Allen: Test Paper of Jee (Main) Examination - 2019
Allen: Test Paper of Jee (Main) Examination - 2019
Allen: Test Paper of Jee (Main) Examination - 2019
(3)
(4)
1 k
2p m
1 3k
2p m
B EN
Ans. (1) Sol.
l
x=
LL
2
K1A1 + K 2 A 2
q K eq =
A1 + A 2
=
( )
K1 pR 2 + K 2 3pR 2 ( )
4pR 2
q
A
K1 + 3K 2
=
Sol. Kx 4
3. A travelling harmonic wave is represented by
l the equation y (x, t) = 10–3 sin (50 t + 2x), where
t = -2Kx cos q x and y are in meter and t is in seconds. Which
2
of the following is a correct statement about the
æ Kl 2 ö wave?
Þ t = ç 2 ÷ q = -Cq The wave is propagating along the
è ø
(1) negative x-axis with speed 25ms–1
Kl 2 (2) The wave is propagating along the positive
1 C 1 2
f= = x-axis with speed 25 ms–1
Þ 2p I 2p Ml 2 (3) The wave is propagating along the positive
12 x-axis with speed 100 ms–1
(4) The wave is propagating along the negative
1 6K
Þ f= x-axis with speed 100 ms–1
2p M Ans. (1)
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JEE ( Main ) Examination–2019/Morning Session/12-01-2019
Sol. y= a sin(wt + kx) (1) 10 V/m (2) 24 V/m
Þ wave is moving along –ve x-axis with speed (3) 30 V/m (4) 6 V/m
Ans. (2)
w 50
n= Þn= = 25m / sec.
K 2 96
Sol. Prefracted = PI
4. A straight rod of length L extends from x = a 100
to x=L + a. The gravitational force is exerts on 96
Þ K 2A t =
2
a point mass 'm' at x = 0, if the mass per unit K1A 2i
100
length of the rod is A + Bx2, is given by:
96
Þ r2 A t =
2
r1A i2
(1) Gm éê A æç 1 - 1 ö÷ - BL ùú 100
ë èa+L aø û
96 1
´ ´ ( 30 )
2
é 1 ù Þ A 2t =
(2) Gm ê A æç -
1 ö 100 3
÷ + BL ú
ë èa a+Lø û 2
é æ 1 1ö ù 64
- ÷ + BL ú ´ ( 30 ) = 24
2
(3) Gm ê A ç At
ë è
é æ1
a
(4) Gm ê A ç -
+ L
1 ö
ë èa a+Lø
a ø
ù
÷ - BL ú
û
û EN 6.
100
The output of the given logic circuit is :
Ans. (2)
Y
LL
B
Sol.
(1) AB
(2) AB
dm= (A + Bx )dx2
(3) AB + AB
GMdm
dF = (4) AB + AB
A
x2
Ans. (2)
a + L GM
= F = òa
x
(2 )
A + Bx 2 dx
A
A
a+ L A+ B A + B =A + B
é A ù
= GM ê- + Bx ú Sol.
Y = (A + B)A
ë x ûa A+B A+ B A
B
é æ1 1 ö ù B
= GM ê A ç - ÷ + BL ú
ë èa a+Lø û Y = (A + B) A
5. A light wave is incident normally on a glass slab
of refractive index 1.5. If 4% of light gets = A + AB
reflected and the amplitude of the electric field
of the incident light is 30V/m, then the ( )
= A AB
amplitude of the electric field for the wave
propogating in the glass medium will be: = A ( A + B)
= A + AB = AB
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JEE ( Main ) Examination–2019/Morning Session/12-01-2019
7. In the figure shown, after the switch 'S' is turned
r2 µ n
from position 'A' to position 'B', the energy
dissipated in the circuit in terms of capacitance 1 2 1
E= kr + mv 2 µ r 2
'C' and total charge 'Q' is: 2 2
A B µn
9. Two electric bulbs, rated at (25 W, 220 V) and
S
(100 W, 220 V), are connected in series across
C 3C a 220 V voltage source. If the 25 W and
100 W bulbs draw powers P 1 and P 2
respectively, then:
3 Q2 3 Q2 1 Q2 5 Q2 (1) P1 = 9 W, P2 = 16 W
(1) (2) (3) (4)
8 C 4 C 8 C 8 C (2) P1 = 4 W, P2 = 16W
Ans. (1) (3) P1 = 16 W, P2 = 4W
(4) P1 16 W, P2 = 9W
1 Ans. (3)
Sol. Vi = CE2
2 220 2
Sol. R1 =
25
( CE )
2
Vf =
1
2
2 ´ 4c
3 3
DE = CE 2 ´ = CE 2
4 8
=
1 CE 2
2 4
EN R2 =
L=
220 2
100
220
R1 + R 2
8. A particle of mass m moves in a circular orbit
P 1 = i2 R 1
1 2 P2 = i2 (R2 = 4W)
in a central potential field U(r) = kr . If Bohr's
LL
2
quantization conditions are applied, radii of 220 2 220 2
= ´
possible orbitals and energy levels vary with æ 220 2 220 2 ö 25
ç + ÷
quantum number n as: è 25 100 ø
1 1
(1) rn µ n2 , En µ (2) rn µ n,E n µ 400
n2 n = = 16W
A
25
(3) rn µ n, En µ n (4) rn µ n , En µ n
10. A satellite of mass M is in a circular orbit of
Ans. (4)
radius R about the centre of the earth. A
dV mv 2 meteorite of the same mass, falling towards the
Sol. F= = kr =
dr r earth, collides with the satellite completely
inelastically. The speeds of the satellite and the
nh meteorite are the same, just before the collision.
