JEE Main & Advanced JEE Main Paper (Held on 09-4-2019 Morning)

  • 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) \[0.01\text{ }kg/{{m}^{3}}\]

    Correct Answer: B

    Solution :

    \[\rho =\frac{m}{\text{v}}\] 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


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