JAMB PHYSICS

Question 1

If a wire $30\text{cm}$ long is extended to $30.5\text{cm}$ by a force of $300\text{N}$. Find the strain energy of the wire.

A. $7.50\text{J}$

B. $750.00\text{J}$

C. $75.00\text{J}$

D. $0.75\text{J}$

Explanation

Original length of wire = $30\text{cm}$
New length of wire after extension = $30.5\text{cm}$
Extension ($e$) = $30.5\text{cm}–30\text{cm}$
Extension ($e$) = $0.5\text{cm}$
Extension ($e$) = $0.005\text{m}$

Force ($F$) = $300\text{N}$

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 $\text{J}$ 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 = $\frac{1}{2}\times 300\times 0.005$
Energy = $0.75\text{J}$

Question 2

A bob of weight $0.1\text{N}$ hangs from a massless string of length $50\text{cm}$.

A variable horizontal force which increases from zero is applied to pull the bob until the string makes an angle of $60°$ with the vertical.
The workdone is...

A. $0.250\text{J}$

B. $0.025\text{J}$

C. $0.050\text{J}$

D. $0.50\text{J}$

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 $0.1\text{N}$ hanging from a string of $50\text{cm}$ or better still, $0.5\text{m}$.



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 $\times$ 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

$\cos \theta =\frac{adj}{hyp}$

$\cos 60°=\frac{y}{0.5}$

$y=0.5\cos 60°$

$y=0.25\text{m}$

Now that we know that $y$ is $0.25\text{m}$, we can easily get $x$.

Recall,
$y+x=0.5$

$0.25+x=0.5$

$x=0.5–0.25$

$x=0.25\text{m}$

Now that we have gotten the value for $x$, we can then finally determine our workdone:
Workdone = $0.1\text{N}\times 0.25\text{m}$
Workdone = $0.025\text{J}$.

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 $\text{P}$ and $\text{Q}$ of lengths $l_1$ and $l_2$ are heated through the same temperature difference. If the ratio of the linear expansivities of $\text{P}$ to $\text{Q}$ is $2:3$, and the ratio of their lengths is $3:4$. What is the ratio of the increase in lengths of $\text{P}$ to $\text{Q}$?

A. $5:7$

B. $2:1$

C. $1:2$

D. $7:5$

Explanation

Length of $\text{P}$ = $l_1$
Length of $\text{Q}$ = $l_2$
Temperature difference is the same, this means that: $∆{\theta }_{\text{P}}$ = $∆{\theta }_{\text{Q}}$
Ratio of linear expansivity of $\text{P}$ to $\text{Q}$ is $\frac{{\alpha }_{\text{P}}}{{\alpha }_{\text{Q}}}$ = $\frac{2}{3}$
Ratio of length of $\text{P}$ to $\text{Q}$ is $\frac{l_{\text{P}}}{l_{\text{Q}}}$ = $\frac{3}{4}$
Ratio of increase in length of $\text{P}$ to $\text{Q}$ is $\frac{∆l_{\text{P}}}{∆l_{\text{Q}}}$ = $?$

Linear expansivity ($\alpha$) = $\frac{∆l}{l_i∆\theta }$

Where $∆l$ = change in Length
               $l_i$ = initial length
            $∆\theta$ = change in temp

${\alpha }_{\text{p}}$ = $\frac{∆l_{\text{p}}}{l_{\text{p}}∆{\theta }_{\text{p}}}$

$∆l_{\text{p}}$ = ${\alpha }_{\text{p}}{l}_{\text{p}}∆{\theta }_{\text{p}}$

From the above ratios:

${\alpha }_{\text{p}}$ = $2$
$l_{\text{p}}$ = $3$

$∆l_{\text{p}}$ = $2\times 3\times ∆{\theta }_{\text{p}}$
$∆l_{\text{p}}$ = $6∆{\theta }_{\text{p}}$

${\alpha }_{\text{Q}}$ = $\frac{∆l_{\text{Q}}}{l_{\text{Q}}∆{\theta }_{\text{Q}}}$

$∆l_{\text{Q}}$ = ${\alpha }_{\text{Q}}{l}_{\text{Q}}∆{\theta }_{\text{Q}}$




${\alpha }_{\text{Q}}$ = $3$
$l_{\text{Q}}$ = $4$

$∆l_{\text{Q}}$ = $3\times 4\times ∆{\theta }_{\text{Q}}$
$∆l_{\text{Q}}$ = $12∆{\theta }_{\text{Q}}$

