Question

When one metre, one $kg$ and one minute are taken as fundamental units, the magnitude of a force is 36 units. What is the value of this force on CGS system?

• A. $10^{3}$ dyne
• B. $10^{5}$ dyne
• C. $10^{6}$ dyne
• D. $10^{7}$ dyne

Answer 1

Answer: (A)

$M_{1} =1 kg$
$L_{1} =1 m$
$ T_{1} =1 min$
$ n_{1} =36$
$M_{2}=1 g$
$L_{2}=1 cm$
$T_{2}=1 s$
$n_{2}=?$
The dimensional formula of force is $\left[ MLT ^{-2}\right]$.
$\therefore a=1, b=1, c=-2$
$n_{2}=n_{1}\left[\frac{M_{1}}{M_{2}}\right]^{a}\left[\frac{L_{1}}{L_{2}}\right]^{b}\left[\frac{T_{1}}{T_{2}}\right]^{c}$
$=36\left[\frac{1 kg }{1 g }\right]^{1}\left[\frac{1 m }{1 cm }\right]^{1}\left[\frac{1 min }{1 s }\right]^{-2}$
$=36\left[\frac{1000 g }{ g }\right]^{1}\left[\frac{100 cm }{1 cm }\right]^{1}\left[\frac{60 s }{1 s }\right]^{-2}$
$=\frac{36 \times 10^{3} \times 10^{2}}{3600}=10^{3}$
Hence, in the CGS system of units the value of given force is $10^{3}$ dyne.