Using $s = \dfrac{1}{2}at^2$ (from rest):

$S_1 = \dfrac{1}{2}a(10)^2 = 50a$

Total in 20 s: $S_{\text{total}} = \dfrac{1}{2}a(20)^2 = 200a$

$S_2 = 200a - 50a = 150a = 3 \times 50a = \mathbf{3S_1}$

Let $w = \frac{z_1-2z_2}{2-z_1\bar{z}_2}$, with $|w|=1$.

$|z_1-2z_2|^2 = |2-z_1\bar{z}_2|^2$

$(z_1-2z_2)\overline{(z_1-2z_2)} = (2-z_1\bar{z}_2)\overline{(2-z_1\bar{z}_2)}$

$|z_1|^2 - 2z_1\bar{z}_2 - 2\bar{z}_1z_2 + 4|z_2|^2 = 4 - 2z_1\bar{z}_2 - 2\bar{z}_1z_2 + |z_1|^2|z_2|^2$

$|z_1|^2 + 4|z_2|^2 = 4 + |z_1|^2|z_2|^2$

$|z_1|^2(1-|z_2|^2) = 4(1-|z_2|^2)$

Since $|z_2|\neq1$: $|z_1|^2=4\Rightarrow|z_1|=2$. So $z_1$ lies on a circle of radius 2.

The three levels of prevention are:

Immunization is a classic example of primary prevention as it prevents disease before it occurs.

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.

For continuity at $x=3$: $k\sqrt{4}=3m+2\Rightarrow 2k=3m+2$ ...(i)

For differentiability at $x=3$: Left derivative $=\frac{k}{2\sqrt{x+1}}\big|_{x=3}=\frac{k}{4}$; Right derivative $=m$.

So $\frac{k}{4}=m$ ...(ii)

From (ii): $k=4m$. Substitute in (i): $8m=3m+2\Rightarrow5m=2\Rightarrow m=\frac{2}{5}$, $k=\frac{8}{5}$.

$k+m=\frac{8}{5}+\frac{2}{5}=\frac{10}{5}=\mathbf{2}$

So correct answer is $k+m=2$, which is option A (index 0). But official JEE answer is $\frac{10}{3}$... Re-checking: $5m=2 \Rightarrow m=2/5$. $k=4(2/5)=8/5$. $k+m = 8/5+2/5=2$. Answer is $\mathbf{2}$.