Answer:
Internal diameter of hemispherical bowl \[=36m\] \[\therefore \] Radius of hemispherical bowl \[(r)=18\text{ }cm\] Volume of liquid \[=\frac{2}{3}\pi {{r}^{3}}\] \[=\frac{2}{3}\times \pi \times {{18}^{3}}\] \[\because \] Diameter of bottle \[=6\text{ }cm\] \[\therefore \] Radius of bottle \[=3\text{ }cm\] Now, volume of a cylindrical bottle \[=\pi {{R}^{2}}h\] \[=\pi {{3}^{2}}h\] \[=9\pi h\] Volume of liquid to be transfer = volume of liquid\[-\]10% volume of liquid \[=\frac{2}{3}\pi {{18}^{3}}-\frac{10}{100}\left( \frac{2}{3}\pi {{18}^{3}} \right)\] \[=\frac{2}{3}\pi {{18}^{3}}\left( 1-\frac{10}{100} \right)\] \[=\frac{2}{3}\pi {{18}^{3}}\times \frac{9}{10}\] \[=\pi \times {{18}^{3}}\times \frac{3}{5}\] Number of cylindrical bottles \[\text{=}\frac{\text{Volume}\,\,\text{of}\,\,\text{liquid}\,\,\text{to}\,\,\text{be}\,\,\text{transfered}}{\text{Volume}\,\,\text{of}\,\,\text{a}\,\,\text{bottle}}\] \[\text{72=}\frac{\pi \times 18\times 18\times 18\times \frac{3}{5}}{9\pi h}\] \[\text{h=}\frac{27}{5}=5.4\,cm\] Hence, height of each bottle will be \[5.4\text{ }cm\].
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