In the density measurement of a cube, the mass and edge length are measured as $(10.00 \pm 0.10)\,\,kg\,$ and $(0.10 \pm 0.01)\,\,m\,$ respectively. The error in the measurement of density is
JEE MAIN 2019, Diffcult
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$\rho=\frac{m}{v}$
Maximum $\%$ error in $\rho$ will be given by
$\frac{\Delta p}{p} \times 100 \%=\left(\frac{\Delta m}{m}\right) \times 100 \%+3\left(\frac{\Delta L}{L}\right) \times 100 \% \ldots(i)$
This is not applicable as error is big.
$\rho_{\min }=\frac{m_{\min }}{v_{\max }}=\frac{9.9}{(0.11)^3}=7438 \ kg / m ^3$
and $\rho_{\max }=\frac{m_{\max }}{v_{\min }}=\frac{10.1}{(0.09)^3}=13854.6 \ kg / m^3$
$\Delta \rho=6416.6 \ kg / m^3$
Maximum $\%$ error in $\rho$ will be given by
$\frac{\Delta p}{p} \times 100 \%=\left(\frac{\Delta m}{m}\right) \times 100 \%+3\left(\frac{\Delta L}{L}\right) \times 100 \% \ldots(i)$
This is not applicable as error is big.
$\rho_{\min }=\frac{m_{\min }}{v_{\max }}=\frac{9.9}{(0.11)^3}=7438 \ kg / m ^3$
and $\rho_{\max }=\frac{m_{\max }}{v_{\min }}=\frac{10.1}{(0.09)^3}=13854.6 \ kg / m^3$
$\Delta \rho=6416.6 \ kg / m^3$
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