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VMC | Paper-1 1 Joint Entrance Exam-2019 | Advanced
SOLUTIONS 2019-Joint Entrance Examination - Advanced | Paper-1
PART-I PHYSICS
SECTION-1
1.(4) (1) The potential is reduced by dq
dv kR
Also, 0 kQ
vR
2
20
( 4 )
(4 )
dv dq R
v Q R 0 dv v
(1) is wrong
(2) The field is reduced by 02
vkdq dvdE
R RR
(2) is wrong
(3) Initial field at centre is zero, so field increases by 02
vkdq dvdE
R RR
(3) is wrong
(4) Potential at centre 0 0(1 ) v dv v
Potential at 0 0 02 (1 2 )2 ( / 2)
R kdqv v dv v
R
Ratio1
1 2
(4) is correct
2.(2) A particle at distance r will experience gravitational force by the sphere of radius r (or mass say M)
2 2
2
2GM mv v K rm M r
r Gm m Gmr
2
2KdM dr
Gm
22
24
Kr dr dr
Gm
2 22
K
r m G
3.(4) 1 2( )0
tkN N e
1 2( )0 0[1 t
Ca Ar KN N N N N e
Given : 1 2
1 2
( )
( )
199 99
t
Ca Art
K
N N e
N e
1 2( ) 1
100 te
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1 2( ) 100 2 10 t n n
10 1010
2(2 3) 4610 9.2 10
5(4 5 0 5) 10
t Years
4.(2) 1/40( ) (1 ) T t T t
4
0
0
( )
T t Tt
BT
Differentiating w.r.t. time
304 4
0
1 ( )4[ ( ) ] 1
dT t
T t TdtT
… (i)
Also dQ CdT dT P
Pdt dt dt C
(C – heat capacity)
Putting in (i) :
3
04 4
0
4 [ ( ) ]P T t TC
T
SECTION-2 1.(134) L mVR
As L and m are dimension less and [R] = L
1[ ] V L
Kinetic energy, 2 21[ ]
2 K mv K L
(3) is correct
Dimension of work is also 2L
2 3[ ] [ ] F L L F L
Power, 3 1 4[ ] [ ][ ] P FV P F v L L L
(2) is wrong
Linear momentum p mv
1[ ] [ ][ ] p m v L (4) is correct
2.(234) For 2h R and ,r R the sphere will completely be inside the cylinder.
enclosedQ Q
0
Q
(3) is correct
For 3
5
Rr and
4
2 5
h R
2 2 2
2 29 86
2 25
h R Rr R
For 3
5
Rr and ,
5
Rh the sphere will completely be outside the cylinder.
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VMC | Paper-1 3 Joint Entrance Exam-2019 | Advanced
0 0 enclosedQ (4) is correct
For 2h R and 3
5
Rr
3
sin5
r
R
Area of curved surface of sphere enclosed by cylinder
22[2 (1 cos ) ] S R
2 44 1
5
S R
2 54
enclosedQ Q
Q SR
05
Q
(2) is correct
Similarly, for 4
2 ,5
R
h R r
4
sin5
2 2 32[2 (1 cos ) ] 4 1
5
S R R
2
2
54enclosed
Q QQ S
R
0
2
5
Q
(1) is wrong
3.(123) In case I, capillary rise in 1T is 3
2 cos0 2(0.075)(1) 757.5
10001000(10)(0.2 10 )
h mgr
cm
(3) is correct
In case II, capillary rise in 2T is 2 cos60
3.75
hgr
cm
(1) is correct
In case I, if joint is 5 cm above water, water will not rise in 2T as the surface tension force
( 2 cos60 ) r can balance the weight of only 3.75 cm water column. Water will rise till 5 cm
and its curvature will adjust at some angle between 0 and 60° to balance the weight of 5 cm water column
(4) is wrong Due to different volume of meniscus, the correct will be different for both cases. (2) is correct.
