JAMB UTME - Physics - 2001
Question 1 Report
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therefore 200 * 103 = I/2 * π√l * 20 * 10-12
therefore (2.0 * 105)2 = I/4 * π2 * L * 20 * 10-12
THEREFORE L = I/4.0 * 1010 * 4 * 10 * 20 * 10-12
(if π2 = 10)
= I/16 * 1011 * 2.0 * 10-11
Therefore L = I/32 H
Question 2 Report
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and I(rms) = I0/√2 = 2.0A/√2
Question 3 Report
In the diagram above, which of the simple pendula will resonate with P when set into oscillation
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Question 4 Report
A cell can supply current of 0.4 A and 0.2 A through a 4.0Ω and 10.0Ω resistors respectively.
This internal resistance of the cell is
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Question 5 Report
If ray travelling in air is incident on a transparent medium as shown in the diagram, the refractive index of the medium is given as
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| From the diagram above, refractive index of the medium = | sin(90- α ) |
| _sin β |
| Reflective index of the medium = | cos α |
| sin β |
Question 6 Report
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W = FD = (PV/L) x L = PV
= 2 x 105
(2 x 10-6 x 6)
= 2 x 105 x 1.2 x 10-5
= 2.4 J
Question 7 Report
I. The earth is not spherical but elliptical in shape
II. Variation in latitude and longitude
III. Rotation of the earth on its axis
IV. Variation in the density of the earthOn which combination of the above does the weight of an object vary on the earth's surface?
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Question 8 Report
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Question 9 Report
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i. Depolarize = CuSO4 solution
ii. Positive electrode = Copper container
iii Negative electrode = zinc rod/plate
Question 10 Report
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by f0 = v/λ, where v = velocity of sound, λ = wave length.
At its fundamental, note, the the length of the string is given l = λ/2, => λ0 = 2l, where λ0 = wave length fundamental note. for the first over tone, λ1 = l.
For the second over tone, λ2 = 21/3 = (2 x 24)/3 = 16cm.
i.e. λ2 = 16cm.
Question 11 Report
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So if θ = 35°, then the reflected ray is rotated through 2 x 10o = 20°
Question 12 Report
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Question 14 Report
The diagram above is a block and tackle pulley system in which an effort of 80N is use to lift a load of 240N. The efficiency of the machine is
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| ∴ efficiency = | M.A |
|
|||
| V.R |
= Load/Effort x 1/V.R x 100/1
= 240/80 x 1/6 x 100/1
= 50%
∴ Efficiency = 50%
Question 15 Report
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Question 16 Report
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Then wavelength = V/F = 342/85.5 = 4m
Then if wavelength = 4m
=> wavelength/2 = 4/2 = 2m = NN,
Thus NA = wavelength/4 = 4/4 = 1m.
Question 17 Report
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Question 19 Report
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Question 20 Report
In the circuit diagram above , the ammeter reads a current of 3A when R is 5Ω and 6A when R is 2Ω. Determine the value of x
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| effective resistance R = | 5X |
| 5+X |
and when R = 2Ω
| effetive resistance R = | 2X |
| 2+X |
Thus since E = I(R+r), and if r = 0 then E = IR
∴(i) E = 3(5X/(5+x)); (ii) E = 6(2X(2+x))
∴3(5X/(5+X)) = 6(2X/(2+X))
=> (5X/(5+X)) = 2(2x/(2+X))
∴ (5X/(5+X)) = (4X/(2+X))
5x(2+x) = 4x(5+x)
10x + 5x2 = 20x + 4x2
5x2 - 4x2 = 20x - 10x
=>x2 = 10x
∴x2 - 10x = 0
x(x - 10) = 0
∴ x = 0 or 10
∴ x = 10Ω
Question 21 Report
The diagram above shows two capacitors P and Q of capacitance 2μF and 4μF respectively connected to a d.c source. The ratio of energy stored in P to Q is
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Question 22 Report
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But as the stone descends, its potential energy decreases, while its kinetic energy increases.
Half way during the fall, (h = 5m), its P.E. = 50J, and consequently, its K.E. should be 50J, such that from the principle of conservation of energy, the total energy of the stone at any instant should be conserved.
