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The density of a material in $SI\, units$ is $128\, kg\, m^{-3}$. In certain units in which the unit of length is $25\, cm$ and the unit of mass $50\, g$, the numerical value of density of the material is
A$40$
B$16$
C$640$
D$410$
(JEE MAIN-2019)
Solution
$\begin{array}{l}
\frac{{128kg}}{{{m^3}}} = \frac{{125\left( {50g} \right)\left( {20} \right)}}{{{{\left( {25cm} \right)}^3}{{\left( 4 \right)}^3}}}\\
= \frac{{128}}{{64}}\left( {20} \right)units\\
= 40units
\end{array}$
\frac{{128kg}}{{{m^3}}} = \frac{{125\left( {50g} \right)\left( {20} \right)}}{{{{\left( {25cm} \right)}^3}{{\left( 4 \right)}^3}}}\\
= \frac{{128}}{{64}}\left( {20} \right)units\\
= 40units
\end{array}$
Standard 11
Physics
