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Size of object, h = 7.5cm
radius of curvature of convex mirror, R = +25cm
focal length, f = R/2 = 25/2 cm
object distance, u = -40cm
image distance = v

we know that
 \frac{1}{v} + \frac{1}{u}= \frac{1}{f} \\ \\ \Rightarrow  \frac{1}{v}= \frac{1}{f} -  \frac{1}{u}\\ \\ \Rightarrow  \frac{1}{v}= \frac{1}{25/2} -  \frac{1}{40}\\ \\ \Rightarrow  \frac{1}{v}= \frac{2}{25} -  \frac{1}{40}\\ \\ \Rightarrow  \frac{1}{v}= \frac{16-5}{200}\\ \\ \Rightarrow  \frac{1}{v}= \frac{11}{200}\\ \\ \Rightarrow  v= \frac{200}{11}cm

magnification =  \frac{h'}{h} =- \frac{v}{u}
h' = height of image

\frac{h'}{h} =- \frac{v}{u}\\ \\ h' = -h \times \frac{v}{u}=-7.5 \times \frac{200}{11 \times (-40)}= \frac{75}{22}\ cm=3.41cm

Size of image is 3.41cm.