4.(134) The x-comp of dl is parallel to velocity so, emf is induced only due to its y-component
0 0 01
yd Bv dy B v dy
L
2
0 0
0
1
yB v dy
L
1
0 0 0 01 1
1( 1) 1
LB v L B v L
L
As the integration is independent of the shape of wire so we can replace by a straight wire with same initial and final y-coordinates,
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VMC | Paper-1 4 Joint Entrance Exam-2019 | Advanced
(1) is correct
For 0 00, 2 B v L
(2) is wrong
For 0 04
2,3
B v L
(3) is correct
As L (4) is correct
5.(124) When 1 2n n n
1 1 1 2( 1)
( 1)n
nf R R R
… (i)
When 1 2,n n n n n
1 2
1 1 1 1 1 1 1( 1) ( 1)n n n
f f f f R R
1 (2 2)
n n
f f R … (ii)
( ) ( )i ii
2( 1) 2 2
12 2 2 2 2 2
f f n f n n
f n n f n n n n
As ( 1)n n
1
12( 1)2 1
df n n
f n nn
As the ratio of independent of sign of radius of curvature,
(1) is correct
f
f and
n
n
are opposite in sign (as 1)n
(2) is correct
Also 1 1 1
2 2 1 12
nn n
f n
f n
(3) is wrong
For 31.5, 10 , 20n n t cm
3(10 )
(20) 0.022(1.5 1)
f cm (4) is correct
6.(3, 4) Just after 1S is closed, capacitors behave as zero resistance wires (as they are initially uncharged)
5
25100 30 70
I A mA
(3) is correct
After 1S is closed for a long time current will be zero. Assuming total charge circulation to be q in
anticlockwise direction, and applying KVL,
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5 010 80 80
q q q
40q C
Voltage across 140
4 volts10
C
(4) is correct
Also 1
4 voltsP Q CV V V
(1) is wrong
Now immediately after 2S is also closed, charges remain same
1
404
10CV V
2
0CV
3
400.5
80CV V
4
400.5
80CV V
Applying KVL in two loops 4 30 10 70( ) 0x x y
100 70 6x y . . . . (i)
4 0.5 5 30 0.5 100 70( ) 0y y x y
200 70 0y x . . . . . (ii)
Solving (i) and (ii), 0.0769x A
7.(2, 4) 3 6100 10 2 10 VR 350 10 50VR k
(3) is wrong
16 32 10 10 1 10 AR 320 10 0.02AR
(2) is correct
Reading of (A) :
V
AV
IR R
RR R
Reading of (V) : V
V
R RV I
R R
Measured value of (50000) (1000)
980.4(51000)
V
V
R RVR
I R R
(4) is correct
If the ideal cell replaced by a cell of internal resistance 5 , reading of ammeter (I) and voltmeter
V
V
R RV I
R R
will change, but V
I remains same.
(1) is wrong.
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8.(3, 4) Process 1 2 and 3 4 are isobaric (as )V T
Process 2 3 and 4 1 are isochoric (as V is constant)
(1) is wrong 1 2 0( )P PQ nC T nC T
03 4 2P P
TQ nC T nC
1 2
3 4
2Q
Q
(2) is wrong 1 2 0( )W P V nR T nR T
2 3 0W
3 4 0W
Total work, 0 0
2 2
nRT RTW
(3) is correct 1 2 0( )P PQ nC T nC T
2 3 0( )V VQ nC T nC T
1 2
2 3
5
3P
V
Q C
Q C
(4) is correct
SECTION-3 1.(8.12 or 8.13)
1330 10
(120)330
f
2330 10cos53
(120)330 30cos37
f
330 6(120)
330 24
2 1bf f f 336 340
(120) 8.128306 330
2.(0.75) ˆ ˆ( 2 )F yi xj N
ˆ ˆ ˆ ˆ. ( 2 ) .( )dw F dr yi xj dxi dyj
2ydx xdy For AB : dy = 0, y = 1
1
0
ABw dx
For BC : dx = 0, x = 1
0.5
1
2 2 ( 0.5)BCw dy
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For CD : dy = 0, y = 0.5 0.5
1
(0.5) (0.5 ) ( 0.5) 0.25CDw dx
For DE : dx = 0, x = 0.5 0
0.5
2 (0.5) ( 0.5) 0.5DEw dy
For EF : dy = 0, y = 0 0FEw
For FA : dx = 0, x = 0 0FAw
Total work ( ) ( 0.25 ) ( 0.5 ) 0 0 0.75 0.75( 1) 0.75
3.(50.00) Time taken to cross the structure ( sinV is the velocity along the length)
sin
Lt
V
For max ,sint should be minimum
So, should be minimum angle for TIR
1.44
sin sin1.5C C
max 11.44
1.5
Lt
V
Also, 8
8
1
3 102 10
1.5
CV
n
8 9max 8
9.6 1.55 10 50 10
2 10 1.44t s s
4.(1.00) For very large N, is very small. Let it be dx
d N
x m m d
N x
Capacitance of elemental capacitor
0 01mK A AxdC K
d dx
Such capacitors are connected in series
0 00
1 12
dd dx d
nC dC K A d x K A
0 12
K AC
d n
5.(2.00) 1 2sin 60 sin30T T
2 13T T 1 2cos60 cos30T T mg
11 1 1
33 100 2 100 50
2 2
TT T T N
2 13 50 3T T N
T
AY
1 1 2
2 2 1
1 3(2) 2
13
T Yc
s T Y
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6.(270.00) 5( ) (80 30) 5 (110 80)s L C , where C is heat capacity of calorimeter
250 5 30s L C
80( ) (50 30) (80 50)s C
1600 30s c 250 5 1600s L s 5 1350 270L
L ss
PART-II CHEMISTRY
SECTION-1
1.(4) 2 6 4HC C , CH CH and p - MeO C H acts as electron withdrawing group while
3 2CH CH acts as electron releasing group. Presence of electron withdrawing group increases
acidic strength while presence of electron releasing group decreases acidic strength.
