Question 1
If a wire \(30\textnormal{cm}\) long is extended to \(30.5\textnormal{cm}\) by a force of \(300\textnormal{N}\). Find the strain energy of the wire.A. \(7.50 \textnormal{J}\)
B. \(750.00 \textnormal{J}\)
C. \(75.00 \textnormal{J}\)
D. \(0.75 \textnormal{J}\)
Explanation
Original length of wire = \(30 \textnormal{cm}\)
New length of wire after extension = \(30.5 \textnormal{cm}\)
Extension (\(e\)) = \(30.5 \textnormal{cm}\, – \, 30 \textnormal{cm}\)
Extension (\(e\)) = \(0.5 \textnormal{cm}\)
Extension (\(e\)) = \(0.005 \textnormal{m}\)
Force (\(F\)) = \(300 \textnormal{N}\)
Energy = \(\frac{1}{2}Fe\)
NOTE! you may want to mistaken tensile strain for strain energy; tensile strain which is \(\frac{e}{l}\) has no unit, but energy has unit \(\textnormal{J}\) which is the units of all the options. This systematically means that we should find the energy in the stretched wire, not tensile strain.
Energy = \(0.75\textnormal{J}\)
Question 2
A bob of weight \(0.1\textnormal{N}\) hangs from a massless string of length \(50\textnormal{cm}\).The workdone is...
A. \(0.250 \textnormal{J}\)
B. \(0.025 \textnormal{J}\)
C. \(0.050 \textnormal{J}\)
D. \(0.50 \textnormal{J}\)
Explanation
For this kind of question we will need to plot a diagram to understand how we will solve it.
Our first step is to draw a bob of weight \(0.1\textnormal{N}\) hanging from a string of \(50 \textnormal{cm}\) or better still, \(0.5 \textnormal{m}\).
Secondly, we draw the horizontal displacement of the bob from its rest point, \(60°\) to the vertical.
From this diagram, we can see that during the oscillation of the bob, there would be a vertical change in length which is the vertical distance \(x\) the bob travels, and we shall be using this to calculate for workdone.
Workdone = force \(×\) distance
From the image above, we need to find the distance \(x\).
We do this by first getting \(y\).
\(y\) can easily be gotten by trigonometry
\(\cos \theta = \frac{adj}{hyp}\)
\(\cos 60° = \frac{y}{0.5}\)
\(y = 0.5 \cos60°\)
\(y = 0.25\textnormal{m}\)
Now that we know that \(y\) is \(0.25 \textnormal{m}\), we can easily get \(x\).
\(y + x = 0.5\)
\(0.25 + x = 0.5\)
\(x = 0.5 – 0.25\)
\(x = 0.25\textnormal{m}\)
Workdone = \(0.1\textnormal{N} × 0.25\textnormal{m}\)
Workdone = \(0.025\textnormal{J}\).
NOTE! if you do not understand this question you don't need to stress yourself, it's very difficult and doesn't always come out.
Question 3
Two metals \(\textnormal{P}\) and \(\textnormal{Q}\) of lengths \(l_1\) and \(l_2\) are heated through the same temperature difference. If the ratio of the linear expansivities of \(\textnormal{P}\) to \(\textnormal{Q}\) is \(2:3\), and the ratio of their lengths is \(3:4\). What is the ratio of the increase in lengths of \(\textnormal{P}\) to \(\textnormal{Q}\)?A. \(5:7\)
B. \(2:1\)
C. \(1:2\)
D. \(7:5\)
Explanation
Length of \(\textnormal{P}\) = \(l_1\)
