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Question Number 218778 by Spillover last updated on 15/Apr/25

Answered by mr W last updated on 15/Apr/25

Commented by mr W last updated on 17/Apr/25

$$V_{cone}=\frac{\pi R^{2}h}{3}$$$$V_o={\left(\frac{\tan \varphi -\tan \theta }{\tan \varphi +\tan \theta }\right)}^{\frac{3}{2}}{V}_{cone}$$$$$$$$seeQ44017$$

Commented by mr W last updated on 15/Apr/25

Commented by mr W last updated on 15/Apr/25

Commented by mr W last updated on 18/Apr/25

$$\frac{h_1+H}{h_1}=\frac{R}{r}\Rightarrow h_1=\frac{H}{\frac{R}{r}-1}$$$$\tan \varphi =\frac{H}{R-r}$$$$\tan \theta =\frac{H}{R+r}$$$$\frac{\tan \varphi -\tan \theta }{\tan \varphi +\tan \theta }=\frac{\frac{H}{R-r}-\frac{H}{R+r}}{\frac{H}{R-r}+\frac{H}{R+r}}=\frac{r}{R}$$$$V_1=\frac{\pi r^2{h}_1}{3}=\frac{\pi r^{3}H}{3(R-r)}$$$$V_{Total}={\left(\frac{R}{r}\right)}^3{V}_1$$$$V_W+V_1={\left(\frac{\tan \varphi -\tan \theta }{\tan \varphi +\tan \theta }\right)}^{\frac{3}{2}}{V}_{Total}$$$$V_W+V_1={\left(\frac{r}{R}\right)}^{\frac{3}{2}}{\left(\frac{R}{r}\right)}^3{V}_1={\left(\frac{R}{r}\right)}^{\frac{3}{2}}{V}_1$$$$V_W=[{\left(\frac{R}{r}\right)}^{\frac{3}{2}}-1]V_1$$$$V_W=\frac{\pi r^{3}H}{3(R-r)}[{\left(\frac{R}{r}\right)}^{\frac{3}{2}}-1]$$$$$$$$V_W+V_1={\left(\frac{h+h_1}{h_1}\right)}^3{V}_1$$$$V_W=[{(\frac{h}{h_1}+1)}^3-1]V_1$$$$[{\left(\frac{R}{r}\right)}^{\frac{3}{2}}-1]V_1=[{(\frac{h}{h_1}+1)}^3-1]V_1$$$${\left(\frac{R}{r}\right)}^{\frac{3}{2}}={(\frac{h}{h_1}+1)}^3$$$$\frac{h}{h_1}=\sqrt{\frac{R}{r}}-1$$$$\Rightarrow h=\frac{H}{\frac{R}{r}-1}(\sqrt{\frac{R}{r}}-1)=\frac{H}{\sqrt{\frac{R}{r}}+1}✓$$

Commented by Spillover last updated on 15/Apr/25

$$Wonderfullsolution.thankyou$$

Answered by Spillover last updated on 15/Apr/25

Answered by Spillover last updated on 15/Apr/25

Answered by Spillover last updated on 17/Apr/25

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