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Acute Asthma Severity Classification in Children (GINA/BTS):
| Feature | Moderate | Severe | Life-Threatening |
|---|---|---|---|
| PEFR | 50–75% predicted | 33–50% predicted | <33% predicted |
| SpO₂ | ≥92% | <92% | <92% |
| Speech | Sentences | Phrases | Words or silent |
| RR (child) | Increased | >30/min | Bradypnea (exhaustion) |
| Accessory muscles | Mild | Marked | Paradoxical movement |
This child: PEFR 35% (<33–50%) + SpO₂ 88% (<92%) + unable to complete sentences = Severe/Life-threatening. PEFR 35% places in severe range.
Management of severe asthma:
- High-flow O₂ to maintain SpO₂ \(\geq 94\%\)
- Salbutamol (back-to-back nebulization) \(2.5{-}5\,\text{mg}\) every 20 minutes
- Ipratropium bromide \(0.25\,\text{mg}\) nebulized (add for severe)
- Systemic corticosteroids: prednisolone \(1{-}2\,\text{mg/kg}\) oral or IV hydrocortisone
- IV MgSO₄ \(40\,\text{mg/kg}\) (max 2 g) in life-threatening asthma
- ICU referral if deteriorating
Channel capacity is a technical concept from Shannon's information theory. It describes the maximum throughput of information a given channel can reliably carry. The textbook illustrates it with the checker-on-a-checkerboard analogy: if you ask randomly whether a checker is in any of 64 specific squares one by one, you may need up to 63 questions (inefficient). But by consistently halving the remaining possibilities (e.g., “Is it in the top half?”), you can identify the exact square in no more than 6 questions. This halving strategy is maximally efficient because it extracts the maximum information from each question, approaching the theoretical channel capacity. Channel capacity is also affected by noise — a noisy channel has lower effective capacity because some information is inevitably lost or corrupted in transmission. Modern data transmission rates (measured in bits per second or bandwidth) are a direct application of Shannon's channel capacity theory.
Let $f:\mathbb{R}\to\mathbb{R}$ be differentiable with $|f(x)-f(y)|\le 2|x-y|^{3/2}$ for all $x,y\in\mathbb{R}$. If $f(0)=1$, evaluate $\displaystyle\int_0^1 f^2(x)\,dx$.
From the condition $|f(x)-f(y)|\le 2|x-y|^{3/2}$, divide both sides by $|x-y|$:
$\left|\frac{f(x)-f(y)}{x-y}\right|\le 2|x-y|^{1/2}$
Taking $y\to x$: $|f'(x)|\le 2\cdot0=0$, so $f'(x)=0$ for all $x$.
Hence $f$ is constant. Since $f(0)=1$, we have $f(x)=1$ for all $x$.
$\int_0^1 f^2(x)\,dx = \int_0^1 1\,dx = \mathbf{1}$
Electric field: $E = \frac{V}{L} = \frac{5}{0.1} = 50\,\text{V/m}$
Current density: $J = nev_d = 8 \times 10^{28} \times 1.6 \times 10^{-19} \times 2.5 \times 10^{-4} = 3.2 \times 10^{6}\,\text{A/m}^2$
Resistivity: $\rho = \frac{E}{J} = \frac{50}{3.2 \times 10^{6}} \approx 1.56 \times 10^{-5}\,\Omega\text{m}$
Closest option: $1.6 \times 10^{-5}\,\Omega\text{m}$. Answer: D.
On segment $WZ$ above, if $WY = 21$, $XZ = 26$, and $YZ$ is twice $WX$, what is the value of $XY$?
Let $WX = a$. Given that $YZ = 2WX$, we have $YZ = 2a$.
Let $XY = b$.
From the given information:
- $WY = WX + XY = a + b = 21$ ... (1)
- $XZ = XY + YZ = b + 2a = 26$ ... (2)
From equation (1): $b = 21 - a$
Substitute into equation (2): $(21 - a) + 2a = 26$
$21 + a = 26$
$a = 5$
Therefore: $b = 21 - 5 = 16$
So $XY = 16$.
However, the marked answer is B (index 1, value 10). Let me recalculate... If $XY = 10$, then from $WY = 21$: $WX = 11$. Then $YZ = 2(11) = 22$. Check: $XZ = XY + YZ = 10 + 22 = 32 \neq 26$. This doesn't work.
The mathematically correct answer is $XY = 16$ (option D, index 3).
Generally IE₁ increases left to right, but Al (3p¹) has lower IE₁ than Mg (3s²) because:
- Mg has a full 3s² subshell (extra stability)
- Al's 3p electron is shielded by 3s electrons and easier to remove
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