Q5. A concave mirror has a radius of curvature 12 cm. At what distance from the mirror should an object be placed to give a virtual image three times the object size?

The focal length $f$ of a spherical mirror is half its radius of curvature $R$.

$$f = \frac{R}{2}$$

Given $R = 12 \text{ cm}$.

$$f = \frac{12 \text{ cm}}{2}$$ $$f = 6 \text{ cm}$$

For a concave mirror, the focal length is considered positive in the mirror formula.

The image is virtual and three times the object size. For a virtual image, the magnification $m$ is positive.

$$m = +3$$

The magnification formula is $m = -\frac{v}{u}$, where $v$ is the image distance and $u$ is the object distance.

$$3 = -\frac{v}{u}$$ $$v = -3u$$

The negative sign for $v$ indicates that the image is virtual and formed behind the mirror.

The mirror formula is $\frac{1}{f} = \frac{1}{u} + \frac{1}{v}$.
Substitute $f = 6 \text{ cm}$ and $v = -3u$ into the formula.

$$\frac{1}{6 \text{ cm}} = \frac{1}{u} + \frac{1}{-3u}$$ $$\frac{1}{6 \text{ cm}} = \frac{1}{u} - \frac{1}{3u}$$

To combine the terms on the right side, find a common denominator, which is $3u$.

$$\frac{1}{6 \text{ cm}} = \frac{3}{3u} - \frac{1}{3u}$$ $$\frac{1}{6 \text{ cm}} = \frac{3 - 1}{3u}$$ $$\frac{1}{6 \text{ cm}} = \frac{2}{3u}$$

Cross-multiply to solve for $u$.

$$3u = 2 \times 6 \text{ cm}$$ $$3u = 12 \text{ cm}$$ $$u = \frac{12 \text{ cm}}{3}$$ $$u = 4 \text{ cm}$$

The object should be placed at a distance of $\boxed{\text{4 cm}}$ from the mirror.