JAMB UTME - Physics - 1998
Frage 1 Bericht
Electricity is supplied to a school along a cable of total resistance \(0.5\,\Omega\) with the maximum current drawn from the mains as \(100\text{ A}\). The maximum energy dissipated as heat for \(1\text{ hr}\) is
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Frage 2 Bericht
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Frage 3 Bericht
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3 x 2 x 10-5 =V2−1515×(100−0)
V2 = 15.09cm3
Frage 4 Bericht
The force of repulsion between two point positive charges \(5\,\mu\text{C}\) and \(8\,\mu\text{C}\) separated at a distance of \(0.02\) map art is. \(\left[\dfrac{1}{4\pi\varepsilon_o}=9 \times 10^9\text{ Nm}^2\text{C}^{-2}\right]\)
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F =5×10×8×100.002×0.002 x 9 x 10
F = 900N
Frage 5 Bericht
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Frage 6 Bericht
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= 1 x (3 x 108)2 x 106
= 9 x 1010J
Frage 7 Bericht
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Frage 8 Bericht
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Frage 9 Bericht
An object is placed in front of two plane mirrors inclined at an angle of \(0^\circ\). If the total number of images formed is 7, Find the value of \(\theta\)
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7 = 360θ -1
thus,θ = 45o
Frage 10 Bericht
A platinum resistance thermometer records \(3.0\,\Omega\) at \(0^\circ\mathrm{C}\) and \(8.0\,\Omega\) at \(100^\circ\mathrm{C}\). If it records \(6.0\,\Omega\) in a certain environment, The temperature of the medium is
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6Ω .....θ oC
3Ω .....0oC
6−38−3 = θ−0100−0
therefore, θ = 60oC
Frage 11 Bericht
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Frage 12 Bericht
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thus; 0.875 = 0.21upthrust in H2O
upthrust in H2O = 0.24N
Frage 13 Bericht
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Frage 15 Bericht
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Frage 16 Bericht
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V1V2=√T1T2
therefore, 360V2=√27+273127+273
V2=240√3 m/s
Frage 17 Bericht
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Total pressure = (1.3 x 106) + (1 x 105)
= 1.4 x 106Nm-2
Frage 19 Bericht
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Frage 20 Bericht
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Frage 21 Bericht
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Frage 22 Bericht
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mcθ = mL
m x 4200 x 17 = 42 x 3.4 x 105
therefore, m = 200g
Frage 24 Bericht
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Frage 25 Bericht
A body on the ground is acted on by a force of 10N at a point P as shown in the fig above. What force is needed to stop the body from moving eastwards
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Frage 26 Bericht
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Frage 27 Bericht
A 500W heater is used to heat 0.6kg of water from \(25^\circ\mathrm{C}\) to \(100^\circ\mathrm{C}\) in \(t_2\) seconds. If another 1000W heater is used to heat 0.2kg of water from \(10^\circ\mathrm{C}\) to \(100^\circ\mathrm{C}\) in t, seconds, find \(\frac{t_1}{t_2}\)
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Heat energy = Electrical energy
mcθ = power x time
0.6 x c x (100 - 25) = 500 x t1
and, 0.2 x c x (100 - 10)= 100 x t2
t1t2
= 5
Frage 28 Bericht
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Frage 29 Bericht
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Frage 30 Bericht
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= 0.15(80-50)
= 4.5N/s
Frage 31 Bericht
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Frage 33 Bericht
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Frage 36 Bericht
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Frage 37 Bericht
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Frage 38 Bericht
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Frage 39 Bericht
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Frage 40 Bericht
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Frage 41 Bericht
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Frage 42 Bericht
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Frage 44 Bericht
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Frage 46 Bericht
A parallel plate capacitor has a common plate area of \(5 \times 10^{-8}\,\mathrm{m}^2\) and plate separation of \(2 \times 10^3\,\mathrm{m}\), Assuming free space, What is the capacitance? [\(\epsilon_o = 9.0 \times 10^{-12}\,\mathrm{C}^2\mathrm{N}^{-1}\mathrm{m}^{-2}\)]
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= 5×10−8×10−12×92×10−3
= 22.5 x 10-17F
