Physics 2007 Jamb Examination

Inductance:
From the diagram: $L=0.10\ \text{H}$, $f=\dfrac{50}{\pi}\ \text{Hz}$, supply voltage $V_s=75\ \text{V}$ and current $I=1.5\ \text{A}$.1. Inductive reactance: $$X_L = 2\pi f L = 2\pi\left(\dfrac{50}{\pi}\right)(0.10)=2\times50\times0.10=10\ \Omega.$$2. Magnitude of the total impedance:$$Z=\dfrac{V_s}{I}=\dfrac{75}{1.5}=50\ \Omega.$$Resistance $R$ from $Z^2=R^2+X_L^2$:$$R=\sqrt{Z^2-X_L^2}=\sqrt{50^2-10^2}$$  $$=\sqrt{2500-100}=\sqrt{2400}=48.99\ \Omega\approx49\ \Omega.$$

For a stretched string \(f\propto\sqrt{T}\) at fixed length and mass per unit length. Doubling frequency requires quadrupling tension.\[T_2 = T_1\left(\dfrac{f_2}{f_1}\right)^2 = 400\left(\dfrac{500}{250}\right)^2 = 400\times4 = 1600\;\text{N}\]

Three identical cells connected in parallel have the same terminal voltage as one cell while their internal resistances combine like resistors in parallel. For internal resistance \(r=2.5\,\Omega\) and three identical parallel branches:\[r_{eq}=\dfrac{r}{3}=\dfrac{2.5}{3}\approx0.83\,\Omega\]The net e.m.f. equals the single cell e.m.f. = 1.5 V.

Use the vector relation \(\mathbf{F}=I\mathbf{L}\times\mathbf{B}\). With magnetic field into the page and force upwards, the current direction must be towards the left so that \(\mathbf{L}\times\mathbf{B}\) points upward (use right-hand rule for cross product with conventional current).

One quarter period corresponds to \(\omega t=\pi/2\). Since \(\sin(\pi/2)=1\):\[\varepsilon=\varepsilon_0\sin\left(\dfrac{\pi}{2}\right)=\varepsilon_0\]

The bob attains maximum speed at the equilibrium position where potential energy is minimum and kinetic energy is maximum. The labelled point at the centre (equilibrium) is where speed is highest.

We solve it using the principle of moments.

The rod is uniform, so its weight acts at its midpoint (25 cm from either end).

From the diagram (given in your question set), the forces balance the rod on the fulcrum. One force acts at 20 cm from F, the other at 30 cm from F.

Taking moments about F:

W × 25= (40 × 20) + (40 × 30)= 800 + 1200 = 2000$$W = \dfrac{2000}{25} = 80 \,\text{N}$$

Only the horizontal component of the force does work along the floor.
Horizontal component = \(2\cos60^\circ=2\times0.5=1\,\text{N}\). Work = force × distance = \(1\times3=3\,\text{J}\).

For isotropic solids \(\beta\approx3\alpha\) where \(\alpha\) is linear expansivity. So \(\alpha=\beta/3=6.3\times10^{-6}/3=2.1\times10^{-6}\).
Areal expansivity ≈ \(2\alpha=2\times2.1\times10^{-6}=4.2\times10^{-6}\).

For small temperature change \(\Delta T\), length increase \(e=\alpha l\Delta T\).
Hence \(\dfrac{e_1}{e_2}=\dfrac{\alpha_1 l_1}{\alpha_2 l_2}\).
Rearranging for the length ratio: \[\dfrac{l_1}{l_2}=\dfrac{e_1}{e_2}\cdot\dfrac{\alpha_2}{\alpha_1}=\dfrac{1}{2}\cdot\dfrac{4}{3}=\dfrac{2}{3}.\]