Ratio of $\frac{∆l_{\text{P}}}{∆l_{\text{Q}}}$ will finally become;

$\frac{∆l_{\text{P}}}{∆l_{\text{Q}}}=\frac{6∆{\theta }_{\text{p}}}{12∆{\theta }_{\text{Q}}}$

$\frac{∆l_{\text{P}}}{∆l_{\text{Q}}}=\frac{6\cancel{∆{\theta }_{\text{p}}}}{12\cancel{∆{\theta }_{\text{Q}}}}$[Since $∆{\theta }_{\text{P}}$ = $∆{\theta }_{\text{Q}}$]

$\frac{∆l_{\text{P}}}{∆l_{\text{Q}}}=\frac{6}{12}$

$\frac{∆l_{\text{P}}}{∆l_{\text{Q}}}=\frac{1}{2}$

$∆l_{\text{P}}:∆l_{\text{Q}}=1:2$

Question 4

The density of a certain oil on frying becomes $0.4\text{kgm}{}^{–3}$ with a volume of $20{\text{m}}^3$. What will be its initial volume when its initial density is $0.8\text{kgm}{}^{–3}$ assuming no loss of oil due to spillage.

A. $10{\text{m}}^3$

B. $5{\text{m}}^3$

C. $8{\text{m}}^3$

D. $12{\text{m}}^3$

Explanation

Density of frying oil ($\rho$) = $0.4\text{kgm}{}^{–3}$
Volume of frying oil ($\text{V}$) = $20{\text{m}}^3$

Initial volume of oil (${\text{V}}_i$) = $?$
Initial density of oil (${\rho }_i$) = $0.8\text{kgm}{}^{–3}$

$\rho$ = $\frac{m}{\text{V}}$

Where $\rho$ = density
            $m$ = mass
            $\text{V}$ = volume

Since there is no loss of oil, this means that mass of oil remain unchanged before frying and during frying $=$ mass of frying oil is equal to initial mass of oil.

We can get the mass of frying oil by;

$0.4$ = $\frac{m}{20}$

$m$ = $0.4\times 20$

$m$ = $8\text{kg}$

Since mass was unchanged, we know that initial mass of oil will be $8\text{kg}$
We can thus get the initial volume of the oil from the density formula:

$0.8$ = $\frac{8}{\text{V}}$

$0.8\times \text{V}$ = $8$

$\text{V}$ = $\frac{8}{0.8}$

$\text{V}$ = $10{\text{m}}^3$

Question 5

The pressure of one mole of an ideal gas of volume ${10}^{–2}{\text{m}}^3$ at a temperature of $27°\text{C}$ is ...[Molar gas constant = $8.3{\text{Jmol}}^{–1}{\text{K}}^{–1}$]

A. $2.24\times {10}^4{\text{Nm}}^{–1}$

B. $2.24\times {10}^5{\text{Nm}}^{–1}$

C. $2.49\times {10}^5{\text{Nm}}^{-1}$

D. $2.49\times {10}^4{\text{Nm}}^{-1}$

Explanation

Number of moles ($\text{n}$) = $1\text{mol}$
Volume ($\text{V}$)= ${10}^{–2}{\text{m}}^3$
Temperature ($\text{T}$) = $27°\text{C}$ = $(27+273)\text{K}$ = $300\text{K}$
Pressure ($\text{P}$) = $?$
Molar gas constant ($\text{R}$) = $8.3{\text{Jmol}}^{–1}{\text{K}}^{–1}$

To Calculate for pressure, we use;

$\text{PV}$ = $\text{nRT}$

$\text{P}\times {10}^{-2}$ = $1\times 8.3\times 300$

$\text{P}$ = $\frac{1\times 8.3\times 300}{{10}^{-2}}$

$\text{P}$ = $2.49\times {10}^5{\text{Nm}}^{-1}$

Question 6

The wavelength of a wave travelling with a velocity of $420{\text{ms}}^{-1}$ is $42\text{m}$. What is its period$?$

