Here, h = 5 cm = height of an object

R = Radius of curvature = 30 cm

∴

f = R/2

f = 30/2

f = 15 cm

v = -20 cm

v = ?, h' = ?

Using mirror formula,

1/f = 1/u + 1/v

∴ 1/v = 1/f - 1/u

1/v = 1/15 + 1/20

1/v = 7/60

⇒ v = 60/7

v = 8.57 cm

‍

The distance of image is 8.57 cm behind the mirror.

Magnification m = h'/h = -(v/u)

∴ h' = -(v/u) x h

h' = - 60/7 x 1/(-20) x 5

h' = 15/7

h' = 2.17

h' = 2.2 cm

Image formed is small, virtual and erect in nature.