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(A) Given: Focal length $f = -10 	ext{ cm}$ (for a concave mirror),object distance $u = -30 	ext{ cm}$.Using the mirror formula $\frac{1}{v} + \frac

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$3$

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(A) Given: Focal length $f = -10 	ext{ cm}$ (for a concave mirror),object distance $u = -30 	ext{ cm}$.Using the mirror formula $\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$,we have $\frac{1}{v} = \frac{1}{-10} - \frac{1}{-30} = \frac{-3+1}{30} = -\frac{2}{30} = -\frac{1}{15}$.Thus,$v = -15 	ext{ cm}$.The velocity of the image $v_I$ with respect to the mirror $v_m$ is given by $v_{I/m} = -\left(\frac{v}{u}ight)^2 v_{o/m}$.Here,$v_{o/m} = v_o - v_m$ and $v_{I/m} = v_I - v_m$.Given that the image is at rest with respect to the laboratory frame,$v_I = 0$.So,$0 - v_m = -\left(\frac{-15}{-30}ight)^2 (v_o - v_m)$.$-v_m = -\left(\frac{1}{2}ight)^2 (v_o - v_m) = -\frac{1}{4} (v_o - v_m)$.$v_m = \frac{1}{4} v_o - \frac{1}{4} v_m$.$\frac{5}{4} v_m = \frac{1}{4} v_o$.$v_m = \frac{v_o}{5} = \frac{15}{5} = 3 	ext{ cm s}^{-1}$.Therefore,the magnitude of $V_m$ is $3 	ext{ cm s}^{-1}$.

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