li CO Li O CO
. Other carbonates are thermally stable.
26.(D) As the size of anion increases covalent character increases.
27.(C)
Cl alc.
KOH
NBS alc.
KOH
NBS
(A)
Br
(B) (C) (D) Br
alc.
KOH
(E)
28.(D) As XeF4 has square planar shape the dipole moments of Xe−F bonds compensate each other and hence it is
non-polar.
29.(D) 2NO2 = N2O4
Mavg = 2 × VD = 2 × 27.6
Let number of moles of 2NO and 2 4N O be m, n respectively
2 × 27.6 m 46 n 92
m n
m
0.8m n
30.(C) Experimental dipole moment = 1.03 D
Expected dipole moment assuming 100% ionic
Character e bond length
10 84.8 10 1.275 10 esu.cm
184.8 1.275 10 esucom
6.12D
% ionic character 1.03
100 16.67%6.12
17 %
Vidyamandir Classes
VMC/Final Step [Part-A] 5 HPT/Class XIth/JEE-2015
HPT-1 Physics
31.(C) 2 21
2mv ks
2
KV S
M
2
2 2
r t
t r
v dva ,a
R dt
a a a
32.(C) 0 w w k
0 d
w kdt
0 00
01
t
kt
ddt
w k
we
k
33.(B) Since every body has an acceleration of g sin on
the inclined plane force on nth block is zero.
34.(A) 1 1 1T m a
2 12T T
2 2 2 2 M g T M a
1 22a a
Solving above equations
2
21 24
M ga
M M
35.(A)
2
0180 37 180 37
Mg N
sin sin
2 N Mg
36.(A)
2
3
2
xW x dx , 3 6 18 yW ,
net x yW W W
2 23 18 x yW | x | W
4 9 xW
5 xW
So 13netW
37.(B) 2 21 2
2 2
l lt , t
g sin g sin cos
38.(D) P KE
39.(C)
21
2
2 0
0
VR =
VCos 45
R
= 2 2 R
40.(D) 2
01
4
330
Ve
V Cos
01 2 30V V Cos
41.(C) Initially there is no sliding between two blocks so
they move together with same acceleration but
when sliding starts B moves with constant
acceleration while of A increases with time.
42.(A) 2 22 12
2mR MR
t t
By using conservation of angular momentum.
43.(B)
andR m r Mm M m M
They rotate about their com,
2
2
GMm mv
r
On solving
2GMv
M m
44.(D) Cyclic process 5 AB BC CAW Q W W W
5 10 1 0 CAW
5CAW
45.(B) 21
3 2
GMm GMP.E, T .E. mv
R R h
Vidyamandir Classes
VMC/Final Step [Part-A] 6 HPT/Class XIth/JEE-2015
46.(A) 1 2
2At H H
a g
According to given data 1 2 0H h,H
2A
t ha g
--- (1) further
22
At' h
a g ----(2)
From (1) and (2) 2t ' t
47.(D) 2 2' . 2 4 [(3 ) ]W T A T r
2 264 8 8 8r T . r T W
48.(B) Tangential velocity of the rope, v r
Tangential acceleration, ( )dv d
a rdt dt
dr
dt
Here 2 n and dr
nddt
22a n d
221 1
a n dT w w
g g
49.(C) 2
5 40 8
5 20
realmdd .
m M
The distance of the dog from the shore is
10 – 3.2 = 6.8 m
50.(D) 2 2 2fw wi
20 2
0 24
……………..(1)
203
24
20
0 24
'
………………(2)
3'
,
3'
, So
3
n rotations
51.(B) 1A OP P gh
2B OP P gh
A BP P p.l.a
1 2
102
10
ah h l.
g = 2 cm
52.(B)
2
2
u sin y
g
; 2u gh
2
22
2
gh.sin y h sin
g
53.(A) Tension produced due to cooling yA
y.A. .
Net force mg yA
Stress mg
yA
Strainmg
Ay
Change in length mg
Ay
54.(B) As water is powered down,
water level increases first
than becomes constant. So
length of air column
decreases first and then
becomes constant.
55.(D) 0
s
v vn n
v v
;
0 5 1 53
0 5 0 5
v . v . vn n n n
v . v . v
Change in frequency 2n
56.(B) 0 05 20 50y . sin x t
1 (20 50 0)x
20 20 5 100x ;
2 [20 (5 25) 50 5]
600 250 350
Change in phase = 250
57.(A) Heat released if temp of water decreases from
30ºC to 0ºC
1 5 1 30 150cal Q
Heat required to increase the temperature of ice
Vidyamandir Classes
VMC/Final Step [Part-A] 7 HPT/Class XIth/JEE-2015
2 5 0.5 20 50cal Q
Heat required to melt the ice completely
3 5 80 400cal Q
Heat available would be less than heat required to
melt.
So final temperature 0ºC and mixture will be
heterogeneous.
58.(B)
22 2 2
2 1
3[1 ] [1 ]
2 4
l lCD l
2 22
2 1
3[1 2 ] [1 2 ]
4 4
l ll
2 22
2 1
3 3 12 ( )
4 4 4
l ll
1 24
59.(A)
2left rightn n
1 21 2
2 2
T T
1 24T T
Taking torque about com.
1 2. ( ) T x T l x , 2 24 . ( ) T x T l x ,
5
lx
60.(D) 2 Re ev g
3
2 e
kTgR
m,
2
3 emgR
Tk
HPT-1 Mathematics
61.(B) Put 2|x 2| t t t 2 0
t 1, 2
If t = 1 or if t 2 cannot be possible as | x 2| 2 not possible.
x 2 1 x 3, 1 Sum of roots be 4
62.(D) For P: 4
0 0 a , 4 5,5
For Q : a 4
0 0 a 4, 5a 5
For R: and20
D 0, 0, a a , 4 5,9
(1) TrueP Q (2) TrueR P
(3) P Q R' 4, 5 All are true.
Vidyamandir Classes
VMC/Final Step [Part-A] 8 HPT/Class XIth/JEE-2015
63.(D) Statement-1: Clearly y1 = | sin | x | | and y2 = x + | x |
has three pt. of intersection in 2 2 ,
Statement-2: True:
64.(B) Graph of 1 0 0 1| f |x| | x , , and Graph of 1 0 0 1| f x | x , ,
Graph of g (x) = | f |x|| | f x | is:
Graph of [ | g (x) |] is:
Range of is {0, 1}
65.(B) Case – I: 3 0 1log x x
2
3 32 0log x log x a
For Real solution : 2
1 4 2 0. .a 1
8a …(i)
Case-2: 3 0log x x < 1
2
3 32 0log x log x a
For real solution: (1)2 – 4.2.a > 0
1
8a …(ii)
3
1 1 8
4
alog x
Here log3 x < 0
1 1 89
04
1 1 8 0a
1 8 1a (+ve sign)
a > 0 …(iii)
Hence from (i), (ii) and (iii)
1
08
a
66.(A) 1 is a root of f (x), then
f (x) = (x + 1) (ax + b)
f (1) + f (2) = 0 2(a + b) + 3 (2a + b) = 0
8a + 5b = 0
8
5
b
a
= other root
2
O
2
-1 1
-1 1 O
-1 1
O 1 –1
2
O
2
Vidyamandir Classes
VMC/Final Step [Part-A] 9 HPT/Class XIth/JEE-2015
67.(D) 2 2 2 2 0x y gx fy c Put y = 0
2 2 0x gx c
1 2 2x x g
1 2x x C
Similarly put 0x
2 2 0 y fy c
1 2 1 22y y f y y c
Area of quadrilateral = 1 2 1 21
2x x y y
= 2 2
1 2 1 2 1 2 1 21
4 42
x x x x y y y y
= 2 214 4 4 4
2g c f c = 2 22 g c f c
68.(B) 2 2 1
8 2 2cos
Now other root is conjugate of this: 2 1
2 2
Sum of roots = 1b 1b
Product of root = 1
8c
11
8b, c ,
69.(C) Statement-1: 3 2 1tan x x 4
x
General solution is 4
n
But for this value, tan 2x = N.D, which does not satisfy the given equation as it reduces to Indeterminate form.
Statement is false!
Statement-2: sin 2x + cos 4 x = 2
sin 2x = 1, cos 4x = 1
cos 4x = 1 21 2 2 1sin x
1 2 1 not possible No solutions (False)
70.(C) We know that logax is defined when 0 0 1x , a , a
x
|x|log
x is defined if: 0 0 and 1
|x|, x x
x
Case-I: 0|x|
x x > 0 …(i)
Case-II: 0x 1x , …(ii)
Case-III: 1x 1 2x , …(iii)
From (i), (ii) and (iii), we get 2x ,
(0, y1)
(x, 0)
(0, y1)
(x2, 0)
Vidyamandir Classes
VMC/Final Step [Part-A] 10 HPT/Class XIth/JEE-2015
For 2x , we find that 1 0x x
|x|log log
x
1 0 for all 22
f x cos x ,
Domain 2 and range2
f x ,
71. (B) As 04
, , tan cot
Since 1 1tan and cot
1 and 1cot tan
tan cot
4 1t t which only holds in (B)
72.(B) Area of quadrilateral
= Area of (ABMD) + Area of BMC rectangle
= 1
2 22
r . x . r x
= 3 xr
But 3xr = 18 xr = 6 …(i)
OBE OBK
EOB KOB
OCN OCK
90NOC KOC
and In 2
90x r
ONC tanr
2x r
cotr
…(ii)
In x r
OEB tanr
…(iii)
From (ii) and (iii)
2
x r r
r x r
2 3 0x x r
30
2
rx x …(iv)
From (iii) and (iv) we get : r = 2
73.(C) Since is isosceles, hence centroid is desired point.
Coordinates are 4
33
,
74.(A) As 1 2 3 4 5 1 4 6 7 10 0a a a a a
x = 1 is root of a1x4 + a2x
3 + a3x
2 + a4x + a5 = 0
Maximum roots = 4
and complex root are in pair form
Hence the given equation has at least two real roots.
A
D C
B
x
x – r
2x – r
2x
O
r
r
r
M
E
90
N
K
r
P (3, 4)
O
(0, 0) (6, 0)
Q
R
Vidyamandir Classes
VMC/Final Step [Part-A] 11 HPT/Class XIth/JEE-2015
75.(B) Since AP PQ QB
The co-ordinate of P are (a, 0) and of Q are 2 0a, , equation of the circles an AP, PQ and QB as diameter are
2 0 x x a y , 22 0 x a x a y and 22 3 0 x a x a y
If h,k be any pt. on the locus, then
2 2 2 23 9 8 h k ah a b 2 2 2 23 9 8 0x y ax a b
76.(B) We have 2 21 x x x x
If f x is divisible by 2 21 0 0 x x , f , f
3 3 0 1 1 0 P Q P Q ………………… (1)
6 2 6 20 1 1 0 P Q P Q …………………. (2)
From (1) and (2) we get,
1 0 1 0 P , Q
3 P x and 3Q x are divisible by 3 1x , Hence by 1x
Since 3 3 f x P x xQ x , We get f x is divisible by 1x
77.(C)
1
131 2
13
S Choice 2 22 1 2 20 25 0 5 0 5 4
log S log /A . . .
Choice 5 5
25
3 1 2
3
10 008 0 2 8
5
log S log
logB . .
/
78.(D) 1
2 2 12 6 6
n
sin A B sin A B n
……………. (i)
Also A B C and 2 B A C 33
B B
From (i) 1n
5 5
26 4 12
A B A ,C
79.(A) 1 1 1
1 2 2 3 11 1 1
n n
d d dtan tan ............. tan
a a a a a a
1 1 13 2 12 1
1 2 2 3 11 1 1
n n
n n
a a a aa atan tan .......... tan
a a a a a a
1 1 1 1 1 1
2 1 3 2 1
n ntan a tan a tan a tan a ............ tan a tan a
1 1
1 ntan a tan a
1C
2C 3
C
P Q
A
B
3a a 2a
x-axis
0 0,
Vidyamandir Classes
VMC/Final Step [Part-A] 12 HPT/Class XIth/JEE-2015
80.(B)
R1
R2
R3
Case No. of Balls
R1 R2 R3
No. of different ways
I 4 1 1 26 2 1
4 1 1 4 2 C C C
II 3 2 1 26 3 1 4
3 2 1 3 3 2 C C C C
III 3 1 2 26 3 1 4
3 2 1 3 3 2 C C C C
IV 2 2 1 26 4 2 4
2 2 2 2 2 2 C C C C
Total = 26 6
81. (C) 2 3
1 1 1
3 3 39 3
.....
xtill
; 2 3
1 1 1
3 3 34
..........
ytill
= (4)1/4
4
1 1
12
ri /
r
r r
eZ i =
4
4
2
12
i /
i /
e
e= i
1 4
3 4 3 2 /
x y z i i
Argument = 1 2
3
tan
Choice: (C)
82. (A) 3 2 1 0 1 0 1 0 f x x bx cx , f f b cand
1 0 1 0 ' ' b / w i.e.lies to
1 1 1 11
tan tan tan cot = 2 2 / /
83. (C) 1 1 1 312
2
tan tan A tan cot A tan cot A
= 3
1 14
21
2 1
tan A cot A cot Atan tan cot A
cot A= 1 1
2 21 1
tan A cot Atan tan
tan A cot A
= 1 12 21 1
tan A tan Atan tan
tan A tan A= 14 4 1
4
. tan
Vidyamandir Classes
VMC/Final Step [Part-A] 13 HPT/Class XIth/JEE-2015
84. (A) 5 4 44
4 5
2 5 2 80 onr shoe outof pairseach pair
C
We need to select only one shoe out of any one pair. So we choose out of spairs & then choose one shoe out of each.
