Question and Answers Forum

All Questions      Topic List

Geometry Questions

Previous in All Question      Next in All Question      

Previous in Geometry      Next in Geometry      

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  .

$${Minimum}\:{Volume}\:{of}\:{water}\:{so}\:\: \\ $$$${that}\:{topmost}\:{ball}\:{remains}\: \\ $$$${submerged}\:{is}\:\:=\:\pi\left(\mathrm{2}{R}\right)^{\mathrm{2}} {h}−\frac{\mathrm{4}×\mathrm{4}}{\mathrm{3}}\pi{R}^{\mathrm{3}} \\ $$$$\:\:\:\:\boldsymbol{{h}}\:=\:\boldsymbol{{R}}+\boldsymbol{{x}}+\boldsymbol{{R}} \\ $$$$\:\:\:\:\:\:\:=\:\mathrm{2}{R}+{R}\sqrt{\mathrm{2}} \\ $$$$\:\:{V}_{{min}} =\:\pi\left(\mathrm{25}\right)\left(\mathrm{5}+\frac{\mathrm{5}\sqrt{\mathrm{2}}}{\mathrm{2}}\right){cm}^{\mathrm{3}} −\frac{\mathrm{2}\pi}{\mathrm{3}}×\mathrm{125}{cm}^{\mathrm{3}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:=\:\mathrm{125}\pi\left[\frac{\left(\mathrm{2}+\sqrt{\mathrm{2}}\right)}{\mathrm{2}}−\frac{\mathrm{2}}{\mathrm{3}}\right]\:{cm}^{\mathrm{3}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\approx\:\mathrm{408}.\mathrm{58}\:{cm}^{\mathrm{3}} \:. \\ $$

Commented by MrW3 last updated on 09/Sep/18

answer is correct sir. thanks.

$${answer}\:{is}\:{correct}\:{sir}.\:{thanks}. \\ $$

Answered by MrW3 last updated on 09/Sep/18

Commented by MrW3 last updated on 09/Sep/18

R=radius of ball=2.5 cm  Centers of 4 balls build a regular  tetrahedon with edge length a=2R.  the distance between two opposite  edges is x=(a/(√2))=(√2)R.    height from bottum of glass to top  of the balls is h=x+2R=((√2)+2)R.    water needed just to cover all balls is  V_(min) =π(2R)^2 h−4×((4πR^3 )/3)  =((√2)+2−(4/3))4πR^3 ≈408.58 cm^3

$${R}={radius}\:{of}\:{ball}=\mathrm{2}.\mathrm{5}\:{cm} \\ $$$${Centers}\:{of}\:\mathrm{4}\:{balls}\:{build}\:{a}\:{regular} \\ $$$${tetrahedon}\:{with}\:{edge}\:{length}\:{a}=\mathrm{2}{R}. \\ $$$${the}\:{distance}\:{between}\:{two}\:{opposite} \\ $$$${edges}\:{is}\:{x}=\frac{{a}}{\sqrt{\mathrm{2}}}=\sqrt{\mathrm{2}}{R}. \\ $$$$ \\ $$$${height}\:{from}\:{bottum}\:{of}\:{glass}\:{to}\:{top} \\ $$$${of}\:{the}\:{balls}\:{is}\:{h}={x}+\mathrm{2}{R}=\left(\sqrt{\mathrm{2}}+\mathrm{2}\right){R}. \\ $$$$ \\ $$$${water}\:{needed}\:{just}\:{to}\:{cover}\:{all}\:{balls}\:{is} \\ $$$${V}_{{min}} =\pi\left(\mathrm{2}{R}\right)^{\mathrm{2}} {h}−\mathrm{4}×\frac{\mathrm{4}\pi{R}^{\mathrm{3}} }{\mathrm{3}} \\ $$$$=\left(\sqrt{\mathrm{2}}+\mathrm{2}−\frac{\mathrm{4}}{\mathrm{3}}\right)\mathrm{4}\pi{R}^{\mathrm{3}} \approx\mathrm{408}.\mathrm{58}\:{cm}^{\mathrm{3}} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com