A. $1.0\text{s}$

B. $0.1\text{s}$

C. $0.5\text{s}$

D. $1.2\text{s}$

Explanation

Velocity ($v$) = $420{\text{ms}}^{–1}$
Wavelength ($\lambda$) = $42\text{m}$
Period ($\text{T}$) = $?$

$v$ = $f\lambda$

But $f$ = $\frac{1}{T}$,

velocity is therefore  $v$ = $\frac{\lambda }{T}$

$420$ = $\frac{42}{T}$

$420\times T$ = $42$

$T$ = $\frac{42}{420}$

$T$ = $0.1\text{s}$

Question 7

The velocity of sound in air at $16°\text{C}$ is $340{\text{ms}}^{–1}$. What will it be when the pressure is doubled and its temperature raised to $127°\text{C}$$?$

A. $4,000{\text{ms}}^{–1}$

B. $160,000{\text{ms}}^{–1}$

C. $8,000{\text{ms}}^{–1}$

D. $400{\text{ms}}^{–1}$

Explanation

${\text{T}}_1$ = $16°\text{C}$ = $(16+273)\text{K}$ = $289\text{K}$
$v_1$ = $340{\text{ms}}^{–1}$

${\text{T}}_2$ = $127°\text{C}$ = $(127+273)\text{K}$ = $400\text{K}$
$v_2$ = $?$

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);

$v\alpha \sqrt{\text{T}}$

$v$ = $\text{K}\sqrt{\text{T}}$[Where $\text{K}$ is constant of proportionality]

$\frac{v}{\sqrt{\text{T}}}$ = $\text{K}$

$\frac{v_1}{\sqrt{{\text{T}}_1}}$ = $\frac{v_2}{\sqrt{{\text{T}}_2}}$

Inputting the known values, we have:

$\frac{340}{\sqrt{289}}$ = $\frac{v_2}{\sqrt{400}}$

$\frac{340}{17}$ = $\frac{v_2}{20}$

$20$ = $\frac{v_2}{20}$

$20\times 20$ = $v_2$

$v_2$ = $400{\text{ms}}^{–1}$

Question 8

An object $4\text{cm}$ high is placed $15\text{cm}$ from a concave mirror of focal length $5\text{cm}$. The size of the image is

A. $3\text{cm}$

B. $5\text{cm}$

C. $4\text{cm}$

D. $2\text{cm}$

Explanation

Object height (${\text{h}}_o$) = $4\text{cm}$
Object distance from mirror ($u$) = $15\text{cm}$
Focal length of mirror ($f$) = $5\text{cm}$
Size of image (${\text{h}}_i$) = $?$

To solve for this kind of question, we can only use magnification formula, which is:

$m$ = $\frac{v}{u}$ = $\frac{{\text{h}}_i}{{\text{h}}_o}$

Where $m$ = magnification
      $v$ = distance of image from mirror
     $u$ = distance of object from mirror
      ${\text{h}}_i$ = height of image
      ${\text{h}}_o$ = height of object

But we cannot directly use this formula because we have two unknowns which are $v$ and what we are to get, ${\text{h}}_i$.

This means that we have to calculate for $v$ first.

We calculate for $v$ by using the mirror formula;

$\frac{1}{f}$ = $\frac{1}{v}+\frac{1}{u}$

$\frac{1}{5}$ = $\frac{1}{v}+\frac{1}{15}$

$\frac{1}{5}–\frac{1}{15}$ = $\frac{1}{v}$

$\frac{3–1}{15}$ = $\frac{1}{v}$

$\frac{2}{15}$ = $\frac{1}{v}$

$v$ = $\frac{15}{2}$

Now that we have gotten the value for $v$, we can then recall our magnification formula:

$\frac{v}{u}$ = $\frac{{\text{h}}_i}{{\text{h}}_o}$

$\frac{\frac{15}{2}}{15}$ = $\frac{{\text{h}}_i}{4}$

$\frac{15}{2}÷15$ = $\frac{{\text{h}}_i}{4}$

$\frac{15}{2}\times \frac{1}{15}$ = $\frac{{\text{h}}_i}{4}$

$\frac{\cancel{15}}{2}\times \frac{1}{\cancel{15}}$ = $\frac{{\text{h}}_i}{4}$

$\frac{1}{2}\times 1$ = $\frac{{\text{h}}_i}{4}$

$1\times 4$ = $2\times {\text{h}}_i$

$\frac{4}{2}$ = ${\text{h}}_i$

${\text{h}}_i$ = $2\text{cm}$

Question 9

An object is embedded in a block of ice, $10\text{cm}$ below the plane surface. If the refractive index of the ice is $1.50$, the apparent depth of the object below the surface is

