JAMB UTME - Physics - 2007
Tambaya 1 Rahoto
Bayanin Amsa
Tambaya 2 Rahoto
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?
Bayanin Amsa
Tambaya 3 Rahoto
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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Tambaya 4 Rahoto
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Tambaya 5 Rahoto
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Tambaya 6 Rahoto
Bayanin Amsa
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 |
Tambaya 7 Rahoto
Bayanin Amsa
Thus, the correct option is p-type and n-type respectively
Tambaya 8 Rahoto
[h = 6.63 x 10-34js, c = 3 x 108 ms-1]
Bayanin Amsa
Tambaya 9 Rahoto
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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Tambaya 10 Rahoto
Bayanin Amsa
∴ 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
Tambaya 11 Rahoto
Bayanin Amsa
Tambaya 13 Rahoto
Bayanin Amsa
Tambaya 14 Rahoto
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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Tambaya 15 Rahoto
Bayanin Amsa
Tambaya 16 Rahoto
Bayanin Amsa
Resistor = 50A
power loss= I2R = 502 x 2
= 2500 x 2 = 5.0 x 103W
Tambaya 18 Rahoto
Bayanin Amsa
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
Tambaya 19 Rahoto
Bayanin Amsa
5 x 10-5
Tambaya 20 Rahoto
.
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
Bayanin Amsa
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
Tambaya 21 Rahoto
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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Tambaya 22 Rahoto
Bayanin Amsa
Tambaya 23 Rahoto
Bayanin Amsa
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
Tambaya 24 Rahoto
Bayanin Amsa
= (1500 x 40 x40)/50
= 48000 = 4.8 x 104 N
Tambaya 25 Rahoto
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
Bayanin Amsa
Tambaya 26 Rahoto
[c = 3.0 x 8 ms-1]
Bayanin Amsa
(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
Tambaya 27 Rahoto
The figure above shows 4 forces 3N, 10N, 3√3N and 6N acting on a particle P. The resultant of he four forces is
Bayanin Amsa
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
Tambaya 28 Rahoto
From the diagram above, the bob of the pendulum has the fastest speed at
Bayanin Amsa
Tambaya 29 Rahoto
Bayanin Amsa
Tambaya 30 Rahoto
Bayanin Amsa
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
Tambaya 31 Rahoto
Bayanin Amsa
Tambaya 32 Rahoto
Bayanin Amsa
=> 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
Tambaya 33 Rahoto
I. Mass
II. Density
III. Temperature
IV. Nature of substance
Which of the above affect diffusion?
Bayanin Amsa
Tambaya 34 Rahoto
Bayanin Amsa
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.
Tambaya 35 Rahoto
[g = 10ms-2]
Bayanin Amsa
= 850 x 10 x 500
= 4250000
= 4.25 x 106 NM-2
Tambaya 36 Rahoto
Bayanin Amsa
Tambaya 37 Rahoto
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Tambaya 38 Rahoto
Bayanin Amsa
Tambaya 39 Rahoto
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Tambaya 40 Rahoto
During the nuclear reaction described by
| 235 | W → | 235 | X→ | 231 | Y |
| 92 | 93 | 91 |
The particles emitted are respectively
Bayanin Amsa
Thus , the emission is β and α particles respectively
Tambaya 41 Rahoto
Bayanin Amsa
Tambaya 42 Rahoto
Bayanin Amsa
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
Tambaya 43 Rahoto
From the diagram above, the inductive reactance and the resistance R are respectively
Bayanin Amsa
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
Tambaya 44 Rahoto
Bayanin Amsa
Tambaya 45 Rahoto
Bayanin Amsa
Work done by this force : 60
F cos 60 x X
= 2 x 0.5 x 3
= 3J
Tambaya 46 Rahoto
Bayanin Amsa
Tambaya 47 Rahoto
[G = 6.67 X 10-11 Nm2 Kg2]
Bayanin Amsa
= 3.71 X 10-8N.
Tambaya 48 Rahoto
A gramophone record takes 5s to reach its constant angular velocity of \(4\pi\ \text{rads}^{-1}\) from rest. Find its constant angular acceleration.
Bayanin Amsa
Tambaya 49 Rahoto
Bayanin Amsa
Tambaya 50 Rahoto
[specific heat capacity of water = 4200 jkg -1k-1]
