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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>]
$${quick}\:{short}\:{Q} \\ $$$$\mathrm{2}\:{things}\:{are}\:{mixed}\: \\ $$$$\left.\mathrm{1}\right){both}\:{same}\:{volume} \\ $$$$\left.\mathrm{2}\right){both}\:{same}\:{mass} \\ $$$${density}\Rightarrow{d}_{{v}} \:{and}\:{d}_{{m}} \\ $$$${d}_{{v}} \:?\:{d}_{{m}} \left[=,<\:{or}>\right] \\ $$
Commented by mahdipoor last updated on 04/Nov/25
V_1 =V_2    and  m_1 =m_2    ⇒ρ_1 =(m_1 /V_1 )=(m_2 /V_2 )=ρ_2
$$\mathrm{V}_{\mathrm{1}} =\mathrm{V}_{\mathrm{2}} \:\:\:\mathrm{and}\:\:\mathrm{m}_{\mathrm{1}} =\mathrm{m}_{\mathrm{2}} \:\:\:\Rightarrow\rho_{\mathrm{1}} =\frac{\mathrm{m}_{\mathrm{1}} }{\mathrm{V}_{\mathrm{1}} }=\frac{\mathrm{m}_{\mathrm{2}} }{\mathrm{V}_{\mathrm{2}} }=\rho_{\mathrm{2}} \\ $$
Commented by fantastic last updated on 04/Nov/25
no no sir  they are different cases
$${no}\:{no}\:{sir} \\ $$$${they}\:{are}\:{different}\:{cases} \\ $$
Commented by mahdipoor last updated on 04/Nov/25
case 1   d_V =((d_1 V+d_2 V)/(2V))=((d_1 +d_2 )/2)  case 2  d_m =((m+m)/((m/d_1 )+(m/d_2 )))=((2d_1 d_2 )/(d_1 +d_2 ))  d_m  ? d_V  ⇒ ((2d_1 d_2 )/(d_1 +d_2 )) ? ((d_1 +d_2 )/2) ⇒ 4d_1 d_2  ? (d_1 +d_2 )^2   ⇒ 0 ? (d_1 −d_2 )^2   ⇒ ?≡≤ ⇒ d_m ≤d_V    { ((if   d_1 =d_2   ⇒ d_m =d_V )),((else  d_m <d_V )) :}
$$\mathrm{case}\:\mathrm{1}\: \\ $$$$\mathrm{d}_{\mathrm{V}} =\frac{\mathrm{d}_{\mathrm{1}} \mathrm{V}+\mathrm{d}_{\mathrm{2}} \mathrm{V}}{\mathrm{2V}}=\frac{\mathrm{d}_{\mathrm{1}} +\mathrm{d}_{\mathrm{2}} }{\mathrm{2}} \\ $$$$\mathrm{case}\:\mathrm{2} \\ $$$$\mathrm{d}_{\mathrm{m}} =\frac{\mathrm{m}+\mathrm{m}}{\frac{\mathrm{m}}{\mathrm{d}_{\mathrm{1}} }+\frac{\mathrm{m}}{\mathrm{d}_{\mathrm{2}} }}=\frac{\mathrm{2d}_{\mathrm{1}} \mathrm{d}_{\mathrm{2}} }{\mathrm{d}_{\mathrm{1}} +\mathrm{d}_{\mathrm{2}} } \\ $$$$\mathrm{d}_{\mathrm{m}} \:?\:\mathrm{d}_{\mathrm{V}} \:\Rightarrow\:\frac{\mathrm{2d}_{\mathrm{1}} \mathrm{d}_{\mathrm{2}} }{\mathrm{d}_{\mathrm{1}} +\mathrm{d}_{\mathrm{2}} }\:?\:\frac{\mathrm{d}_{\mathrm{1}} +\mathrm{d}_{\mathrm{2}} }{\mathrm{2}}\:\Rightarrow\:\mathrm{4d}_{\mathrm{1}} \mathrm{d}_{\mathrm{2}} \:?\:\left(\mathrm{d}_{\mathrm{1}} +\mathrm{d}_{\mathrm{2}} \right)^{\mathrm{2}} \\ $$$$\Rightarrow\:\mathrm{0}\:?\:\left(\mathrm{d}_{\mathrm{1}} −\mathrm{d}_{\mathrm{2}} \right)^{\mathrm{2}} \:\:\Rightarrow\:?\equiv\leqslant\:\Rightarrow\:\mathrm{d}_{\mathrm{m}} \leqslant\mathrm{d}_{\mathrm{V}} \\ $$$$\begin{cases}{\mathrm{if}\:\:\:\mathrm{d}_{\mathrm{1}} =\mathrm{d}_{\mathrm{2}} \:\:\Rightarrow\:\mathrm{d}_{\mathrm{m}} =\mathrm{d}_{\mathrm{V}} }\\{\mathrm{else}\:\:\mathrm{d}_{\mathrm{m}} <\mathrm{d}_{\mathrm{V}} }\end{cases} \\ $$
Commented by fantastic last updated on 05/Nov/25
correct sir
$${correct}\:{sir} \\ $$

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