As depicted in the figure, a U-shaped wire is situated in front of...

For the concave mirror, the focal length is $f = -\dfrac{R}{2} = -\dfrac{20}{2} = -10\text{ cm}$.

The U-shaped wire consists of three segments: two vertical and one horizontal.

1. For the right vertical segment:
Object distance, $u_1 = -30\text{ cm}$.
Using the mirror formula, $\dfrac{1}{v_1} + \dfrac{1}{u_1} = \dfrac{1}{f}$
$\dfrac{1}{v_1} - \dfrac{1}{30} = -\dfrac{1}{10} \Rightarrow \dfrac{1}{v_1} = \dfrac{1}{30} - \dfrac{1}{10} = -\dfrac{2}{30}$
$v_1 = -15\text{ cm}$
Magnification, $m_1 = -\dfrac{v_1}{u_1} = -\dfrac{-15}{-30} = -\dfrac{1}{2}$
Length of its image, $L_1 = |m_1| \times h_o = \dfrac{1}{2} \times 10 = 5\text{ cm}$.

2. For the left vertical segment:
Object distance, $u_2 = -(30 + 10) = -40\text{ cm}$.
$\dfrac{1}{v_2} - \dfrac{1}{40} = -\dfrac{1}{10} \Rightarrow \dfrac{1}{v_2} = \dfrac{1}{40} - \dfrac{1}{10} = -\dfrac{3}{40}$
$v_2 = -\dfrac{40}{3}\text{ cm}$
Magnification, $m_2 = -\dfrac{v_2}{u_2} = -\dfrac{-40/3}{-40} = -\dfrac{1}{3}$
Length of its image, $L_2 = |m_2| \times h_o = \dfrac{1}{3} \times 10 = \dfrac{10}{3}\text{ cm}$.

3. For the horizontal segment:
This segment lies along the principal axis from $u_1 = -30\text{ cm}$ to $u_2 = -40\text{ cm}$.
Its image lies along the principal axis from $v_1 = -15\text{ cm}$ to $v_2 = -\dfrac{40}{3}\text{ cm}$.
Length of its image, $L_3 = |v_1 - v_2| = \left|-15 - \left(-\dfrac{40}{3}\right)\right| = \left|-15 + \dfrac{40}{3}\right| = \dfrac{5}{3}\text{ cm}$.

Total length of the image is the sum of the lengths of these three segments:
$L = L_1 + L_2 + L_3 = 5 + \dfrac{10}{3} + \dfrac{5}{3} = 5 + \dfrac{15}{3} = 5 + 5 = 10\text{ cm}$.