2001 - JAMB Physics Past Questions and Answers - page 5
41
A
3/8
B
9/11
C
11/9
D
8/3
correct option: b
| I = | _E |
| R+r |
For parallel of 5Ω and 10Ω
1/R = 1/10 + 1/5 = 1+2/10 = 3/10
∴R = 10/3
Total R = 31/3 + 3 x 1 = 71/2
∴I = 6/71/3 = 6/1 x 3/22 = 18/22 = 9/11
∴I = 9/11
42
A
4:1
B
2:1
C
1:4
D
1:2
correct option: d
Energy = 1/2CV2
∴Energy of P = 1/2 x 2 x V2 = V2
Energy of Q = 1/2 x 4 x V2 = 2V2
∴ P/Q = V2/2V2 = 1/2 => P:Q = 1:2
Users' Answers & Comments∴Energy of P = 1/2 x 2 x V2 = V2
Energy of Q = 1/2 x 4 x V2 = 2V2
∴ P/Q = V2/2V2 = 1/2 => P:Q = 1:2
43
A
4V, 6V and 2V
B
2V, 6V and 4V
C
6V, 4V and 2V
D
6V, 2V and 4V
correct option: d
C = Q/V => Q = CV : ∴V = Q/C
∴V1 = Q/C1; V2 = Q/C;
V3 = Q/C3
for series
1/C = 1/C1 + 1/C2 + 1/C3
= 1/2 + 1/6 + 1/3
= 6/6 = 1
∴ C = 1μF = 1.0 x 10-6F
∴ Q = CV = 1.0 x 10-6 x 12
= 12.0 x 10-6C
∴V1 = Q/C1
= 6V
V2 = Q/C2
= 2V
V3 = Q/C3
∴ 6V, 2V and 4V
Users' Answers & Comments∴V1 = Q/C1; V2 = Q/C;
V3 = Q/C3
for series
1/C = 1/C1 + 1/C2 + 1/C3
= 1/2 + 1/6 + 1/3
| = | 3+1+2 |
| _6 |
= 6/6 = 1
∴ C = 1μF = 1.0 x 10-6F
∴ Q = CV = 1.0 x 10-6 x 12
= 12.0 x 10-6C
∴V1 = Q/C1
| = | 12.0x10-6 |
| 2 x 10-6 |
= 6V
V2 = Q/C2
| = | 12.0x10-6 |
| 6 x 10-6 |
= 2V
V3 = Q/C3
| = | 12.0x10-6 |
| 3 x 10-6 |
∴ 6V, 2V and 4V
44
A
0.5 E
B
2.0 E
C
0.2 E
D
0.0 E
correct option: d
0.0 E since one of the electric fields points in the opposite direction (left)
Users' Answers & Comments45
A
8Ω
B
2Ω
C
10Ω
D
4Ω
correct option: c
for a parallel R and X, when R = 5Ω,
and when R = 2Ω
Thus since E = I(R+r), and if r = 0 then E = IR
∴(i) E = 3(5X/(5+x)); (ii) E = 6(2X(2+x))
∴3(5X/(5+X)) = 6(2X/(2+X))
=> (5X/(5+X)) = 2(2x/(2+X))
∴ (5X/(5+X)) = (4X/(2+X))
5x(2+x) = 4x(5+x)
10x + 5x2 = 20x + 4x2
5x2 - 4x2 = 20x - 10x
=>x2 = 10x
∴x2 - 10x = 0
x(x - 10) = 0
∴ x = 0 or 10
∴ x = 10Ω
Users' Answers & Comments| effective resistance R = | 5X |
| 5+X |
and when R = 2Ω
| effetive resistance R = | 2X |
| 2+X |
Thus since E = I(R+r), and if r = 0 then E = IR
∴(i) E = 3(5X/(5+x)); (ii) E = 6(2X(2+x))
∴3(5X/(5+X)) = 6(2X/(2+X))
=> (5X/(5+X)) = 2(2x/(2+X))
∴ (5X/(5+X)) = (4X/(2+X))
5x(2+x) = 4x(5+x)
10x + 5x2 = 20x + 4x2
5x2 - 4x2 = 20x - 10x
=>x2 = 10x
∴x2 - 10x = 0
x(x - 10) = 0
∴ x = 0 or 10
∴ x = 10Ω
46
A
31A
B
48A
C
60A
D
80A
correct option: b
Ir.m.s = Vr.m.s/Z
But Z =
√ R2 + XL2
=
√ 42 + 32
=
√ 16 + 9
=
√ 25
∴Ir.m.s = 240/5
= 48.0A
Users' Answers & CommentsBut Z =
√ R2 + XL2
=
√ 42 + 32
=
√ 16 + 9
=
√ 25
∴Ir.m.s = 240/5
= 48.0A
47
Find the frequencies of the first three harmonics of a piano string of length 1.5m, if the velocity of the wave on the string is 120ms-1
A
40Hz, 80Hz, 120Hz
B
80Hz, 160Hz, 240Hz
C
180Hz, 360Hz, 540Hz
D
369Hz, 180Hz, 90Hz
correct option: a
in vibrations in string, the fundamental frequency happens to be the first harmonic, and is given as
f0 = V/2l.
Where V = Velocity of the sound wave, I = Length of the string.
The second harmonic is the first overtone given by
f1 = V/l = 2f0 ;
and the third harmonic happens to be the second overtone, given by
f2 = 3V/2l = 2f0.
Thus if V = 120m/s and l = 1.5m
Thus f1 = 2f0 = 2 x 40 = 80Hz
Thus f2 = 3f0 = 3 x 40 = 120Hz
Thus we have in order, 40Hz, 80Hz, 120Hz
Users' Answers & Commentsf0 = V/2l.
Where V = Velocity of the sound wave, I = Length of the string.
The second harmonic is the first overtone given by
f1 = V/l = 2f0 ;
and the third harmonic happens to be the second overtone, given by
f2 = 3V/2l = 2f0.
Thus if V = 120m/s and l = 1.5m
Thus f1 = 2f0 = 2 x 40 = 80Hz
Thus f2 = 3f0 = 3 x 40 = 120Hz
Thus we have in order, 40Hz, 80Hz, 120Hz
48
A bread toaster uses a current of 4A when plugged in a 24oV line. it takes 1 minute to toast slices of bread. what is the energy consumed by the toaster
A
5.76 x 104J
B
1.60 x 104J
C
3.60 x 103J
D
1.60 x 102J
correct option: a
Electrical Energy = IVt
= 4 x 240 x (1 x 60)
= 57600J
= 5.76 x 104J
Users' Answers & Comments= 4 x 240 x (1 x 60)
= 57600J
= 5.76 x 104J
49
What is the angle of dip at the magnetic equator
A
\( 45^{\circ} \)
B
\( 0^{\circ} \)
C
\( 90^{\circ} \)
D
\( 180^{\circ} \)
correct option: b
The angle of dip dip is the angle which the resultant earth's magnetic field makes with the horizontal.
At the magnetic equator, which is entirely along the horizontal this angle is zero
Users' Answers & CommentsAt the magnetic equator, which is entirely along the horizontal this angle is zero