Mirror is moving towards the particles (negative  x-axis) with speed  2cm/sec. Particles A and B are moving with speeds of $10\sqrt{2}\text{\hspace{0.17em}cm}/\text{sec}$ and $5\text{\hspace{0.17em}cm}/\text{sec}$ respectively in the direction shown in figure. Magnitude of velocity of image of the particle B with respect to image of A is 

 
Velocity of image normal to the plane of mirror is  $V_i=2V_m-V_{\text{object}}$

Velocity of image parallel to the plane of mirror is  $V_i=V_{\text{object}}$
Velocity of particle image B ,  $V_{Bi}=2\left(-2\widehat{i}\right)-\left(-5\widehat{i}\right)=\widehat{i}$
Velocity of particle A is  $=10\sqrt{2}\cos {45}^{\circ }\widehat{i}+10\sqrt{2}\sin {45}^{\circ }\widehat{j}$
Velocity of image of particle A perpendicular to the mirror,  $V_{Aix}=2\left(-2\hat{i}\right)-\left(10\hat{i}\right)=14\left(-\widehat{i}\right)$
Velocity of image of particle A parallel to mirror
$V_{\text{Aiy}}=10\widehat{j}$
$\therefore V_{Ai}=-14\hat{i}+10\hat{j}$
Relative velocity of image  $V_{BAi}=V_{Bi}-V_{Ai}=\widehat{i}-\left[14\left(-\widehat{i}\right)+10\widehat{j}\right]=15\widehat{i}-10\widehat{j}$
$\therefore V_{BAi}=\sqrt{{15}^2+{10}^2}=\sqrt{325}cm/\sec$
Therefore, the correct answer is  18.03