85.(A) 4 3 2 2 22
2 12 3 2 1 2 3
n n n n n n n
n n=
22 1 1
1 2n n nn n
=
22 1
1
n nn
2008 2008
1 1
1 1 11 1
1 1
n n
n n n n=
1 1 20082008 2008
1 2009 2009
86.(B) Let roots be: a, a d , a d then sum of roots = 3 1
B
aA
As a = 1/3 is a root so put in original equation 9 27
1 27 1 9 3 0 12 2
/ / /
For any 1 , x yline
Definitely passes then 9 27
2 2
, .
87.(B) 1 2 2 3 1 3
1 2 3 1 2 3 2
1
Required root at are 1 2 3
1 1 1
, ,
1 1 y x
x yor (Replace)
=
3 21 1
1 0
ay y
= 3 1 0 by ay = 3 1 0 bx ax
88.(D) This equation can be written as sin x . cos x [sin2 x + cos
2 x + sin x . cos x] = 1
1 1 sin x . cos x sin x . cos x 2 2 2 4 sin x sin x
= 2 4 16
2 1 52
sin x which is not possible.
89.(C)
90.(C) nP cos cos n
n
nP cos Re cos i sin 21
n
nP x Re x i x
2 1n
nZ cos i sin cos n i sin n x x
2 1
nn
Z cos i sin cos n i sin x x
2 22 1 1
n n
Z Z cos n x x x x = 2Pn(x)
Vidyamandir Classes
VMC/Final Step [Part-A] 14 HPT/Class XIth/JEE-2015
HPT-2 Chemistry
1.(B) 3CaCO will dissociate to give CaO and 2CO until 2P COK P 1.5atm
Use PV = nRTfor extra CO2 produced.Find ‘n’ which will equal to net of moles of CaO.
PV 1 1 1
nRT 0.0821 292 24
2.(A) 2 2 n
n n 2
N N XeF
XeF XeF N
r P M
r P M
2 2 n
n 2
n
N F XeF
XeN N
XeF
Molar volume
r t t 57
Molar volumer t 38
t
nXeFM57 0.8
38 1.6 28
n
2 2
XeF
57 1.6M 28
38 0.8
=252
Xe nF 252
131 n 19 252 = n = 6
Molecular formula = XeF6
3.(A) PV = nRT W
RTM
WP RT
V M
RT
M
A
3P RT
M
;
BP RT
2M
A
B
3RT
P 3 2MM
P MRT
2M
= 6
4.(C)
5.(C)
2H
HH
nr 0.529 A
Z
3
3
2
Be
BeBe
nr 0529
Z
3
3 3
22H H Be
2 2HBe Be
Zr n 2 4
r Z n 1 3
3
0
Be
a 16
r 9
30
Be
9ar
16
Vidyamandir Classes
VMC/Final Step [Part-A] 15 HPT/Class XIth/JEE-2015
6.(A) C(s) + CO2 (g) 2CO (g) ; …(i)
1
14pK 10 atm
2CO(g) + 2Cl2 (g) 2COCl2 (g) ; …(ii)
2
3 2 2pK (6 10 ) atm
Add (i) and (ii)
C(s) + CO2 (g) + 2Cl2(g) 2COCl2 (g) ;
Kp = 14 610 36 10
= 36 × 108
For given reaction ng = – 1
c pK K (RT)
= 36 × 108 × 0.0821 × 1120
11 1
cK 3.31 10 M
7.(A) Solubility of CuS and HgS (AB type) is spK ; solubility of 2 2Ag S (A B type) sp
3K
4
8.(D)
CH2CH2OH
H+
H2O CH2CH2
H
+
CHCH3 +
+
H
CH3 CH3
+ H2O
H+
OH
CH3
9.(B)
OH CH3CH=CH H+
OH CH3CH2CH
Br
(more stable)
OH CH3CH2CHBr
(major)
+
10.(B) The blue colouration is due to solvated electrons 11.(A)
12.(D) 3NH N g 3H g 1
1H x Kcal.mol
2 4N H 2N g 4H g 1
2H yKcal.mol
1 N HH 3 x …(1)
2 N H N NH 4 y …(2)
From (1) &(2) N N
xy 4.
3
1
N N
4x 3y 4xy Kcal.mol
3 3
13.(D) Factual
Vidyamandir Classes
VMC/Final Step [Part-A] 16 HPT/Class XIth/JEE-2015
14.(C) In the hydrides of group 15 and group 16 (except NH3 and H2O) the energy difference between 3s and 3p orbitals is
quite high. Hybridization increases the energy of 3s orbital so much that lone pair rather prefers to occupy
unhybridizeds orbital. For example in PH3, 600 kJ mol-1
of energy is required to hybridize the central atom. So to
avoid such energy demanding hybridization P forms bonds with unhybridizedp orbitals leaving the lone pair in the
spherical s orbital which leads to a bond angle close to 90°.
15.(D) [H+]1 = 10
–2
[H+]2 = 10
–6
Factor = 41
2
H10
H
16.(B) Peroxide effect is only shown by HBr 17.(A)
C=C H
Br
H
Br ; C=C
H
Br H
Br ; C=C
Br H
Br H
18.(B) Assuming mass of CH4 and H2 are 1 gram of each
4 2CH H
1 1Mole Mole
16 2
2Hp Mole fraction Ptotal
2H total
8p P
9
19.(B) Vacant d orbitals available only in sulphur.
20.(D) CH2=CH2 Cl2, H2O H2SO4
alc. KOH
CH2ClCH2OH ClCH2CH2OCH2CH2Cl
(B)
CH2=CHOCH=CH2
(C)
Heat
21.(D) Oxygen exist +ve oxidation state only with 'F' In F2O, 'F' in -1 oxidation state and 'O' has +2 oxidation state
22.(D) 4 2 4 2 2 4 2 4 4 2 22KMnO 3H SO 5H C O K SO 2MnsO 10CO 8H O
COOH
1 1 1 4 2 2 2 COOH
M V n KMnO M V n |
4 2
110 V 5 10 0.5 2 , 1V 20L
23.(B)
X 2Y, Z P Q
Initial mol 1 0 1 0 0
At equilibrium 1- 2 1-
1
2
12Y
PX
1
2P
P 1K
1PP
1
; 2
2 2P q
Pz
2
P PP P 1 1
K1P
P1
Vidyamandir Classes
VMC/Final Step [Part-A] 17 HPT/Class XIth/JEE-2015
1
21
P 2
4 PK
1
2
22
P 2
PK
1
Given is 1
2
P
P
K 1
K 9
Substituting values from equation (i) & (ii) into (iii) we get :
21
2 1 1
22 22
2
4 P
1 4P 1 P 11
9 P 9 P 36P
1
24.(D) For acidic buffer,
a
ApH pK log
HA
When the acid is 50% ionized, [A ] [HA] or apH pK log1 or apH pK
Given apK 4.5
pH 4.5
pOH 14 4.5 9.5
25.(C) maximum number of orbital’s in a sub-orbit = maximum number of electrons with parallel spin 2l 1
26.(D) 4HClO is a strong acid.
27.(B) Borazole is polar due to BN bond; benzene is nonpolar due to CC bonds.
28.(D)
Anhydrous
+ CH2Cl2 AlCl3
CH2Cl
Anhydrous AlCl3
CH2
29.(B) 3
vapVap
H 30 10S T 400K
T 75
30.(C) Three replaceable H so molarity 3 = Normality
HPT-2 Physics
31.(D) 0n
cos 0mg T
cosT mg
2sinT mrw
1
2Max
gP
h
Vidyamandir Classes
VMC/Final Step [Part-A] 18 HPT/Class XIth/JEE-2015
32.(D)
0 08 30 6 300 7
10
rod
sin sin. rad / s
33.(D) ( ) ( 1)T n mg m g a a n g
Max. possible retardation g
2
12( 1) .v n g h
2
22v gh
& 1 2 h h h
On solving 2 ( 1)gh n
vn
34.(A) N Fsin 60º 3g
Fcos60º [Fsin 60º 3g]
35.(B) Let masses on two sides be ' 'm and m m , where '2
mm
' ' 1m g T m a ' ' 2T m m g m m a
2 4 1
m'F T m' g mg
m
36.(B) 2
cos 1mv
T mgl
;
212
2 2
mglmgl mv
On solving T = 1.5 mg
37.(B) 1 cm cmy y
1 1
10 7 30 301
40 40
y y y
1 .y cm
38.(C) Applying conservation of momentum
0.50 5 0
2.5
v
v
Vidyamandir Classes
VMC/Final Step [Part-A] 19 HPT/Class XIth/JEE-2015
39.(A) 20
0
w= 10+0.50x 21
40.(C) cos cos2
v u
cos
cos / 2
uv
21 1cos
2 2gravityW K mv m u
2 2 21cos tan
2 2mu
.
41.(B) 2
5
R
T
K
K
5
7TK Mg R r
21 5
2 7mv mg R r
v
wR r
10
7( )
g
R r
42.(D) F
MR
F f Ma MR F
0f
43.(D) 21
2 2 2e
GMm GMmmv
l / l /
2
2e
GMv
l
44.(D) The buoyant force pushed body in upwards direction for which cube will rise
When 0.2 kg is removed, cube rises by 2 cm
So, ( ) mg Buoyancy force
2 2 30.2 ( 2 10 )10 g a g
2 210a
110a m or 10 cm
45.(A) According to perpendicular axis theorem
1 2 1 2and2
II I I I I
Vidyamandir Classes
VMC/Final Step [Part-A] 20 HPT/Class XIth/JEE-2015
Similarly 3 4 3 4and2
II I I I I
So, 1 3I I
46.(D) Using conservation of energy 21 1
2 2FX mv
2 11 120
2 2
YAXX mv v ms
L
47.(C)
0
1/201/2
0
2
w t
w
dw dwKw K dt w Kt
dt w
The body comes to stop in time 0
0
2 wt
K , further
2
02
Ktw w
0
20 00 0
0 2
0
2
12 2
t
t
wdtK w wKwK
w wKK
dt
0 00 0
3 3
w ww w
48.(A) 10 kx B g
10 kx kx g
10 5 5 10 1
2 500 10
g gx m
k k
10x cm
49.(D) Force by curved surface
= Fbottom – mg =
22 .