Question 23 Report
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= power out put/ power + that due * 100/1
to internal resistance
=> 75/1 = 12R *100/1
12(R+r)
i.e 75/100 = 6/6+r => 75 (6+r) = 100/*6
i.e 450 + 75r = 600
75r =600 - 450
i.e r = 150/75 = 2Ω
i.e r = 2Ω
Question 24 Report
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Question 25 Report
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Question 26 Report
The diagram above shows a closed side 0.5m in a uniform electric field E in the direction shown by the arrows. What is the flux Φ for the box?
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Question 27 Report
The diagram above shows the capacitors C1, C2 and C3 2μF, 6μF and 3μF respectively. The potential difference across C1, C2 and C3 respectively are
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∴V1 = Q/C1; V2 = Q/C;
V3 = Q/C3
for series
1/C = 1/C1 + 1/C2 + 1/C3
= 1/2 + 1/6 + 1/3
| = | 3+1+2 |
| _6 |
= 6/6 = 1
∴ C = 1μF = 1.0 x 10-6F
∴ Q = CV = 1.0 x 10-6 x 12
= 12.0 x 10-6C
∴V1 = Q/C1
| = | 12.0x10-6 |
| 2 x 10-6 |
= 6V
V2 = Q/C2
| = | 12.0x10-6 |
| 6 x 10-6 |
= 2V
V3 = Q/C3
| = | 12.0x10-6 |
| 3 x 10-6 |
∴ 6V, 2V and 4V
Question 28 Report
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Question 29 Report
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T1 = 127 + 273; T2 = -73 + 273
= 400K; = 200K
Let the initial volume V1; and the final volume = V2
? P1V1/T1 = P2V2/T2
P1V1T2 = P2V2T1
? V2/V1 = P1T2/P2T1
? V2/V1 = (300 x 200)/(120 x 400) = 5/4
? V2 : V1 = 5 : 4
Question 30 Report
I. Low pressure.
II. High pressure.
III. High p.d.
IV. Low p.d.
Which combination of the above is true of the conduction of electricity through gases?
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Question 31 Report
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Question 33 Report
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Thus if r = 1m = 100cm
f = 1/2 = -50
∴ if U = 4cm = 400cm
f = -50cm
∴1/u + 1/v = 1/f
1/400 + 1/v = -1/50
∴ -1/v = 1/400 + 1/50
-1/v = (1 + 8)/400 = 9/400
∴ v = -400/9
v = 4/9
Question 34 Report
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Question 35 Report
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Question 36 Report
A resistance R is connected across the terminal of an electric cell of internal resistance 2Ω and the voltage was reduced to 3/5 of its nominal value.
The value of R is
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V = 3/5 E
=> 3E/5R = E/R+2
i.e 3E(R+2) = 5RE
3RE + 6E = 5RE
i.e 3R + 6 = 5R
6 = 5R - 3R
=> 2r = 6 i.e R = 6/2 = 3Ω
Question 37 Report
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Question 38 Report
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Question 39 Report
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Question 40 Report
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Question 41 Report
The particles emitted when \( {}^{39}_{19}\mathrm{K} \) decay to \( {}^{39}_{19}\mathrm{K} \) is
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Since neither the mass number (39), nor the atomic number (19) is affected by the emission,the particles emitted can only be gamma ray, which is not a charge particle, but simply a radiation.
Question 42 Report
In the diagram above, determine the r.m.s current
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√ R2 + XL2 =
√ 42 + 32 =
√ 16 + 9 =
√ 25
∴Ir.m.s = 240/5
= 48.0A
Question 43 Report
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Question 44 Report
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Question 45 Report
In the diagram above , the current I is
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Question 46 Report
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This volume of water in the cylinder = πr2l, where r = radius of the cylinder, and l = level of water rose in the cylinder.
Therefore, π x r2 x l = 4/3 x π x 33
Then r2 = 36
r = √36
r = 6cm
Question 47 Report
If a ray traveling in air is incident on a transparent medium as shown in the diagram, the refractive index of the medium is given as
Question 48 Report
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∴ (p - 760)/-760 = (20 - 0)/80
80(p-760) = -760(20-0)
∴80p - 60800 = -15200
80p = 60800 - 15200
80p = 45600
∴ p = 45600/80
p = 570mmHg
Question 49 Report
