2005 - JAMB Physics Past Questions and Answers - page 5
41
Two liquids L1 and L2 are contained in a U - tube. The height and the density of L1 are 8 cm and 103 kgm-3 respectively. If the density of L2 is 800 kgm-3, its height measured from the same level is
A
8 cm
B
10 cm
C
16 cm
D
12 cm
correct option: b
Generally, from Hares Experiment, density is inversely proportional to height
i.e h ∝1/d; → h1d1 = h2d2
∴ 8 x 103 = h2 x 800
= 10cm
Users' Answers & Commentsi.e h ∝1/d; → h1d1 = h2d2
∴ 8 x 103 = h2 x 800
| ∴ h2 = | 8 x 103 |
| _800 |
= 10cm
42
106J of heat is required to boil off completely 2kg of a certain liquid. Neglecting heat lost to the surroundings, the latent heat of vaporization of the liquid is
A
2.0 x 106J kg-1
B
5.0 x 106J kg-1
C
2.0 x 105J kg-1
D
5.0 x 105J kg-1
correct option: d
From the heat exchange involved, the latent heat of vaporization, L can be deduced from the relation:
ML = 106, where M = mass of the liquid.
∴ 2 x L = 106
=> L = 106/2 = 5.0 x 105J kg-1
Users' Answers & CommentsML = 106, where M = mass of the liquid.
∴ 2 x L = 106
=> L = 106/2 = 5.0 x 105J kg-1
43
A
4V
B
12V
C
2V
D
6V
correct option: c
The effective capacitance in the circuit is given by:
1/C = 1/2 + 1/3 + 1/6 = 6/6
therefore C = 1μF or
1.0 x 10-6F.
Since C = Q/V, its implies that the charges Q, flowing in the circuit is given by Q = CV
= 1.0 x 10-6 x 12
= 12.0 x 10-6 Coulomb.
∴ P.d across the 6μF,
= 2.0 volts
Users' Answers & Comments1/C = 1/2 + 1/3 + 1/6 = 6/6
therefore C = 1μF or
1.0 x 10-6F.
Since C = Q/V, its implies that the charges Q, flowing in the circuit is given by Q = CV
= 1.0 x 10-6 x 12
= 12.0 x 10-6 Coulomb.
∴ P.d across the 6μF,
| V = Q/C = | 12.0 x 10-6 |
| 6.0 x 10-6F |
= 2.0 volts
44
If the distance between two point charges its increased by a factor of four, the magnitude of the electrostatic force between them will be
A
1/4 of its former value
B
1/16 of its former value
C
1/2 of its former value
D
4 times its former value
correct option: b
For the two point charges, the magnitude of the electrostatic force between them is given by the equation:
F ∝ 1/r2, and if the distance is increased by a factor of 4, that is new distance = 4r,
then, new force F ∝ 1/(4r)2
= ∝ 1/16r2
Thus it will be 1/16 of the former value
Users' Answers & CommentsF ∝ 1/r2, and if the distance is increased by a factor of 4, that is new distance = 4r,
then, new force F ∝ 1/(4r)2
= ∝ 1/16r2
Thus it will be 1/16 of the former value
45
The velocity of sound in air will be doubled if it's absolute temperature is
A
Quadrupled
B
Constant
C
Halved
D
Doulbed
correct option: a
In general, the velocity of sound in air varies directly as the square root of temperature measured in kelvin.
That V \( \propto \sqrt{T} \implies V^2 \propto T. \\
\text{Therefore} \frac{V^2_1}{T_1} = \frac{V^2_2}{T_2} \\
\text{Thus Let } V_1 = 4m/s \\ T_1 = 10K \\ \text{Therefore } V_2 = 2V_1 = 8m/s \\
\implies \frac{4^2}{10} = \frac{8^2}{T_2} \\
T_2 = \frac{64 \times 10}{16} = 40K\\
T_2 = 4T_1 \)
Thus when the velocity of sound in air is doubled, it's absolute temperature will be quadrupled.
