A) \[66\frac{2}{3}%\]
B) \[78\frac{1}{2}%\]
C) 100%
D) None of these
Correct Answer: C
Solution :
Let radius of hemispherical bowl and cylindrical vessel be r. Also, r = 50% more than h \[\Rightarrow \]r = 50% of h + h \[\Rightarrow \]\[r=\frac{3h}{2}\] ? (i) Now, volume of bowl \[({{V}_{1}})\,=\frac{2}{3}\pi {{r}^{3}}\] ?(ii) and volume of vessel \[({{V}_{2}})=\pi {{r}^{2}}h\] ?(iii) Dividing eqn. (iii) by (ii), we get \[\frac{{{V}_{1}}}{{{V}_{2}}}=\frac{\frac{2}{3}\pi {{r}^{3}}}{\pi {{r}^{2}}h}=\frac{2r}{3h}=\frac{2}{3h}\left( \frac{3h}{2} \right)\] [by (i)] = 1 \[\Rightarrow \] \[{{V}_{1}}={{V}_{2}}\] \[\therefore \]Volume of bowl = Volume of vessel The amount of beverage that can be poured into vessel is 100%
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