A diverging lens has a focal length of 30 cm. At _____ cm distance an object of height 5 cm from the optical centre of the lens should be placed so that its image is formed 15 cm away from the lens. The size of the image is  _______ cm.

In the case of a diverging lens, both the object distance and the image distance from the optical center are negative.
Given: 

Focal length f = -30 cm.
Distance of the image from the lens v = -15 cm.
The height of the object is h₁=5 cm. 

We will write the lens formula to find the distance of the object from the lens.

$\frac{1}{\mathrm{f}}=\frac{1}{\mathrm{v}}-\frac{1}{\mathrm{u}}$

=>$\frac{1}{\mathrm{u}}=\frac{1}{\mathrm{v}}-\frac{1}{\mathrm{f}}$

Here u is the distance of the object from the optical center.

$\frac{1}{\mathrm{u}}=\frac{1}{-15}-\frac{1}{-30}$

u = - 30 cm

We know that magnification is given by:

$\mathrm{m}=\frac{\mathrm{v}}{\mathrm{u}}=\frac{{\mathrm{h}}_2}{{\mathrm{h}}_1}$
$\Rightarrow \frac{\mathrm{v}}{\mathrm{u}}=\frac{{\mathrm{h}}_2}{{\mathrm{h}}_1}$

Substitute, v = -15 cm, u = -30 cm and h₁ = 5 cm to find the value of h₂.

$\frac{-15 \mathrm{cm}}{-30 \mathrm{cm}}=\frac{{\mathrm{h}}_2}{5 \mathrm{cm}}$

h₂ = 2.5 cm.

As a result, the object should be located 30 cm from the optical center, and the size of the image is 2.5 cm.
Hence, the correct answer is 30cm, 2.5 cm.