Users' Answers & CommentsThat V \( \propto \sqrt{T} \implies V^2 \propto T. \\
\text{Therefore} \frac{V^2_1}{T_1} = \frac{V^2_2}{T_2} \\
\text{Thus Let } V_1 = 4m/s \\ T_1 = 10K \\ \text{Therefore } V_2 = 2V_1 = 8m/s \\
\implies \frac{4^2}{10} = \frac{8^2}{T_2} \\
T_2 = \frac{64 \times 10}{16} = 40K\\
T_2 = 4T_1 \)
Thus when the velocity of sound in air is doubled, it's absolute temperature will be quadrupled.
46
A force of 200N acts between two objects at a certain distance apart. The value of the force when the distance is halved is
A
400N
B
200N
C
100N
D
800N
correct option: d
let the masses of the two objects be M1 and M2
and their distance apart = r
Therefore the force acting F1 = \( \frac{GM1M2}{r^2} \)
for a new distance of r/2, we have a force of
F2 = \( \frac{GM2M2}{r/2} \\
\text{Thus } F1 \times r^2 = GM1M1 \text{and} \\ F2 \times (r/2)^2 = GM1M1 \\
\text{Since } GM1M1 = GM1M1 \\
\implies F1 \times r^2 = F2 \times (r/2)^2 \\
\text{Therefore } 200 \times r^2 = F2 \times r^2/4 \\
F2 = \frac{4 \times 200 \times r^2}{r^2} = 800N \)
Users' Answers & Commentsand their distance apart = r
Therefore the force acting F1 = \( \frac{GM1M2}{r^2} \)
for a new distance of r/2, we have a force of
F2 = \( \frac{GM2M2}{r/2} \\
\text{Thus } F1 \times r^2 = GM1M1 \text{and} \\ F2 \times (r/2)^2 = GM1M1 \\
\text{Since } GM1M1 = GM1M1 \\
\implies F1 \times r^2 = F2 \times (r/2)^2 \\
\text{Therefore } 200 \times r^2 = F2 \times r^2/4 \\
F2 = \frac{4 \times 200 \times r^2}{r^2} = 800N \)
47
If two inductors of inductances 3H and 6H are arranged in series, the total inductance is
A
9.0H
B
18.0H
C
0.5H
D
2.0H
48
Am elastic material has a length of 36cm when a load of 40N is hung on it and a length of 45cm when a load of 60N is hung on it. The original length of the string is
A
12cm
B
20cm
C
18cm
D
15cm
correct option: c
For Hooke's law F = Ke
\( \implies F/e = K \\
f1/e1=f2/e2 \\
\text{Let the original length} = t_0 \\
\text{therefore} e1 = (36 - t_0)cm ; e2 = (46 - t_0)cm \\ \text{if f1} = 40N \text{f2} = 60N \\ \text{Then } \frac{40}{36 - t_0} = \frac{60}{45 - t_0} \\ \implies 40(45 - t_0) = 60(36 - t_0) \\ \text{therefore } 1800 - 40t_0 \\
2160 - 60t_0 \\ 60t_0 - 40t_0 = 2160 - 1800 \\ 20t_0 = 360 \
t_0 = 18cm \)
Users' Answers & Comments\( \implies F/e = K \\
f1/e1=f2/e2 \\
\text{Let the original length} = t_0 \\
\text{therefore} e1 = (36 - t_0)cm ; e2 = (46 - t_0)cm \\ \text{if f1} = 40N \text{f2} = 60N \\ \text{Then } \frac{40}{36 - t_0} = \frac{60}{45 - t_0} \\ \implies 40(45 - t_0) = 60(36 - t_0) \\ \text{therefore } 1800 - 40t_0 \\
2160 - 60t_0 \\ 60t_0 - 40t_0 = 2160 - 1800 \\ 20t_0 = 360 \
t_0 = 18cm \)