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A particle is projected vertically upward with $u = 10\text{ m/s}$. Air exerts a resistive force $F = -0.2v^2$ on it ($m = 2\text{ kg}$, $g = 10\text{ m/s}^2$). The maximum height attained is:
Net force (taking up as positive): $F_{net} = -mg - 0.2v^2 = -20 - 0.2v^2$
Using $ma = v\dfrac{dv}{dx}$: $2v\dfrac{dv}{dx} = -20 - 0.2v^2$
Separating variables: $\dfrac{v\,dv}{20+0.2v^2} = -\dfrac{dx}{2}$
Integrating from $v = 10$ to $v = 0$: let $k = 20 + 0.2v^2$, $dk = 0.4v\,dv$
$\dfrac{1}{0.4}\big[\ln(20) - \ln(40)\big] = -\dfrac{H}{2} \Rightarrow H = \dfrac{2\ln 2}{0.4} = \mathbf{5\ln 2}$
Key properties:
- Wide hysteresis loop (A): High retentivity and high coercivity — hard magnetic material. Good for permanent magnets (used in electric generators).
- Narrow hysteresis loop (B): Low retentivity, low coercivity, low energy loss per cycle — soft magnetic material. Ideal for transformers and electromagnets (needs to magnetise/demagnetise repeatedly with low loss).
Electric generators use permanent magnets (wide loop, material A). Transformers and electromagnets use soft material (B). So Option C is correct.
This is a disproportionation reaction. The balanced equation in alkaline medium is:
$S_{8} + 12OH^{-} \rightarrow 4S^{2-} + 2S_{2}O_{3}^{2-} + 6H_{2}O$
Therefore, the value of the coefficient 'a' is $12$.
[cite: 13]\(\text{Cl}_2 > \text{Br}_2 > \text{F}_2 > \text{I}_2\)
F₂ has an anomalously low bond energy (158 kJ/mol) due to lone pair repulsion in the small F atom, making \(\text{Cl}_2\) stronger. So the order \(\text{F}_2 > \text{Cl}_2\) for bond dissociation energy is incorrect.
All other options are correct: oxidising power, acid strength of HX in water, and electronegativity all follow the stated orders.
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