2.(4) 32 3 2 3Cr (salt) B O Cr(BO )
Green beads of chromium metaborate will be formed.
3.(2) Calamine 3ZnCO Malachite 3 2CuCO Cu(OH)
Magnetite 3 4Fe O Cryolite 3 6Na AlF
4.(3) Molar conductivity decreases with concentration till CMC & further decreases rapidly beyond
CMC.
SECTION-2
1.(3, 4) Here 2Q SnCl
2SnCl Cl 3(X)Sn Cl 3(sp hybrid, pyramidal)
2 3 3 2(Y)
SnCl Me N Me N SnCl (Coordination compound)
2 2 4(Z)
SnCl 2CuCl SnCl 2CuCl
2.(3, 4) Standard formation of enthalpy is the enthalpy involved in the reaction when one mole of a
substance is formed from its constituent elements in their standard elemental state.
8 2 2
2 3
1S (s) O (g) SO (g)
8 formation reactions3
O (g) O (g)2
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3.(1, 2, 3) 2 2 2 4 2(W)
1MnO 2KOH O K MnO H O
2
(W) 6Mn state, paramagnetic, green coloured and tetrahedral shape of the anion 24MnO
Structure bonds between (d)(p)
O Mn
Electrolyric4Oxidation
W KMnO (X)
(X) 7Mn , diamagnetic, purple coloured and tetrahedral shape of the anion 14MnO
Structure
(d)(p)
O Mn
4.(3, 4) (3) 2 3 3 3NO , NH , POCl , CH Cl (polar molecule)
(4) 2 6 5 2 5SO , C H Cl, H Se, BrF (polar molecule)
5.(1, 2, 4) rms3RT
M rms
1
M
av3
kT2
independent of molar mass
rms T
rms is doubled when temperature is increased four times
6.(2, 3)
7.(1, 3, 4) Sucrose 2(leavo)(dextro)
H O glucose + fructose
Glucose [O] gluconic acid
Monosaccharides are simplest carbohydrates and can't be hydrolysed
Cyclic hemiacetal pyranose (six membered) forms of glucose are called as anomers.
