Question Number 225625 by fantastic last updated on 04/Nov/25
![quick short Q 2 things are mixed 1)both same volume 2)both same mass density⇒d_v and d_m d_v ? d_m [=,< or>]](https://www.tinkutara.com/question/Q225625.png)
$${quick}\:{short}\:{Q} \\ $$$$\mathrm{2}\:{things}\:{are}\:{mixed}\: \\ $$$$\left.\mathrm{1}\right){both}\:{same}\:{volume} \\ $$$$\left.\mathrm{2}\right){both}\:{same}\:{mass} \\ $$$${density}\Rightarrow{d}_{{v}} \:{and}\:{d}_{{m}} \\ $$$${d}_{{v}} \:?\:{d}_{{m}} \left[=,<\:{or}>\right] \\ $$
Commented by mahdipoor last updated on 04/Nov/25

$$\mathrm{V}_{\mathrm{1}} =\mathrm{V}_{\mathrm{2}} \:\:\:\mathrm{and}\:\:\mathrm{m}_{\mathrm{1}} =\mathrm{m}_{\mathrm{2}} \:\:\:\Rightarrow\rho_{\mathrm{1}} =\frac{\mathrm{m}_{\mathrm{1}} }{\mathrm{V}_{\mathrm{1}} }=\frac{\mathrm{m}_{\mathrm{2}} }{\mathrm{V}_{\mathrm{2}} }=\rho_{\mathrm{2}} \\ $$
Commented by fantastic last updated on 04/Nov/25

$${no}\:{no}\:{sir} \\ $$$${they}\:{are}\:{different}\:{cases} \\ $$
Commented by mahdipoor last updated on 04/Nov/25

$$\mathrm{case}\:\mathrm{1}\: \\ $$$$\mathrm{d}_{\mathrm{V}} =\frac{\mathrm{d}_{\mathrm{1}} \mathrm{V}+\mathrm{d}_{\mathrm{2}} \mathrm{V}}{\mathrm{2V}}=\frac{\mathrm{d}_{\mathrm{1}} +\mathrm{d}_{\mathrm{2}} }{\mathrm{2}} \\ $$$$\mathrm{case}\:\mathrm{2} \\ $$$$\mathrm{d}_{\mathrm{m}} =\frac{\mathrm{m}+\mathrm{m}}{\frac{\mathrm{m}}{\mathrm{d}_{\mathrm{1}} }+\frac{\mathrm{m}}{\mathrm{d}_{\mathrm{2}} }}=\frac{\mathrm{2d}_{\mathrm{1}} \mathrm{d}_{\mathrm{2}} }{\mathrm{d}_{\mathrm{1}} +\mathrm{d}_{\mathrm{2}} } \\ $$$$\mathrm{d}_{\mathrm{m}} \:?\:\mathrm{d}_{\mathrm{V}} \:\Rightarrow\:\frac{\mathrm{2d}_{\mathrm{1}} \mathrm{d}_{\mathrm{2}} }{\mathrm{d}_{\mathrm{1}} +\mathrm{d}_{\mathrm{2}} }\:?\:\frac{\mathrm{d}_{\mathrm{1}} +\mathrm{d}_{\mathrm{2}} }{\mathrm{2}}\:\Rightarrow\:\mathrm{4d}_{\mathrm{1}} \mathrm{d}_{\mathrm{2}} \:?\:\left(\mathrm{d}_{\mathrm{1}} +\mathrm{d}_{\mathrm{2}} \right)^{\mathrm{2}} \\ $$$$\Rightarrow\:\mathrm{0}\:?\:\left(\mathrm{d}_{\mathrm{1}} −\mathrm{d}_{\mathrm{2}} \right)^{\mathrm{2}} \:\:\Rightarrow\:?\equiv\leqslant\:\Rightarrow\:\mathrm{d}_{\mathrm{m}} \leqslant\mathrm{d}_{\mathrm{V}} \\ $$$$\begin{cases}{\mathrm{if}\:\:\:\mathrm{d}_{\mathrm{1}} =\mathrm{d}_{\mathrm{2}} \:\:\Rightarrow\:\mathrm{d}_{\mathrm{m}} =\mathrm{d}_{\mathrm{V}} }\\{\mathrm{else}\:\:\mathrm{d}_{\mathrm{m}} <\mathrm{d}_{\mathrm{V}} }\end{cases} \\ $$
Commented by fantastic last updated on 05/Nov/25

$${correct}\:{sir} \\ $$