Length of \(\textnormal{Q}\) = \(l_2\)
Temperature difference is the same, this means that: \(∆\theta_{\small{\textnormal{P}}}\) = \(∆\theta_{\small{\textnormal{Q}}}\)
Ratio of linear expansivity of \(\textnormal{P}\) to \(\textnormal{Q}\) is \(\Large{\frac{\alpha_{\small{\textnormal{P}}}}{\alpha_{\small{\textnormal{Q}}}}}\) = \(\large{\frac{2}{3}}\)
Ratio of length of \(\textnormal{P}\) to \(\textnormal{Q}\) is \(\Large{\frac{l_{\small{\textnormal{P}}}}{l_{\small{\textnormal{Q}}}}}\) = \(\large{\frac{3}{4}}\)
Ratio of increase in length of \(\textnormal{P}\) to \(\textnormal{Q}\) is \(\Large{\frac{∆l_{\small{\textnormal{P}}}}{∆l_{\small{\textnormal{Q}}}}}\) = \(?\)
Linear expansivity (\(\alpha\)) = \(\Large{\frac{∆l}{l_i∆\theta}}\)
Where \(∆l\) = change in Length
\(l_i\) = initial length
\(∆\theta\) = change in temp
Where \(∆l\) = change in Length
\(l_i\) = initial length
\(∆\theta\) = change in temp
\(\alpha_{\textnormal{p}}\) = \(\Large{\frac{∆l_{\textnormal{p}}}{l_\textnormal{p}∆\theta_{\textnormal{p}}}}\)
\(∆l_{\textnormal{p}}\) = \(\alpha_{\textnormal{p}}l_{\textnormal{p}}∆\!\!\theta_{\textnormal{p}}\)
From the above ratios:
\(\alpha_{\textnormal{p}}\) = \(2\)
\(l_{\textnormal{p}}\) = \(3\)
\(∆l_{\textnormal{p}}\) = \(2 × 3 ×\!\! ∆\!\!\theta_{\textnormal{p}}\)
\(∆l_{\textnormal{p}}\) = \(6\! ∆\!\!\theta_{\textnormal{p}}\)
\(∆l_{\textnormal{p}}\) = \(\alpha_{\textnormal{p}}l_{\textnormal{p}}∆\!\!\theta_{\textnormal{p}}\)
From the above ratios:
\(\alpha_{\textnormal{p}}\) = \(2\)
\(l_{\textnormal{p}}\) = \(3\)
\(∆l_{\textnormal{p}}\) = \(2 × 3 ×\!\! ∆\!\!\theta_{\textnormal{p}}\)
\(∆l_{\textnormal{p}}\) = \(6\! ∆\!\!\theta_{\textnormal{p}}\)
\(\alpha_{\small{\textnormal{Q}}}\) = \(\Large{\frac{∆l_{\small{\textnormal{Q}}}}{l_{\small{\textnormal{Q}}}∆\theta_{\small{\textnormal{Q}}}}}\)
\(∆l_{\small{\textnormal{Q}}}\) = \(\alpha_{\small{\textnormal{Q}}}l_{\small{\textnormal{Q}}}∆\!\!\theta_{\small{\textnormal{Q}}}\)
\(\alpha_{\small{\textnormal{Q}}}\) = \(3\)
\(l_{\small{\textnormal{Q}}}\) = \(4\)
\(∆l_{\small{\textnormal{Q}}}\) = \(3 × 4 ×\!\! ∆\!\!\theta_{\small{\textnormal{Q}}}\)
\(∆l_{\small{\textnormal{Q}}}\) = \(12\! ∆\!\!\theta_{\small{\textnormal{Q}}}\)
\(∆l_{\small{\textnormal{Q}}}\) = \(\alpha_{\small{\textnormal{Q}}}l_{\small{\textnormal{Q}}}∆\!\!\theta_{\small{\textnormal{Q}}}\)
\(\alpha_{\small{\textnormal{Q}}}\) = \(3\)
\(l_{\small{\textnormal{Q}}}\) = \(4\)
\(∆l_{\small{\textnormal{Q}}}\) = \(3 × 4 ×\!\! ∆\!\!\theta_{\small{\textnormal{Q}}}\)
\(∆l_{\small{\textnormal{Q}}}\) = \(12\! ∆\!\!\theta_{\small{\textnormal{Q}}}\)
\(\frac{∆l_{\small{\textnormal{P}}}}{∆l_{\small{\textnormal{Q}}}} = \frac{6∆\theta_{\textnormal{p}}}{12∆\theta_{\small{\textnormal{Q}}}}\)
\(\frac{∆l_{\small{\textnormal{P}}}}{∆l_{\small{\textnormal{Q}}}} = \frac{6\cancel{∆\theta_{\textnormal{p}}}}{12\cancel{∆\theta_{\small{\textnormal{Q}}}}}\)[Since \(∆\theta_{\small{\textnormal{P}}}\) = \(∆\theta_{\small{\textnormal{Q}}}\)]
