Can u please solve number 9

Dear Student ,
Here in this case the given data is : object distance = u = 30 cm and focal length = f = 15 cm .
So from the equation of spherical mirror we can write that ,
$$\begin{gathered} \frac{1}{v}+\frac{1}{u}=\frac{1}{f} \\ \Rightarrow \frac{1}{v}=\frac{1}{f}-\frac{1}{u}=\frac{1}{15}-\frac{1}{30} \\ \Rightarrow \frac{1}{v}=\frac{1}{30} \\ \Rightarrow v=30 cm \\ So the image will form 30 cm in front of the mirror . \\ Now , the magnification , m=\frac{image dis\tan ce}{object dis\tan ce}=\frac{30}{30}=1 \\ Now again we know that , magnification , m=\frac{image length}{object length}=\frac{image length}{4} \\ So , the image length=object length\times 1=4\times 1=4 cm \end{gathered}$$
Regards