Final JEE - Main Exam September, 2020/02-09-2020/Evening Session
FINAL JEE–MAIN EXAMINATION – SEPTEMBER, 2020 (Held On Wednesday 02nd SEPTEMBER, 2020)
TEST PAPER WITH ANSWER & SOLUTION
PHYSICS 1.
The figure shows a region of length 'l' with a
2.
The displacement time graph of a particle
uniform magnetic field of 0.3 T in it and a
executing S.H.M. is given in figure :
proton entering the region with velocity 4 × 105 ms–1
(sketch is schematic and not to scale) displacement
making an angle 60° with the field. If the proton completes 10 revolution by the time it cross the
O
region shown, 'l' is close to (mass of proton = 1.67 × 10 –27 kg, charge of the proton = 1.6 × 10–19 C)
l
(1) 0.11 m
(2) 0.22 m
(3) 0.44 m
(4) 0.88 m
Official Ans. by NTA (3)
total time t = 10 T
V COS 60° =
V 2
l
V t 2
l=
V 2pm 10 ´ 2 qB
= 4 ´ 10 5 ´ 10 ´
= 0.439
(4) (A), (B) and (C)
at
3T displacement zero (x = 0), so a = 0 4
F=0 (B) at t = T displacement (x) = A x maximum, So acceleration is maximum. (C) V = w A 2 – x 2 Vmax at x = 0 Vmax = Aw
Kinematics
l=
(3) (A) and (D)
Official Ans. by NTA (4) Sol. (A) F = ma a = –w2x
A 60°
time (s)
(2) (B), (C) and (D)
LL
2 pm T= qB
3T T 5T 4 4
T 4
EN
60°
2T 4
Which of the following statements is/are true for this motion ? 3T (A) The force is zero t = 4 (B) The acceleration is maximum at t = T T (C) The speed is maximum at t = 4 (D) The P.E. is equal to K.E. of the oscillation T at t = 2 (1) (A), (B) and (D)
B
Sol.
TIME : 3 PM to 6 PM
at t =
T , x = 0, So Vmax. 4
(D) KE = PE 3.14 ´ 1.67 ´ 10 -27 1.6 ´ 10 -19 ´ 0.3
\ at x = at t =
A . 2
T x = -A 2
(So not possible) 1