If there is an increase in linear dimensions of the objects, the associated de-Broglie wavelength
A) Increases
B) Decreases
C) Remains unchanged
D) Depends on density

Hint: According to the dual nature of radiation or wave-particle duality, the De Broglie wavelength is the wavelength associated with a particle in relation to its mass and momentum. This wavelength is usually inversely proportional to its force. Similar to light, a particle, too, can exhibit dual behaviour and the De Broglie wavelength is a characteristic of its wave nature.

Formulae used:
 $\lambda = \dfrac{h}{p}$
$ \Rightarrow \lambda = \dfrac{h}{{mv}}$

$\lambda $ is the De Broglie wavelength of the particle, $h$ is the Planck’s constant and $p$ is the momentum of the particle.

Complete step by step solution:
Louis de Broglie, a physicist, was the one to suggest that particles might exhibit both wave properties and particle properties. These have been proven experimentally.
As his experiments progressed, De Broglie derived the equation for the De Broglie wavelength in a schematic manner.
$E = hv$ for a photon and $\lambda v = c$ for an electromagnetic wave.
$E = m{c^2}$, means, $\lambda = \dfrac{h}{{mc}}$ which the is equivalent of $\lambda = \dfrac{h}{p}$.
Here $m$ is the relativistic mass of the photon, and not the rest mass; since the rest mass of a photon is zero.
Now, if a particle is moving with a velocity $v$ , the momentum $p = mv$ and hence $\lambda = \dfrac{h}{{mv}}$
Therefore, the de Broglie wavelength formula is expressed as;
$ \Rightarrow \lambda = \dfrac{h}{{mv}}$

As this De Broglie wavelength is inversely proportional to mass, as in;
$\lambda \propto \dfrac{1}{m}$
Therefore an increase in linear dimensions, which will inevitably cause an increase in mass (density remaining constant), will lead to a decrease in the De Broglie wavelength.

Therefore, Option (B) is the correct option.

Note: It is important to note that an increase in linear dimension need not necessarily lead to an increase in mass, such as the case of stretching of a wire. Here it has been assumed that density remains constant in the process of increment and has led to an increase in mass, thereby a decrease in wavelength.