A thin cylindrical rod of length 10cm is placed horizontally on the...

As per sign conventions, object distance ${u}_{A} = - 45cm$, focal length $f = - 20cm$.

From the mirror formula

$\dfrac{1}{{v}_{A}} + \dfrac{1}{{u}_{A}} = \dfrac{1}{f} ⇒ \dfrac{1}{{v}_{A}} + \dfrac{1}{- 45} = \dfrac{1}{- 20} ⇒ {v}_{A} = \dfrac{- 45 \times ( - 20 )}{- 45 - ( - 20 )} = \dfrac{- 900}{25} ⇒ {v}_{A} = - 36cm$

And ${u}_{B} = - 35cm$

$\dfrac{1}{{v}_{B}} + \dfrac{1}{{u}_{B}} = \dfrac{1}{f} ⇒ \dfrac{1}{{v}_{B}} + \dfrac{1}{- 35} = \dfrac{1}{- 20} ⇒ {v}_{B} = \dfrac{- 35 \times ( - 20 )}{- 35 - ( - 20 )} = \dfrac{700}{- 15} ⇒ {v}_{B} = - \dfrac{140}{3}cm$

Now, length of image$= {v}_{A} - {v}_{B}$

$= \left(- 36 + \dfrac{140}{3}\right)cm$

$= \dfrac{- 108 + 140}{3}cm$

$= \dfrac{32}{3}cm$

$∴ x = 32$