\(\frac{∆l_{\small{\textnormal{P}}}}{∆l_{\small{\textnormal{Q}}}} = \frac{6}{12}\)
\(\frac{∆l_{\small{\textnormal{P}}}}{∆l_{\small{\textnormal{Q}}}} = \frac{1}{2}\)
\(∆l_{\small{\textnormal{P}}} : ∆l_{\small{\textnormal{Q}}} = 1:2\)
Question 4
The density of a certain oil on frying becomes \(0.4\, \textnormal{kgm}\!^{–3}\) with a volume of \(20\,\textnormal{m}^3\). What will be its initial volume when its initial density is \(0.8\, \textnormal{kgm}\!^{–3}\) assuming no loss of oil due to spillage.A. \(10 \,\textnormal{m}^3\)
B. \(5 \,\textnormal{m}^3\)
C. \(8 \,\textnormal{m}^3\)
D. \(12 \,\textnormal{m}^3\)
Explanation
Density of frying oil (\(\rho\)) = \(0.4 \, \textnormal{kgm}\!^{–3}\)
Volume of frying oil (\(\textnormal{V}\)) = \(20\, \textnormal{m}^3\)
Initial volume of oil (\(\textnormal{V}_i\)) = \(?\)
Initial density of oil (\(\rho_i\)) = \(0.8\, \textnormal{kgm}\!^{–3}\)
\(\rho\) = \(\Large{\frac{m}{\textnormal{V}}}\)
Where \(\rho\) = density
\(m\) = mass
\(\textnormal{V}\) = volume
Where \(\rho\) = density
\(m\) = mass
\(\textnormal{V}\) = volume
We can get the mass of frying oil by;
\(0.4\) = \(\Large{\frac{m}{20}}\)
\(m\) = \(0.4 × 20\)
\(m\) = \(8\,\textnormal{kg}\)
Since mass was unchanged, we know that initial mass of oil will be \(8\,\textnormal{kg}\)
We can thus get the initial volume of the oil from the density formula:
\(0.8\) = \(\Large{\frac{8}{\textnormal{V}}}\)
\(0.8 × \textnormal{V}\) = \(8\)
\(\textnormal{V}\) = \(\Large{\frac{8}{0.8}}\)
\(\textnormal{V}\) = \(10\,\textnormal{m}^3\)
Question 5
The pressure of one mole of an ideal gas of volume \(10^{–2}\,\textnormal{m}^3\) at a temperature of \(27°\textnormal{C}\) is ...[Molar gas constant = \(8.3\,\textnormal{Jmol}^{–1}\textnormal{K}^{–1}\)]A. \(2.24 × 10^4 \,\textnormal{Nm}^{–1}\)
B. \(2.24 × 10^5 \, \textnormal{Nm}^{–1}\)
C. \(2.49 × 10^5 \, \textnormal{Nm}^{-1}\)
D. \(2.49 × 10^4 \, \textnormal{Nm}^{-1}\)
Explanation
Number of moles (\(\textnormal{n}\)) = \(1 \,\textnormal{mol}\)
Volume (\(\textnormal{V}\))= \(10^{–2}\,\textnormal{m}^3\)
Temperature (\(\textnormal{T}\)) = \(27°\textnormal{C}\) = \(\left(27 + 273\right)\textnormal{K}\) = \(300\,\textnormal{K}\)
Pressure (\(\textnormal{P}\)) = \(?\)
Molar gas constant (\(\textnormal{R}\)) = \(8.3\,\textnormal{Jmol}^{–1}\textnormal{K}^{–1}\)
To Calculate for pressure, we use;
\(\textnormal{PV}\) = \(\textnormal{nRT}\)
\(\textnormal{P} × 10^{-2}\) = \(1 × 8.3 × 300\)
\(\textnormal{P}\) = \(\frac{\Large{1 × 8.3 × 300}}{\Large{10^{-2}}}\)
\(\textnormal{P}\) = \(2.49 × 10^5 \,\textnormal{Nm}^{-1}\)
Question 6
The wavelength of a wave travelling with a velocity of \(420 \,\textnormal{ms}^{-1}\) is \(42\,\textnormal{m}\). What is its period\(?\)A. \(1.0 \,\textnormal{s}\)
B. \(0.1 \,\textnormal{s}\)
C. \(0.5 \,\textnormal{s}\)
D. \(1.2 \,\textnormal{s}\)
Explanation
