Question

Using dimension show that $$ 1 \;Jaule = 10^{2} erg $$

Answer 1

Work $$ = [mL^{2}T^{-2}] $$

$$ 1 \;Jaule = (\frac{m_{1}}{m_{2}})^{1} (\frac{L_{1}}{L_{2}})^{2} (\frac{T_{1}}{T_{2}})^{-2} \times erg $$

$$ = (\frac{kg}{g}) (\frac{m}{cm})^{2} (\frac{sec}{sec} )^{-2} erg $$

$$ = ( \frac{l \omega o}{l}g) (\frac{l \omega}{l}cm)^{2} erg $$

$$ L = 10^{3} \times 10^{4} erg $$

$$ 1 \;Joule's = 10^{2} erg $$.