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Equal volumes of two substances of densities \mathrm{\rho_1 } and \mathrm{\rho_2} are mixed together. The density of the mixture would be

Option: 1

\mathrm{\frac{1}{2}\left(\rho_1+\rho_2\right)}


Option: 2

\mathrm{\left(\rho_1+\rho_2\right)}


Option: 3

\mathrm{\sqrt{\rho_1 \rho_2}}


Option: 4

\mathrm{\frac{\rho_1 \rho_2}{\left(\rho_1+\rho_2\right)}}


Answers (1)

best_answer

The density of the mixture is

                 \mathrm{ \begin{aligned} \rho & =\frac{m_1+m_2}{V_1+V_2} \\\\ & =\frac{\rho_1 V_1+\rho_2 V_2}{V_1+V_2} \end{aligned} }

where\mathrm{ V_1 \, \, and \, \, V_2 } are the volumes of the two substances.

Given \mathrm{ V_1=V_2=V}. Therefore, \mathrm{ \rho=\frac{1}{2}\left(\rho_1+\rho_2\right).}

Hence the correct choice is (a).

Posted by

Divya Prakash Singh

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