a) The mirror's radius of curvature will be 2.7 cm
B) The magnification of the image will be 4.0.
When a hollow spherical is divided into pieces and the exterior surface of each cut portion is painted, it forms a mirror, with the inner surface reflecting the light.
From the mirror equation;
[tex]\frac{1}{p}+\frac{1}{q} =\frac{1}{f} \\\\ \frac{1}{1}+\frac{1}{-10} =\frac{1}{f} \\\\ \frac{1}{f} = 1-\frac{1}{10} \\\\ \frac{1}{f} = \frac{9}{10} \\\\ f= 1.1111[/tex]
Hence,
The radius of curvature is twice of the focal length;
[tex]\rm R = 2f \\\\ \rm R = 2\times 1.111 \\\\ R=2.2222 \ cm[/tex]
Hence, the mirror's radius of curvature will be 2.7 cm.
The magnification factor is found as;
[tex]\rm m = \frac{-q}{p} \\\\ m = \frac{-(-10)}{1} \\\\ m = 10[/tex]
Hence, the magnification of the image will be 4.0.
To learn more about the concave mirror, refer to the link;
https://brainly.com/question/25937699
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