Velocity (\(v\)) = \(420 \, \textnormal{ms}^{–1}\)
Wavelength (\(\lambda\)) = \(42\,\textnormal{m}\)
Period (\(\textnormal{T}\)) = \(?\)
\(v\) = \(f\lambda\)
But \(f\) = \(\frac{1}{T}\),
velocity is therefore \(v\) = \(\frac{\lambda}{T}\)
\(420\) = \(\frac{42}{T}\)
\(420 × T\) = \(42\)
\(T\) = \(\frac{42}{420}\)
\(T\) = \(0.1 \, \textnormal{s}\)
Question 7
The velocity of sound in air at \(16°\textnormal{C}\) is \(340\,\textnormal{ms}^{–1}\). What will it be when the pressure is doubled and its temperature raised to \(127°\textnormal{C}\)\(?\)A. \(4,000 \,\textnormal{ms}^{–1}\)
B. \(160,000 \,\textnormal{ms}^{–1}\)
C. \(8,000 \,\textnormal{ms}^{–1}\)
D. \(400 \,\textnormal{ms}^{–1}\)
Explanation
\(\textnormal{T}_1\) = \(16°\textnormal{C}\) = \(\left(16 + 273\right) \,\textnormal{K}\) = \(289\,\textnormal{K}\)
\(v_1\) = \(340 \,\textnormal{ms}^{–1}\)
\(\textnormal{T}_2\) = \(127°\textnormal{C}\) = \(\left(127 + 273\right) \,\textnormal{K}\) = \(400\,\textnormal{K}\)
\(v_2\) = \(?\)
Generally, we know that the velocity of sound in air is directly proportional to the square root of air temperature (velocity of sound is independent of pressure of air);
\(v \:\:\large{{\alpha}}\:\: \sqrt{\textnormal{T}}\)
\(\frac{\large{v}}{\large{\sqrt{\textnormal{T}}}}\) = \(\textnormal{K}\)
\(\frac{\large{v_1}}{\large{\sqrt{\textnormal{T}_1}}}\) = \(\frac{\large{v_2}}{\large{\sqrt{\textnormal{T}_2}}}\)
Inputting the known values, we have:
\(\frac{\large{340}}{\large{\sqrt{289}}}\) = \(\frac{\large{v_2}}{\large{\sqrt{400}}}\)
\(\frac{\large{340}}{\large{17}}\) = \(\frac{\large{v_2}}{\large{20}}\)
\(20\) = \(\frac{\large{v_2}}{\large{20}}\)
\(20 × 20\) = \(v_2\)
\(v_2\) = \(400 \,\textnormal{ms}^{–1}\)
Question 8
An object \(4\,\textnormal{cm}\) high is placed \(15\,\textnormal{cm}\) from a concave mirror of focal length \(5\,\textnormal{cm}\). The size of the image isA. \(3\,\textnormal{cm}\)
B. \(5\,\textnormal{cm}\)
C. \(4\,\textnormal{cm}\)
D. \(2\,\textnormal{cm}\)
Explanation
Object height (\(\textnormal{h}_o\)) = \(4\,\textnormal{cm}\)
Object distance from mirror (\(u\)) = \(15\,\textnormal{cm}\)
Focal length of mirror (\(f\)) = \(5\,\textnormal{cm}\)
Size of image (\(\textnormal{h}_i\)) = \(?\)
To solve for this kind of question, we can only use magnification formula, which is:
\(m\) = \(\frac{\large{v}}{\large{u}}\) = \(\frac{\large{\textnormal{h}_i}}{\large{\textnormal{h}_o}}\)
Where \(m\) = magnification
\(v\) = distance of image from mirror
\(u\) = distance of object from mirror
\(\textnormal{h}_i\) = height of image
\(\textnormal{h}_o\) = height of object
Where \(m\) = magnification
\(v\) = distance of image from mirror
\(u\) = distance of object from mirror
\(\textnormal{h}_i\) = height of image
\(\textnormal{h}_o\) = height of object
But we cannot directly use this formula because we have two unknowns which are \(v\) and what we are to get, \(\textnormal{h}_i\).
