JAMB UTME - Physics - 2007
Pregunta 1 Informe
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Pregunta 2 Informe
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Pregunta 3 Informe
[c = 3.0 x 8 ms-1]
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(1.5)/1 = (3.0 x 108)/V
V = (3.0 x 108)/1.5
V = 2.0 x 108 ms-1
But x = Vt
t = (x)/Vt
t = (x)/v = (9 x 10)4/2.0x 108 = 4.5 x 10-2S
∴ Time = 4.5 x 10-12S
Pregunta 4 Informe
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Pregunta 5 Informe
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= (1500 x 40 x40)/50
= 48000 = 4.8 x 104 N
Pregunta 6 Informe
I. Increase the melting point of the liquid.
II. Increase the boiling point of the liquid.
III. Decrease the melting point of the liquid.
IV. Decrease the boiling point of the liquid.
Which of the statement above about the effect on increase in pressure in a liquid are correct?
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Pregunta 7 Informe
[G = 6.67 X 10-11 Nm2 Kg2]
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= 3.71 X 10-8N.
Pregunta 8 Informe
I. They should be identical
II. They should originate from the same source.
III. They should be coherent.
IV. They should be monochromatic.
From the statements above, the conditions for two waves to interfere are?
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Pregunta 9 Informe
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Pregunta 10 Informe
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Pregunta 11 Informe
The figure above shows 4 forces 3N, 10N, 3√3N and 6N acting on a particle P. The resultant of he four forces is
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Total vertical upward force:
= 3√3 + 10 cos 60o
= 3√3 + 5.
Total downward force
= 6 cos 30o
= 6 x √(3/2) = 3√3
Total horizontal forces towards right;
= 3N + cos 30o
= 3N + 10 x √(3/2)
= 3N + 5√3
Total horizontal forces towards left:
= 6 cos 60o
= 6 x 0.5
= 3.0N
Summary:
Vertical resultant forces upward:
= 3√3 + 5 - 3√3
= 5N
Horizontal resultant forces towards right:
= 3N + 5√3 - 3√3
= 5N
Horizontal resultant force towards right
= 3N + 5√3 - 3N
= 5√3N
∴ R2 = 52 + (5√3)2
= 25 + 25 x 3
= 100
∴ R = √100
= 10N
Pregunta 12 Informe
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Pregunta 13 Informe
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Pregunta 14 Informe
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R = (V2)/60 = (120 X 120)/60 = 240 Ω, the fuse wire that should ensure current does not exceed this value should be rated 200 Ω
Pregunta 15 Informe
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Work done by this force : 60
F cos 60 x X
= 2 x 0.5 x 3
= 3J
Pregunta 16 Informe
[g = 10ms-2]
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= 850 x 10 x 500
= 4250000
= 4.25 x 106 NM-2
Pregunta 17 Informe
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And for incline plane
V.R =1Sinθ=V.R=1Sin30=10.5=2therefore 60120=M.A2=M.A=120100=1.2but M.A =loadeffort=120Etherefore120E=1.2=E=1201.2=100N
Therefore Effort up the plane = 100N
Pregunta 18 Informe
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Resistor = 50A
power loss= I2R = 502 x 2
= 2500 x 2 = 5.0 x 103W
Pregunta 19 Informe
I. Mass
II. Density
III. Temperature
IV. Nature of substance
Which of the above affect diffusion?
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Pregunta 20 Informe
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∴ 40,000 = 1 x 800
1 = (40.000)/800 = 50A. tune the current through resistor = 50A.
∴ power loss = 12R = 502 X 2 = 2500 X 2 = 5.0 X 103W
Pregunta 21 Informe
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Pregunta 22 Informe
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=> area expansivity β= 2∝.
And cubic expansivity δ = 3∝.
=> 6.3 x 10-6 = 3∝
∴ ∝ =
| ∴ ∝ = | 6.3 x 10-6 | = 2.1x 10-6 |
| __3 |
∴ Area = 2∝ = 2 x 2.1x10-6
= 4.2 x 10-6K-1
Pregunta 23 Informe
A gramophone record takes 5s to reach its constant angular velocity of \(4\pi\ \text{rads}^{-1}\) from rest. Find its constant angular acceleration.
