Physics - Geometrical Optics Question with Solution | TestHub

To produce an upright and magnified image, a concave mirror must be used. We are given that the object distance $u=2$ cm and the magnification $m=3$.

The magnification formula is:

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

where $v$ is the image distance.

Since $m=3$ and $u=2$ cm, we have:

$$3=-\frac{v}{2}$$

$$v=-6cm$$

The image is virtual (negative sign) and located 6 cm behind the mirror.

Now, we use the mirror formula:

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

where $f$ is the focal length of the mirror.

Plugging in the values, we get:

$$\frac{1}{f}=\frac{1}{2}+\frac{1}{-6}$$

$$\frac{1}{f}=\frac{1}{2}-\frac{1}{6}=\frac{3-1}{6}=\frac{2}{6}=\frac{1}{3}$$

$$f=3cm$$

The radius of curvature $r$ is related to the focal length by:

$$r=2f$$

$$r=2(3)=6cm$$

Therefore, the mirror must be a concave mirror with a radius of curvature of 6 cm.

Final Answer: The final answer is $\boxed{A}$