A rod of length 10 cm lies along the principle axis of a concave...

For the given concave mirror, the focal length is $f = -10$ cm.

The rod lies along the principal axis. From the figure, the near end of the rod is at a distance of $20$ cm from the pole of the mirror. Since the length of the rod is $10$ cm, the far end is at a distance of $20 + 10 = 30$ cm from the pole.

Let the near end be $A$ and the far end be $B$.
Object distance for end $A$, $u_A = -20$ cm.
Object distance for end $B$, $u_B = -30$ cm.

Using the mirror formula, $\dfrac{1}{v} + \dfrac{1}{u} = \dfrac{1}{f}$

For end $A$:
$\dfrac{1}{v_A} + \dfrac{1}{-20} = \dfrac{1}{-10}$
$\dfrac{1}{v_A} = \dfrac{1}{20} - \dfrac{1}{10} = -\dfrac{1}{20}$
$v_A = -20$ cm

For end $B$:
$\dfrac{1}{v_B} + \dfrac{1}{-30} = \dfrac{1}{-10}$
$\dfrac{1}{v_B} = \dfrac{1}{30} - \dfrac{1}{10} = \dfrac{1 - 3}{30} = -\dfrac{2}{30} = -\dfrac{1}{15}$
$v_B = -15$ cm

The length of the image is the distance between the images of the two ends:
$\text{Length of image} = |v_A - v_B| = |-20 - (-15)| = |-5| = 5$ cm.

Answer: $5$