• # question_answer In the density measurement of a cube, the mass and edge length are measured as $\left( 10.00\pm 0.10 \right)\text{ }kg$and$\left( 0.10\pm 0.01 \right)\text{ }m,$respectively. The error in the measurement of density is:         [JEE Main 9-4-2019 Morning] A) $0.10\text{ }kg/{{m}^{3}}$B)   $0.31\text{ }kg/{{m}^{3}}$C) $0.07\text{ }kg/{{m}^{3}}$           D) $0.01\text{ }kg/{{m}^{3}}$

$\rho =\frac{m}{\text{v}}$ maximum % error in S will be given by $\frac{\Delta \rho }{\rho }\times 100%=\left( \frac{\Delta m}{m} \right)\times 100%+3\left( \frac{\Delta L}{L} \right)\times 100%$                                                     ...(i) which is only possible when error is small which is not the case in this question. Yet if we apply equation (i), we get $\Delta \rho =3100kg/{{m}^{3}}$ Now, we will calculate error, without using approximation. $\rho {{ & }_{\min }}=\frac{{{m}_{\min }}}{{{\text{v}}_{\max }}}=\frac{9.9}{{{(0.11)}^{3}}}=7438kg/{{m}^{3}}$ $\And \,\rho {{ & }_{\max }}=\frac{{{m}_{\max }}}{{{\text{v}}_{\min }}}=\frac{10.1}{{{(0.09)}^{3}}}=13854.6kg/{{m}^{3}}$ $\Delta \,\rho =6416.6kg/{{m}^{3}}$ No option is matching. Therefore this question should be awarded bonus