Force acting on a particle is $10\text{N}$. If units of length and time are doubled and the unit of mass is halved, then find the numerical value of force in the new system of units.

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

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

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

$f=10\text{N}=10$$\times \left[\frac{\frac{2{\text{kg}}^{{}^{\prime }}}{1}\times \frac{1{\text{m}}^{{}^{\prime }}}{2}}{{\frac{1{\text{s}}^{{}^{\prime }}}{2^2}}^2}\right]$

$\Rightarrow f=40\text{units}$

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