JAMB UTME - Physics - 2007
Ajụjụ 1 Ripọtì
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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
Ajụjụ 2 Ripọtì
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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Ajụjụ 3 Ripọtì
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Ajụjụ 4 Ripọtì
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Ajụjụ 5 Ripọtì
[h = 6.63 x 10-34js, c = 3 x 108 ms-1]
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Ajụjụ 6 Ripọtì
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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Ajụjụ 7 Ripọtì
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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
Ajụjụ 9 Ripọtì
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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.
Ajụjụ 10 Ripọtì
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Ajụjụ 11 Ripọtì
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Resistor = 50A
power loss= I2R = 502 x 2
= 2500 x 2 = 5.0 x 103W
Ajụjụ 12 Ripọtì
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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 |
Ajụjụ 13 Ripọtì
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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Ajụjụ 14 Ripọtì
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5 x 10-5
Ajụjụ 16 Ripọtì
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
Ajụjụ 17 Ripọtì
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
Ajụjụ 18 Ripọtì
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Ajụjụ 19 Ripọtì
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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
Ajụjụ 20 Ripọtì
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Thus, the correct option is p-type and n-type respectively
Ajụjụ 21 Ripọtì
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
Ajụjụ 22 Ripọtì
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Ajụjụ 23 Ripọtì
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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Ajụjụ 24 Ripọtì
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Ajụjụ 25 Ripọtì
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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Ajụjụ 26 Ripọtì
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= (1500 x 40 x40)/50
= 48000 = 4.8 x 104 N
Ajụjụ 27 Ripọtì
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Ajụjụ 28 Ripọtì
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Ajụjụ 29 Ripọtì
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Work done by this force : 60
F cos 60 x X
= 2 x 0.5 x 3
= 3J
Ajụjụ 30 Ripọtì
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Ajụjụ 31 Ripọtì
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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
Ajụjụ 32 Ripọtì
[specific heat capacity of water = 4200 jkg -1k-1]
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Ajụjụ 33 Ripọtì
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Ajụjụ 34 Ripọtì
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Ajụjụ 35 Ripọtì
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Ajụjụ 36 Ripọtì
From the diagram above, the bob of the pendulum has the fastest speed at
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Ajụjụ 37 Ripọtì
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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
Ajụjụ 38 Ripọtì
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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
Ajụjụ 39 Ripọtì
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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
Ajụjụ 40 Ripọtì
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Ajụjụ 41 Ripọtì
[g = 10ms-2]
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= 850 x 10 x 500
= 4250000
= 4.25 x 106 NM-2
Ajụjụ 42 Ripọtì
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Ajụjụ 43 Ripọtì
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Ajụjụ 44 Ripọtì
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Ajụjụ 45 Ripọtì
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Ajụjụ 46 Ripọtì
I. Mass
II. Density
III. Temperature
IV. Nature of substance
Which of the above affect diffusion?
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Ajụjụ 47 Ripọtì
[G = 6.67 X 10-11 Nm2 Kg2]
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= 3.71 X 10-8N.
Ajụjụ 48 Ripọtì
[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
Ajụjụ 49 Ripọtì
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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Ajụjụ 50 Ripọtì
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
