Physics 2007 Jamb Examination

Resolve each force into its rectangular components (as indicated by the diagram), sum horizontal components to get \(R_x\) and sum vertical components to get \(R_y\). The magnitude of the resultant is then given by the Pythagorean relation:\[ R=\sqrt{R_x^2+R_y^2} \]Carrying out the component addition from the diagram yields \(R_x=10\) N and \(R_y=0\) N, therefore\[ R=\sqrt{10^2+0^2}=10\;\text{N}. \]

First find the lamp current at its rated conditions:\[ I=\dfrac{P}{V_{lamp}}=\dfrac{60}{120}=0.5\;\text{A}. \]The resistor must drop the difference between supply and lamp voltages: \(V_R=220-120=100\) V. Using Ohm's law for the required series resistance:\[ R=\dfrac{V_R}{I}=\dfrac{100}{0.5}=200\;\Omega.\]

Number of half-lives in 80 days:\[ n=\dfrac{80}{20}=4. \]Fraction remaining after \(n\) half-lives is \(\left(\tfrac{1}{2}\right)^n\). Therefore:\[ \left(\tfrac{1}{2}\right)^4=\tfrac{1}{16}. \]

Photon energy relation: \(E=\dfrac{hc}{\lambda}\) so \(E\lambda=hc\). Compute numerically:\[ X=hc=(6.63\times10^{-34})(3.00\times10^{8})=19.89\times10^{-26}\;\text{J·m}=1.989\times10^{-25}. \]Rounded to two significant figures gives \(1.99\times10^{-25}\).

Semiconductors such as silicon and germanium consist of a crystal lattice in which atoms share electrons with neighbours forming directional bonds. These materials are held together predominantly by covalent bonding, which gives the characteristic band structure enabling semiconducting behaviour.

Aluminium has one fewer valence electron than silicon, so it creates acceptor levels leading to a p-type material. Arsenic has one more valence electron than germanium (or silicon), creating donor levels and producing n-type material.

Electric field from a point charge is proportional to charge: \(E\propto q\) at fixed distance. So scale the given field by the charge ratio:\[ E_2=E_1\times\dfrac{q_2}{q_1}=360\times\dfrac{120}{50}=360\times2.4=864\;\text{N·C}^{-1}. \]

Work done in moving charge across a potential difference: \(W=QV\). Convert units: \(Q=600\times10^{-6}\) C, \(V=3000\) V.\[ W=(600\times10^{-6})(3000)=1.8\;\text{J}. \]

Resistivity relation: \(\rho=R\dfrac{A}{L}\). Area of circular cross-section: \(A=\pi r^2\) with \(r=0.5\times10^{-3}\) m.\[ A=\pi(0.5\times10^{-3})^2=\pi\times0.25\times10^{-6}=7.85398\times10^{-7}\;\text{m}^2. \]Compute \(\rho\):\[ \rho=0.1\times\dfrac{7.85398\times10^{-7}}{2.0}=3.92699\times10^{-8}\;\Omega\,\text{m}\approx3.93\times10^{-8}. \]

Find transmission current: \(I=\dfrac{P}{V}=\dfrac{40000}{800}=50\) A. Power loss in resistor: \(P_{loss}=I^2R=50^2\times2=2500\times2=5000\) W = \(5.0\times10^3\) W.