At what distance from a concave mirror focal length 10 cm should an object 2 cm long be placed in order to get an erect image 6 cm tall?

​Height of the object 'h_o' = 2 cm

Focal length of the mirror 'f' = −10 cm
Height of the image 'h_i' = 6 cm
We have to find the distance of the object from the mirror 'u'.
Using the magnification formula, we get
$$\begin{gathered} m=\frac{h_i}{h_o}=-\frac{v}{u} \\ m=\frac{6}{2}=-\frac{v}{u} \\ m=3=-\frac{v}{u} \\ \mathrm{thus}v=-3u \end{gathered}$$

Now, using the mirror formula, we get

$$\begin{gathered} \frac{1}{f}=\frac{1}{v}+\frac{1}{u} \\ \frac{1}{-10}=\frac{1}{-3u}+\frac{1}{u} \\ \frac{1}{10}=\frac{1}{3u}-\frac{3}{3u}=\frac{-2}{3u} \\ u=\frac{-2\times 10}{3} \\ u=-6.67\mathrm{cm} \end{gathered}$$

Thus, the distance of the object from the mirror 'u' is −6.67 cm.