8.(1, 2, 4) 31 2 4XX X X238 234 234 234 23092 90 91 90U Th Pa Z Th
1X : particle 2X : particle
3X : particle 4X : particle
Z : 23492 Z (isotope of uranium)
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SECTION-3
1.(8.92) 2 2Fe (aq) S (aq) FeS(s) 17CK 1.6 10
0.03 M 0.1 M 0.03 x 0.1 x
Since CK 1 x 0.03
2[S ] 0.1 0.03 0.07M
2 217
1[Fe ][S ]
1.6 10
2 17 17 171[Fe ] 10 8.928 10 y 10
1.6 0.07
y 8.93 (Round off)
or y 8.92 (After Truncation)
2.(6.75) x y zr k[A] [B] [C]
5 x y z6 10 k(0.2) (0.1) (0.1) ........ (i)
5 x y z6 10 k(0.2) (0.2) (0.1) ........ (ii)
4 x y z1.2 10 k(0.2) (0.1) (0.2) ........ (iii)
5 x y z9 10 k(0.3) (0.1) (0.1) ........ (iv) From (i) & (ii) y 0
From (i) & (iii) z 1
From (i) & (iv) x 1
r k[A][C]
From equation (i)
56 10 k(0.2)(0.1)
5
32
6 10k 3 10
2 10
3r 3 10 [A][C]
For 3 3 3[A] 0.15mol dm , [B] 0.25mol dm & [C] 0.15mol dm
3r 3 10 0.15 0.15
5r 6.75 10
y 6.75
3.(4)
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VMC | Paper-1 11 Joint Entrance Exam-2019 | Advanced
4.(4)
5.(1.03) 650 640 0.5 78i
640 M 39
f fT i K m
650 640 M 39 0.5 10005.12
640 0.5 78 M 39
10 10005.12 1.025 1.03
640 78
6.(19) 4 2 2 6 2XeF O F XeF O
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VMC | Paper-1 12 Joint Entrance Exam-2019 | Advanced
PART-III MATHEMATICS
SECTION-1
1.(3) 1 11
2 1
y
x
1 2 y x 1 x y
1 y x
0 0
0 0
4 4 2
2 ( ) ( ) 2
z z x
z z i x iy x iy i 4 2 4 2 4 2 1
2 2 2 ( 1) 2 (2 )
x x x
iiy i i y i x i
2.(3) 1 M I M
2 0 M M I ( ) rT M 4 4 1sin cos , 1
2
4 4 2 2| | ( (1 )(1 )) M S C S C 4 4 2 2 2 2( 2), S C S C S C t
2 1( 2) 0,
4
t t t
* 1
2
* * *37 29
16 16
3.(3) Line with slope 1
mpassing through (3, –2)
3 2 n my m
( 1) (3 2 ) x m mx m
2
3 3 3
51
m
xm
215 15 3 3 m m
23 15 18 2 m m
2 5 6 0 m m 2,3m
4.(4)
2 8
2
1 2
87 x dx dx
x
78 4 7
3 n 14
16 23
n sq. units
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VMC | Paper-1 13 Joint Entrance Exam-2019 | Advanced
SECTION-2 1.(2, 4) 1B = Event of selection of bag 1B (5 Red, 5 green)
13
( )10
P B
2B = Event of selection of Bag 2B (3 Red, 5 Green)
13
( )10
P B
3 B Event of selection of Bag 2B (5 Red, 3 Green)
34
( )10
P B
R Event of selection of Red Ball
G Event of selection of a Green Ball.
1 1 2 2 3 3( ) ( / ) ( ) ( / ) ( ) ( / ) ( ) P G P G B P B P G B P B P G B P B
5 3 5 3 3 4 39
10 10 8 10 8 10 80
3 3 33
( ) ( / ) ( )( / )
( ) ( )
P B G P G B P BP B G
P G P G
12 12 480 39 1339
80
33
( / )8
P G B
2.(1, 3, 4) 1 2(1 , 2 , 2 ), (2 , , 2 ) L L
D.Rs of 1 2L L (1 2 , 2 2 2 )
1 21ˆˆ( 2 2 ) 0 9 1 09
L L i j k
1 22ˆˆ ˆ(2 2 ) 0 9 29
L L i j k
1 28 2 2 4 2 4
, , , , ,9 9 9 9 9 9
N N
DRs of 1 2N N (2, 2, 1)
3.(2, 3) 6 cos 4 sin 12 sin 2 A
1 max 12, sin 2 14
R A
2 23 2
,2 2
a b
1 1
2,
2 2
n nn n
a b
Eccentricity 2
2
4 51 1
9 3 n
nn
be n
a
2 36, 3.... R R1 1
1224
11
2
N
n nn n
R R
Latus rectum
2
29
