An object of mass 5.0 kg is suspended by an aluminum wire of length 4.0 m and diameter, d = 2.0 × 10−3 m. Calculate the extension produced in the wire. [Young's modulus of aluminium = 7.0 × 1010 Nm−2, π = 3.14, g = 10.0 m s−2]

The force $F$ is the weight of the object.

$$F = mg$$

Given $m = 5.0 \text{ kg}$ and $g = 10.0 \text{ m s}^{-2}$.

$$F = 5.0 \text{ kg} \times 10.0 \text{ m s}^{-2}$$ $$F = 50.0 \text{ N}$$

The wire has a circular cross-section, so its area $A$ is given by $A = \pi r^2$ or $A = \pi \left(\frac{d}{2}\right)^2$.
Given $d = 2.0 \times 10^{-3} \text{ m}$ and $\pi = 3.14$.

$$A = 3.14 \times \left(\frac{2.0 \times 10^{-3} \text{ m}}{2}\right)^2$$ $$A = 3.14 \times (1.0 \times 10^{-3} \text{ m})^2$$ $$A = 3.14 \times 1.0 \times 10^{-6} \text{ m}^2$$ $$A = 3.14 \times 10^{-6} \text{ m}^2$$

Young's Modulus $Y$ is given by the formula:

$$Y = \frac{\text{Stress}}{\text{Strain}} = \frac{F/A}{\Delta L/L}$$

We need to find the extension $\Delta L$. Rearrange the formula:

$$\Delta L = \frac{F L}{A Y}$$

Substitute the calculated values and given values:

$$L = 4.0 \text{ m}$$ $$Y = 7.0 \times 10^{10} \text{ N m}^{-2}$$ $$\Delta L = \frac{50.0 \text{ N} \times 4.0 \text{ m}}{(3.14 \times 10^{-6} \text{ m}^2) \times (7.0 \times 10^{10} \text{ N m}^{-2})}$$ $$\Delta L = \frac{200.0 \text{ N m}}{21.98 \times 10^{4} \text{ N}}$$ $$\Delta L = \frac{200.0}{219800} \text{ m}$$ $$\Delta L \approx 0.0009099 \text{ m}$$

Given values have 2 or 3 significant figures. Let's use 3 significant figures.

$$\Delta L \approx 9.10 \times 10^{-4} \text{ m}$$

The extension produced in the wire is $\boxed{\text{9.10} \times 10^{-4} \text{ m}}$.

a) The waveform of the output voltage for a half-wave rectifier will show only the positive (or negative) half-cycles of the input sinusoidal voltage, with the other half-cycles being blocked, resulting in gaps between the pulses.
(A sketch would show a series of positive half-sine waves, followed by a flat line at zero, then another positive half-sine wave, and so on.)

b) Two disadvantages of half-wave rectification are:

• It has a low efficiency because only half of the input AC power is converted to DC output.
• It produces a high ripple factor, meaning the output DC voltage is not smooth and contains a significant AC component, requiring extensive filtering.

a) A diamagnetic substance will orient itself perpendicular to a strong magnetic field.

b) One factor that affects the magnetism of paramagnetic substances is temperature. As temperature increases, the random thermal motion of atoms disrupts the alignment of magnetic dipoles, reducing the overall magnetism.

c) One use of ferromagnetic materials is in the cores of transformers to concentrate magnetic flux and improve efficiency.