A. $6.67\text{cm}$

B. $7.63\text{cm}$

C. $7.50\text{cm}$

D. $2.50\text{cm}$

Explanation

Real depth of object below ice = $10\text{cm}$
Refractive index of ice = $1.50$
Apparent depth of object = $?$

$\mathsf{Refractive}\mathsf{index}$ = $\frac{\mathsf{Real}\mathsf{depth}}{\mathsf{Apparent}\mathsf{depth}}$

$1.50$ = $\frac{10}{\mathsf{A}\mathsf{.}\mathsf{d}}$

$1.50\times \mathsf{A}\mathsf{.}\mathsf{d}$ = $10$

$\mathsf{A}\mathsf{.}\mathsf{d}$ = $\frac{10}{1.50}$

$\mathsf{A}\mathsf{.}\mathsf{d}$ = $6.67\text{cm}$

Question 10

Three capacitors of capacitance $2\mu \text{F}$, $4\mu \text{F}$, $8\mu \text{F}$ are connected in parallel and a p.d of $6\text{V}$ is maintained across each capacitor. The total energy stored is

A. $2.52\times {10}^{–6}\text{J}$

B. $6.90\times {10}^{–6}\text{J}$

C. $2.52\times {10}^{–4}\text{J}$

D. $6.90\times {10}^{–4}\text{J}$

Explanation

For three capacitors that are connected in parallel, the effective capacitance is    ${\text{C}}_1+{\text{C}}_2+{\text{C}}_3$

Effective capacitance = $2\mu \text{F}+4\mu \text{F}+8\mu \text{F}$

Effective capacitance = $14\mu \text{F}$
Effective capacitance = $14\times {10}^{–6}\text{F}$[$\mu$ = ${10}^{–6}$]

$\text{E}$ = $\frac{1}{2}{\text{CV}}^2$

Where $\text{E}$ = energy
          $\text{C}$ = capacitance
          $\text{V}$ = voltage

$\text{E}$ = $\frac{1}{2}\times 14\times {10}^{–6}\times 6^2$

$\text{E}$ = $7\times 36\times {10}^{–6}$

$\text{E}$ = $252\times {10}^{–6}$

$\text{E}$ = $2.52\times {10}^{–4}\text{J}$

Question 11

Three $4\mathrm{\Omega }$ resistors connected in parallel have a potential difference of $16\text{V}$ applied across them. What is the total current in the circuit

A. $12\text{A}$

B. $8\text{A}$

C. $10\text{A}$

D. $14\text{A}$

Explanation

For three resistors connected in parallel, effective resistance is

$\frac{1}{{\text{R}}_{\text{total}}}$ = $\frac{1}{{\text{R}}_1}+\frac{1}{{\text{R}}_2}+\frac{1}{{\text{R}}_3}$

$\frac{1}{{\text{R}}_{\text{total}}}$ = $3\times \frac{1}{4}$

$\frac{1}{{\text{R}}_{\text{total}}}$ = $\frac{3}{4}$

${\text{R}}_{\text{total}}$ = $\frac{4}{3}$


Now we can calculate for the total current in the circuit by :

$\text{V}$ = $\text{IR}$

$16$ = $\text{I}\times \frac{4}{3}$

$3\times 16$ = $4\text{I}$

$\text{I}$ = $\frac{3\times 16}{4}$

$\text{I}$ = $12\text{A}$

Question 12

The diagram above shows a balanced metre bridge. The value of x is

A. $66.7\text{cm}$

B. $25.0\text{cm}$

C. $33.3\text{cm}$

D. $75.0\text{cm}$

Explanation

In a balanced metre bridge, the resistance is directly proportional to the length of the conductor;

$\text{R}\alpha \text{l}$

$\text{R}$ = $\text{Kl}$ [where $\text{K}$ is the constant of proportionality]

$\frac{\text{R}}{\text{l}}$ = $\text{K}$

$\frac{{\text{R}}_1}{{\text{l}}_1}$ = $\frac{{\text{R}}_2}{{\text{l}}_2}$

${\text{R}}_1$ = effective resistance of two $8\mathrm{\Omega }$ resistors connected in parallel.