This means that we have to calculate for \(v\) first.
We calculate for \(v\) by using the mirror formula;
\(\frac{\large{1}}{\large{f}}\) = \(\frac{\large{1}}{\large{v}} + \frac{\large{1}}{\large{u}}\)
\(\frac{\large{1}}{\large{5}}\) = \(\frac{\large{1}}{\large{v}} + \frac{\large{1}}{\large{15}}\)
\(\frac{\large{1}}{\large{5}} – \frac{\large{1}}{\large{15}}\) = \(\frac{\large{1}}{\large{v}}\)
\(\frac{\large{3\,–\,1}}{\large{15}}\) = \(\frac{\large{1}}{\large{v}}\)
\(\frac{\large{2}}{\large{15}}\) = \(\frac{\large{1}}{\large{v}}\)
\(v\) = \(\frac{\large{15}}{\large{2}}\)
Now that we have gotten the value for \(v\), we can then recall our magnification formula:
\(\frac{\large{v}}{\large{u}}\) = \(\frac{\large{\textnormal{h}_i}}{\large{\textnormal{h}_o}}\)
\(\frac{\large{\frac{15}{2}}}{\large{15}}\) = \(\frac{\large{\textnormal{h}_i}}{\large{4}}\)
\(\frac{15}{2} ÷ 15\) = \(\frac{\large{\textnormal{h}_i}}{\large{4}}\)
\(\frac{15}{2}×\frac{1}{15}\) = \(\frac{\large{\textnormal{h}_i}}{\large{4}}\)
\(\frac{\cancel{15}}{2}×\frac{1}{\cancel{15}}\) = \(\frac{\large{\textnormal{h}_i}}{\large{4}}\)
\(\frac{1}{2}×1\) = \(\frac{\large{\textnormal{h}_i}}{\large{4}}\)
\(1 × 4\) = \(2 × \textnormal{h}_i\)
\(\frac{\large{4}}{\large{2}}\) = \(\textnormal{h}_i\)
\(\textnormal{h}_i\) = \(2 \, \textnormal{cm}\)
Question 9
An object is embedded in a block of ice, \(10\,\textnormal{cm}\) below the plane surface. If the refractive index of the ice is \(1.50\), the apparent depth of the object below the surface isA. \(6.67 \, \textnormal{cm}\)
B. \(7.63 \,\textnormal{cm}\)
C. \(7.50 \, \textnormal{cm}\)
D. \(2.50 \, \textnormal{cm}\)
Explanation
Real depth of object below ice = \(10\, \textnormal{cm}\)
Refractive index of ice = \(1.50\)
Apparent depth of object = \(?\)
\(\mathsf{Refractive \,\, index}\) = \(\frac{\large{\mathsf{Real \,depth}}}{\large{\mathsf{Apparent\, depth}}}\)
\(1.50\) = \(\frac{\large{10}}{\mathsf{A.d}}\)
\(1.50 × \mathsf{A.d}\) = \(10\)
\(\mathsf{A.d}\) = \(\frac{\large{10}}{\large{1.50}}\)
\(\mathsf{A.d}\) = \(6.67 \, \textnormal{cm}\)
Question 10
Three capacitors of capacitance \(2μ\textnormal{F}\), \(4μ\textnormal{F}\), \(8μ\textnormal{F}\) are connected in parallel and a p.d of \(6\textnormal{V}\) is maintained across each capacitor. The total energy stored isA. \(2.52 × 10^{–6}\,\textnormal{J}\)
B. \(6.90 × 10^{–6}\,\textnormal{J}\)
C. \(2.52 × 10^{–4}\,\textnormal{J}\)
D. \(6.90 × 10^{–4}\,\textnormal{J}\)
Explanation
For three capacitors that are connected in parallel, the effective capacitance is \(\textnormal{C}_1 \,+ \,\textnormal{C}_2\,+ \, \textnormal{C}_3\)