98
22 8
2 23
2
b
a
8 1
3 24 6
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VMC | Paper-1 14 Joint Entrance Exam-2019 | Advanced
4.(2, 3) ( ) dy
Y y X xdx
0,
pdy
Y y xdx
2
2 2 1
pdy
PY x xdx
21
'
dy x
ydx x
2' 1 0 xy x
21
, sin
xdy dx c x
x
cos dx d
2cos
sin
y d c | cos cos | cos n ec c
2
21 11
x
y n xx
5.(2, 3, 4) 2 1 0 x x
1 5
2
x
1 5 1 5, , ( )
2 2
5, 1, 1 2 1 21 0 0 n n n 2 1 21 0 0 n n n 1 1 2 2( ) ( ) ( ) 0 n n n n n n 1 2 1 20 n n n n n na a a a a a or 2 1 n n na a a
1 1 n n nb a a1 1 1 1 1 2 1 2( 1) ( 1)
n n n n n n
1 1( 2) ( 2)
n n1 15 5 5 5
2 2
5
n n
n nnb
1 1 1
1
10 1010 5
n n
nn
n n n
a 1 1 5 1 5 1 40 5 10
356 895 19 5 19 5 5
1 1 1
1
5
n n nr r
rr r r
a
1 1 1 1 1 11 1
1 1 1 15 5
n n n n
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1 1
15 5
n n n n
1 1 n na a 2 1 na
1 1 1
1 5 1 5 48 12
10 10 10 10 356 8910 19 5 19 5
n n
nn n n
bn
6.(1, 2, 4) 3 12 2
sin sinR
P Q
3 1sin , sin
2 2P Q
60 or 120 ; 30 or 150 P Q
60 , 30 , 90P Q R (Right triangle)
60 , 150 180P Q P Q (Not valid)
120 , 30 30P Q R (Valid)
120 , 150P Q (Not valid)
120 , 30 1P Q R q r
Slope of 1 / 4 1
3 3 / 4 3 3
RS
Equation: 3
3 32
x y
1 1
6 6k OE
27 1 28 7
16 16 4 2RS
Area
0 0 1
1 1 30 1
2 6 48
3 11
4 4
SOE
1 1 33 3(2 3)2 2 2
5 23 2 31
2
r
7.(1, 2, 3) 2adj. (adj. ) | |M M a = 2, b = 1
0 2 4 0 1 2
2 4 6 2 1 2 3 | | 2
6 2 2 3 1 1
M
2 1 1 2
2 3 2 2 2 3 2 2 4 3 3 3 3 1 2 3 2 2
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1 1 2 1 1 1 1 1 3
1 1 1 2(adj. ) adj. 2(adj. )
| |M M M M M
M
22 4 4adj. adj. | | ( 2) 16M M M
8.(1, 2, 3)
4 3 2
4
2
5 20 30 20 30
5( 1) 20 1
' 2 11 3
2 8 73
ln ( 2) 1
x x x xx
xx
f xx
x xx
x
3 220 60 60 20 02 0 1
" 4 8 1 31 3
2
x x x x
xf x x
xx
''(1 ) 4 0, ''(1 ) 2 0f f is continuous inf R ( )f ( )f is ontof
SECTION-3
1.(10.00) 2
1
2rAC
BC r
2 2 2AC BC AC AB
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2 4 AC AB AD16 18 23
4 1010
2.(0.75)
1 1
( , ,0) , ,02 2
ˆˆ
1 1 10
2 2 21 1 1
6 6 3
i j k
1 1 1 1 1 1 3ˆˆ 32 6 6 6 2 6 12
i j k
2 3 3(6 ) 36 0.75
144 4
3.(157.00) (1,3) {1,4,7,10,....}AP
(2,5) {2,7,12,17,....}AP
(3,7) {3,10,17,24,....}AP
(1,3) (2,5) (7,15) {7,22,37,52,...} AP AP AP
(2,5) (3,7) (17,35) {17,52,...}A AP AP
(1,3) (2,5) (3,7) (52,105) 157 AP AP AP AP a d
4.(4.00) /4 sin
sin/4
2 12
(1 )(2 cos2 )
x
x
eI dx
e x
/4 /4
2 20 0
2 2
2 cos 1 2sin
dx dxI
x x
/4 2
20
2 sec
1 3tan
x dx
x
1
20
2
1 3
dt
t
11
0
2 3tan
3 1/ 3
t
2 3
3 3
2 3
9I
2 4.327 27 4
81 I
5.(3.00) 22 2 2( )( ) a b c a b c a b c 2 3 2 a b c ab bc ca
2 2 21 1(( ) ( ) ( ) ) (1 1 4) 3
2 2 a b b c c a
Vidyamandir Classes
VMC | Paper-1 18 Joint Entrance Exam-2019 | Advanced
6.(0.50) 2 E seven1, two zeros
92 2
9.8( ) 36
2 m E C
1 21 2
( ) 18 1( / )
36 36 2
n E EP E E
Two zero’s in same row/column
1 0 0
1 1 1 0
1 1 1
3 6 18
Two zeros not in same row/column
1 1 0
1 0 1 1
1 1 1