Physics 2006 Jamb Examination

Wave number $k=\dfrac{\pi}{4}\;\text{rad per unit length}$. Phase difference $\Delta\phi=kx$. Set $\Delta\phi=45^{\circ}=\dfrac{\pi}{4}$, so $\dfrac{\pi}{4}x=\dfrac{\pi}{4}\Rightarrow x=1$ (cm).\[ Y = \sin \pi (150\pi t - \pi x/4) \] Comparing the above with the standard wave equation: \[ Y = A \sin \pi \left(\dfrac{2vt}{\lambda} - \dfrac{2x}{\lambda}\right) \] \[ \Rightarrow \text{the phase difference at } x/4 = \dfrac{2x}{\lambda} \] \[ \therefore \; 14 = 2\lambda \quad \Rightarrow \quad \lambda = 8 \, \text{cm} \] \[ \therefore \; \dfrac{2\pi x}{8} = 45^\circ \] Since \(\pi = 180^\circ\): \[ 45^\circ \times 8 = 2 \times 180^\circ \times x \] \[ \therefore \; x = 1 \, \text{cm} \]

From the diagram one complete wave (one period) spans 4 s (reading time-axis divisions). Frequency $f=\dfrac{1}{T}=\dfrac{1}{4}=0.25\;\text{Hz}$.

Field lines for unlike (north–south) poles emerge from the north and enter the south, showing continuous lines connecting the two poles concentrated between them. 

An n–p–n transistor has n-type emitter and collector with a p-type base; the emitter arrow points outward (conventional current leaving the emitter). The figure that shows n–p–n with the arrow pointing out is option C.

With base current zero, the transistor receives no bias to allow collector–emitter current; it is in the cut-off region (no conduction).

Pressure of the Gas, P = A + h = 76 + 10 86 cm Hg

Vertical equilibrium: $T\sin30^{\circ}=75\;\text{N}$ and $\sin30^{\circ}=0.5$. Thus $T=\dfrac{75}{0.5}=150\;\text{N}$.

From the ideal gas relation $PV=\text{constant (}=nRT\text{)}$, plotting $V$ against $1/P$ gives slope $=nRT$ whose unit is pressure$\times$volume = energy, i.e. work.

For projectile motion: $R=\dfrac{u^2\sin2\theta}{g}$ and $H=\dfrac{u^2\sin^2\theta}{2g}$. Requirement $R=3H$ gives $\sin2\theta=\tfrac{3}{2}\sin^2\theta$. Simplifying yields $\tan\theta=\tfrac{4}{3}$, so $\theta\approx53^{\circ}$.

Resistive force $F=\dfrac{P}{v}=\dfrac{80\times10^{3}}{32}=2500\;\text{N}=2.50\times10^{3}\;\text{N}$.