Calculating for ${\text{R}}_1$:

$\frac{1}{R_1}$ = $\frac{1}{8}+\frac{1}{8}$

$\frac{1}{R_1}$ = $\frac{1+1}{8}$

$\frac{1}{R_1}$ = $\frac{2}{8}$

$\frac{1}{R_1}$ = $\frac{1}{4}$

$R_1$ = $4\mathrm{\Omega }$

Now that we have ${\text{R}}_1$, we can go ahead with our calculations;

${\text{l}}_1$ = $x\text{cm}$
${\text{R}}_2$ = $8\mathrm{\Omega }$
${\text{l}}_2$ = $100–x\text{cm}$

Substituting known values into the formula:

$\frac{4}{x}$ = $\frac{8}{100–x}$

$4(100–x)$ = $8x$

$400–4x$ = $8x$

$400$ = $8x+4x$

$400$ = $12x$

$\frac{400}{12}$ = $x$

$33.3$ = $x$

$x$ = $33.3\text{cm}$

Question 13

In the diagram above, a $200\text{W}$ bulb is lighted by a $240\text{V}$ a.c mains supply. If $1\text{kWh}$ is sold at ₦$40$, the cost of keeping the bulb lighted for a day is.

A. ₦$192.00$

B. ₦$1.92$

C. ₦$19.20$

D. ₦$1920.00$

Explanation

Power of light bulb ($\text{P}$) = $200\text{W}$
Voltage from a.c mains supply ($\text{V}$) = $240\text{V}$
$1\text{kWh}$ = ₦$40$
Time ($\text{t}$) = $1$ day = ($1\times 24$)$\text{h}$ = $24\text{h}$

Amount in $\text{Wh}$ that will be used to run the light bulb for a day is $\text{power}\times \text{time}$ = $200\times 24$ = $4800\text{Wh}$

Conversion of $\text{Wh}$ to $\text{kWh}$:

= $\frac{4800}{1000}$

= $4.8\text{kWh}$

Recall,

$1\text{kWh}$ = ₦$40$
$4.8\text{kWh}$ = ₦$x$[by cross multiplication]

$x\times 1$ = $4.8\times 40$

$x$ = $192$

Question 14

Two inductors of inductances $5m\text{H}$ and $15m\text{H}$ are connected in series and a current of $5\text{A}$ flows through them. The total energy stored in the inductors is

A. $250.0\text{J}$

B. $50.0\text{J}$

C. $62.5\text{J}$

D. $500.0\text{J}$

Explanation

For two inductors connected in series, the effective inductance is ${\text{L}}_1+{\text{L}}_2$

Effective inductance:
= $5m\text{H}+15m\text{H}$
= $20m\text{H}$
= $20\times {10}^{–3}\text{H}$

Total energy stored in the inductor is given by:

$\text{E}$ = $\frac{1}{2}{\text{LI}}^2$

Where $\text{E}$ = energy
         $\text{L}$ = inductance
         $\text{I}$ = current

$\text{E}$ = $\frac{1}{2}\times 20\times {10}^{–3}\times 5^2$

$\text{E}$ = $0.25\text{J}$

Question 15

The ground state energy for a hydrogen atom is $5.44\times {10}^{–19}\text{J}$. If an electron drops from zero to ground state, calculate the frequency of the emitted radiation.
[ $\text{h}$ = $6.6\times {10}^{–34}\text{Js}$ ]

A. $2.0\times {10}^{16}\text{Hz}$

B. $2.0\times {10}^{15}\text{Hz}$

C. $5.0\times {10}^{15}\text{Hz}$

D. $5.0\times {10}^{16}\text{Hz}$

Explanation

Energy ($\text{E}$) = $5.44\times {10}^{–19}\text{J}$
Frequency = $?$
$\text{h}$ = $6.6\times {10}^{–34}\text{Js}$

$\text{E}$ = $\text{h}f$

$5.44\times {10}^{–19}$ = $6.6\times {10}^{–34}\times f$

$\frac{5.44\times {10}^{–19}}{6.6\times {10}^{–34}}$ = $f$

$8.24\times {10}^{14}$ = $f$

$f$ = $8.24\times {10}^{14}\text{Hz}$