Determine the period of the pendulum.

The time taken to move from the mean position to a maximum displacement is one-quarter of a full oscillation.

$$\frac{T}{4} = 0.5 \text{ s}$$ $$T = 4 \times 0.5 \text{ s}$$ $$T = 2.0 \text{ s}$$

Frequency ($f$) is the reciprocal of the period ($T$).

$$f = \frac{1}{T}$$ $$f = \frac{1}{2.0 \text{ s}}$$ $$f = \boxed{\text{0.5 Hz}}$$

6)

The formula for apparent depth ($d'$) is $d' = \frac{d}{n}$, where $d$ is the real depth and $n$ is the refractive index.
For the transparent material:

$$d'_1 = \frac{12 \text{ cm}}{6/5}$$ $$d'_1 = 12 \text{ cm} \times \frac{5}{6}$$ $$d'_1 = 10 \text{ cm}$$

For the glass block:

$$d'_2 = \frac{9 \text{ cm}}{3/2}$$ $$d'_2 = 9 \text{ cm} \times \frac{2}{3}$$ $$d'_2 = 6 \text{ cm}$$

Vertical displacement is the difference between the total real depth and the total apparent depth.

$$\text{Vertical displacement} = d_{\text{total}} - d'_{\text{total}}$$ $$\text{Vertical displacement} = 21 \text{ cm} - 16 \text{ cm}$$ $$\text{Vertical displacement} = \boxed{\text{5 cm}}$$

7)
To reverse bias a diode, the positive terminal of the dry cell is connected to the cathode (the bar side of the diode symbol), and the negative terminal is connected to the anode (the triangle side of the diode symbol).

\begin{tikzpicture}[scale=0.8] % Dry cell \draw (0,0) to (0,0.5); \draw[line width=1.5pt] (0,0.5) to (0,0.8); % Positive terminal \draw (0,0) to (0,-0.5); \draw[line width=0.5pt] (0,-0.5) to (0,-0.8); % Negative terminal \node at (0.5, 0.65) {$+$}; \node at (0.5, -0.65) {$-$}; % Wires \draw (0,0.8) -- (2,0.8); \draw (0,-0.8) -- (2,-0.8); % Diode (reverse biased) \draw (2,0.8) -- (3,0.8); \draw (3,-0.8) -- (2,-0.8); \draw (3,0.8) -- (3,-0.8); % Cathode bar \draw[fill=white] (3,0) -- (2,0.8) -- (2,-0.8) -- cycle; % Anode triangle \node at (2.5, 0) {$\blacktriangleright$}; % Arrow for diode direction % Connecting wires \draw (3,0.8) -- (4,0.8); \draw (3,-0.8) -- (4,-0.8); \end{tikzpicture}

8)

The image is real and half the size of the object, so the magnification ($M$) is $-\frac{1}{2}$ (negative for a real, inverted image).

$$M = \frac{v}{u} = -\frac{1}{2}$$

For a real image formed by a converging lens, $u$ and $v$ are positive distances. The negative sign in magnification indicates inversion.
So, the magnitude of the image distance is half the magnitude of the object distance:

$$v = \frac{u}{2}$$

The distance between the object and the image is $450 \text{ mm}$. For a real image, the object and image are on opposite sides of the lens, so the distance between them is $u+v$.

$$u + v = 450 \text{ mm}$$

Substitute $v = \frac{u}{2}$ into the equation:

$$u + \frac{u}{2} = 450 \text{ mm}$$ $$\frac{3u}{2} = 450 \text{ mm}$$ $$u = \frac{450 \times 2}{3} \text{ mm}$$ $$u = 300 \text{ mm}$$

Now find $v$:

$$v = \frac{u}{2} = \frac{300 \text{ mm}}{2}$$ $$v = 150 \text{ mm}$$

For a converging lens forming a real image, the lens formula is:

$$\frac{1}{f} = \frac{1}{u} + \frac{1}{v}$$ $$\frac{1}{f} = \frac{1}{300 \text{ mm}} + \frac{1}{150 \text{ mm}}$$

Find a common denominator:

$$\frac{1}{f} = \frac{1}{300 \text{ mm}} + \frac{2}{300 \text{ mm}}$$ $$\frac{1}{f} = \frac{3}{300 \text{ mm}}$$ $$\frac{1}{f} = \frac{1}{100 \text{ mm}}$$ $$f = \boxed{\text{100 mm}}$$