A rod of length 10 cm lies along the principal axis of a concave mirror of focal length 10 cm in such a way that the end closer to the pole is 20 cm away from it. Find the length of the image (in cm)

The situation is shown in fig. The radius of curvature of the mirror is $r=2f=20cm$  . Thus,   the nearer and B of the rod AB is at the center of the curvature and hence, its image will be formed at B itself. We shall now locate the image of A here $u=-30cmandf=-10cm$  . We have 
$\frac{1}{u}+\frac{1}{v}=\frac{1}{f}or\frac{1}{v}=\frac{1}{f}-\frac{1}{u}$
  $=\frac{1}{-10cm}-\frac{1}{-30cm}$     
   Or ,  $v=-15cm$
   Thus, the image of A is formed at 125 cm from the pole. The length of the image is, therefore, 5.0 cm