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If $0 < st < 1$, which of the following could be true?
We need $0 < st < 1$, meaning $st$ is a small positive number. This requires $s$ and $t$ to have the same sign.
- Option A: $s < -1, t > 0$ β $st < 0$. Not possible.
- Option B: $s < -1, t < -1$ β both negative, so $st > 0$. For example $s = -2, t = -0.3$: $st = 0.6$, which satisfies $0 < st < 1$. β
- Option C: $s > -1, t < -1$ β could give $st$ negative or positive depending on sign of $s$. If $s = 0.5, t = -2$: $st = -1 < 0$. Not valid for all cases.
- Option D: $s > 1, t < -1$ β $st < -1 < 0$. Not possible.
Option B is the one that can satisfy the condition.
Consider the capacitor as strips of width $dx$ parallel to the shorter side. At horizontal position $x$ from the left, the gap has dielectric filling of height $\frac{d \cdot x}{a}$ (triangular region). The local gap (from bottom) filled with dielectric is $\frac{xd}{a}$ and air gap is $d - \frac{xd}{a} = \frac{d(a-x)}{a}$.
Each strip is two capacitors in series (dielectric + air). Integrating across the width gives:
$C = \int_0^a \frac{\varepsilon_0 K \cdot a \, dx}{d[(K-1)x/a + 1] \cdot a} \Rightarrow C = \frac{K\varepsilon_0 a^2}{d(K-1)}\ln K$
(Select the best word. Note: two words produce sentences alike in meaning β choose the most accurate.)
The ordinal level of measurement allows data to be ranked or ordered (e.g., first, second, third; or high, medium, low), but the intervals between ranks are not necessarily equal. Examples include class rankings, Likert scale ratings, and educational grades as categories.
Correct Answer: B — Ordinal.
1. Calculate Refractive Index ($\mu$):
$\mu = \frac{c}{v} = \frac{3 \times 10^8}{2 \times 10^8} = 1.5$.
2. Determine Radius of Curvature ($R$):
For a lens with radius of aperture $r = 3 \text{ cm}$ and thickness $t = 0.3 \text{ cm}$, the radius of curvature is given by $R \approx \frac{r^2}{2t}$ (for small thickness).
$R = \frac{3^2}{2 \times 0.3} = \frac{9}{0.6} = 15 \text{ cm}$.
3. Lens Maker's Formula:
$\frac{1}{f} = (\mu - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)$. For a plano-convex lens, $R_1 = R$ and $R_2 = \infty$.
$\frac{1}{f} = (1.5 - 1) \left( \frac{1}{15} \right) = 0.5 \times \frac{1}{15} = \frac{1}{30}$.
Thus, $f = 30 \text{ cm}$.
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