mvr =
2p The subsequent motion of the combined body
will be :
r2 µ n
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JEE ( Main ) Examination–2019/Morning Session/12-01-2019
(1) in a circular orbit of a different radius
1 f
(2) in the same circular orbit of radius R Sol. Energy = nRT = PV
2 2
(3) in an elliptical orbit
(4) such that it escapes to infinity f
Ans. (3) =
2
( )
3 ´ 10 6 ( 2 )
is:
(1) 3 × 102 (2) 108 J
(3) 6 × 104 J (4) 9 × 106 J K1
Ans. (4)
(1) 12W (2) 25W (3) 5W (4) 22W
Ans. (4)
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JEE ( Main ) Examination–2019/Morning Session/12-01-2019
AB = Vp × t
E
Sol. case I ig = = C q0 ..(i) BC = Vt
220 + R g
Case II AB
cos60° =
BC
æ ö
ç ÷ 1 Vp ´ t
E 5 Cq0 =
ig = ç ÷´ = 2 Vt
ç 5R g ÷ ( R g + 5) 5 ..(ii)
çç 220 + ÷÷
è 5 + Rg ø V
VP =
2
200Rg = 4400
Rg = 22W 2mqDV
r=
Ans. – 4 qB
LL
16. A person standing on an open ground hears the
sound of a jet aeroplane, coming from north at m
rµ
an angle 60° with ground level. But he finds q
the aeroplane right vertically above his position.
If u is the speed of sound, speed of the plane rp 1
=
is : rµ 2
A
V
d ·S
L
2L
C
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JEE ( Main ) Examination–2019/Morning Session/12-01-2019
d
(1) 3d (2)
2
(3) d (4) 2d
Ans. (1)
Sol.
Sol. 3d
I S 2 Before colision After collision
L vm
L L 3d v1 M
m
2
v = 2gl (1 - cos q0 ) v1 = 2gl (1 - cos q1 )
By momentum conservation
1 - cos q0 - 1
l and a bob of mass m, is released from a small
angle q0. It strikes a block of mass M, kept on By componendo divided
a horizontal surface at its lowest point of
oscillations, elastically. It bounces back and æq ö
sin ç 1 ÷
m -M 1 - cos q1 è2ø
goes up to an angle q1. Then M is given by : = =
m+M 1 - cos q0 sin æ q0 ö
ç ÷
m æ q0 - q1 ö m æ q0 + q1 ö è 2ø
(1) 2 ç q + q ÷ (2) 2 ç q - q ÷
è 0 1ø è 0 1ø
M q0 - q1 q0 - q1
Þ m = q +q Þ M = q +q
æ q0 + q1 ö æ q0 - q1 ö 0 1 0 1
(3) m ç q - q ÷ (4) m ç q + q ÷
è 0 1ø è 0 1ø
Ans. (3)
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JEE ( Main ) Examination–2019/Morning Session/12-01-2019
(1) 6A (2) 7.5A
21. What is the position and nature of image
(3) 5.5A (4) 3A
formed by lens combination shown in figure?