[ ] . .3
R RgR R g
33
3
g Rg R
32
3g R
50.(B) Force required to separate the plates 2TA
d
51.(B) 2 L a bt
Suppose 2ˆ ˆ L ai bt j
ˆ2t btj
Angle between L and t is 45º
So, 2a bt
a
tb
2a
t bb
Vidyamandir Classes
VMC/Final Step [Part-A] 21 HPT/Class XIth/JEE-2015
52.(D) So F > 15, then acceleration of two block together
15
15
Fa
15
15 515
Ff
Max. value of f = 25
15
25 153
F
30 15 F
45F
53.(D) Less/Gain in time 1
( )2
time 511.2 10 10 86400 7
2
36.28
54.(C) 3
rms
RTV
m
21 3
2 2rmsK mV KT
55.(A)
0
0
. . t
Q m s dt0
2
0
( ) t
m at bt c dt
3 20 0
03 2
mat bt
Q Ct
56.(A) 2 2AD (68) (53) 42.6
Speed of Car = 42.6
2130.2
57.(B) 0
1
vn
4( e)
; fundamental mode
0
2
3vn
4( e)
; first overtone
By above equations
1 23 3e e
2 13e 0.025m
2
58. 2 4x ay
Slope = tan1
.24 2
dy xx
dx a a
Restoring force sin tan2
mgxmg mg
a
2
2
g
a
Vidyamandir Classes
VMC/Final Step [Part-A] 22 HPT/Class XIth/JEE-2015
59.(A) 1 1 1
1 5[ ] . . . 100 125
2 2
RQ n cv T R
2
1 3[ ] . 100 37.5
4 2
RQ R
162.5 162.5 2 Q R
60.(B)
Area under P V curve with volume
Axis = work done
HPT-2 Mathematics
61.(B) are roots of 3 2 0 x px qx r p, q, r
2 2 2 2 2 2
2 2 2 2 2 2
1 1 1
2 2 2
2 2 2
2
2
2
2
q rp
r
62.(A) 2
2
2 1 9 40
8 32
ax a x af x
x x
2
2
2 1 9 40
4 16
ax a x a
x
2 2 1 9 4 0 ax a x a x R 2 22 1 9 4 0 a a a a and 0a
28 2 1 0 a a and 0a 28 4 2 1 0 a a a and 0a
4 2 1 1 2 1 0 a a a and 0a 2 1 4 1 0 a a and 0a
1 2 a / or 1 4 / and 0a
So, 1
2
a ,
63.(B) S1 : 2 0 ax bx c 0a
If 0 a b c then one root is 1 product of roots c
a
Vidyamandir Classes
VMC/Final Step [Part-A] 23 HPT/Class XIth/JEE-2015
So another root c
a
S2 : 2 f x ax bx c 0a
It has finite minimum value of graph is upward parabola roots are opposite sign
So 0 0f
S3 : is repeated root of 2 0 ax bx c
So 22 ax bx c a x
Sn : 2 0 ax bx c 0a
Irrational roots occur in conjugate pair only if a, b, c are rational.
64.(D)
2 4 6 200
1 3 5 199
100
a a a ..... a
a a a ...... a
...............................................
d
100
d
65.(C) 1 2 3 x x 3 4 12 x x
1 2 x x A 3 4 12 x x
x1 x2 x3 x4 G.P.
a ar ar2 ar
3 1r
1 3 a r
2 1 12 ar r
……………………………………..
2 2 r , Hence r =2
a = 1
2 2 A a.ar a r
2 5 52 32 B a r
66.(C) 2 2 2
2 1
1 2
r
rT
.....r
6 2 1 6
1 2 1 1
r
r r r r r
1 16
1
r r
1 1
61
n nS T
r r
1 6
6 11 1
n
nS
n n
6nn
S lim S
Vidyamandir Classes
VMC/Final Step [Part-A] 24 HPT/Class XIth/JEE-2015
67.(B) 3 3
32 182 2
cos x sin x sin x
19 56 9 7cos x cos x sin x
18 9sin x cos x sin x cos x cos x sin x 18 9 9cos x sin x sin x bcos x
27a b
68.(C) 2001 2001 2001
2001 20012 3 4 6
cos cot sec tan cos ec
667
667 1 0 1 1 1 22 4 2
cos cot sec tan cos ec
69.(A) 22 2
tan tan
2
2
2
2
123 5
13 5 2
5 31
25 3
12
tan
tancos
costan
tan
2 2
2 2
3 1 5 12 2
5 1 3 12 2
tan tan
tan tan
2 2 2
2 2 2
8 2 8 8 12 2 2
8 2 8 8 12 2 2
tan tan tan
cos
tan tan tan
70.(B) 2
2 12 2 2 4
2cos x cos x cos x
2 211 2 2 2 2 2 1
2cos x cos x cos x 2 2 1 3
1 2 2 2 2 2 22 2
cos x cos x cos x cos x
71.(B) 2 0x bx c
2
8cos
is the roots of equation
21
1 14 18 2 2 2
cos
cos
21 1 1 1
1 1 02 2 2 2
b c
1 1 1 1
1 2 1 04 2 2 2
b c
1 1
1 2 04 2 2 2 2
b bc
Since b, c Q so comparing rational and irrational parts with zero we get : 1
02 2 2 2
b and
30
2 8
bc
1b ; 1
8c
Vidyamandir Classes
VMC/Final Step [Part-A] 25 HPT/Class XIth/JEE-2015
72.(A) Domain of 1x x is 1 1x 1x x ….(1)
Case I For 1 0x , equation (i) holds.
Case II 0 1x ,
1x x or, 2 1x x
or 2 1 0x x or 1 5
02
x ,
1 5
12
x ,
73.(D) 3
2x iy
cos i sin
. . . .(i)
3
2x iy
cos i sin
. . . .(ii)
Multiplying (i) and (ii)
2 2 9
5 4x y
cos
Adding (i) and (ii)
3 2
5 4
cosx
cos
2 2 24 12 94
5 4
cosx x y
cos
2 2 5 5 4
45 4
cosx x y
cos
2 24 3x x y
74.(A) 5 5 3 1
4 46 6 2 2
z cos i sin z i
2 3 2z i
75.(A) 21 1 2 3x . x a a
2
1 1 1 4 1 3x x a a a
For solution
1 3 0a a At 1 3, ,
76.(A) 1 2
1 2
31
3
z z
z z
1 2 1 2
1 2 1 2
3 31
3 3
z z z z.
z z z z
2
1 1 1 2 1 2 23 9z z z z z z z 2 2
1 2 1 2 1 29 3 z z z z z z
2 2 2 2
1 2 1 29 9z z z z 2 21 29 1 0z z
But 2 1z 1 3z
Vidyamandir Classes
VMC/Final Step [Part-A] 26 HPT/Class XIth/JEE-2015
77.(A) 1 2 4x x 1 2 2y y
2 3 10x x 2 3 6y y
1 3 6x x 3 1 14y y
1 2 3 10x x x 1 2 3 11y y y
2 4x 2 3y
4 3B ,
Equation of medium 10 37 0x y
78.(B) 3 4 3 3 4 3 0a x y x y
1 2 0L L
3 4 3 0x y
3 4 3 0x y
Point is 1 0,
79.(A) 2 2 14 10 151 0x y x y . . . .(i) ;
Centre 7 5 15, , r
Let 2 7P ,
1 4 49 28 70 151 56 0S
maxD r PS 2 2
7 2 5 7 15 13 15 28 and minD PT ST PS r PS
15 13 2
80.(B) Equation of line (L) is
3
7 5 2 52
2 10 3 15y x
3 2 5x y
Point E 2 3 15 0x y 3 2 5 0x y
E (3, 2)
Point C 5
03
,
Area of circle CEB 1 1 10 46 230 5 3 3 0 0
120 0 2 4 02 2 3 6 3
/
81.(D) Let 1 1 2 2 3 3A r , r , B r , r , C r , r
1 2 32 2OA . OB . OC r r r
On putting point (r, r) in the curve
3 3 23 30 72 55 0r r r r
3 24 30 72 55 0r r r
It has roots 1 2 3r , r , r
Vidyamandir Classes
VMC/Final Step [Part-A] 27 HPT/Class XIth/JEE-2015
1 2 3
55
4r r r
55
2OA.OB.OC
4 24
55
OA.OB.OC
82.(C) Equation of line passing through A is 1 1 0px qy qx py
1 1 0px qy qx py
Passed (p, q) 2 2 1 2 1 0 p q pq
2 2 1 2 1 0p q pq
2 2 1
1 2
p q
pq
83.(D) Let the centre the (h, h)
3 6 10 h h 9 10h 10 9h /
line 2 3 x y is tangent to the circle 2 3
4 55
h h
3 3 20 h 20
13
h
20
13
h 23
3h
So centre is 23 23
3 3
,
Equation of circle is
2 223 23
803 3
x y
84.(A) Since given triangle is a right angled isosceles triangle
Angle between them is 1 2
1 2
45 451
m mtan
m m
. . . . .(i)
Where 1 2
1andm m
k
From (i)
1
2
12
11
1 1 11 1
m
mk
km
m
85.(A) 3 4 2 9 0y x , y bx are parallel 2
63 1
bb
4 3 10y x, ay x are perpendicular 1
3 1 3 3 6 18a . aba
86.(A) Let 1 1P x , y be a point in the locus.
Let OAB . Then 1 1y xsin , cos
b a
2 2
2 2 1 12 2
1 1x y
cos sina b
Locus is 2 2
2 21
x y
a b A
B
O
a
b
x1
x1
y1
P(x1, y1)
Vidyamandir Classes
VMC/Final Step [Part-A] 28 HPT/Class XIth/JEE-2015
87.(A) 3 3 3 3 3 30 3 0 3
a b c
b c a abc a b c a b c abc
c a b
G.E = 3 3 3
2 2 2 32 2 2 8a b c / abc
a / bc b / ca c / ab
88.(A)
10 10
10 10
f xe
x xe f x log
x x
222
2 2
2
20010
200 100 20 10100 2200 10100 100 2010
100
e
x
x x x xxf log log log f xx xx x x
x
2 2
200 1 200 12 0 5
2 2100 100
x xf f x f x f k .
x x
89.(B) If 0x . Then 2 3f x x x x . If x < 0. Then 2f x x x x
If 0x . Then 2 3 2 3 2 2f x f x f x x x x x x
If x < 0. Then 2 2 3 2 2f x f x f x x x x x x
90.(A) 1 2fofof
0 33fofof
1 2fofof
HPT-3 Chemistry
1.(C) In xyd both lobes is in xy plane so yz and zx are the nodal plane
2.(B) No of hybrid orbitals equal to number of atoms attached + lone pair = 5 + 1 = 6
So, hybridization is 3 2sp d
3.(D) meg of 2 4H SO 200 0.1 2 40
Meg of KOH 200 0.2 1 40 meq
Heat involved 40
57.3 2.292kJ1000
2.3k J
4.(C) 2 1Cl g 2Cl g ; H 242.3 kJ / mol
2 2I g 2I g ; H 151 kJ / mol ; 3ICl g I g Cl g ; H 211.3 kJ / mol
2 2 4I s I g ; H 62.8 kJ / mol
Required equation : 2 2
1 1I g Cl g ICl g ; H ?
2 2
62.8 151 242.3
H 211.32
=16.75 kJ/mol
Vidyamandir Classes
VMC/Final Step [Part-A] 29 HPT/Class XIth/JEE-2015
5.(B) 2 2 1s
C O CO x 1
2 2 22 g
1H O H O x 2
2
4 2 2 2 3CH 2O CO 2H O x 3
By (1) + 2 × (2) – (3) we get
(g) 2 4(g)C 2H (g) CH
1 2 3H x 2x x
6.(A) Change in 5 2
oxidation state (Mn) (C)
Coefficients 2 5
7.(C)
Fe
Cl2
Cl Cl Cl
Cl Cl
8.(D) 2H A H HA
51
2
H HAK 1 10
H A
2HA H A
2
102
2
H AK 5 10
H A
Overall dissociation constant = 1 2K=K K
5 10
15
1 10 5 10
5 10
9.(B) H+ + HOCl H2O + Cl
(Electrophile)
10.(D) The solubility of alkaline earth metal chlorides decreases down the group.
11.(B) Nitrogen is more electronegative than ‘H’.
12.(B)
CCl3CH=CH2
OH2
Cl2 CCl3CHCH2
Cl+
.. CCl3CHCH2
HO+H
Cl H+
CCl3CHCH2
OH
Cl
13.(C) The enol form is aromatic in nature.
Vidyamandir Classes
VMC/Final Step [Part-A] 30 HPT/Class XIth/JEE-2015
14.(D) 3.6 M solution means 3.6 moles of H2SO4 is present in 1000 ml of solution.
Mass of 3.6 moles of 2 4H SO 3.6 98g 352.8g
ass of 2 4H SO in 1000 ml of solution =352.8 g
Given 29 g of 2 4H SO in present in 100 g of solution
2 429% H SO by mass
352.8 g of 2 4H SO is present in
100352.8 1216g
29 of solution
Now, density Mass 1216
1.216 g/mlVolume 1000
=1.22 g/ml
15.(A) Milli eq. of Na2S2O3 = 24.35 1/10 = 2.435
This would be the milli eq. of I2 and therefore that of Cl2 (which liberates I2 from KI).
Millieq. of Cl2 in 500 ml. = 2.435 500/25 = 48.7
Meq.of Cl2 = meq of bleaching powder = Meq of available Cl2 in the bleaching powder.
% of chlorine = 48.7 35.5
100 30.33%1000 5.7
16.(B) 2)Y(32
)X(
NMgNMg3 ;
Mg3N2 + 6H2O 3Mg(OH)2 + 2NH3 (Y)
CuSO4 Blue colour
17.(A) 18.(B)
Note: chiral centre are marked.
.
.
19.(A) Because solution does not contain Cl+ ion and hence the cyclic chloronium ion is not formed.
