Question Number 43264 by MrW3 last updated on 09/Sep/18
Commented by ajfour last updated on 09/Sep/18
Answered by ajfour last updated on 09/Sep/18
![Minimum Volume of water so that topmost ball remains submerged is = π(2R)^2 h−((4×4)/3)πR^3 h = R+x+R = 2R+R(√2) V_(min) = π(25)(5+((5(√2))/2))cm^3 −((2π)/3)×125cm^3 = 125π[(((2+(√2)))/2)−(2/3)] cm^3 ≈ 408.58 cm^3 .](https://www.tinkutara.com/question/Q43290.png)
$${Minimum}\:{Volume}\:{of}\:{water}\:{so}\:\: \\ $$$${that}\:{topmost}\:{ball}\:{remains}\: \\ $$$${submerged}\:{is}\:\:=\:\pi\left(\mathrm{2}{R}\right)^{\mathrm{2}} {h}−\frac{\mathrm{4}×\mathrm{4}}{\mathrm{3}}\pi{R}^{\mathrm{3}} \\ $$$$\:\:\:\:\boldsymbol{{h}}\:=\:\boldsymbol{{R}}+\boldsymbol{{x}}+\boldsymbol{{R}} \\ $$$$\:\:\:\:\:\:\:=\:\mathrm{2}{R}+{R}\sqrt{\mathrm{2}} \\ $$$$\:\:{V}_{{min}} =\:\pi\left(\mathrm{25}\right)\left(\mathrm{5}+\frac{\mathrm{5}\sqrt{\mathrm{2}}}{\mathrm{2}}\right){cm}^{\mathrm{3}} −\frac{\mathrm{2}\pi}{\mathrm{3}}×\mathrm{125}{cm}^{\mathrm{3}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:=\:\mathrm{125}\pi\left[\frac{\left(\mathrm{2}+\sqrt{\mathrm{2}}\right)}{\mathrm{2}}−\frac{\mathrm{2}}{\mathrm{3}}\right]\:{cm}^{\mathrm{3}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\approx\:\mathrm{408}.\mathrm{58}\:{cm}^{\mathrm{3}} \:. \\ $$
Commented by MrW3 last updated on 09/Sep/18
$${answer}\:{is}\:{correct}\:{sir}.\:{thanks}. \\ $$
Answered by MrW3 last updated on 09/Sep/18
Commented by MrW3 last updated on 09/Sep/18
$${R}={radius}\:{of}\:{ball}=\mathrm{2}.\mathrm{5}\:{cm} \\ $$$${Centers}\:{of}\:\mathrm{4}\:{balls}\:{build}\:{a}\:{regular} \\ $$$${tetrahedon}\:{with}\:{edge}\:{length}\:{a}=\mathrm{2}{R}. \\ $$$${the}\:{distance}\:{between}\:{two}\:{opposite} \\ $$$${edges}\:{is}\:{x}=\frac{{a}}{\:\sqrt{\mathrm{2}}}=\sqrt{\mathrm{2}}{R}. \\ $$$$ \\ $$$${height}\:{from}\:{bottum}\:{of}\:{glass}\:{to}\:{top} \\ $$$${of}\:{the}\:{balls}\:{is}\:{h}={x}+\mathrm{2}{R}=\left(\sqrt{\mathrm{2}}+\mathrm{2}\right){R}. \\ $$$$ \\ $$$${water}\:{needed}\:{just}\:{to}\:{cover}\:{all}\:{balls}\:{is} \\ $$$${V}_{{min}} =\pi\left(\mathrm{2}{R}\right)^{\mathrm{2}} {h}−\mathrm{4}×\frac{\mathrm{4}\pi{R}^{\mathrm{3}} }{\mathrm{3}} \\ $$$$=\left(\sqrt{\mathrm{2}}+\mathrm{2}−\frac{\mathrm{4}}{\mathrm{3}}\right)\mathrm{4}\pi{R}^{\mathrm{3}} \approx\mathrm{408}.\mathrm{58}\:{cm}^{\mathrm{3}} \\ $$