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Using $n_1 u_1 = n_2 u_2$:
$128 \times [\text{kg} \cdot \text{m}^{-3}] = n_2 \times [(50 \text{ g}) \cdot (25 \text{ cm})^{-3}]$
$n_2 = 128 \times \left( \frac{1000 \text{ g}}{50 \text{ g}} \right) \times \left( \frac{25 \text{ cm}}{100 \text{ cm}} \right)^3$
$n_2 = 128 \times 20 \times \left( \frac{1}{4} \right)^3 = \frac{128 \times 20}{64} = 2 \times 20 = 40$.
Enterobiasis (Pinworm/Threadworm) caused by Enterobius vermicularis โ most common helminth infection in school-age children worldwide.
Key features:
- Perianal/perineal pruritus at night (female migrates to perianal region at night to deposit eggs)
- Restless sleep, irritability
- Vaginal pruritus in girls (retrograde migration)
- Diagnosis: Scotch tape/adhesive tape test โ applied perianally early morning before bathing or defecation โ microscopy reveals ova
Treatment:
- Mebendazole 100 mg single dose OR Albendazole 400 mg single dose
- Repeat dose after 2 weeks (kills hatched worms from eggs not affected by first dose)
- Treat ALL household members simultaneously (highly contagious โ fecal-oral + auto-inoculation)
Hygiene measures:
- Short fingernails, frequent handwashing
- Wash underwear/bedlinen in hot water
- Avoid nail-biting
Albendazole dosing in helminths: \(400\,\text{mg}\) single dose (Ascaris, hookworm, Enterobius); \(400\,\text{mg}\) daily ร 3 days (Trichuris)
If $\displaystyle\int_0^{\pi/3}\dfrac{\tan\theta}{\sqrt{2k\sec\theta}}\,d\theta = 1-\dfrac{1}{\sqrt{2}}$ for $k>0$, find $k$.
$\displaystyle\int_0^{\pi/3}\dfrac{\tan\theta}{\sqrt{2k}\cdot\sqrt{\sec\theta}}\,d\theta = \dfrac{1}{\sqrt{2k}}\int_0^{\pi/3}\dfrac{\sin\theta}{\cos\theta}\cdot\sqrt{\cos\theta}\,d\theta=\dfrac{1}{\sqrt{2k}}\int_0^{\pi/3}\sin\theta\cdot(\cos\theta)^{-1/2}d\theta$
Let $u=\cos\theta$, $du=-\sin\theta\,d\theta$; limits: $1$ to $1/2$:
$=\dfrac{1}{\sqrt{2k}}\int_1^{1/2}(-u^{-1/2})du=\dfrac{1}{\sqrt{2k}}\left[2\sqrt{u}\right]_1^{1/2}... $ Wait: $\int_1^{1/2}(-u^{-1/2})du=\int_{1/2}^1 u^{-1/2}du=[2\sqrt{u}]_{1/2}^1=2-\sqrt{2}=2(1-1/\sqrt{2})$
$\dfrac{2(1-1/\sqrt{2})}{\sqrt{2k}}=1-\dfrac{1}{\sqrt{2}} \Rightarrow \dfrac{2}{\sqrt{2k}}=1 \Rightarrow \sqrt{2k}=2 \Rightarrow 2k=4 \Rightarrow k=\mathbf{2}$
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