2010 - JAMB Physics Past Questions and Answers - page 4
pumping out of the gas in the discharge tube reduces the pressure inside the tube, therefore, electricity is conducted at low pressure
Users' Answers & Comments[h = 6.6 x 10-34 Js]
Correct Answer: Option E
∆E = hf <br />
f = (∆E)/h = (5.44 x 10-19)/6.6 x 10-34 = 8.24 x 1014
Users' Answers & Commentstransistor serve the same purpose as triode valve, which is amplification of signals.
Users' Answers & CommentsA semiconductor is a material whose conductivity is between good conductor and insulator. Example of pure semi conductors are silicon and germanium
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Above are the V-t graphs for
I Uniform acceleration
II uniform retardation
III Uniform velocity
Thus the car has no acceleration between point F and G since that portion of the graph represents uniform velocity
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A bod y is said to be in stable equilibrium if after a slight displacement, the body returns to its former position. Thus, position III, VI and XI represents position of stable equilibrium
Users' Answers & CommentsLength of the string
L = y + x = 0.5m
∴ x = L - y
= distance moved in the direction of force (which is the weight) when the body is released But cos 60o = y/0.5
∴ y = 0.5cos60 = 0.5 x 1/2 = 0.25
∴ x = l - y = 0.5 - 0.25 = 0.25m
Work done = force x distance moved in the direction of force.
= 0.1 x 0.25 = 0.0252J
Users' Answers & CommentsMetals P, Q
original length lP, lQ
Increase in length ΔlP, ΔlQ
Linear expansivities αP, αQ
∴ αP/αQ = 2/3 ; and lP/lQ = 3/4
But increase in length, Δl = αlΔθ
∴ΔlP = αPlPΔθ
ΔlQ = αQlQΔθ
∴ | ΔlP | = | αPlPΔθ | = | 2 | x | 3 | = | 1 |
ΔlQ | αQlQΔθ | 3 | 4 | 2 |
∴ Ratio of increase in length of P to Q = 1:2
Users' Answers & CommentsThe velocity of sound in air, V∝√T, where T is absolute temperature
Thus V = K√T or K = V/T
V is independent of pressure and density at a fixed temperature.
∴ V1/√T1 = V2/√T2
T1 = 16 + 273 = 289K, V1 = 340ms-1
T2 = 127 + 273 = 400K, V2 = ?
340 | = | V2 |
√289 | √400 |
∴ V2 = | 340√400 |
_√289 |
= 400m/s
Users' Answers & CommentsC = C1 + C2 + C3
= 2 + 4 + 8 = 14μF
V = 6V
Energy = 1/2CV2
1/2 x 14 x 10-6 x 62
= 2.52 x 10-4 J
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