The origin of $x$ and $y$ coordinates is the pole of a concave mirror of focal length $20\text{cm}$. The $\text{x-axis}$ is the optical axis with $x>0$ being the rear side of mirror. A point object at the point $(30\text{cm},1\text{cm})$ is moving with a velocity $10\text{cm/s}$ in positive $\text{x-}$direction. The velocity of the image in $\text{cm/s}$ is approximately

The correct option is C $-40\hat{i}+2\hat{j}$
Given,
Focal length, $f=-20\text{cm}$
Object distance, $u=30\text{cm}$
Initial height, $h_0=1\text{cm}$
Velocity of image along positive $\text{x-axis}$,
$\frac{du}{dt}=10\text{cm/sec}$

If $v$ is the image distance, then

$\frac{1}{v}+\frac{1}{u}=\frac{1}{f}$

Substituting the values, we get

$\frac{1}{v}-\frac{1}{30}=-\frac{1}{20}$

$\Rightarrow v=-60\text{cm}$

For the velocity of image along the $\text{x-axis}$ we can use,

$\frac{dv}{dt}=-\frac{v^2}{u^2}\left(\frac{du}{dt}\right)$

Substituting the values,
$v_x=-{\left(\frac{60}{30}\right)}^2\times 10$

$\Rightarrow v_x=-40\text{cm/sec}$

Linear magnification

$m=\frac{f}{f-u}$

$\Rightarrow \frac{dm}{dt}=-\left[\frac{f}{(f-u{)}^2}\right]\left(\frac{du}{dt}\right)$

Substituting the values,

$\Rightarrow \frac{dm}{dt}=-\frac{\left(20\right)}{(-20+30{)}^2}\times 10$

$\Rightarrow \frac{dm}{dt}=-2$

Velocity of image along $\text{y-axis}$

$v_y=\frac{d{h}_i}{dt}=-\left(\frac{dm}{dt}\right)h_0$

$\Rightarrow v_y=(-1)\times (-2)\times 1=2\text{cm/s}$

Negative sign indicates that $h_i$ decreases with time.

Velocity of the image

$v_I=v_x\hat{i}+v_y\hat{j}$

$\therefore v_I=-40\hat{i}+2\hat{j}$