Ans. (1)
(f1, f2 are focal lengths)
Sol. Ideal inductor will behave like zero
resistance long time after switch is closed
2 cm
A B
R
O e R
+q
l
–2q
l
l
x
+q
V -20 5
ˆi + ˆj ˆj - ˆi
20 (1) (ql) (2) 3ql
V= 2 2
3
LL
For second lens (3) - 3q l ˆj (4) 2ql ˆj
20 14 Ans. (3)
V= -2 =
3 3
y
1 1 1
- =
A
V 14 -5
3 –2q
V = 70cm Sol. P1 P2
22. In the figure shown, a circuit contains two
x
identical resistors with resistance R = 5W and +q +q
an inductance with L = 2mH. An ideal battery
of 15 V is connected in the circuit. What will |P1| = q(d)
be the current through the battery long after the |P2| = qd
switch is closed? |Resultant| = 2 P cos30º
S æ 3ö
L
2 qd çç 2 ÷÷ = 3 qd
è ø
R
15 V
R
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JEE ( Main ) Examination–2019/Morning Session/12-01-2019
y
®
24. The position vector of the centre of mass r cm
S
of an symmetric uniform bar of negligible area O
of cross-section as shown in figure is : L P Q
M x
N
(1) 40 A, perpendicular into the page
L
(2) 40 A, perpendicular out of the page
L 2L 3L
(3) 20 A, perpendicular out of the page
® 13 5
(1) r cm = Lxˆ + Lyˆ (4) 20 A, perpendicular into the page
8 8
Ans. (4)
® 11 3
(2) r cm = L xˆ + Lyˆ Sol. Magnetic field at ‘O’ will be done to ‘PS’ and
8 8 ‘QN’ only
® 3 11 i.e. B0 = BPS + BQN ® Both inwards
(3) r cm = L xˆ + Lyˆ
8 8 Let current in each wire = i
® 5 13 µ 0i µ 0i
(4) r cm = L xˆ + Lyˆ \ B0 = +
Ans. (1)
8 8 EN or
\
10–4 =
2pd
i = 20 A
4pd 4pd
µ 0i
=
2 ´10-7 ´ i
4 ´10-2
2m (L,L)
26. In a meter bridge, the wire of length 1 m has
m 2L, L a non-uniform cross-section such that, the
Sol. 2 5L , 0
variation dR of its resistance R with length l
LL
2
2L m 3L dl
5mL dR 1
2mL + 2mL + is µ . Two equal resistances are
2 = 13 L dl l
X cm =
4m 8 connected as shown in the figure. The
æ Lö galvanometer has zero deflection when the
A
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JEE ( Main ) Examination–2019/Morning Session/12-01-2019
2
is accelerated by a potential difference of 2500
1 B
1 2 3 4 5 lA
3 V. The ratio of de-Broglie wavelengths l is
V(m ) B
(1) 1 J (2) 5 J
close to :
(3) 10 J (4) 30 J
(1) 10.00 (2) 14.14 (3) 4.47 (4) 0.07
Ans. (3)
Ans. (2)
Sol. Since P–V indicator diagram is given, so work
Sol. K.E. acquired by charge = K = qV
done by gas is area under the cyclic diagram.
h h h
1 l= p= =
\ DW = Work done by gas = ×4×5J 2mK 2mqV
2
= 10 J
lA 2m Bq B VB 4m.q.2500
\ l = =
m.q.50
= 2 50
B 2m A q A VA
= 2 × 7.07 = 14.14
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JEE ( Main ) Examination–2019/Morning Session/12-01-2019
30. There is a uniform spherically symmetric Sol. At any instant 't'
surface charge density at a distance R0 from the Total energy of charge distribution is constant
origin. The charge distribution is initially at rest
1 KQ 2 KQ 2
and starts expanding because of mutual i.e. mV 2
+ = 0 +
2 2R 2R 0
repulsion. The figure that represents best the
speed V(R(t)) of the distribution as a function
1 KQ2 KQ2
of its instantaneous radius R (t) is : \ mV 2
= -
2 2R 0 2R
V(R(t)) V(R(t))
V0
2 KQ2 æ 1 1 ö
\ V= .ç - ÷
m 2 è R0 R ø
(1) (2)
R0 R(t)
R(t) R0 KQ 2 æ 1 1 ö 1 1
\ V= ç - ÷ =C -
V(R(t)) m è R0 R ø R0 R
V(R(t))
Also the slope of v-s curve will go on
decreasing
(3)
Ans. (1)
R0 R(t)
(4)
R0
EN
R(t)
\ Graph is correctly shown by option(1)
LL
A
10 E