20.(C) 2XeF and 3I
are linear in shape
21.(C) PV
ZnRT
9 V 0.082 273 0.9
0.9 V 2.24 L / mole1 0.082 273 9
For 1 millimole, the volume is 2.24 mL
22.(C) Factual
23.(C) Initial pair = 750 – 100 = 650mm
1
1 1 2 2 1 2
VP V P V 650 V P
3
Final Pair = 1950
Final Ptotal =1950 +100 = 2050 mm = 2050 torr
24.(A) Mortar is Ca(OH)2 mixed with silica and water.
25.(B)
148
5
Kh Kw 10h 10
C Kb.C 10 0.1
4h 10
2%h 10 0.01%
Vidyamandir Classes
VMC/Final Step [Part-A] 31 HPT/Class XIth/JEE-2015
26.(D) Solubility of salt of AB type is spK
27.(A) Factual
28.(C) h
x.m. v4
2 hx .m x v
4
2 h
x4 m
, h
x4 m
1
2
hx
m
29.(A) For 2A B and 2AB type 3
spK 4s and for 3AB type 3
spK 27s
30.(A) 2 10–4
= 0.5 2
= 2 10–2
[H+] = 0.5 2 10
–2
= 10
–2
Hence, pH = 2
HPT-3 Physics
31.(A) If the breadth of the lake is l and velocity of boat is vb. Time in going and coming back on a quite day
2
Q
b b b
l l lt
v v v .....(i)
Now if va is the velocity of air- current then time taken in going across the lake,
1
b a
lt
v v
[As current helps the motion]
and time taken in coming back 2
b a
lt
v v
[As current opposes the motion]
So 1 2Rt t t 2
2
[1 ( / ) ]b a b
l
v v v
.....(ii)
From equation (i) and (ii)
2
2 2
11 [ 1 1]
[1 ( / ) ]
aR
Q a b b
vtas
t v v v
i.e. R Qt t
i.e. time taken to complete the journey on quite day is lesser than that on rough day.
32.(D) dv dv dx
adt dx dt
2dvv x
dx (Given)
0
02
0
S
v
vdv x dx
0
02 3
02 3
S
v
v x
2 30
2 3
v S
12 303
2
vS
Vidyamandir Classes
VMC/Final Step [Part-A] 32 HPT/Class XIth/JEE-2015
33.(A) In this problem it is assumed that particle although moving in a vertical loop but its speed remain constant.
Tension at lowest point
2
max
mvT mg
r
;
Tension at highest point
2
min
mvT mg
r
2
max
2min
5
3
mvmg
T r
T mvmg
r
By solving we get, 4v gr 4 9.8 2.5 98 /m s
34.(B) As the mass of 10 kg has acceleration 12 m/s2 therefore it apply 120N force on mass 20kg in a backward direction.
Net forward force on 20 kg mass = 200 – 120 = 80N
Acceleration280
4 /20
m s .
35.(C) AB BGF f f
( )AB a BG A Bm g m m g
0.2 100 10 = 0.3(300) 10
200 900 1100 N
36.(C) The total time from A to C
Ac AB BCt t t
( / 4) BCT t
where T = time period of oscillation of spring mass system
BCt can be obtained from, sin(2 / ) BCBC AB T t
Putting 1
2
BC
AB we obtain
12BC
Tt
2
4 12 3AC
T T mt
k
.
37.(D)
2 2 2 2
2 22 2 2
u m u PS
g gm m g
38.(D)
If h is the common height when they are connected, by conservation of mass
1 1 2 2 1 2( )A h A h h A A
1 2( )/2h h h [as 1 2A A A given]
Vidyamandir Classes
VMC/Final Step [Part-A] 33 HPT/Class XIth/JEE-2015
As (h1/2) and (h2/2) are heights of initial centre of gravity of liquid in two vessels., the initial potential energy of the
system
2 21 2 1 2
1 2
( )( ) ( )
2 2 2i
h h h hU h A g h A gA
...(i)
When vessels are connected the height of centre of gravity of liquid in each vessel will be h/2,
i.e. ( 1 2( )
4
h h [as 1 2( )/2]h h h
Final potential energy of the system
1 2 1 2( )
2 4F
h h h hU A g
21 2( )
4
h hA g
…(ii)
Work done by gravity 2 2 21 2 1 2
1[2( ) ( ) ]
4i fW U U gA h h h h
2
1 2
1( )
4 gA h h
39.(A) Velocity of 50 kg. mass after 5 sec of projection v u gt 100 9.8 5 51 /m s
At this instant momentum of body is in upward direction
50 51 2550 /initialP kg m s
After breaking 20 kg piece travels upwards with 150 m/s let the speed of 30 kg mass is V
20 150 30finalP V
By the law of conservation of momentum
initial finalP P
2500 20 150 30 15 /V V m s
i.e. it moves in downward direction.
40.(A) Let mass A moves with velocity v and collides in elastically with mass B, which is at rest.
According to problem mass A moves in a perpendicular direction and let the mass B moves at angle with the
horizontal with velocity v.
Initial horizontal momentum of system
(before collision) = mv ....(i)
Final horizontal momentum of system
(after collision) = mV cos ....(ii)
From the conservation of horizontal linear momentum
mv = mV cos v = V cos ...(iii)
Initial vertical momentum of system (before collision) is zero.
Final vertical momentum of system sin3
mvmV
Vidyamandir Classes
VMC/Final Step [Part-A] 34 HPT/Class XIth/JEE-2015
From the conservation of vertical linear momentum
sin 03
mvmV
sin3
vV
...(iv)
By solving (iii) and (iv)
22 2 2 2(sin cos )
3
vv V
224
3
vV
2
3V v .
41.(B) Let 1 ,m m 2 2 ,m m 3 3 ,m m 4 4m m
1ˆ ˆ0 0r i j
2
3ˆ ˆ ˆcos60 sin602 2
a ar a i a j i j
3
3 3ˆ ˆ ˆ ˆ( cos60) sin602 2
ar a a i a j ai j
4ˆ ˆ0r ai j
by substituting above value in the following formula 1 1 2 2 3 3 4 4
1 2 3 4
3 ˆ0.954
m r m r m r m rr ai a J
m m m m
So the location of centre of mass 3
0.95 ,4
a a
42.(A) For translatory motion the force should be applied on the centre of
mass of the body.
Let mass of rod AB is m so the mass of rod CD will be 2m.
Let y1 is the centre of mass of rod AB and y2 is the centre of mass of
Rod CD. We can consider that whole mass of the rod is placed at
their respective centre of mass i.e., mass m is placed at 1y and mass
2m is placed at 2y .
Taking point ‘C’ at the origin position vector of point 1y and 2y can
be written as 1r = ˆ2l j , 2ˆr l j ,
and 1m m and 2 2m m
Position vector of centre of mass of the system 11 2 2
1 2
cmm r m r
rm m
=
ˆ ˆ2 2
2
m l j ml j
m m
=
ˆ4
3
ml j
mjl ˆ
3
4
Hence the distance of centre of mass from C =4
3l .
43.(A) Due to net force in downward direction and towards
left centre of mass will follow the path as shown in
figure
Vidyamandir Classes
VMC/Final Step [Part-A] 35 HPT/Class XIth/JEE-2015
44.(A) Mass of element .2dm xdx
MI of element 2.2 .dI xdx x
32
R
r
I x dx
4 4
2 22
4( )
M R r
R r
=
2 2( )
2
M R r
45.(A) Weight of the rod = W
Reaction of boy 4
B
WR Reaction of man
3
4M
WR
As the rod is in rotational equilibrium 0
02
B M
LR R x
30
4 2 4
W L Wx
6
Lx
Distance from other end, 2
Ly x
y
2
2 6 6 3
L L L L
46.(B) 2
2
2 2 4
1 311
2
gh ghv gh
K
R
47.(C) . .K E U
21 1 1
2eMV GM M
R R h
…(i)
Also 2
eGMg
R …(ii)
On solving (i) and (ii)
2
21
Rh
gR
V
48.(B) Initial length (circumference) of the ring = 2r
Final length (circumference) of the ring = 2R
Change in length = 2R – 2r.
lengthoriginal
length in changestrain
2 ( )
2
R r
r
R r
r
Now Young's modulus / /
/ ( )/
F A F AE
l L R r r
R r
F AEr
49.(B) Force acting on the base
F P A hdgA 30.4 900 10 2 10 7.2N
Vidyamandir Classes
VMC/Final Step [Part-A] 36 HPT/Class XIth/JEE-2015
50.(D) When the bob is immersed in water its effective weight = 1m
mg g mg
1effg g
'
eff
T g
T g 1 2
mT
K
51.(B) From Bernoulli's theorem,
2 21 1
2 2A A A B B BP dv dgh P dv dgh
Here, A Bh h
2 21 1
2 2A A B BP dv P dv
2 21[ ]
2A B B AP P d v v
Now, 0,A Bv v r and A BP P hdg
2 21
2hdg dr or
2 2
2
rh
g
52.(C) On heating the system; x, r, d all increases, since the expansion of isotropic solids is similar to true photographic
enlargement.
53.(C) It is given that the volume of air in the flask remains the same. This means that the expansion in volume of the vessel
is exactly equal to the volume expansion of mercury.
i.e., g LV V or g g L LV V
6
4
1000 (3 9 10 )150
1.8 10
g g
L
L
VV cc
54.(C) Since in the region AB temperature is constant therefore at this temperature phase of the material changes from solid
to liquid and (H2 – H1) heat will be absorb by the material. This heat is known as the heat of melting of the solid.
Similarly in the region CD temperature is constant therefore at this temperature phase of the material changes from
liquid to gas and (H4 – H3) heat will be absorb by the material. This heat as known as the heat of vaporisation of the
liquid.
55.(C) ( )f iQ U W U U W 30 ( 40) 10fU 60fU J
56.(B) ABW 0
BCW nR T 6R 1400 8400R
CDW 0
DAW 6R 600 3600R
W 4800R 4.8R kJ 4.8 8.31 kJ = 30 kJ
57.(D) Rate of cooling (here it is rate of loss of heat)
( ) ( )l l c c
dQ d dmc W m c m c
dt dt dt
60 55(0.5 2400 0.2 900)
60
dQ
dt
115
sec
J .
Vidyamandir Classes
VMC/Final Step [Part-A] 37 HPT/Class XIth/JEE-2015
58.(C) Frequency of vib. is stretched string 1
2(
Tn
Length) m
When the stone is completely immersed in water, length changes but frequency doesn’t ( unison reestablished)
Hence length T ( 1)
air
water
TL V g
l T V g
(Density of stone = and density of water =1)
1
L
l
2
2 2
L
L l
59.(D) Intensity 22a
Here, 2
1
A
B
a
a and
1
2
A
B
2 22 1 1
1 2 1
A
B
I
I
60.(B) Suppose N resonance occurred before tube coming out.
Hence by using (2 1)
4
N vl
n
(2 1) 3301.5
4 660
N
6N .
HPT-3 Mathematics
61.(D) Since, and are roots of 2 0 px qx r , 0p .
q r
,p p
as p, q and r are in AP. 2 q p r …..(i)
Also, 1 1
4
4
4 4
q r
p p 4 q r
2 4 r p r 9 p r [From (i)]
4 4 4
9 9
q r r
p p r and
1
9 9
r r
p r
2 2
4 16 4 16 36
81 9 81
52
81
213
3
62.(D) Let 32 3 f x x x k
On differentiating w.r.t. x, we get : 26 3 0 f x x , x R
Thus, f (x) is strictly increasing function.
Hence, f (x) = 0 has only one real root, so two roots are not possible.
Vidyamandir Classes
VMC/Final Step [Part-A] 38 HPT/Class XIth/JEE-2015
63.(A) Given equation are
2 2 3 0 x x .......(i)
and 2 0 ax bx c ..…(ii)
Since, equation (i) has imaginary roots.
So, equation (ii) will also have both roots same as equation (i).
Thus, 1 2 3
a b c
Hence, : :a b c is 1: 2 : 3.
64.(C) Let z x iy, given Re(z) = 1
x = 1 1 z iy
Since, the complex roots are conjugate of each other
1 z iy and 1 iy are two roots of 2 0 z z
Product of roots =
1 1 iy iy
21 1y 1,
65.(B) We have, 1 1 z z z i
Clearly, z is the circumcentre of the triangle formed by the
vertices (1, 0) and (0, 1) and 1 0 , which is unique.
Hence, the number of complex number z is one.
