Physics 2007 Jamb Examination

Find the lamp current at its rated conditions first, then the series resistance needed so the lamp sees 120 V.\[ I = \dfrac{P}{V_{lamp}} = \dfrac{60}{120} = 0.5\;\text{A} \]Voltage to be dropped across the series resistor:\[ V_R = 220 - 120 = 100\;\text{V} \] Required resistance:\[ R = \dfrac{V_R}{I} = \dfrac{100}{0.5} = 200\;\Omega \]

Determine the transmission current from the transmitted power and supply voltage, then compute I²R loss.
Transmitted current:\[ I = \dfrac{P_{trans}}{V} = \dfrac{40000}{800} = 50\;\text{A} \]Power loss in resistor:\[ P_{loss} = I^2R = 50^2 \times 2 = 2500 \times 2 = 5000\;\text{W} = 5.0\times10^3\;\text{W} \]

The magnetic force per unit length between two parallel currents is equal in magnitude and opposite in direction on each wire. If the given force per unit length is 5×10^-5 N, the other wire experiences the same magnitude.

Lenz's law states that the induced emf produces a current whose magnetic effect opposes the change in magnetic flux that produced it. This ensures that energy is conserved (the induced current does not create energy but opposes the change that produced it).

For an inclined plane, efficiency η relates ideal effort and actual effort by η = (weight × sinθ) / effort. Solve for effort:\[ \text{ideal effort} = W\sin\theta = 120\times\sin30^\circ = 120\times0.5 = 60\;\text{N} \]\[ \eta = \dfrac{60}{F} = 0.60 \Rightarrow F = \dfrac{60}{0.60} = 100\;\text{N} \]

Under Hooke's law (in the elastic limit) extension is proportional to applied force. Given initial extension e₁ and force F₁:\[ F_1 = mg = 0.4\times9.8 = 3.92\;\text{N} \]\[ e_1 = 0.02\;\text{m} \]For force F₂ = 15 N the extension e₂ = e₁(F₂/F₁):\[ e_2 =0.02\times\dfrac{15}{3.92} \approx 0.07653\;\text{m} \]New length = original length + e₂ = 4.00 + 0.07653 ≈ 4.08 m.

Work done during an isothermal process of an ideal gas relates pressure and volume; Boyle's law (P ∝ 1/V at constant T) underpins the PV integral used to compute work in isothermal processes.

State differences arise from how closely particles are packed (spaces) and the strength of intermolecular forces. These determine structure, compressibility, and flow characteristics of each state.

Given data:\[\text{Area } A = 118.8\ \text{cm}^2 = 118.8\times10^{-4}\ \text{m}^2 = 0.01188\ \text{m}^2,\]\[\text{thickness } t = 10^{-6}\ \text{m},\qquad \rho = 9\times10^{3}\ \text{kg·m}^{-3},\]\[I = 10\ \text{A},\qquad E = 3.3\times10^{-7}\ \text{kg·C}^{-1}.\]Volume of copper deposited:\[V = A\times t = 0.01188\times10^{-6} =1.188\times10^{-8}\ \text{m}^3.\]Mass of copper required:\[m = \rho V = 9\times10^{3}\times 1.188\times10^{-8} = 1.0692\times10^{-4}\ \text{kg}.\]Total charge required (using electrochemical equivalent):\[Q=\dfrac{m}{E}=\dfrac{1.0692\times10^{-4}}{3.3\times10^{-7}} \approx 324\ \text{C}.\]Time at 10 A:\[t=\dfrac{Q}{I}=\dfrac{324}{10}=32.4\ \text{s}\approx 0.54\ \text{min}.\]\[\boxed{t\approx 32.4\ \text{s}\ (\,\approx 0.54\ \text{min}\,).}\]

First step: $^{235}_{92}\text{W} \;\rightarrow\; ^{235}_{93}\text{X}$  
$A$ unchanged, $Z$ increases by 1 $\;\Rightarrow\;$ $\beta^-$-decay (a neutron converts to a proton, emitting $\beta^-$).  

Second step: $^{235}_{93}\text{X} \;\rightarrow\; ^{231}_{91}\text{Y}$  
$A$ decreases by 4, $Z$ decreases by 2 $\;\Rightarrow\;$ $\alpha$-decay (emission of an $\alpha$-particle, $^{4}_{2}\text{He}$).  

Hence, the sequence of emissions is: $\beta^-$ then $\alpha$.