Question-103001

Question Number 103001 by I want to learn more last updated on 12/Jul/20

Answered by ajfour last updated on 12/Jul/20

let new depth be y. V=πR^2 H=πR^2 (y−(h/2))+π((h/2))(R^2 −r^2 ) ⇒ H=y−(h/2)+(h/2)(1−(r^2 /R^2 )) 6=y−(9/2)+(9/2)(1−(1/4)) ⇒ y=6+(9/8) = 7.125 m

$${let}\:{new}\:{depth}\:{be}\:\boldsymbol{{y}}. \\ $$$$\boldsymbol{{V}}=\pi{R}^{\mathrm{2}} {H}=\pi{R}^{\mathrm{2}} \left({y}−\frac{{h}}{\mathrm{2}}\right)+\pi\left(\frac{{h}}{\mathrm{2}}\right)\left({R}^{\mathrm{2}} −{r}^{\mathrm{2}} \right) \\ $$$$\Rightarrow\:{H}={y}−\frac{{h}}{\mathrm{2}}+\frac{{h}}{\mathrm{2}}\left(\mathrm{1}−\frac{{r}^{\mathrm{2}} }{{R}^{\mathrm{2}} }\right) \\ $$$$\:\:\:\:\:\:\mathrm{6}={y}−\frac{\mathrm{9}}{\mathrm{2}}+\frac{\mathrm{9}}{\mathrm{2}}\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{4}}\right) \\ $$$$\Rightarrow\:\:\:\:{y}=\mathrm{6}+\frac{\mathrm{9}}{\mathrm{8}}\:=\:\mathrm{7}.\mathrm{125}\:{m} \\ $$

Commented by I want to learn more last updated on 12/Jul/20

$$\mathrm{I}\:\mathrm{appreciate}\:\mathrm{sir} \\ $$

Commented by I want to learn more last updated on 12/Jul/20

Sir, sorry to disturb. Can you show me how the formular is derived?. Thanks sir.

$$\mathrm{Sir},\:\mathrm{sorry}\:\mathrm{to}\:\mathrm{disturb}.\:\mathrm{Can}\:\mathrm{you}\:\mathrm{show}\:\mathrm{me}\:\mathrm{how}\:\mathrm{the} \\ $$$$\mathrm{formular}\:\mathrm{is}\:\mathrm{derived}?.\:\:\mathrm{Thanks}\:\mathrm{sir}. \\ $$

Commented by ajfour last updated on 12/Jul/20

$${Do}\:{you}\:{have}\:{the}\:{answer},\:{is}\:{it} \\ $$$${correct}? \\ $$

Commented by I want to learn more last updated on 12/Jul/20

I did not have the answer sir. But i will go with your work sir.

$$\mathrm{I}\:\mathrm{did}\:\mathrm{not}\:\mathrm{have}\:\mathrm{the}\:\mathrm{answer}\:\mathrm{sir}.\:\mathrm{But}\:\mathrm{i}\:\mathrm{will}\:\mathrm{go}\:\mathrm{with}\:\mathrm{your}\:\mathrm{work}\:\mathrm{sir}. \\ $$

Commented by mr W last updated on 12/Jul/20

$${answer}\:{correct}! \\ $$

Answered by mr W last updated on 12/Jul/20

Commented by mr W last updated on 12/Jul/20

volume of water in tank: V=πR^2 h_1 with R=12m, h_1 =6m volume of wooden rod: V_w =πr^2 H with r=6m, H=9m half of wooden rod is in water: πR^2 h_2 =V+(V_w /2) πR^2 h_2 =πR^2 h_1 +((πr^2 H)/2) ⇒h_2 =h_1 +((Hr^2 )/(2R^2 ))=6+(9/2)((6/(12)))^2 =7.125 m

$${volume}\:{of}\:{water}\:{in}\:{tank}: \\ $$$${V}=\pi{R}^{\mathrm{2}} {h}_{\mathrm{1}} \:{with}\:{R}=\mathrm{12}{m},\:{h}_{\mathrm{1}} =\mathrm{6}{m} \\ $$$${volume}\:{of}\:{wooden}\:{rod}: \\ $$$${V}_{{w}} =\pi{r}^{\mathrm{2}} {H}\:{with}\:{r}=\mathrm{6}{m},\:{H}=\mathrm{9}{m} \\ $$$$ \\ $$$${half}\:{of}\:{wooden}\:{rod}\:{is}\:{in}\:{water}: \\ $$$$\pi{R}^{\mathrm{2}} {h}_{\mathrm{2}} ={V}+\frac{{V}_{{w}} }{\mathrm{2}} \\ $$$$\pi{R}^{\mathrm{2}} {h}_{\mathrm{2}} =\pi{R}^{\mathrm{2}} {h}_{\mathrm{1}} +\frac{\pi{r}^{\mathrm{2}} {H}}{\mathrm{2}} \\ $$$$\Rightarrow{h}_{\mathrm{2}} ={h}_{\mathrm{1}} +\frac{{Hr}^{\mathrm{2}} }{\mathrm{2}{R}^{\mathrm{2}} }=\mathrm{6}+\frac{\mathrm{9}}{\mathrm{2}}\left(\frac{\mathrm{6}}{\mathrm{12}}\right)^{\mathrm{2}} =\mathrm{7}.\mathrm{125}\:{m} \\ $$

Commented by I want to learn more last updated on 12/Jul/20

$$\mathrm{Wow},\:\mathrm{thanks}\:\mathrm{sir}\:\mathrm{for}\:\mathrm{confirmation}. \\ $$

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