66.(A) 7
1 A B ,
7
2 A B [ 21 ]
14 A B
2 A B 14 12 2 2[ ]
1 A B
On comparing, we get : A = 1, B = 1
67.(D) Let and be the roots of equation Alternate Solution
2 2 1 0 x a x a Since, 2 a and
Then, 2 a and 1 a Let 22 2 2 f a
Now, 22 2 2
22 2 1 a a 2 2 6 a a
22 2 2 2 1a a 2 2 f a a
2 2 2 2 6 a a For maxima or minima, put 0 f a
22 2 1 5 a 2 2 0 a a = 1
Vidyamandir Classes
VMC/Final Step [Part-A] 39 HPT/Class XIth/JEE-2015
The value of 2 2 will be least, if 1 0 a Now, 1 2 0 f
a = 1 So, f (a) is minima at a = 1
68.(C) Given, that 0 z i w z i w
z = i w w i z
and arg(zw) = 2 arg i z
2 arg i arg z
22
arg z
2
arg i
3
4arg z
69.(C) 1 8 2 7 99 910 10 2 11 10 3 11 10 10 11 k. .....
2 911 11 11
1 2 3 1010 10 10
k .... ...(i)
2 9 1011 11 11 11 11
1 2 9 1010 10 10 10 10
k ..... …(ii)
On subtracting equation (ii) from equation (i), we get
2 9 1011 11 11 11 11
1 1 1010 10 10 10 10
k .....
10
10
111
1010 11 1110
1110 101
10
k
10 1011 11
10 10 10 1010 10
k k = 100
70.(A) Since x, y and z are in AP 2 y x z
Also, 1tan x, 1tan y and 1tan z are in AP
1 1 12 tan y tan x tan z
1 1
2
2
11
y x ztan tan
xzy
2 11
x z x z
xzy 2 y xz
Since x, y and z ar in AP as well as in GP. x y z
71.(C) Given 2 4 6 200 a a a ..... a ...(i)
and 1 3 5 199 a a a ....... a ...(ii)
On subtracting equation (ii) from equation (i), we get
2 1 4 3 6 5 200 199 a a a a a a ...... a a
100 d d d ..... times
100 d 100
d
Vidyamandir Classes
VMC/Final Step [Part-A] 40 HPT/Class XIth/JEE-2015
72.(D) Number of notes that the person counts in 10 min 10 150 1500
Since, a10, a11, a12,….. are in AP with common difference 2 .
Let n be the time taken to count remaining 3000 notes, then 2 148 1 2 30002
nn
2 149 3000 0 n n 24 125 0 n n
n = 24, 125
Then, the total time taken by the person to count all notes = 10 24 = 34 min
[neglecting n = 125 because for this value of n, a125 will be negative, which is not possible as currency notes cannot be
negative]
73.(C) Given that,
21 2
21 2
p
q
a a ..... a p
a a ...... a q
21
2
1
2 12
2 12
pa p d
p
q qa q d
[where, d is a common difference of an AP]
1
1
2
2
a d pd p
a d qd q 12 0 a d p q 1
2
da
Now, 6 1
21 1
5
20
a a d
a a d
5112
4120
2
dd
dd
74.(B)
2 3
11! 2! 3!
x x x xe . . . .
1 1 1
2! 3! 4! ... 11 1 1
1 12! 3! 4!
... e
75.(B) Since, 13 31 3 2 4 3 1x x, log , log . are in AP.
1 2
13 3 32 3 2 3 4 3 1x xlog log log . 1
3 33 2 3 4 3 1 x xlog log .
13 2 12 3 3x x.
3
2 12 3tt [Let 3x t ]
212 5 3 0t t 3 1 4 3 0t t
1 3
3 4t ,
33
4
x [Since, 3x cannot be negative]
33
4log x
31 4x log
76.(C) 2 3 4
12 3 4
x x xlog x x
2 31 11 1 1 1 12 3
x x x . . . log x log xe e e x
Vidyamandir Classes
VMC/Final Step [Part-A] 41 HPT/Class XIth/JEE-2015
77.(C) Let coordinate of the point be ,
Since, , lie on 4 2 0ax ay c and 5 2 0bx by d
4 2 02
ca a c
a
. . . .(i)
Also, 5 2 03
db b d
b
. . . .(ii)
From equations (i) and (ii), 2 3
c d
a b
3 2bc ad
78.(D) Let the coordinate of the centre of T be (0, k)
Distance between their centre
2
1 1 1k k 21 1 1 2k k k
21 2 2k k k 2 21 2 2 2k k k k
1
4k
So, the radius of circle T is k i.e., 1
4
79.(B) So, the sides of a triangle will be 2, 2 and 2 22 2 2 2
x-coordinate of incentre
2 0 2 2 0 2 2
2 2 2 2
. .
2 2 2
2 22 2 2 2
80.(A) 2 2 0x y ax and 2 2 2x y c touch each other.
(i) If circles touch internally,
2 2 2 2
a a a ac c
a c
(ii) If circles touch externally,
2 2 2 2
a a a ac c
c = 0, i.e., not possible as c > 0
Hence, the circles should touch internally and a c .
83.(C) Equation of AB is x, y
Y y dy
X x dx
. . . .(i)
x-intercept is 0dy
x y . ,dx
and y-intercept is 0
dy, y x
dx
As P is mid-point of AB,
Vidyamandir Classes
VMC/Final Step [Part-A] 42 HPT/Class XIth/JEE-2015
2dx
x x y .dy
dxy x
dy
0dx dy
x y
Integrating both sides, we get :
log x log y log c
xy c , as it passes through (2, 3).
6 6c xy
82.(C) As, x y a and 1ax y , intersect in I quadrant.
Therefore, x and y intercept are positive.
1
01
ax
a
and
10
1
a ay
a
1 0a and 1 0a a
1a and 1a a . . . .(i)
Case 1 If 1 0a
2 1a [Not possible]
Case 2 If 20 1a a
1a . . . .(ii)
1a or 1a ,
83.(A) 4 7 0sin sin sin
4 7 0sin sin sin 4 2 4 3 0sin sin . cos
4 1 2 3 0sin cos 1
4 0 32
sin , cos
Case I : 4 0sin
As, 0 0 4 4 4 2 3, , …..(i)
Case II : 1
32
cos
As, 0 0 3 3 2 4 5
33 3 3
, ,
…..(ii)
From (i) and (ii) we get : 3 2 4 8
4 2 4 9 9 9, , , , ,
84.(C) 3
2cos cos cos
2 3 0cos cos cos
2 2 2 2 2 22 0cos cos cos sin cos sin cos sin cos
2 2
0sin sin sin cos cos cos
It is possible when, 0sin sin sin and 0cos cos cos
Hence, both statements A and B are true.
Vidyamandir Classes
VMC/Final Step [Part-A] 43 HPT/Class XIth/JEE-2015
85.(B) Given that, 1 1
2
ycos x cos
2
1 21 12 4
xy ycos x
2
21 12 4
xy yx cos
2
22 1 1 24
yx cos xy
On comparing both sides, we get : 2 2
2 2 24 1 4
4 44
x ycos x y xycos
2 2 2 2 2 2 24 4 4 4x y x y cos x y xycos
2 2 24 4 4x xycos y sin
86.(A) Given that,
1 1cot cos tan cos x . . . .(i)
We know that,
1 1
2cot cos tan cos
. . . .(ii) 1 1
2cot x tan x
On adding equations (i) and (ii), we get :
122
cot cos x
4 2
xcos cot
12
12
xcot
cosx
cot
2 2
2 2
x xcos sin
cosx x
cos sin
On squaring both sides, we get :
2 2
2 2
22 2 2 2
22 2 2 2
x x x xcos tan sin cos
cosx x x x
cos tan sin cos
1
1
sin xcos
sin x
2
2
112
11
2
tansin x
sin xtan
Applying componendo and dividendo rule, we get : 2
2sin x tan
87.(C) Given, 2n A and 4n B
8n A B
The number of subsets of A B having 3 or more elements 8 8 83 4 8C C . . . C
8 8 8 80 1 22 C C C 256 1 8 28 219 0 12n n n n
nC C . . . C
88.(B) Given, 3n
nT C
11 3
nnT C
11 3 3 10n n
n nT T C C [Given]
Vidyamandir Classes
VMC/Final Step [Part-A] 44 HPT/Class XIth/JEE-2015
2 3 3 310 10n n n nC C C C
5n
89.(C) The number of ways in which 4 novels can be selected 64 15C
The number of ways in which 1 dictionary can be selected 31 3C
Now, we have 5 places in which middle place is fixed.
4 novels can be arranged in 4! ways.
The total number of ways = 15 4! 3 = 15 24 3 = 1080
90.(B) Total number of ways
10 10 10 101 2 3 4C C C C
10 45 120 210 385
HPT-4 Chemistry
1.(C) Let the initial pressure of HCN in the following reaction be Po atm.
2HCN(g) H2(g) + (CN)2(g)
Initial pressure Po
Pressure at equilib. Pox x/2 x/2
Given that 2
x = 0.5 x = 1
Kp = 2 2H (CN)
2HCN
(P ) (P
(P ) =
2o
0.5 0.5
(P 1)
= 4 × 10
6.
On solving, Po = 251 atm.
2.(D) For the reaction,
N2O4(g) 2NO2(g)
V).aa(
a4K
22
c
…(i)
where ‘a’ is initial mole of N2O4 present in ‘V’L and is its degree of dissociation.
Also, n
cp )RT(KK
On reducing the volume of container to V
L,2
initial concentration of N2O4 becomes 2a
.V
An increase in concentration
leads to more dissociation of N2O4 in order to have Kc constant, a characteristic constant for a given reaction at a
temperature.
3.(B) 2Na + O2 Na2O2 4.(A) Heating of AlCl3 results in Al(OH)3.
5.(B) NO3 NH4
+5
(n = 8)
Equivalent weight of =8
M3NO
= 8
62 = 7.75.
6.(A) Mg burns in air to react with N2 and gives Mg3 N2.
Vidyamandir Classes
VMC/Final Step [Part-A] 45 HPT/Class XIth/JEE-2015
7.(A) 2KOH + CO2 K2CO3 + H2O
m. mol 20 1 0
18 – 1
milli equivalent of acid = milli equivalent of K2CO3 + milli equivalent of KOH = 1 + 18 = 19
N = millequivalent 19
V(ml) 200 0.095 N.
8.(B) MnO2 + 4HCl MnCl2 + Cl2 + H2O
Cl2 + 2KI I2 + 2 KCl
Moles of MnO2 = 1.087
7.8
Moles of Cl2 = 0.1
Moles of I2 = 0.1
Mass of I2 = 0.1 254 = 25.4 gm.
9.(A) 10.(D) Both have same molecular mass.
11.(B)
Mg/ether CH3CN
Br
(A)
MgBr C=NMgBr +
CH3
H3O+
CH3C=O
12.(A) CH3 + D2 O
4
2
NaBD
Hg(OAc)
Actually oxymercuration demercuration reaction occur according to syn addition of D2O without any rearrangement
because of no carbocation formation.
13.(D) 14.(A) 2 4
4
CuCl H / Pd(BaSO )2 2 2
NH Cl2H CH CH C CH CH CH CH CH CH
15.(D)
16.(C) Down the group thermal stability of alkaline earth metal carbonates increases.
17.(B) CaH2 + 2H2O Ca(OH)2 + 2H2
Ca(OH)2 + CO2CaCO3 + H2O
18.(D)
CH3—CH2—CH2—CH2
1CHO
2 3 4 5
(Pentan-1-al)
CH3—CH2—CH—CH3
1CHO
2 3 4
(2-methyl butan-1-al)
19.(B)
CH2—CH==CH—C—H CH2==CH—CH==C—H
O O—H
H
Vidyamandir Classes
VMC/Final Step [Part-A] 46 HPT/Class XIth/JEE-2015
20.(A)
CHO
H
H
CH2OH
OH
HO
(I)
CHO
H OH
HOH2 C
(II)
OH
H
CHO
H First
interchange
OH
HO CH2OH
H
Second
interchange
CHO
H
H
CH2OH
OH
HO
(I)
Similarly
CHO HO
H
CH2OH
HO
(III)
CHO
H After two
interchanges at C
OH
H OH
(III)
H
*
*
CH2OH
(I) and (II) are identical; (I) and (III) are diastereomers.
21.(A) C—H is weakest bond so rate is maximum in C6H6.
22.(D) OH Strongly activating group because of +R effect.
Cl Weakly deactivating group because of I effect.
NO2 Strongly deactivating group because of R effect and I effect.
So, 4 > 1 > 3 > 2.
23.(B) N2 has maximum bond energy due NN bond.
24.(C) Out of N and P, N has higher IE, and out of O and S, O has higher IE and out of N and O, N has higher IE, due to
greater stability of the exactly half-filled 2p-subshell.
