An object $2.5cm$ high is placed at a distance of $10cm$ from a concave mirror of radius of curvature $30cm$. The size of the image is:

The correct option is D $7.5cm$

Given that,

Distance $u=-10cm$

Height $h=2.5cm$

Radius of curvature $R=30cm$

We know that,

$-f=\frac{R}{2}$

$-f=\frac{30}{2}$

$f=-15cm$

Now, using mirror’s formula

$\frac{1}{f}=\frac{1}{v}+\frac{1}{u}$

$-\frac{1}{15}=\frac{1}{v}-\frac{1}{10}$

$\frac{1}{v}=-\frac{1}{15}+\frac{1}{10}$

$v=30cm$

Now, the magnification is

$m=\frac{-v}{u}$

$m=\frac{30}{10}$

$m=3$

We know that,

$m=\frac{h^{\prime }}{h}$

$3=\frac{h^{\prime }}{2.5}$

$h^{\prime }=7.5cm$

Hence, the size of the image is $7.5cm$