JAMB UTME - Physics - 2007
Vraag 1 Verslag
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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Vraag 2 Verslag
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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Vraag 3 Verslag
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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Vraag 4 Verslag
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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
Vraag 5 Verslag
[G = 6.67 X 10-11 Nm2 Kg2]
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= 3.71 X 10-8N.
Vraag 6 Verslag
[specific heat capacity of water = 4200 jkg -1k-1]
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Vraag 7 Verslag
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
Vraag 8 Verslag
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Vraag 9 Verslag
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Vraag 10 Verslag
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Vraag 11 Verslag
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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
Vraag 12 Verslag
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5 x 10-5
Vraag 13 Verslag
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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Vraag 14 Verslag
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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
Vraag 15 Verslag
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Vraag 16 Verslag
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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.
Vraag 17 Verslag
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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
Vraag 18 Verslag
I. Mass
II. Density
III. Temperature
IV. Nature of substance
Which of the above affect diffusion?
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Vraag 19 Verslag
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Work done by this force : 60
F cos 60 x X
= 2 x 0.5 x 3
= 3J
Vraag 20 Verslag
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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Vraag 21 Verslag
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Thus, the correct option is p-type and n-type respectively
Vraag 22 Verslag
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Vraag 23 Verslag
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Vraag 24 Verslag
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Vraag 25 Verslag
[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
Vraag 26 Verslag
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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 |
Vraag 27 Verslag
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Vraag 28 Verslag
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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
Vraag 29 Verslag
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
Vraag 30 Verslag
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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Vraag 31 Verslag
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Vraag 32 Verslag
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Vraag 34 Verslag
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Vraag 35 Verslag
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Vraag 36 Verslag
[h = 6.63 x 10-34js, c = 3 x 108 ms-1]
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Vraag 37 Verslag
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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
Vraag 38 Verslag
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Vraag 39 Verslag
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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Vraag 40 Verslag
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Vraag 41 Verslag
From the diagram above, the bob of the pendulum has the fastest speed at
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Vraag 42 Verslag
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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
Vraag 43 Verslag
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= (1500 x 40 x40)/50
= 48000 = 4.8 x 104 N
Vraag 44 Verslag
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Vraag 45 Verslag
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Resistor = 50A
power loss= I2R = 502 x 2
= 2500 x 2 = 5.0 x 103W
Vraag 46 Verslag
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Vraag 47 Verslag
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Vraag 48 Verslag
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
Vraag 49 Verslag
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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
Vraag 50 Verslag
[g = 10ms-2]
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= 850 x 10 x 500
= 4250000
= 4.25 x 106 NM-2