25.(D) Catalyst
2 2 2 2(steam)
CO H H O CO H
26.(A)
27.(B) Diborane is a dimer of 3BH having 3c-2e bonds, as well as 2c-2e bonds. There are two bridged bonds and four
terminal bonds.
28.(C) E W K.E
K.E E W
12400
4.2 2.0ev2000
19 19K.E 2.0 1.6 10 3.2 10 Joule
29.(D) 30.(A) AB5 has trigonal bypyramidal structure
Vidyamandir Classes
VMC/Final Step [Part-A] 47 HPT/Class XIth/JEE-2015
HPT-4 Physics
31.(A) By adjoining graph 0ABW and JWBC 24010]25[108 34
JWWW BCABAC 2402400
Now, JQQQ BCABAC 800200600
From FLOT ACACAC WUQ
240800 ACU .560 JUAC
32.(B) Volume of the gas is constant V = constant TP
i.e., pressure will be doubled if temperature is doubled
02PP
Now let F be the tension in the wire. Then equilibrium of any one piston gives APAPPAPPF 0000 )2()(
33.(A) The distance covered by the ball during the last t seconds of its upward motion = Distance covered by it in first t
seconds of its downward motion
From 2
2
1tguth
2
2
1tgh [As u = 0 for it downward motion]
34.(C) Both the cylinders are in parallel, for the heat flow from one end as shown.
Hence 1 1 2 2
1 2
eq
K A K AK
A A
; where A1 = Area of cross-section of inner cylinder = R
2 and 2A Area of cross-section
of cylindrical shell 2 2 2{(2 ) ( ) } 3R R R
2 21 2 1 2
2 2
( ) (3 ) 3
43eq
K R K R K KK
R R
35.(A) Since 2 22 1( )
2
Lt x x
k
2 2 ( )( )
( )2 2
L L x y x yt x y
k K
36.(B) For observer note of B will not change due to zero relative motion.
Observed frequency of sound produced by A = (330 30)
660 600330
Hz
No. of beats = 600 – 596 = 4
Vidyamandir Classes
VMC/Final Step [Part-A] 48 HPT/Class XIth/JEE-2015
37.(A) According to problem
1
2 4
T v
L m L …..(i) and
1 8 3
2 4
T v
L m L
..…(ii)
Dividing equation (i) and (ii), 1
18 3
TT N
T
38.(B) Time taken by particle to move from x=0 (mean position) to x =4 (extreme position)1.2
0.34 4
Ts
Let t be the time taken by the particle to move from x=0 to x=2 cm
2
sin 2 4siny a t tT
1 2sin
2 1.2t
20.1
6 1.2t t s
. Hence time to move from x = 2 to x = 4 will be equal to 0.3 – 0.1 = 0.2 s
Hence total time to move from x = 2 to x = 4 and back again sec4.02.02
39.(B) Loss of weight at 27ºC is
= 46 – 30 = 16 = V1 × 1.24 l × g …(i)
Loss of weight at 42ºC is
= 46 – 30.5 = 15.5 = V2× 1.2 l × g …(ii)
Now dividing (i) by (ii), we get 16
15.5 = 1
2
1.24
1.2
V
V
But 2
1
V
V = 1 + 3 (t2 – t1) =
15.5 1.24
16 1.2
= 1.001042
3 (42º – 27º) = 0.001042 = 2.316 × 10–5
/ºC.
40.(C) Since specific heat = 0.6 kcal/gm °C = 0.6 cal/gm°C
From graph it is clear that in a minute, the temperature is raised from 0°C to 50°C.
Heat required for a minute = 50 0.6 50 = 1500 cal.
Also from graph, Boiling point of wax is 200°C.
41.(B) Let specific gravities of concrete and saw dust are 1 and 2 respectively.
According to principle of floatation weight of whole sphere = up thrust on the sphere
3 3 3 3
1 2
4 4 4( ) 1
3 3 3R r g r g R g
3 3 3 3
1 1 2R r r R
3 3
1 1 2( 1) ( )R r
31 2
31 1
R
r
3 3
1 2 13
1
1
1
R r
r
3 3
1 2 13
1 22
( ) 1
1
R r
r
1 0.3 2.44
2.4 1 0.3
Mass of concrete
Mass of saw dust
Vidyamandir Classes
VMC/Final Step [Part-A] 49 HPT/Class XIth/JEE-2015
42.(B) mg N
So c
1.
2
If length decreases by cm/hr then candle moves down by 1 cm/hr.
43.(D)
2L dgl
2Y
2 32
6
(8) 1.5 10 109.6 10 m
2 5 10
44.(B) Let the width of each plate is b and due to surface tension liquid will rise upto
height h then upward force due to surface tension
= 2 cosTb …(i)
Weight of the liquid rises in between the plates
= ( )Vdg bxh dg …(ii)
Equating (i) and (ii) we get , 2 cosT bxhdg 2 cosT
hxdg
45.(B) Let velocities of these masses at r distance from each other be 1v and 2v respectively.
By conservation of momentum
1 1 2 2 0m v m v
1 1 2 2m v m v … (i)
By conservation of energy
Change in P.E.= change in K.E.
2 21 2
1 1 2 2
1 1
2 2
Gm mm v m v
r
2 2 2 21 1 2 2 1 2
1 2
2m v m v Gm m
m m r …(ii)
On solving equation (i) and (ii)
22
1
1 2
2
( )
Gmv
r m m
and
21
2
1 2
2
( )
Gmv
r m m
1 2 1 2
2| | | | ( )app
Gv v v m m
r
46.(B) Linear density of the rod varies with distance dm
dx (Given) dm dx
Position of centre of mass cm
dm xx
dm
3
0
3
0
( )dx x
dx
3
0
3
0
(2 )
(2 )
x xdx
x dx
33
2
0
33
0
3
22
xx
xx
9 9 36 12.
9 21 76
2
m
Vidyamandir Classes
VMC/Final Step [Part-A] 50 HPT/Class XIth/JEE-2015
47.(C) Conservation of angular momentum 1 1 2 2 1 2I I I I
Angular velocity of system 1 1 2 2
1 2
I I
I I
Rotational kinetic energy = 21 2
1
2I I
2 2
1 1 2 21 1 2 21 2
1 2 1 2
1
2 2
I II II I
I I I I
48.(A) By the conservation of energy
P.E. of rod = Rotational K.E.
21
sin2 2
lmg I 2
2
32
1sin
2
mllmg
3 sing
l
But in the problem length of the rod 2L is given
3 sin
2
g
L
49.(B) 2 20.3 0.7(1.4 )I x x
For minimum work moment of inertia of the system should be
minimum i.e. 0dI
dx
dI
dx 0.3 2 0.7 2 (1.4 ) 0x x 0.98x m
50.(B) Let two boys meet at point C after time 't' from the starting. Then AC vt , 1BC v t
2 2 2( ) ( ) ( )AC AB BC
2 2 2 2 21v t a v t
By solving we get
2
2 21
at
v v
51.(D) For upper half
2 2 2 / 2 2( sin ) / 2 sinv u al g l gl
For lower half
20 2 (sin cos )
2
lu g
sin (sin cos )gl gl
cos 2sin 2tan
52.(C) Maximum force by surface when friction works 2 2 2 2 2( ) 1F f R R R R
Minimum force R when there is no friction
Hence ranging from R to 2 1R
We get,
2 1Mg F Mg
Vidyamandir Classes
VMC/Final Step [Part-A] 51 HPT/Class XIth/JEE-2015
53.(C) Tension the string ( )m g a Breaking force 20( ) 25g a g 2/ 4 2.5 /a g m s
54.(A) v u at 2 2 2 2 cosv u a t u at
200 100 2 10 2 10 cos135v 10 /m s
sin 10sin135
tan 1cos 10 2 10cos135
at
u at
45
i.e. resultant velocity is 10 m/s towards East.
55.(B) 2 2 2(3 ) (2 ) 2 3 2 cosR P P P P …(i)
2 2 2(2 ) (6 ) (2 ) 2 6 2 cosR P P P P …(ii)
By solving (i) and (ii), cos 1/ 2 120
56.(C) ˆ ˆ5 2As v t i t j ˆ ˆx ya a i a j ˆ ˆ5 2i j
ˆ ˆ( )x yF ma i m g a j
2 2| | ( ) 26x yF m a g a N
57.(D) Force acting on plate, dp dm
F vdt dt
Mass of water reaching the plate per sec =dm
dt1 2( )Av A v v 1 2
2
( )V
v vv
( 1 2v v v velocity of water coming out of jet w.r.t. plate)
( A Area of cross section of jet
2
V
v ) 1 2 1 2
2
( ) ( )dm V
F v v v v vdt v
2
1 2
2
( )V
v vv
58.(D) Since the maximum tension BT in the string moving in the vertical circle is at the bottom and minimum tension TT is
at the top.
2B
B
mvT mg
L and
2T
T
mvT mg
L
2
2
4
1
B
B
T T
mvmg
T L
T mvmg
L
or
2
2
4
1
B
T
v gL
v gL
or 2 24 4B Tv gL v gL but
2 2 4B Tv v gL
Vidyamandir Classes
VMC/Final Step [Part-A] 52 HPT/Class XIth/JEE-2015
2 24 4 4T Tv gL gL v gL
23 9Tv gL
2 10
3 3 103
Tv g L or 10 / secTv m
59.(B) Angular momentum m(vcos )H
2 2sin 45º. .
22
v vm
g
3
4 2
mv
g
60.(D) At the highest point, vertical component of velocity becomes zero so there will be only horizontal velocity and it is
perpendicular to the acceleration due to gravity.
HPT-4 Mathematics
61.(C) As OA OB OC So O (0, 0) is circumcentre of ABC
1 1
3 3
cos cosG ,
So 1
3 3
cos
1 1cos , sin
Orthocenter 1 1cos , sin
62.(D) 1 1 0L y x i.e. Below the line
2 2 1 0L x y Above and on the line
63.(C) A, B, C lie on the circle of centre 2 3, and of radius 3 and they from equilateral triangle
incentre is 2 3,
64.(B) 1 1 2 21 1 1 8 1 2r , C , , C , , r Two circles are separated to each other.
2y x a, neither touch nor cut circle
1r length of perpendicular
2 1
14 1
a
5 1 5 1a , , . . . .(i)
Similarly 16 1
25
a 15 2 5 15 2 5a , , . . . .(ii)
Also line will be between the circles if their centres 1 2andC C are on the opposite sides of it
2 1 16 1 0a a 15 1a , . . . . (iii)
Now from (i), (ii) and (iii) 15 2 5 5 1a ,
Vidyamandir Classes
VMC/Final Step [Part-A] 53 HPT/Class XIth/JEE-2015
65.(C) Point of intersection of 3 0x y and 3 0x y are (3, 0)
Parametric form of lines 3 0 3 0
4 and 41 1 1 1
2 2 2 2
x y x y
Coordinates of point on lines and circle which are at a distance of 4 unit from (3, 0) are
2 2 3 2 2 and 2 2 3 2 2, , these are diametric and points
2 2 3 2 2 3 2 2 2 2 0x x y y
2 2 2 23 8 8 4 2 0 6 4 2 9 0 x y y x y x y
66.(A) Statement-2 is correct and 1 1 0 6 0 3S negative
67.(A) From given functional equation 2y x
f xy f x f y , x, y R
Putting 1y
2 1x
f x f x f
1010 10
10
1 1
3 3 1 33 3 1
3 1 2
r
r r
f r
68.(D) 4
xyf x y f
Putting ; 0y 0f x f . . . .(i)
Putting ; 4 4 0x , f f 0 4f [Using 4 4f ]
Putting ; 2011x in equation (i) 2011 0 4f f
69.(B) 1 2
3 3 13 3
f x x x x x sin x
1 1 2 2
3 3 13 3 3 3
x x x x x x x sin x
1 2
33 3
sin x x x x
3sin x is periodic with period with period 2
Period of 2 1 2
3 3 3f x LCM ,
70.(B) Graph is upward parabola so 0a
0c
x-co-ordinate pf vertex 02
b
a
0b
0b
2 4 2 0f a b c
2 4 2 0f a b c
Vidyamandir Classes
VMC/Final Step [Part-A] 54 HPT/Class XIth/JEE-2015
71.(D) are root of 3 25 2 0 x x x
Now 2
2
y
2 1
1
y
y
it is root of given equation so
3 2
3 2
8 1 1 2 15 4 2 0
11 1
y y y.
yy y
3 2 2 3
8 1 20 1 1 2 1 1 2 1 0 y y y y y y
3 23 2 9 8 0y y y
Roots of this equation are 2 2 2
2 2 2
, ,
so product of roots
2 2 2 8
2 2 2 3
72.(B) 1 2 60 80
2 10
. cos cos
sin
11 2 2 80
21
2 10
cos
sin
73.(C) 5 3 0 sin x sin x sin x
2 3 2 3 0 sin x cos x sin x 3 2 2 1 0 sin x cos x
3 0sin x or 1
22
cos x
3 0x , 2 3, , .... 2
03 3
x , , , ....
or 2 4 8 10
23 3 3 3
x , , , ,.....
