JAMB UTME - Physics - 2007
Question 1 Report
A gramophone record takes 5s to reach its constant angular velocity of \(4\pi\ \text{rads}^{-1}\) from rest. Find its constant angular acceleration.
Answer Details
Question 2 Report
Answer Details
Question 3 Report
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?
Answer Details
Question 4 Report
Answer Details
Question 5 Report
Answer Details
Question 6 Report
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]
Answer Details
Question 7 Report
Answer Details
Resistor = 50A
power loss= I2R = 502 x 2
= 2500 x 2 = 5.0 x 103W
Question 8 Report
Answer Details
5 x 10-5
Question 9 Report
[c = 3.0 x 8 ms-1]
Answer Details
(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
Question 10 Report
Answer Details
∴ P = (0.1 x 3.14 x 10-6)/2 x 4
= 0.3925 x 10-7 OR -3.930 X 10-8 Ωm
Question 11 Report
Answer Details
Question 13 Report
Answer Details
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 |
Question 14 Report
From the diagram above, the bob of the pendulum has the fastest speed at
Answer Details
Question 15 Report
Answer Details
Question 16 Report
From the diagram above, the inductive reactance and the resistance R are respectively
Answer Details
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
Question 17 Report
Answer Details
Question 18 Report
Answer Details
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
Question 19 Report
Answer Details
Question 20 Report
Answer Details
Work done by this force : 60
F cos 60 x X
= 2 x 0.5 x 3
= 3J
Question 21 Report
Answer Details
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
Question 22 Report
Answer Details
Question 23 Report
Answer Details
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.
Question 24 Report
Answer Details
Thus, the correct option is p-type and n-type respectively
Question 25 Report
[G = 6.67 X 10-11 Nm2 Kg2]
Answer Details
= 3.71 X 10-8N.
Question 26 Report
.
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
Answer Details
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
Question 27 Report
[g = 10ms-2]
Answer Details
= 850 x 10 x 500
= 4250000
= 4.25 x 106 NM-2
Question 28 Report
Answer Details
Question 29 Report
Answer Details
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
Question 30 Report
Answer Details
Question 31 Report
[h = 6.63 x 10-34js, c = 3 x 108 ms-1]
Answer Details
Question 32 Report
Answer Details
Question 33 Report
Answer Details
Question 34 Report
Answer Details
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
Question 35 Report
Answer Details
Question 36 Report
Answer Details
=> 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
Question 37 Report
Answer Details
Question 38 Report
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?
Answer Details
Question 39 Report
The figure above shows 4 forces 3N, 10N, 3√3N and 6N acting on a particle P. The resultant of he four forces is
Answer Details
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
Question 40 Report
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?
Answer Details
Question 41 Report
Answer Details
Question 42 Report
Answer Details
∴ 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
Question 43 Report
[specific heat capacity of water = 4200 jkg -1k-1]
Answer Details
Question 44 Report
During the nuclear reaction described by
| 235 | W → | 235 | X→ | 231 | Y |
| 92 | 93 | 91 |
The particles emitted are respectively
Answer Details
Thus , the emission is β and α particles respectively
Question 45 Report
Answer Details
Question 46 Report
I. Mass
II. Density
III. Temperature
IV. Nature of substance
Which of the above affect diffusion?
Answer Details
Question 47 Report
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
Answer Details
Question 48 Report
Answer Details
= (1500 x 40 x40)/50
= 48000 = 4.8 x 104 N
Question 49 Report
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?
Answer Details
Question 50 Report
