Force acting on a particle is $40\text{N}$. If units of length and time are halved and the unit of mass is doubled, then find the numerical value of force in the new system of units.

The correct option is A $10$
Given:
Force $\left(f\right)=40\text{N}=40\text{kg}\frac{\text{m}}{{\text{s}}^2}$

If units of length and time are halved and unit of mass is doubled,
Let new units be ${\text{kg}}^{{}^{\prime }},{\text{m}}^{{}^{\prime }},{\text{s}}^{{}^{\prime }}$
$1{\text{kg}}^{{}^{\prime }}=2\text{kg}$
$1{\text{m}}^{{}^{\prime }}=0.5\text{m}$
$1{\text{s}}^{{}^{\prime }}=0.5\text{s}$

then the value of force in the new system of units will be

$f=40\text{N}=40$$\times \left[\frac{\frac{{\text{kg}}^{{}^{\prime }}}{2}\times \frac{2{\text{m}}^{{}^{\prime }}}{1}}{{\frac{2^2{\text{s}}^{{}^{\prime }}}{1}}^2}\right]$

$\Rightarrow f=10\text{units}$

Hence, option $\left(\text{A}\right)$ is the right answer.