Effective capacitance = \(2μ\textnormal{F} + 4μ\textnormal{F} + 8μ\textnormal{F}\)
Effective capacitance = \(14μ\textnormal{F}\)
Effective capacitance = \(14 × 10^{–6}\textnormal{F}\)[\(μ\) = \(10^{–6}\)]
\(\textnormal{E}\) = \(\frac{1}{2}\textnormal{CV}^2\)
Where \(\textnormal{E}\) = energy
\(\textnormal{C}\) = capacitance
\(\textnormal{V}\) = voltage
Where \(\textnormal{E}\) = energy
\(\textnormal{C}\) = capacitance
\(\textnormal{V}\) = voltage
\(\textnormal{E}\) = \(\frac{1}{2}× 14×10^{–6} × 6^2\)
\(\textnormal{E}\) = \(7×36×10^{–6}\)
\(\textnormal{E}\) = \(252 × 10^{–6}\)
\(\textnormal{E}\) = \(2.52 × 10^{–4}\,\textnormal{J}\)
Question 11
Three \(4 \Omega\) resistors connected in parallel have a potential difference of \(16\textnormal{V}\) applied across them. What is the total current in the circuitA. \(12\textnormal{A}\)
B. \(8\textnormal{A}\)
C. \(10\textnormal{A}\)
D. \(14\textnormal{A}\)
Explanation
For three resistors connected in parallel, effective resistance is
\(\frac{\large{1}}{\large{\textnormal{R}}_\textnormal{total}}\) = \(\frac{\large{1}}{\large{\textnormal{R}_1}} + \frac{\large{1}}{\large{\textnormal{R}_2}} + \frac{\large{1}}{\large{\textnormal{R}_3}}\)
\(\frac{\large{1}}{\large{\textnormal{R}}_\textnormal{total}}\) = \(3 × \frac{\large{1}}{\large{4}}\)
\(\frac{\large{1}}{\large{\textnormal{R}}_\textnormal{total}}\) = \(\frac{\large{3}}{\large{4}}\)
\(\textnormal{R}_{\textnormal{total}}\) = \(\frac{\large{4}}{\large{3}}\)
Now we can calculate for the total current in the circuit by :
\(\textnormal{V}\) = \(\textnormal{IR}\)
\(16\) = \(\textnormal{I} × \frac{\large{4}}{\large{3}}\)
\(3 × 16\) = \(4\textnormal{I}\)
\(\textnormal{I}\) = \(\frac{\large{3 × 16}}{\large{4}}\)
\(\textnormal{I}\) = \(12\textnormal{A}\)
Question 12
The diagram above shows a balanced metre bridge. The value of x is
A. \(66.7\,\textnormal{cm}\)
B. \(25.0 \,\textnormal{cm}\)
C. \(33.3 \,\textnormal{cm}\)
D. \(75.0 \, \textnormal{cm}\)
Explanation
In a balanced metre bridge, the resistance is directly proportional to the length of the conductor;
\(\textnormal{R}\:\:\large{\alpha}\:\:\textnormal{l}\)
\(\textnormal{R}\) = \(\textnormal{Kl}\) [where \(\textnormal{K}\) is the constant of proportionality]
\(\frac{\large{\textnormal{R}}}{\large{\textnormal{l}}}\) = \(\textnormal{K}\)
\(\frac{\large{\textnormal{R}}_1}{\large{\textnormal{l}}_1}\) = \(\frac{\large{\textnormal{R}}_2}{\large{\textnormal{l}}_2}\)
\(\textnormal{R}_1\) = effective resistance of two \(8\Omega\) resistors connected in parallel.
