If the present units of length, time and mass (m, s, kg) are changed to 100m, 100s, and (1/ 10)kg then

The correct answer is: The new unit of force is decreased 1/1000 times

To understand why this is correct, we can analyze the relation between the new units and the old units using dimensional analysis.

Let the original units of mass, length, and time be denoted by:
Mass: $M=1\text{kg}$
Length: $L=1\text{m}$
Time: $T=1\text{s}$

According to the question, the new units of length, time, and mass (denoted by primes) are defined as:
New unit of length: $L^{\prime }=100L$
New unit of time: $T^{\prime }=100T$
New unit of mass: $M^{\prime }=\frac{1}{10}M$

Now, let us find how the unit of force changes under this new system. The dimensional formula for force is:
$\left[\text{Force}\right]=\left[M^1{L}^1{T}^{-2}\right]$

Therefore, the original unit of force $F$ is expressed as:
$F=ML{T}^{-2}$

The new unit of force $F^{\prime }$ in the new system is given by:
$F^{\prime }=M^{\prime }{L}^{\prime }{\left(T^{\prime }\right)}^{-2}$

Substitute the relations for the new units in terms of the original units into the formula for $F^{\prime }$:
$F^{\prime }=\left(\frac{1}{10}M\right)\times \left(100L\right)\times {\left(100T\right)}^{-2}$

Now, simplify the expression by collecting the numerical factors:
$F^{\prime }=(\frac{1}{10}\times 100\times \frac{1}{{100}^2})\times ML{T}^{-2}$

Simplify the fractions:
$F^{\prime }=(10\times \frac{1}{10000})\times F$

$F^{\prime }=\frac{1}{1000}F$

Thus, the new unit of force is decreased to $\frac{1}{1000}$ times the original unit of force.