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Pregunta 24 Informe
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Pregunta 25 Informe
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Thus, the correct option is p-type and n-type respectively
Pregunta 26 Informe
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5 x 10-5
Pregunta 27 Informe
During the nuclear reaction described by
| 235 | W → | 235 | X→ | 231 | Y |
| 92 | 93 | 91 |
The particles emitted are respectively
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Thus , the emission is β and α particles respectively
Pregunta 28 Informe
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Pregunta 29 Informe
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Pregunta 30 Informe
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Pregunta 31 Informe
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Pregunta 32 Informe
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Pregunta 33 Informe
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∴ P = (0.1 x 3.14 x 10-6)/2 x 4
= 0.3925 x 10-7 OR -3.930 X 10-8 Ωm
Pregunta 34 Informe
[h = 6.63 x 10-34js, c = 3 x 108 ms-1]
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Pregunta 35 Informe
In the diagram above , if the magnetic field points into the paper and the force on a current - carrying conductor points upwards, what is the direction of the current?
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Pregunta 36 Informe
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When heated to the same temperature range, the ratio of their increase in length e1/e2 = 1:2
But the increase in length of
1 = e1 = ∝1l1Δθ and increase i length of
2 = e2 = ∝2l2Δθ
| => | e1 | = | ∝1l1Δθ | = | ∝1l1 | = | 1 |
| e2 | ∝2l2Δθ | ∝2l2 | 2 |
| But | ∝1 | = | 3 | , ∴ | 3 | x | l1 | = | 1 |
| ∝2 | 4 | 4 | l2 | 2 |
=> 6l1 = 4l2
| ∴ | l1 | = | 4 | = | 2 | OR 2 : 3 |
| l2 | 6 | 3 |
Pregunta 37 Informe
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Pregunta 38 Informe
118.8cm2 surface of the copper cathode of a voltameter is to be coated with 10-6m thick copper of density 9 x 103kgm-3. How long will the process run with 10A constant current?
[3.3 x 10-7kgC-1]
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Pregunta 39 Informe
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Pregunta 40 Informe
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Pregunta 41 Informe
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Using E = Fq .
So we have F = Q × E
F = 50 × 10−6 × 360 = 0.018N
When E2 = F/q
0.018120×10−6 = 150NC−1
Pregunta 42 Informe
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Where P = Pressure
and V = Volume
By definition Pressure = FA=FM×M
Again Workdone = force x displacement
thereforePV=FA×V=FM2×M31PV=Fx×x×x×x×x1=F×x
= force x displacement
Pregunta 43 Informe
From the diagram above, the inductive reactance and the resistance R are respectively
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Xl = 2πfl
= 2 x 50/π x 0.1
= 10Ω
Again from ohms law,
R = V/I = 75/1.5 = 50Ω
∴ Thus the inductive reactance and the resistance are respectively 10Ω, and 50Ω
Pregunta 44 Informe
A 120V, 60W lamp is to be operated on 220V ac supply mains. calculate the value of non inductive resistance that would be required to ensure that the lamp is run on correct value
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Pregunta 45 Informe
From the diagram above, the bob of the pendulum has the fastest speed at
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Pregunta 46 Informe
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wt. in the liquid = 200N
loss in wt. in the liquid = 250 - 200 = 50N
∴ Wt. the liquid of equal vol. of the iron = 50N
∴If the density of the liquid is 1000 lgm-3
then the volume of the liquid = (N)/D = V = (25kg)/1000 =0.025m3 = 2.5 x 10-2 m8.
Pregunta 47 Informe
.
Three cells each of e.m.f 1.5 V and internal resistance 2.5Ω are connected as shown in the diagram above. Find the net e.m.f and the internal resistance
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However the effective internal resistance should be added as parallel resistance.
=>1/v + 1/25 + 1/25 + 1/25 = 1.2
∴ v = 0.83Ω
∴ Effective e.m.f and internal resistance are respectively = 1.5V and 0.83Ω
Pregunta 48 Informe
I. Use a liquid with a high melting point.
II. use a liquid of high volume expansivity.
III. use a capillary tube of large diameter.
Which of the above best describes how the sensitivity of a liquid-in-glass thermometer can be enhanced?
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Pregunta 49 Informe
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Pregunta 50 Informe
[specific heat capacity of water = 4200 jkg -1k-1]