2 4
3 3 3x , ,
But given 02
x ,
3
x
74.(B) (B)
5 5
4
1 1x x
x x
a af x f x f x
a a
Hence f x is neither even nor odd
(C) 2 1
1 1 1 12 2 21 1 1
x
x x x
x x x x ef x
e e e
2
x odd function
1
1
x
x
e
e
odd function
1 even function
f x odd function x odd function + even function = even function.
(D)
5
g x g xf x f x
f x is odd function
Vidyamandir Classes
VMC/Final Step [Part-A] 55 HPT/Class XIth/JEE-2015
75.(B) (i) 2 1f x 1f x
(ii)
10
2
f x
f x
1 2f x , ,
(iii)
1 1 7 77
f x, f x
(iv) 0cos sin f x f x R
Hence value of f x for which g x is defined 7 1 1 1 2 7, , ,
76.(C) Since for 0 1f x x ,
(A) 1 1
10 1 0 44 4
sin x sin xf , sin x
1
2 2sin x ,
102
sin x
0 1x ,
(B) 2
2 2
2 3
2
x xf log
x
2
2 2
2 30 1
2
x xlog
x
2
2
2 31 2
2
x x
x
1
2x ,
(C) 0 1f x x x
(D) f x f sgn x 0 1 0 1x sgn x
x R x x
77.(D) 1 2 11 1nnx x x x ..... x
Log 1 2 11 1nnx log x log x log x ..... log x
Differentiate w.r.t. x
1
1 2 1
1 1 1 1
11
n
nn
nx......
x x x xx
Put x = 2
11
1 2 1
2 2 11 1 1 21
2 2 2 2 1 2 1
nn
n nn
nn........
78.(B) 71 2 61 1x x x x ....... x
Put x = 3
71 2 63 1 2 3 3 3....... a
But 1 63 3 , 2 53 3 & 3 43 3
So 7
1 3 53 1
3 3 3 10932
Vidyamandir Classes
VMC/Final Step [Part-A] 56 HPT/Class XIth/JEE-2015
79.(A) 2 2 2 2 2 21 2 2 3 1
1 1 1
n n
......a a a a a a
2 2
12
1n
k kk a a
2 2 4 22
1
1
n
kk a r r
2 2 2 4
2
1 1
1
n
kka r r
12
2 2 2
1 11
1 1 1
nr
a r r
/
/
2 2 2
22 2 2 2
1
1
n
n
r r
a r r
80.(D)
2
2 2
4 4 2
10 6 5 33f
sin sinsin cos
So 21
14
f 1
12
f 1
12
max minf & f
81.(C) 1 1 1 1 1rT r n r
1r n r
21rT n r r
2
1 1 1
1 1 1 2 11
2 6
n n n
n r
n r r
n n n n n nS T n r r
1 2 1 1 21
2 3 6
n n n n n nn
82.(B) 1 1 1
12 3
nH ....n
50
1 1 12 1 2 2 2
2 3 50S ......
50
1 1 12 50 1
2 3 50S . .....
50 50100S H
83.(A)
2 3 1
4 7 10 3 21
5 5 5 5n
nS .......
…….(i)
2 3 1
1 1 4 7 3 5 3 2
5 5 5 5 5 5n n
n nS .......
…….(ii)
Now subtract (i) and (ii)
2
4 3 3 3 21 1
5 5 5 5n
nS ........ ...... n term
13 1
15 54 3 2
115 5
15
n
n
nS
1
4 3 1 3 21 1
5 4 5 5n n
nS
1 1
5 15 1 3 21
4 16 5 4 5n n
nS
.
84.(A) 2b a c & 2c ab
To prove that : c, a, b are in HP
2bc c a c
ab c b c
2ac c
b c
ac baa
b c
Vidyamandir Classes
VMC/Final Step [Part-A] 57 HPT/Class XIth/JEE-2015
Hence c, a, b are in HP
(2) Eliminating a from the two equation
2
2c
c bb 2 22c bc b 2 22 0b bc c
2 22 2 0b bc bc c 2 0b b c c b c
2c b or b = c which is not possible
2b a c
4b a
: : 4 : : 2a b c b b b
4 :1: 2
85.(B) 8 4
3 2
8 7 6 4 3336
6 2C C
86.(B) Total number of ways such that no row remains empty
= Total number of ways – those ways in which R2 is empty
– those ways in which R3 is empty
86 6! 6! 6!C .
866! 2 26 6!C .
87.(D) Total number of triangles that can be made by excluding R in the following ways.
(i) Take any two points lying on line 1, and are point lying on 2.
or
(ii) Take any one point on line 1, and two points on line 2.
Hence total number of ways = 1 22nC . nC 2 1n n
88.(B) Point of intersection of 8 unequal circles = 822 C =
8 72
2
= 56
Point of intersection of 4 unequal circles = 422 C = 12
Point of intersection of 4 non coincident = 421 C = 6
Point of intersection of 4 circles and 4 lines = 4 41 12 C C = 32
36
89.(A) x, 1, 2 are boys and y, 3, 4 are girls
No of arrangement 2! 2! 4
90.(B) Since, there are 14 stations in total.
Hence the total number of ways different tickets can be generated is 14
C2 = 91.
Since, these are 75 persons, hence total number way of distribution = 91
C75.
Vidyamandir Classes
VMC/Final Step [Part-A] 58 HPT/Class XIth/JEE-2015
HPT-5 Chemistry
1.(B) Acidic solution of Na2S2O3 is unstable due to different oxidation states of sulphur.
2.(A) Iron has more affinity towards oxygen therefore it releases more heat.
3.(B) In isoelectronic species anion is bigger than cation. So Cl
> K+.
4.(C) 2 2 4Be C 2H O 2BeO CH ; 2 2 2 2 2CaC 2H O Ca(OH) C H
5.(B) –CO group shows –R effect and –CH2 group shows +R effect
6.(D) In 13th
group, on moving down the stability of +1 oxidation state increases due to inert pair effect.
7.(A) This one is the Diel’s Alder reaction involving the addition of conjugated alkene with an alkene (dienophile) to give
cyclic product.
NBS/hv
CCl4
Br
A B 8.(A) Anti addition.
9.(B) Peroxide effect means reaction will occur according to free radical mechanism.
10.(A) 2 +6 0 +3
N2H4 + K2CrO4 N2 + Cr(OH)4
nfactor 4 3
Moles of K2CrO4 reacted = 24
194
Equivalents of K2CrO4 reacted = 24 3
194
=
72
194
Equivalents of N2H4oxidised = 72
194
Moles of N2H4oxidised = 72
194 4 =
18
194
Mass of N2H4oxidised = 18 32
194
= 2.97 g.
11.(A) Percentage of metal in its oxide = 60%.
Number of parts of metal that combines with 40 parts of oxygen = 60.
Number of parts of metal that combines with 8 parts of oxygen = 60 8
40
= 12
Equivalent weight of metal = 12.
12.(B) Due to presence of lone pair of electron on nitrogen atom 13.(C)
14.(B) H
ST
H S T (60.01 38.20)273
15.(D) When 2 2H O acts as oxidizing agent 2O is not evolved.
16.(C) In CO2, ‘C’ having it’s maximum oxidation state (4), so it cannot act as a reducing agent.
Vidyamandir Classes
VMC/Final Step [Part-A] 59 HPT/Class XIth/JEE-2015
17.(A) meq of acid = meq of Base i.e. M × V × n-factor = M × V × n-factor
(Acid) (Base)
100 0.01 2 = v 0.2 2
v = 5 ml
18.(A) sK Ag X ,
Where X Cl ,Br ,or I
sK
[Ag ][X ]
For iodide :
17 2168.5 10 M
8.5 10 M0.1M
For bromide:
13 2125.0 10 M
5.0 10 M0.1
19.(B) Initial temperature = 127°C = 400 K
Final temperature = 527°C = 800 K
Since KE T, when temperature is doubled, KE is doubled.
20.(B) For photoelectric effect 2
oh h 1/ 2mv
2
01/ 2mv h h
211/ 2mv = 2 hf0 – hf0 = hf0 …(1)
221/ 2mv = 5hf0 – hf0 = 4hf0 …(2)
2n1 1
n2 2
Eq (1) v 1 v 1
v 4 v 2Eq (2)
21.(B) Since entropy is a state function A B A C C D D BS S S S 50 30 20 60eu
B D D BS 20, S 20
22.(B) (s) (aq)NaCl Na (aq) Cl , H 1 kcal/mol
Heat of dissolution = Lattice energy Absorbed + hydration energy evolved
Let heat of hydration of Na and Cl the 6x and 5x respectively 1 kcal) mole 180 ( 6x) ( 5x)
1/ x 179
179
x11
6 179
6x 97.63kcal / mol11
23.(A) Bond length is inversely proportional to bond order
24.(B) 3 3 2 3 2 5CH COOH CH CH OH CH COOC H
Mass 60 gm 46 gm
(t = 0) moles 1 mole 1 mole
(At ) 0.5 mole 0.5 mole 44
gm 0.588
mole
C
0.5
5K 10
0.5 0.5
5 5
Vidyamandir Classes
VMC/Final Step [Part-A] 60 HPT/Class XIth/JEE-2015
3 2 5 3 2 5CH COOH C H OH CH COOC H
At t = 0 2moles 1 mole –
2 – x 1 – x x
C
x / 5K 10
(2 x) (1 x)
5 5
x
2(2 x)(1 x)
2x 4 6x 2x
22x 7x 4 0
x 0.72
Mass of 3 2 5CH COOHC H formed
0.72 88 63.3 gm
25.(D) Equation of 2 3Na CO eq. of NaOH eq. of HCl
20 0.1 20 0.15 25 N
N 0.2 and molarity = 0.2
26.(B) Mass of aluminium = 54 gm
Moles of aluminium = 54
27 = 2 moles i.e. A2 N atoms of Al
Mass of same number of Mg atoms = 24 × 2 = 48 gm
27.(C) 2BOH HCl BCl H O
2B H O BOH H
(t = 0) c – –
(At ) c(1 – h) chch
For titration: acid acid base baseN V N V
2 2
V 2.5 V volumeof HCl15 5
V 7.5 ml
In resulting Solution: 2 / 5 2.5
BCl 0.110
2w
b
ch K
1 h K
Or
14w
12b
K 10 1h
K C 1010 0.1
Now, H ch
21H 0.1 3.16 10 M
10
28.(B) 2 2 4
4 4 2
O O CH
CH CH O
r n M
r n M
3 16 16 3
2 32 32 4 2
Vidyamandir Classes
VMC/Final Step [Part-A] 61 HPT/Class XIth/JEE-2015
29.(D) H
He
1 1R
y 1 9 He1;
1 1 4y 5 H4R 1
4 9 9
When H x
He
9x
5
30. Error
HPT-5 Physics
31.(C)
23
2 3 2
FW t dt
M
24
3 43
F
W t dtM
2 22 3
2 23 4
3 2 5
74 3
W
W
32.(C) Stress
Strainy
Since stress = 0, so strain = 0
33.(B) 21
cos2
mgR mgR mv
2 2 (1 cosθ) v Rg … (i)
2
cosθ mv
mgR
… (ii)
34.(D) Work done = Change in KE 21
02
mgh F h mv
1
1 10 5 1 6 62
h h
15 18h
1.20h m
35.(C) Fundamental frequency
2
V
n
Beats2 2( )
V V
x
2 ( )
V x
x
22
Vx
36.(A) Stopping distance
2 2
2 2
v vS
a g
37.(C) ω2
v
R
2 2
1
1 1ω
2 2 K I mv
22 2
2
1 12
2 24
vmR mv
R
Vidyamandir Classes
VMC/Final Step [Part-A] 62 HPT/Class XIth/JEE-2015
2 22
1
3
4 2 4
mv mvK mv
Velocity of block ω V R3
2 2
V VV
2 2
2
1 9 9
2 4 8
v mvK m
21
22
3 4 3 8 2
4 9 39 8
K mv
K mv
38.(B) Linear momentum conservation
0 mv mv … (1)
Angular momentum conservation
2
0 ω12
mL
mv x … (2)
Point A is at rest
ω
2
LV
0 0 12
2
mv mv xL
M MLx
6
L
x
39.(B) P P C CK U K V
21
022
C
GMm GMmmv
aa,
2 2
2C
GM GMV
a a
40.(B) 2 3T r Kepler’s III
rd law
321
22
2 1
8 64
T R
RT
1 2: : :1:8T T
41.(C) Acc of earth = 2g, acc of mass = g, Relative acc = 3g
2
3
Ht
g
42.(C) Point of contact position will remain unchanged as friction is sufficient.
