Physics 2008 Jamb Examination
Physics 2008 Jamb Examination
Question 11
The resultant of two forces 12 N and 5 N is 13 N. What is the angle between the two forces?
Using the law of cosines for the resultant of two forces:$$R^2 = F_1^2 + F_2^2 + 2F_1F_2 \cos \theta$$Substitute values:$$13^2 = 12^2 + 5^2 + 2(12)(5)\cos \theta$$$$169 = 144 + 25 + 120\cos \theta$$$$169 = 169 + 120\cos \theta$$$$120\cos \theta = 0$$$$\cos \theta = 0$$$$\theta = 90^\circ$$
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Question 12
A force $F$ required to keep a 5 kg mass moving round a circle of radius 3.5 m at a speed of 7 mĀ·sā»Ā¹. What is the speed if the force is tripled?
Centripetal force isĀ \[
F = \frac{mv^2}{r}.\]
For the initial state:\[ \begin{aligned}
F_0 &= \frac{5 \times 7^2}{3.5} \\
Ā Ā &= \frac{245}{3.5} \\
Ā Ā &= 70\ \text{N}.
\end{aligned}\]
If the force is tripled:\[ F' = 3 F_0 = 210\ \text{N}.\]
Solve for new speed \(v'\):\[ \begin{aligned}
v' &= \sqrt{\frac{F' r}{m}} \\
Ā Ā &= \sqrt{\frac{210 \times 3.5}{5}} \\
Ā Ā &= \sqrt{147} \\
Ā Ā &\approx 12.12\ \text{mĀ·s}^{-1}.
\end{aligned}\]
For the initial state:\[ \begin{aligned}
F_0 &= \frac{5 \times 7^2}{3.5} \\
Ā Ā &= \frac{245}{3.5} \\
Ā Ā &= 70\ \text{N}.
\end{aligned}\]
If the force is tripled:\[ F' = 3 F_0 = 210\ \text{N}.\]
Solve for new speed \(v'\):\[ \begin{aligned}
v' &= \sqrt{\frac{F' r}{m}} \\
Ā Ā &= \sqrt{\frac{210 \times 3.5}{5}} \\
Ā Ā &= \sqrt{147} \\
Ā Ā &\approx 12.12\ \text{mĀ·s}^{-1}.
\end{aligned}\]
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Question 13
A body of mass 4 kg resting on a smooth horizontal plane is simultaneously acted upon by two perpendicular forces 6 N and 8 N. Calculate the acceleration of the motion.
Resultant force magnitude for perpendicular forces is $$F=\sqrt{6^2+8^2}=\sqrt{36+64}=\sqrt{100}=10\ \text{N}.$$ Acceleration: $$a=\dfrac{F}{m}=\dfrac{10}{4}=2.5\ \text{mĀ·s}^{-2}.$$
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Question 14
A block of aluminium is heated electrically by a 30 W heater. If the temperature rises by 10Ā°C in 5 minutes, the heat capacity of the aluminium is
Energy supplied: $$Q=Pt=30\times(5\times60)=30\times300=9000\ \text{J}.$$ Heat capacity $C$ satisfies $$Q=C\Delta T\Rightarrow C=\dfrac{Q}{\Delta T}=\dfrac{9000}{10}=900\ \text{JĀ·K}^{-1}.$$
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Question 15
An object 4 cm high is placed 15 cm from a concave mirror of focal length 5 cm. The size of the image is
Mirror formula: $$\dfrac{1}{f}=\dfrac{1}{v}+\dfrac{1}{u},\quad f=5\ \text{cm},\ u=15\ \text{cm}.$$ So $$\dfrac{1}{v}=\dfrac{1}{f}-\dfrac{1}{u}=\dfrac{1}{5}-\dfrac{1}{15}=\dfrac{3-1}{15}=\dfrac{2}{15}$$ giving $$v=7.5\ \text{cm}.$$ Linear magnification $$m=\dfrac{v}{u}=\dfrac{7.5}{15}=0.5.$$ Image height = $4\times0.5=2\ \text{cm}$.
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Question 16
I. for current amplification.
II. for voltage stabilization.
III. for power amplification.
IV. as a switch.
Which of the above are the uses of transistor?
II. for voltage stabilization.
III. for power amplification.
IV. as a switch.
Which of the above are the uses of transistor?
Transistors are commonly used for current amplification (I), power amplification (III) and as electronic switches (IV). While they can be used within voltage-regulator circuits, the primary list in many standard options excludes II ā so the correct combination here is I, III and IV.
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Question 17
The d.c generator has essentially the same components as the a.c generator except the presence of?
A DC generator requires a commutator (split ring) to convert the alternating emf induced in the armature winding into a unidirectional output. AC generators use slip-rings instead. Thus the distinguishing component is the split ring (commutator).
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Question 18
A step-down transformer has a power output of 50 W and efficiency of 80%. If the mains supply voltage is 200 V, calculate the primary current of the transformer.
Input power $$P_{in}=\dfrac{P_{out}}{\eta}=\dfrac{50}{0.80}=62.5\ \text{W}.$$ Primary current $$I_p=\dfrac{P_{in}}{V_{p}}=\dfrac{62.5}{200}=0.3125\ \text{A}\approx0.31\ \text{A}.$$
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Question 19
Caesium has a work function of $3\times10^{-19}\ \text{J}$. The maximum energy of liberated electrons when it is illuminated by light of frequency $6.7\times10^{14}\ \text{Hz}$ is (take $h=6.6\times10^{-34}\ \text{JĀ·s}$):
\[
\begin{aligned}
E_{ph} &= hf \\
Ā Ā Ā Ā &= (6.6\times10^{-34})(6.7\times10^{14}) \\
Ā Ā Ā Ā &= 4.422\times10^{-19}\ \text{J}
\end{aligned}\] \[ \begin{aligned}
K_{\max} &= E_{ph}-\phi \\
Ā Ā Ā Ā Ā &= 4.422\times10^{-19}-3.0\times10^{-19} \\
Ā Ā Ā Ā Ā &= 1.422\times10^{-19}\ \text{J} \approx 1.42\times10^{-19}\ \text{J}
\end{aligned}\]
E_{ph} &= hf \\
Ā Ā Ā Ā &= (6.6\times10^{-34})(6.7\times10^{14}) \\
Ā Ā Ā Ā &= 4.422\times10^{-19}\ \text{J}
\end{aligned}\] \[ \begin{aligned}
K_{\max} &= E_{ph}-\phi \\
Ā Ā Ā Ā Ā &= 4.422\times10^{-19}-3.0\times10^{-19} \\
Ā Ā Ā Ā Ā &= 1.422\times10^{-19}\ \text{J} \approx 1.42\times10^{-19}\ \text{J}
\end{aligned}\]
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Question 20
Zener diode is used for?
A Zener diode is intentionally operated in its breakdown region to provide a stable reference voltage and thus is widely used for voltage regulation and over-voltage protection.
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