JAMB UTME - Physics - 2005
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The figure above shows a plot of angles of deviation through a glass prism when light is incident at θ degrees on the prism. The incident angle that produces the minimum deviation is
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In the figure above, the coefficient of static friction is 0.5. If the tension in the string is T, the acceleration of the system in motion is
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coefficient of friction , μ = 0.5
∴ acting frictional force = μR = 30 x 0.5
Pulling/sliding force F = 6kg x 10 = 60N.
∴ Net downward force = ma = 60 - 15.0
=> (3 + 6)a = 45
∴ 9a = 45
a = 5m/s2
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In the figure above, the work done by the force of 100N inclined at an angle of 60o to the object dragged horizontally to a distance of 8m is
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⟹F/e=Kf1/e1=f2/e2Let the original length=t0thereforee1=(36−t0)cm;e2=(46−t0)cmif f1=40Nf2=60NThen 4036−t0=6045−t0⟹40(45−t0)=60(36−t0)therefore 1800−40t02160−60t060t0−40t0=2160−180020t0=360t0=18cm
Question 14 Report
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thus for lens of power of 4.0 dioptres we have that: (100)/(f) = +4.0
f= (100)/(4.0) = 25cm
= 0.25cm
Question 15 Report
[g = 10ms2]
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U = upthrust Mg = weight of the body.
∴ 36 x 10 - 260 - U = 0
∴ U = 360 - 260 = 100N
Upthrust on the body = 100N.
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The velocity of sound in air will be doubled if it's absolute temperature is
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In general, the velocity of sound in air varies directly as the square root of temperature measured in kelvin.
That V∝√T⟹V2∝T.ThereforeV21T1=V22T2Thus Let V1=4m/sT1=10KTherefore V2=2V1=8m/s⟹4210=82T2T2=64×1016=40KT2=4T1
Thus when the velocity of sound in air is doubled, it's absolute temperature will be quadrupled.
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The value of F in the figure above when in equilibrium is
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= 20N
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In the circuit diagram of a transistor above, the terminal P, Q and R represent
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If the system above is in equilibrium, the tension in the string Q is
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(1)/(8) + (1)/(24) = (1)/(f)
therefore (3+1)/(24) = (1)/(f)
4/24=1/f;therefore 4f = 24
f = 6cm
Question 31 Report
The figure above shows three capacitors, 2μF, 3μF and 6μF connected in series. If the p.d across the system is 12V, the p.d, across the 6μF capacitor is
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1/C = 1/2 + 1/3 + 1/6 = 6/6
therefore C = 1μF or
1.0 x 10-6F.
Since C = Q/V, its implies that the charges Q, flowing in the circuit is given by Q = CV
= 1.0 x 10-6 x 12
= 12.0 x 10-6 Coulomb.
∴ P.d across the 6μF,
| V = Q/C = | 12.0 x 10-6 |
| 6.0 x 10-6F |
= 2.0 volts
Question 32 Report
\[ {}^{235}_{92}\mathrm{U} + {}^{1}_{0}\mathrm{n} \rightarrow {}^{144}_{56}\mathrm{Ba} + {}^{90}_{36}\mathrm{Kr} + 2\mathrm{X} \]
In the reaction above, X is
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In the reaction above, balancing the left with the right, show that X is neutron
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Question 39 Report
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i.e h ∝1/d; → h1d1 = h2d2
∴ 8 x 103 = h2 x 800
| ∴ h2 = | 8 x 103 |
| _800 |
= 10cm
Question 40 Report
[Resistivity of the wire = 4 x 10-7 Ω m]
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= (4 x 10-7 x 20)/8.0 x 10-6
R = 1.0 Ω
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