Vidyamandir Classes VMC | JEE Main-2020 1 Solutions |7 th January Morning SOLUTIONS JEE Main – 2020 | 7 th January 2020 (Morning) PHYSICS SECTION – 1 1.(2) A logic gate is reversible if we can recover input data from the output. 2.(1) 2000kg m = 4000 r f N = ? = v 4000 20000 T = + , 24000 T N = . P Tv = ; 60 746 24000 u = , 1.9 m/s u = 3.(1) 2 0 0 I It It = − 0 IA = = BA n 0 0 (1 2) =− = − − R d V nAI t dt 1 0 at sec 2 R V t = = 2 0 0 0 0 2 (1 2) 2 R R nI R V I t R R − = = − 4.(4) 0 0 0 ˆ ˆ cos 60 sin 60 2 2 2 E x y − = + − 0 3 1 ˆ ˆ 1 2 2 2 y x = − − 5.(4) 8 6000 10 cm − = For 2 nd minimum 2 sin 2 d = 3 4 d = So, for 1 st minimum, 1 sin d = 1 3 sin 4 d = = 1 25.65 (from sin table) = , 1 25 6.(1) Energy conservation : 2 2 1 1 2 2 2 GMm GMm mu mv R R − − + = + 2 GM V u R = − …(i) Momentum conservation: 9 10 10 2 2 T orbital m m GM GM V V R R = =
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Vidyamandir Classes
VMC | JEE Main-2020 1 Solutions |7th January Morning
SOLUTIONS
JEE Main – 2020 | 7th January 2020 (Morning)
PHYSICS
SECTION – 1
1.(2) A logic gate is reversible if we can recover input data from the output.
2.(1) 2000kgm=
4000rf N=
?=v
4000 20000T = + , 24000T N=
.P T v= ;
60 746 24000 u = , 1.9 m/su =
3.(1) 20 0I I t I t= −
0 IA = = BA n
0 0 (1 2 )
= − = − −R
dV nAI t
dt
1
0 at sec2
RV t= =
2
0 0
0 0
2(1 2 )
2
RR
nI RVI t
R R
− = = −
4.(4) 0 0 0
ˆ ˆcos60 sin 602 2 2
E x y −
= + −
0
3 1ˆ ˆ1
2 2 2y x
= − −
5.(4) 86000 10 cm− =
For 2nd minimum 2sin 2d = 3
4d
=
So, for 1st minimum, 1sind = 1
3sin
4d
= =
1 25.65 (from sin table) = , 1 25
6.(1) Energy conservation :
2 21 1
2 2 2
GMm GMmmu mv
R R
−− + = +
2 GM
V uR
= − …(i)
Momentum conservation:
9
10 10 2 2T orbital
m m GM GMV V
R R
= =
Vidyamandir Classes
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2 [By (1)]10
r
m GMV m u
R= −
Kinetic energy ( )2 2 21 81100 100 (1)
2 10 20 2T r
m m GM GMV V u
R R
= + = + − =
7.(4) 1 2
1 2
1 2mixmix
mix 1 2
P Pp
V V V
n C n CC
C n C n C
+ = =
+
1 21 2
1 2mix
1 1 2
1 2
1 1
1 1
R Rn n
n R n R
+
− − =
+ − −
On rearranging, we get, 1 2 1 2
mix 1 21 1 1
n n n n+= +
− − −
mix
5 3 2
1 1/ 3 2 / 3= +
− mix mix
5 171 1.42
12 12 − = = =
8.(1) Time-period 2 r
Tv
= =
2
0
1
2 a n nT
V Z Z
= ;
30
21
2 a nT
V Z
=
3T n
3
1 12 13
2 2
18
8
T nT T
T n= = = ; 16
2 12.8 10 secT −=
142 16
17.8 10
12.8 10f
−=
9.(2) Loss in P.E. = Gain in K.E.
2 21 1
2 2mgh mv Iw= +
Put, (no slipping)v wr=
2
2 2 21 1
2 2 2
mrmgh mw r w= +
2 23 1 4
4 3= =
ghmgh mw r w
r
10.(3) Equal number of magnetic field lines enters the circular region & comes out of infinite plane excluding
circular area. So, magnetic flux are equal in magnitude but opposite in direction.
0i = −
11.(1) For damped oscillation : 0ma bv kx+ + =
2
20
d x dxm b kx
dtdt+ + = …(i)
For LCR series circuit 0di q
iR Ldt C
− − − =
2
2
10
d q dqL R q
dt Cdt+ + = …(ii)
Comparing equation (i) and (ii), 1
, ,L m C R bk
12.(4) 0
1e
L Dm
f f
= +
, if final image is least distance of distinct vision
150 25
375 15 ef
= +
Vidyamandir Classes
VMC | JEE Main-2020 3 Solutions |7th January Morning
750
2.17 cm 21.7 mm 22 mm345
ef = = =
Also, 0 e
L Dm
f f
=
if final image is at infinity.
150 25
3755 ef
=
, 22 mmef =
13.(4) 0 0 ,E B C C= = speed of light in vacuum
8 80 3 10 3 10 V/mE −=
0 9 V/mE = 3 10 ˆ9sin(1.6 10 48 10 ) V/mE x t k= +
14.(1) 2AB CDI I md= +
2 2
12 16
ml ml= +
27
48
ml=
Radius of gyration 27 7
48 48
ABI ll
m= = =
15.(4) 1
0.42.5
I A A= =
1 0.22
II A= =
16.(4) Take origin at 1 kg mass y = vertical, x = horizontal