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At the magnetic equator, the angle of dip (or inclination) is zero because the magnetic field lines are horizontal.

At a neutral point, the magnetic field is weak or zero, so the magnetic lines of force are not closely packed together. Instead, they are sparse in that region.

Iron is used because it has high magnetic permeability, which enhances the magnetic field produced by the electromagnet.

To calculate the energy stored in the inductor at resonance, we can use the formula for the energy stored in an inductor: Energy stored in an inductor = \(\frac{1}{2}\) x L x I\(^2\) = 0.5 x 0.1 x 1x1 = 0.050J

\(g_1 = \frac{Gm_1m_2}{r^2_1}\) \(m_1m_2 = \frac{gr^2_1}{G}\) \(g_2 = \frac{Ggr^2_1}{Gr^2_2} = \frac{10 \times 0.1^2}{0.2^2}\) = 2.5N

As the plates of a charged variable capacitor are moved closer together, the potential difference between them decreases. This is because the capacitance increases as the distance between the plates decreases. For a constant charge, the potential difference (V = \(\frac{\text{Q}}{\text{C}}\)) will decrease when capacitance (C) increases.

\(\frac{1}{c} = \frac{1}{6} + \frac{1}{6} + \frac{1}{6}\) = \(\frac{1 + 1 + 1}{6} = \frac{3}{6} = \frac{1}{2}\) = 2\(\mu F\)

Energy = \(\frac{1}{2}CV^2 = \frac{1}{2} \times 2 \times 10^{-6} \times 100^2\) = 1.0 x 10-2J

P = IV I = \(\frac{P}{V} = \frac{3 \times 100}{250} = 1.2A\)

Electrons passing through crystals are diffracted due to their wave properties. This wave-like behavior allows them to interact with the periodic arrangement of atoms in the crystal, leading to diffraction patterns similar to those seen with light waves. Thus, the correct answer is C. possesses wave properties.

\(X_L = 2 \pi fL = 2 \times \pi \times 60 \times 0.7 = 264\Omega \)

The production of X-rays in an X-ray tube begins with thermionic emission from the hot cathode.

Table of values L(cm)\(^{3}\) Ls(cm) Lp(cm) \(R_1\) 100 5.3 94.7 17.90 90 6.5 93.5 14.40 80 7.0 93.0 13.30 70 8.5 91.5 10.80 60 9.5 90.5 9.50 L(cm) Is\(^{1}\) Lg \(R_2\) 100 21.60 78.4 3.60 90 22.1 77.9 3.50 80 23.4 76.6 3.30 70 24.3 75.7 3.10 60 25.3 74.7 2.95 L(cm) Ls(cm) Lp(cm) \(R_1\) Ls\(^{1}\)cm\(^{3}\) Lq R\(^{2}\) 100 5.3 94.7 17.90 21.6 78.4 3.60 90 6.5 93.5 14.40 22.1 77.9 3.50 80 7.0 93.0 13.30 23.4 76.6 3.30 70 8.5 91.5 10.80 24.3 75.7 3.10 60 9.5 90.5 9.50 25.3 74.7 2.95 Slope (s) = of the graph = \(\frac{DR_{2}}{DR_{1}} = \frac{1.4}{8.0}\) = 0.175 K = \(\sqrt 5\) = 0.17555 = 0.42 Precautions: Key was opened where readings were not being taken. Tight connections were ensured ParallaX errors were avoided in reading the metre rule. Zero error was noted in reading the metre rule. The jockey was not rolled on or allowed to scratch the wire. (b) The resistivity of a material is defined as the resistance of a unit length of the material/wire of a uni cross-sectional area. OR P = \(\frac{AR}{L}\) where A = cross-sectional area R = eletrical resistance L = length of the wire V = Vg + \(V_R\) = IRg + IR = I (Rg + R) \(\therefore\) \(\frac{V}{I}\) = Rg+ R = \(\frac{1.5}{1.5 \times 10^{-3}}\) = 50 + R = 10\(^{3}\) = 50 + R \(\therefore\) R = 950\(\Omega\)

in vibrations in string, the fundamental frequency happens to be the first harmonic, and is given as f\(_0\) = \(\frac{V}{2l}\) Where V = Velocity of the sound wave, I = Length of the string. The second harmonic is the first overtone given by f1 = \(\frac{V}{l}\) = 2f0 ; and the third harmonic happens to be the second overtone, given by f2 = \(\frac{3V}{2l}\) = 2f0. Thus if V = 120m/s and l = 1.5m Thus f1 = 2f0 = 2 x 40 = 80Hz Thus f2 = 3f0 = 3 x 40 = 120Hz Thus we have in order, 40Hz, 80Hz, 120Hz

Electrical Energy = IVt = 4 x 240 x (1 x 60) = 57600J = 5.76 x 104J