43.(C) Force of reaction at A is 2A
A A
dPF AV
dt
Force of reaction at B 2B
B B
dPF AV
dt
Net force 2 2( )net B A B AF F F A V V
Applying Bernouli’s equation at A & B, we get 2 2
2 2
1 1( )
2 2 a A a BP V h h g P V gh
2 2 2 2B AV V gh F A gh
Vidyamandir Classes
VMC/Final Step [Part-A] 63 HPT/Class XIth/JEE-2015
44.(A) Suppose V’ volume of shell of volume V is hollow, then 1
2B mg
1
ω ( ')5 ω2
V P g V V P g
3
'5
V V
45.(A) Given 5sinπ( 4) y t
5sin 2π 22
ty
Standard equation sin 2πλ
t xy A
y
So 5A
2T
46.(B) At point O wave A. Shows mean position while B shows positive extreme and C negative Extreme
47. Acceleration of system
P
m M
Force on block
MPMa
m M
48.(B) If 1 2&VV are volumes of ball in the upper and lower liquid respectively 1 2VV V
1 1 2 2g V gV g V
Fraction of volume in lower liquid is 2v V
49.(B) Fraction of the ice inside water = 0.9
Fraction of the ice outside water = 0.1
(by laws of flotation).
Minimum length of rope required = (0.1) 10 = 1m
50.(A) The lower plate will rise if spring force on it greater than its weight
2 2kx m g .
Here 2x maximum extension
For this it should be compressed by 1' 'x
2 21 2
1 1 2( )2 2
kx kxm g x x
Hence , 1 21
2m g m gx
k k
F force given to upper plate in downward direction
1 1F m g kx
1 2 F m g m g
Vidyamandir Classes
VMC/Final Step [Part-A] 64 HPT/Class XIth/JEE-2015
51.(A) sinR r l
2 2( sin )
tanm R r l
mg g
tan
sin
g
r l
52.(C) Δ
Δ /
PB
V V
Δ Δ
V P
V B
Δ Δ
100 100
P P
P B
9
Δ 1000.1
2 10
P
6Δ 2 10 P
53.(B) Tension at the midpoint of the lower wire = T1 = 1
10.2
2m g a
= 2.9 0.1 9.8 0.2 30N
54.(D) 2
3x t
x = 2 6 9t t
v = 2 6ds
tdt
When v = 0, 2t – 6 = 0, t = 3 sec.
Then v = 0, displacement x = (3 – 3 ) = 0
When t = 0, 1 2 1 0 6 6 /V t m s
When t = 6, 2 2 6 6 6V
L =
2 22 1
1 1
2 2mv mv =
2 216 6
2m
= 0
55.(B) 1 1 2 2
1 2cm
m x m xX
m m
2 2 22
2 2 20
b p.c. a b p.x
a b p
2
2 2 2
cbX
a b
56.(C) For the bead to reach point ‘B’ after it leaves point ‘A; the horizontal path which it travels should be 2R sin .
Then
2 22
u sinR sin
g
Where ‘u’ velocity at ‘A’
Let V velocity at “O”
According to law of conservation of energy 2 2 1 cosV u gR
12 2V gr cos
cos
Vidyamandir Classes
VMC/Final Step [Part-A] 65 HPT/Class XIth/JEE-2015
57.(A) This can be assumed as a pure rotation about point of contact, say O, with angular velocity = R
, where R is the
radius of hoop. Speed of P will be P = OP
2R Sin 2
or P= 2R Sin 2
or P= 2 Sin
2
58.(D) The momentum of third part will be equal and opposite to the
resultant of momentum of rest two equal parts
Let V is the velocity of third part.
By the conservation of linear momentum
3 12 2m V m 4 2 /V m s
59.(C) Tension at distance 3
4
L from lower end
1
3
4T
Stress 1
3/
4S
60.(A) Slope is irrelevant hence
1/2
22
MT
K
HPT-5 Mathematics
61.(D) 1
2 2sin cosec sinsin
2 2 1 0sin sin 1sin cosec
Given expression 2 2 4 .
62.(B) log x k y z , log y k z x , log z k x y
x y z x y zlog x y z log x log y log z . . .(i)
= xlog x ylog y zlog z 0k x y z y z x z x y
0 1x y z x y zlog x y z x y z
63.(C) 1sin cos
x xcos sin sin cos
and
2 21 1 cos siny cos
cos cos cos
2sin ycos
Now 2
2
2 2 3
1 1sinx y .
cossin cos cos
and
32
3
sinxy
cos
2 3 2 32 2 1x y xy
O
)
R
= v/R
Vidyamandir Classes
VMC/Final Step [Part-A] 66 HPT/Class XIth/JEE-2015
64.(A)
1 2 3
2 1 3 25 0
5 5 4
D System of equations have exactly one solution.
65.(C) 0x x I and 2 4 0x x
2 4 0x x x I
0 4x ,
66. (B) 1
1
1
2 11
11 2
r
r r
r rtanr r
.r r
1 1
1
1
2 1r
r rtan tan
r r
1 1 1 1 1 12 1 3 2 1
3 2 4 3 2 1n
n nlim tan tan tan tan . . . tan tan
n n
1 1
4 2tan
1 1 2 1
1 1 2 3
tan
67.(A) 21 12 1 1sin sin x sin sin x
LHS is defined only for 1
2x But 1 1 1
12
sin sin sin sin
68.(B) 1 11 1S cosec x sec y
x y
0xy let x < 0 and y > 0
11 1
2 06
x cosec x ,x x
and
12y
y 1 1
3 2sec y ,
y
1 11 1
6 2cosec x sec y ,
x y
If x > 0, y < 0
Then 1 1
06
cosec x ,x
and 1 1 2
2 3sec y ,
y
1 11 1 5
2 6cosec x sec y ,
x y
Range = 5
6 2 2 6, ,
which contains the integers 1 and 2.
69.(B) (A) 1 1 2sin x
1cos cos x x when 0 x
(B) 1 2 2 1 2sin sin cos x sin x sin sin cos x
2 1 2 1cos x, cos x
1
2 2sin sin x x ; x
Vidyamandir Classes
VMC/Final Step [Part-A] 67 HPT/Class XIth/JEE-2015
(C) 21 2 2x x
1
2 2tan tan x x ; x
Option Not possible
(D) 1 1sin x
3 3 3sin x
1 3 3tan tan sin x sin x (Not possible)
70-71. 70.(C) 71.(A)
70. 10 10 10 7 7 710 75 5P P
7 3 7 3 7 35 5 5a 3 3 35 5 5, ,
8 8 88P
71. 6
1r
n n n
6 2
3
1
6 7 6 75 6 5
2 2r
. .n n .
441 21 30 390
72.(A) 3 7 4P , , AB
4 2AC
Let equation of AC
3 7
2 245 45
x y
cos sin
3 2 7 2 5 9 1 5x, y , , , ,
Similarly equation of BD
3 7
2 2135 135
x y
cos sin
3 2 7 2 1 9 5 5x, y , , , ,
73.(B) Equation of a line through (1, 1) and parallel to 4x y and 2x y
Now area of 1
22
PAB . AB
2 2
3 2 1 1t t 4 2t ,
4 2B , or 2 4,
74-75. 74.(B) 75.(D)
74. Point from which length of tangents to these circle is same as radical centre :
1 2 0 4 4 4 1 0S S x y x y
2 3 0 6 14 10 0S S x y
3 7 5 0x y
3 3 3 0
4 8 0 2 3
x y
y y , x
75. If circle he drawn taking radical centre as centre and length of tangents from radical centre to any circle s radius will
cut all the three circles orthogonally.
Vidyamandir Classes
VMC/Final Step [Part-A] 68 HPT/Class XIth/JEE-2015
Length of tangent 9 4 9 4 1 1 27S
Equation of circle 2 2
3 2 27x y
2 24 : 6 4 14 0S x y x y
76.(B) 6 3 9f x f x f x f x . . . .(i)
In equation (i) replace x by (x + 3)
3 9 6 12f x f x f x f x . . . .(ii)
(i) and (ii) 12f x f x
f x is a periodic function having period 12
12 12
0 0
13 1
3f x dx f x dx
12
0
12 4f x dx
77.(C) π 3π 5π 7π
1 cos 1 cos 1 cos 1 cos8 8 8 8
2 2π 3π
1 cos 1 cos8 8
2 2
1 1π 3π1 11 cos 1 cos
π 3π 14 4 2 2sin sin
8 8 2 4 8
78.(C) α,β are roots of 2 0x ax a b
2 2α α and β βa a b a a b
2 2
1 1 2 1 1 20
α α β β a b a b a b a ba a
79.(C) We have 1 1 1 1 1
12 2 3 3 4
S . . . .
S = 1
Given 1 2 1 3 25 2 4 3 3cos x cos x x x
1 2 5cos x and 1 3 22 4 3 2cos x x x
2 5 1x and 3 2 4 3 1x x x 2 4x and 3 2 4 4 0x x x
2 2x , and 1 2 2 0x x x
Finally 2 2x ,
Hence sum 2 2 0
80.(A) 22f x f x x x . . . .(i)
Replace x x
Vidyamandir Classes
VMC/Final Step [Part-A] 69 HPT/Class XIth/JEE-2015
22f x f x x x . . . .(ii)
Solving (i) and (ii) we get : 2 3
3
x xf x
23
03
f x f xf f x
0f x or 3f x
2 3
03
x x or
2 33
3
x x 0 3x ; or 2 3 9 0x x (No real roots)
Sum = –3
81.(D) RHS =
20072 4 21 1 1 1 2 1 1
1
x x x x . . . .
x
RHS = 20072 2 21 1 1 1x x . . . . x
=
2 2 20072 2 21 1 1 1x x . . . . x
Hence
2008
20082
2 11 1
11
x
g x . xx
2008
20082 122 2 1 2 1 1g .
2008
2008
2
22 2
2 2 2 22
g . x g
82.(B)
2
21 1
5 6 3 2 2 22
1 4 45 4
n n
n nn n
k k
k k k k nS lim S lim
k k nk k
83.(A) As 1, z , 1z , z cos i sin , 0, .
1 1
1 1
z cos i sin
2 2
1 1
2 11
cos i sin cos i sin
coscos sin
1 1
2 2 2 2 2 2 2
i icot tan
This shows that lies on the line x = 1/2 and 2 2 / arg / , 0Arg . The maximum value of
Arg is never attained.
84.(A) We can choose one denomination in 131C ways, then 3 cards of this denomination can be chosen in 4
3C ways and one
remaining card can be chosen in 481C ways. Thus, the total number of choices is
13 4 481 3 1C C C 13 4 48 2496 .
85.(D) We can permute M, I, I, I, I, P, P in 7!
4! 2!ways. Corresponding to each arrangement of these seven letters, we have 8
places where S can be arranged as shown below with X.
X X X X X X X
We can choose 4 places out of 8 in 84C ways.
Thus, the required number of ways 8 8 64 4 4
7!7
4! 2!
C C C
Vidyamandir Classes
VMC/Final Step [Part-A] 70 HPT/Class XIth/JEE-2015
86.(A) 2
: 4 : 5
abab
a b
2 4
5
ab
a b
5
2
a b
ab
5
2
a b
b a
12
2
a,
b
: 4 :1a b or 1: 4
87.(B) From the given equation we have 1 cos a x
sin y
and
1 cos a x
sin y
Multiplying we get : 2 2 2
2 2 2
2 2
1 cos a xx y a
sin y
88.(A) Let A (1, 1) and B(2, 4)
If P(x, y) divides the line segment AB in the ratio 3 : 2, then
3 2 2 1 3 4 2 1
5 5x , y
8 14
5 5P ,
As the line 2x y k passes through P
8 14
2 65 5
k k
89.(D) Since 2
1 5 5 1f
2
2 20 5 2f
2
3 45 5 3f 25f x x for 1 2 3x , ,
25 0f x x has 1, 2, 3 as roots 25 1 2 3f x x a x x x
21 2 3 5f x a x x x x
Now, 2 23 2 2f x x f x x f x x f x x
Product of roots of 2 23 2 0f x x f x x
= (Product of roots, 0f x x ) (Product of roots of 2 0f x x )
= 1 2 3 1 2 3 36
90.(D) Two lines are to a common line if these two lines are parallel.
22
2
2
11
1
pp p
p
21 1 0 1p p p