In C.G.S. system the magnitude of the force is 100 dynes. In another system where the fundamental physical quantities are kilogram, metre and minute, the magnitude of the force is

The correct option is 3.6.

To find the magnitude of the force in the new system of units, we use the method of dimensions. The dimension of force is given by:
$[F]\; =\; [M^\{1\}\; L^\{1\}\; T^\{-2\}]$
where M represents mass, L represents length, and T represents time.

The relationship between the numerical value and the unit of a physical quantity is constant:
$n\_\{1\}\; u\_\{1\}\; =\; n\_\{2\}\; u\_\{2\}$
where $n\_\{1\}$ and $n\_\{2\}$ are the numerical values in the two systems, and $u\_\{1\}$ and $u\_\{2\}$ are the respective units.

Using the dimensional formula of force, we can write:
$n\_\{2\}\; =\; n\_\{1\}\; [M\_\{1\}/M\_\{2\}]^\{1\}\; [L\_\{1\}/L\_\{2\}]^\{1\}\; [T\_\{1\}/T\_\{2\}]^\{-2\}$

Here, the first system is the C.G.S. system:
$n\_\{1\}\; =\; 100$
$M\_\{1\}\; =\; 1\backslash text\{\; g\}$
$L\_\{1\}\; =\; 1\backslash text\{\; cm\}$
$T\_\{1\}\; =\; 1\backslash text\{\; s\}$

The second system has the following fundamental units:
$M\_\{2\}\; =\; 1\backslash text\{\; kg\}\; =\; 1000\backslash text\{\; g\}$
$L\_\{2\}\; =\; 1\backslash text\{\; m\}\; =\; 100\backslash text\{\; cm\}$
$T\_\{2\}\; =\; 1\backslash text\{\; minute\}\; =\; 60\backslash text\{\; s\}$

Substituting these values into the conversion formula:
$n\_\{2\}\; =\; 100\; \backslash times\; [1/1000]^\{1\}\; \backslash times\; [1/100]^\{1\}\; \backslash times\; [1/60]^\{-2\}$

Simplifying the expression:
$n\_\{2\}\; =\; 100\; \backslash times\; (1/1000)\; \backslash times\; (1/100)\; \backslash times\; (60)^\{2\}$
$n\_\{2\}\; =\; 100\; \backslash times\; (1/100000)\; \backslash times\; 3600$
$n\_\{2\}\; =\; 360000\; /\; 100000$
$n\_\{2\}\; =\; 3.6$

Therefore, the magnitude of the force in the new system is 3.6.