In the CGS system, the magnitude of the force | vedclass

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(C) The formula for conversion between two systems is $n_2 = n_1 \left[ \left( \frac{M_1}{M_2} \right)^a \left( \frac{L_1}{L_2} \right)^b \left( \frac

Vedclass · Accepted Answer

$3.6$

Vedclass · Answer

(C) The formula for conversion between two systems is $n_2 = n_1 \left[ \left( \frac{M_1}{M_2} ight)^a \left( \frac{L_1}{L_2} ight)^b \left( \frac{T_1}{T_2} ight)^c ight]$.The dimensions of force are $[M L T^{-2}]$, so $a=1, b=1, c=-2$.Given $n_1 = 100$, $M_1 = 1 \ g$, $L_1 = 1 \ cm$, $T_1 = 1 \ s$.In the new system, $M_2 = 1 \ kg = 1000 \ g$, $L_2 = 1 \ m = 100 \ cm$, $T_2 = 1 \ min = 60 \ s$.Substituting these values:$n_2 = 100 \left[ \left( \frac{1 \ g}{1000 \ g} ight)^1 \left( \frac{1 \ cm}{100 \ cm} ight)^1 \left( \frac{1 \ s}{60 \ s} ight)^{-2} ight]$$n_2 = 100 \left[ \frac{1}{1000} 	imes \frac{1}{100} 	imes (60)^2 ight]$$n_2 = 100 	imes \frac{1}{1000} 	imes \frac{1}{100} 	imes 3600$$n_2 = 3.6$.

In the $CGS$ system, the magnitude of the force is $100 \ dynes$. In another system where the fundamental physical quantities are $kilogram$, $meter$, and $minute$, the magnitude of the force is:

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