The conversion of 1 ;MW power in a new system | vedclass

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Question

$[P] = [M{L^2}{T^{ - 3}}]$
${n_2} = {n_1}{\left[ {\frac{{{M_1}}}{{{M_2}}}} \right]^x}\,{\left[ {\frac{{{L_1}}}{{{L_2}}}} \right]^y}\,{\left[ {\frac{{

Vedclass · Accepted Answer

$2.16 	imes {10^{12}}unit$

Vedclass · Answer

$[P] = [M{L^2}{T^{ - 3}}]$
${n_2} = {n_1}{\left[ {\frac{{{M_1}}}{{{M_2}}}} ight]^x}\,{\left[ {\frac{{{L_1}}}{{{L_2}}}} ight]^y}\,{\left[ {\frac{{{T_1}}}{{{T_2}}}} ight]^z}$$ = 1 	imes {10^6}\,{\left[ {\frac{{1\,kg}}{{10\,kg}}} ight]^1}\,{\left[ {\frac{{1m}}{{1\,dm}}} ight]^2}\,{\left[ {\frac{{1s}}{{1\min }}} ight]^{ - 3}}$ [$\,1MW = {10^6}W$]
$ = {10^6}\left[ {\frac{{1kg}}{{10kg}}} ight]\,{\left[ {\frac{{10\,dm}}{{1\,dm}}} ight]^2}\,{\left[ {\frac{{1sec}}{{60sec}}} ight]^{ - 3}}$$ = 2.16 	imes {10^{12}}$unit

The conversion of $1 \;MW$ power in a new system having basic unit of mass, length and time as $10 \;kg , 1 \;dm$ and $1 \;minute$ respectively is:

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