In the C.G.S. system,the magnitude of the for | vedclass

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(C) The dimensional formula for force is $[M^1 L^1 T^{-2}]$.Using the conversion formula $n_2 = n_1 \left( \frac{M_1}{M_2} \right)^1 \left( \frac{L_1}

Vedclass · Accepted Answer

$3.6$

Vedclass · Answer

(C) The dimensional formula for force is $[M^1 L^1 T^{-2}]$.Using the conversion formula $n_2 = n_1 \left( \frac{M_1}{M_2} ight)^1 \left( \frac{L_1}{L_2} ight)^1 \left( \frac{T_1}{T_2} ight)^{-2}$.Here,$n_1 = 100$,$M_1 = 1 \ g$,$L_1 = 1 \ cm$,$T_1 = 1 \ s$.In the new system,$M_2 = 1 \ kg = 10^3 \ g$,$L_2 = 1 \ m = 10^2 \ cm$,$T_2 = 1 \ min = 60 \ s$.Substituting these values:$n_2 = 100 \left( \frac{1 \ g}{10^3 \ g} ight)^1 \left( \frac{1 \ cm}{10^2 \ cm} ight)^1 \left( \frac{1 \ s}{60 \ s} ight)^{-2}$$n_2 = 100 	imes \frac{1}{10^3} 	imes \frac{1}{10^2} 	imes (60)^2$$n_2 = 100 	imes 10^{-5} 	imes 3600$$n_2 = 3600 	imes 10^{-3} = 3.6$.

In the $C.G.S.$ 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:

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