Calculating for \(\textnormal{R}_1\):
\(\frac{\large{1}}{\large{R}_1}\) = \(\frac{\large{1}}{\large{8}} + \frac{\large{1}}{\large{8}}\)
\(\frac{\large{1}}{\large{R}_1}\) = \(\frac{\large{1+1}}{\large{8}}\)
\(\frac{\large{1}}{\large{R}_1}\) = \(\frac{\large{2}}{\large{8}}\)
\(\frac{\large{1}}{\large{R}_1}\) = \(\frac{\large{1}}{\large{4}}\)
\(R_1\) = \(4\Omega\)
Now that we have \(\textnormal{R}_1\), we can go ahead with our calculations;
\(\textnormal{l}_1\) = \(x \,\textnormal{cm}\)
\(\textnormal{R}_2\) = \(8\Omega\)
\(\textnormal{l}_2\) = \(100–x \,\textnormal{cm}\)
Substituting known values into the formula:
\(\frac{\large{4}}{\large{x}}\) = \(\frac{\large{8}}{\large{100–x}}\)
\(4(100–x)\) = \(8x\)
\(400 – 4x\) = \(8x\)
\(400\) = \(8x + 4x\)
\(400\) = \(12x\)
\(\frac{\large{400}}{\large{12}}\) = \(x\)
\(33.3\) = \(x\)
\(x\) = \(33.3 \,\textnormal{cm}\)
Question 13
A. ₦\(192.00\)
B. ₦\(1.92\)
C. ₦\(19.20\)
D. ₦\(1920.00\)
Explanation
Power of light bulb (\(\textnormal{P}\)) = \(200\,\textnormal{W}\)
Voltage from a.c mains supply (\(\textnormal{V}\)) = \(240\textnormal{V}\)
\(1\,\textnormal{kWh}\) = ₦\(40\)
Time (\(\textnormal{t}\)) = \(1\) day = (\(1 × 24\))\(\, \textnormal{h}\) = \(24\,\textnormal{h}\)
Amount in \(\textnormal{Wh}\) that will be used to run the light bulb for a day is \(\textnormal{power} × \textnormal{time}\) = \(200 × 24\) = \(4800 \textnormal{Wh}\)
Conversion of \(\textnormal{Wh}\) to \(\textnormal{kWh}\):
= \(\frac{\large{4800}}{\large{1000}}\)
= \(4.8 \textnormal{kWh}\)
Recall,
\(1\,\textnormal{kWh}\) = ₦\(40\)
\(4.8\, \textnormal{kWh}\) = ₦\(x\)[by cross multiplication]
\(x × 1\) = \(4.8 × 40\)
\(x\) = \(192\)
Question 14
Two inductors of inductances \(5\,m\textnormal{H}\) and \(15\,m\textnormal{H}\) are connected in series and a current of \(5\textnormal{A}\) flows through them. The total energy stored in the inductors isA. \(250.0 \,\textnormal{J}\)
B. \(50.0 \, \textnormal{J}\)
C. \(62.5 \, \textnormal{J}\)
D. \(500.0 \, \textnormal{J}\)
Explanation
For two inductors connected in series, the effective inductance is \(\textnormal{L}_1 + \textnormal{L}_2\)
Effective inductance:
= \(5\,m\textnormal{H} + 15\,m\textnormal{H}\)
= \(20\,m\textnormal{H}\)
= \(20 × 10^{–3} \,\textnormal{H}\)
Total energy stored in the inductor is given by:
\(\textnormal{E}\) = \(\frac{1}{2}\textnormal{LI}^2\)
Where \(\textnormal{E}\) = energy
\(\textnormal{L}\) = inductance
\(\textnormal{I}\) = current
Where \(\textnormal{E}\) = energy
\(\textnormal{L}\) = inductance
\(\textnormal{I}\) = current
\(\textnormal{E}\) = \(\frac{1}{2} × 20 × 10^{–3} × 5^2\)
\(\textnormal{E}\) = \(0.25 \,\textnormal{J}\)
Question 15
The ground state energy for a hydrogen atom is \(5.44 × 10^{–19}\,\textnormal{J}\). If an electron drops from zero to ground state, calculate the frequency of the emitted radiation.[ \(\textnormal{h}\) = \(6.6 × 10^{–34} \,\textnormal{Js}\) ]
A. \(2.0 × 10^{16}\textnormal{Hz}\)
B. \(2.0 × 10^{15}\textnormal{Hz}\)
D. \(5.0 × 10^{16}\textnormal{Hz}\)
Explanation
Energy (\(\textnormal{E}\)) = \(5.44 × 10^{–19}\,\textnormal{J}\)
Frequency = \(?\)
\(\textnormal{h}\) = \(6.6 × 10^{–34} \,\textnormal{Js}\)
\(\textnormal{E}\) = \(\textnormal{h}f\)
\(5.44 × 10^{–19}\) = \(6.6 × 10^{–34} × f\)
\(\frac{\large{5.44 × 10^{–19}}}{\large{6.6 × 10^{–34}}}\) = \(f\)
\(8.24 × 10^{14}\) = \(f\)
\(f\) = \(8.24 × 10^{14}\textnormal{Hz}\)
