WAEC SSCE - Physics - 1988 (Essay)

(a) Moment of a force about a point

The moment of a force is its turning effect about a pivot or point. It is given by:

\[ \text{Moment} = F \times d \]

where \(F\) is the force and \(d\) is the perpendicular distance from the pivot to the line of action of the force. The SI unit is newton metre, \(\text{N m}\).

The distance used is not simply any distance from the pivot to the force: it must be measured at right angles to the force's line of action.

(b) Conditions for equilibrium of coplanar parallel forces

• The total force in one direction equals the total force in the opposite direction, so the resultant force is zero.
• The sum of clockwise moments about any point equals the sum of anticlockwise moments about that point, so the resultant moment is zero.

Metre rule calculation

Since the metre rule balances by itself at the \(48\,\text{cm}\) mark, its weight acts at the \(48\,\text{cm}\) mark.

When the \(60\,\text{g}\) body is at \(6\,\text{cm}\), the pivot is at \(30\,\text{cm}\).

• Distance of the \(60\,\text{g}\) mass from the pivot: \(30-6=24\,\text{cm}\).
• Distance of the rule's weight from the pivot: \(48-30=18\,\text{cm}\).

Taking moments about the balance point:

\[ 60 \times 24 = M \times 18 \]

\[ M=\frac{60\times24}{18}=80\,\text{g} \]

(i) Mass of the metre rule \(=80\,\text{g}\).

For the body at the \(13\,\text{cm}\) mark, let the new balance point be \(x\,\text{cm}\) from the zero end. The body is to the left of the pivot and the rule's weight acts to the right:

\[ 60(x-13)=80(48-x) \]

\[ 60x-780=3840-80x \]

\[ 140x=4620 \]

\[ x=33\,\text{cm} \]

(ii) The balance point is \(33\,\text{cm}\) from the zero end.

(c) Relationship between efficiency, mechanical advantage and velocity ratio

Efficiency is:

\[ E=\frac{\text{work output}}{\text{work input}}\times100\% \]

If \(L\) is the load, \(P\) is the effort, \(d_L\) is the distance moved by the load, and \(d_P\) is the distance moved by the effort:

\[ E=\frac{L\times d_L}{P\times d_P}\times100\% \]

Rearranging the factors:

\[ E=\frac{L}{P}\times\frac{d_L}{d_P}\times100\% \]

But:

\[ M.A=\frac{L}{P} \qquad \text{and} \qquad V.R=\frac{d_P}{d_L} \]

Therefore:

\[ \frac{d_L}{d_P}=\frac{1}{V.R} \]

Hence:

\[ \boxed{E=\frac{M.A}{V.R}\times100\%} \]

For the machine, the useful work done on the load is:

\[ W_{\text{output}}=mgh=200\times10\times3.0=6000\,\text{J} \]

Since the efficiency is \(80\%=0.80\):

\[ 0.80=\frac{W_{\text{output}}}{W_{\text{input}}} \]

\[ W_{\text{input}}=\frac{6000}{0.80}=7500\,\text{J} \]

The work done by the person is \(7500\,\text{J}\) (or \(7.5\,\text{kJ}\)).

(a)(i) Electric field intensity

Electric field intensity, \(E\), at a point is the force experienced per unit positive test charge placed at that point:

\[E=\frac{F}{q}\]

It is a vector quantity. Its direction is the direction of the force on a positive test charge. Its SI unit is \(\text{N C}^{-1}\), equivalently \(\text{V m}^{-1}\).

(a)(ii) Electric lines of force

Electric lines of force are imaginary lines used to show an electric field pattern. The tangent to a line at any point gives the direction of the electric field at that point. Lines leave positive charges and enter negative charges. They never cross, because the field cannot have two different directions at one point. Where the lines are closer together, the electric field is stronger.

(b) Resultant electric field intensity at \(P\)

The distances shown are \(0.03\ \text{m}\) from \(+q\) and \(0.05\ \text{m}\) from \(-q\). Since these add to \(0.08\ \text{m}\), point \(P\) lies between the charges.

The field due to a positive charge points away from it. At \(P\), this is towards the negative charge. The field due to a negative charge points towards it, which is also towards the negative charge. Therefore, the two fields act in the same direction and must be added.

For the positive charge:

\[ E_{+}=\frac{1}{4\pi\varepsilon_0}\frac{q}{r^2} =\frac{(9.0\times10^9)(5.0\times10^{-8})}{(0.03)^2} =5.0\times10^5\ \text{N C}^{-1} \]

For the negative charge:

\[ E_{-}=\frac{(9.0\times10^9)(5.0\times10^{-8})}{(0.05)^2} =1.8\times10^5\ \text{N C}^{-1} \]

\[ E=E_{+}+E_{-} =5.0\times10^5+1.8\times10^5 =6.8\times10^5\ \text{N C}^{-1} \]

Result: \(\boxed{6.8\times10^5\ \text{N C}^{-1}}\), directed from \(+q\) towards \(-q\).

The supplied reference answer subtracts the two field magnitudes. That would only be appropriate if the fields were in opposite directions. Here \(P\) is between opposite charges, so both fields point towards \(-q\).

(c) Series \(R\)-\(C\) circuit

Using the circuit values shown in the supplied reference material: \(R=1000\ \Omega\), \(C=10\times10^{-6}\ \text{F}\), \(f=25\ \text{Hz}\), and \(V=90\ \text{V}_{\rm rms}\).

1. Capacitive reactance

   \[ X_C=\frac{1}{2\pi fC} =\frac{1}{2\pi(25)(10\times10^{-6})} =\frac{2000}{\pi} =636.6\ \Omega \]

2. Impedance

   In a series \(R\)-\(C\) circuit, resistance and capacitive reactance are at right angles in the phasor diagram:

   \[ Z=\sqrt{R^2+X_C^2} =\sqrt{(1000)^2+(636.6)^2} =1.19\times10^3\ \Omega \]

3. Current

   \[ I=\frac{V}{Z} =\frac{90}{1185.5} =0.0759\ \text{A}_{\rm rms} \]

4. Voltage across the capacitor

   \[ V_C=IX_C =(0.0759)(636.6) =48.3\ \text{V}_{\rm rms} \]

5. Average power used

   Only the resistor dissipates average power. An ideal capacitor stores and returns energy, so its average power is zero.

   \[ P=I^2R =(0.0759)^2(1000) =5.76\ \text{W} \]

The value \(I^2Z=6.83\) is not the average power in watts; it is associated with apparent power, \(VI\), measured in volt-amperes. For average power in an \(R\)-\(C\) circuit, use \(I^2R\) or \(VI\cos\phi\).

Examination reminder: decide whether electric-field vectors point in the same or opposite directions before adding their magnitudes. In an a.c. circuit containing a capacitor, use \(R\), not \(Z\), when calculating real average power.