SOLUTIONS & ANSWERS FOR JEE MAINS-2021 27th July Shift 2

[Pages:47]SOLUTIONS & ANSWERS FOR JEE MAINS-2021 27th July Shift 2

[PHYSICS, CHEMISTRY & MATHEMATICS]

PART ? A ? PHYSICS

Section A

Ans: at x < x1, KE is smallest and particle is moving at the slowest speed

Sol:

Emech = 8 J At x < x1, U = 8 J

KE = Emech ? U = 8 ? 8 = 0 J (particle is at rest)

at x = x3, U = 4 J KE = Emech ? U = 8 ? 4 = 4 J

at x = x3, U = 0 KE = Emech ? U = 8 J

at x > x1, U = 6 J KE = Emech ? U = 8 ? 6 = 2 J

Ans: 0.003C-1

Sol: RT = R0 [1 + (T ? T0)] 16 = R0 [1 + (15 ? T0)] ---------- (1) 20 = R0 [1 + (100 ? T0)] --------- (2) T0 reference temperature assuming T0 = 0C Using equations (1) and (2) 16 = 1+ 15 20 1+ 100

16 + ?100 ?16 = 20 + ? 15 ? 20 Solving, = 0.003C-1

Ans: 70.3 H

Sol: C = 0.1 F f0 = 60 Hz f0 = 1 2 LC

L

=

1 42f02C

L=

1 42 602 0.110-6

= 70.3 H

Ans: 0.62 J

Sol: T = 2 m K

0.2 = 2 0.5 K

K = 50 2 500 x = A sin (t + )

= 50 cm sin T + 0 4

= 50 cm sin 2

= 5 cm

= 2

T

PE = 1 Kx2 2

( ) = 1 (500) 5 10-2 2 2 = 0.0255

Ans: 6.00 1023 kg

Sol: T = 2 r3 GM

M

=

42r3 GT2

( ( ) ) M

=

6 1011 9 103 103 729 106 2

3

7 hrs, 30 min = 729 ? 106 = 6 ? 1023

Ans: Sol: Basic concept

Ans: 1.41 eV

Sol: Initial energy of electron = +3 eV

In 2nd excited state, energy of electron =

-13.6 eV 32

=-1.51 eV

Loss in energy is emitted as photon, photon energy = hc = 4.51eV

Photoelectric equation,

hc

= + KEmax

4.51 eV =

hc 0

+ KEmax

0

KEmax =

4.51

eV

-

12400 eV A

0

4000 A

= 1.41 eV

Ans: 4F0T / 3 M

Sol:

Given F = F0

1 +

t

-

T

2

T

Acceleration a = F = F0 - F0 t - T 2 M M M T

a = dv dt

dw

=

F0

- F0

t - T 2

dt M M T

v dv

0

= 2T F0 t=0 M

- F0 M

(t

- T)2

T2

dt

=

F0 M

t02T

-

F0 MT2

t3 3

+ T2t -

2t2 2

2T T

0

=

2F0T M

-

F0 MT2

8T3 3

+ 2T3

-

4T3

=

2F0 T M

-

F0T M

8 3

-

2

=

2F0T M

-

F0T M

2 3

= 4F0T 3M

Ans: (a)-(iii), (b)-(ii), (c)-(iv), (d)-(i)

Sol:

(a) Capacitance M-1 L-2 T4 A2 (iii)

(b) Permittivity of free space, 0 M-1L-3T4A2 (ii) (c) Permeability of free space, 0 M1L1T-2A-2 (iv) (d) Electric field, E M1L1T-3A-1 (i)

( ) Ans:

1 40

q (22 )

2

2 -1

Sol: A (-q)

C 2q

(+q)

D

E1

(+q) B

2q E

E2

E'1

O

E3

F

2q G

(-q)

H

Let E1 is the resultant field at point O due to +q at D and 2q at E

E1 =

K2q Kq 22 - 22

=K

q 22

E2 is the resultant filed at point O due to 2q at G and q at B

E2

=

K2q 2

-

Kq 2

=

Kq 2

=

E3

Where E is the resultant field at point O due to 2q at C and q at F Let E1' be the resultant of E2 and E3

E1' =

Kq 2

2

+

Kq 2

2

=

2

Kq 2

E1 and E1' are opposite to each other

E" =resultant of E1 and E1'

( ) 2

Kq 2

-

Kq 22

Kq = 22

2

2 -1

Field at point O due to ?q at A and ?q at H get cancelled

( ) Net

field

=

E1"

=

Kq 22

2

2 -1 K = 1 40

Ans:

2 3

T1

+

1 3

T3

Sol:

WA

= 1- Q2 Q1

= 1- T T1

Q2 Q1

=

T T1

WB

= 1- Q3 Q2

= 1-

T3 T

2Q3 Q2

=

T3 T

2

WA = WB

Q1 ? Q2 =

Q2 2

- Q3

2Q1 + 2Q3 = 3 Q2 Q2

2T1 + T3 = 3 TT

2T1 + T3 = T 33

T1

Q1

A

WA

T

Q2

B

WB

Q3 T3

1

Ans:

8P

23

t2

9m

Sol: Power, P = constant P = FV = (ma) V

P = m dv c dt

t

v

P dt = vdv m

0

0

P t = v2 m2

1

1

v2 =

2P t v = 2P t

2 = 2P

21

t2

m

m m

1

dx

= 2P

21

t2

dt m

x

dx

=

2P

1

2

1 21

t 2dt

0

m 0

1

x = 2P 2 m

3

t2 3

2

1

= 2P 2 m

2

3

t2

3

x

=

8P

1 2

3

t2

9m

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