AIEEE Mock Test 6

AIEEE Mock Test 6
Question:-
Option (A)
2
Option (B)
9
Option (C)
3
Option (D)
11
Correct Option:
D
Question Solution:
Let P(n) = 2.42n+1 + 33n+1
P(1) = 128+81=209 (divisible by 11 only)
Question:-
Option (A)
2.56 H
Option (B)
0.32 H
Option (C)
0.64 H
Option (D)
1.28 H
Correct Option:
D
Question Solution:
Rbilb = (V2/P) = (60x60)/10 = 360 ohm
I = (60 / 10) = (1/6) A
on 100 A.C. Line, (1/6) = (100/Z)
Z = 600 ohm
L2w2 + R2 = (600)2
or L = 1.28H
Question:-
Option (A)
at the pole
Option (B)
at equator
Option (C)
remains stationary
Option (D)
no emf is induced at all
Correct Option:
A
Question Solution:
The car cuts the vertical component of earth’s magnetic field at equator V =0
\ no emf is induced but at the poles as V is maximum, therefore maximum emf is induced.
Question:-
Option (A)
0.1/e6
Option (B)
0.3/e6
Option (C)
0.5/e6
Option (D)
0.7/e8
Correct Option:
B
Question Solution:
I = (dq/dt) = (d/dt) [0.1e-3t]
= 0.1 x (3e-3t) = 0.3e-3t
Current in the wire at t = 2 sec
(I) = 0.3e3x2 = 0.3/e6
Question:-
Option (A)
mg/pr2B
Option (B)
mg/prB
Option (C)
mg/2prB
Option (D)
peB/mg
Correct Option:
C
Question Solution:
The normal to the plane of coil makes 90° angle with the field direction. Therefore the torque on the coil
t = BiA = Bipr2
Since only one edge lifts from the table the torque required is = (mg/2)r
\ Bipr2 = (mg/2)r Þ i = (mg / 2peB)
Question:-
Option (A)
75 mH
Option (B)
74 mH
Option (C)
76 mH
Option (D)
77 mH
Correct Option:
A
Question Solution:
108 = K(600)2, L = K(500)2
(L/108) = (25/36) Þ L = ((108 x 25)/36)
L = 75 mH