Empty The Punchbowl

Since the liquid just goes from one container to other containers, the total volume of the liquid remains the same.

Let $n$ be the required no. of cylindrical bottles.

Volume of total liquid in hemispherical bowl $=\frac{2}{3} \pi r^3 = \frac{2}{3} \pi (9)^3$ $cm^3$

Volume of liquid in one bottle $=\pi r^2 h = \pi (\frac{3}{2})^2 \times 4$ $cm^3$

We have, $\frac{2}{3} \pi (9)^3=n \pi (\frac{3}{2})^2 \times 4$

$\implies \frac{2}{3} \times 729=n(\frac{9}{4}) \times 4$

$\implies 2\times 243=9n \implies n=\boxed{54}$

So, no. of bottles required $=n=\boxed{54}$

V. of hemisphere=2/3 22/7 9 9 9 (i) V.of cylinder=22/7 3/2 3/2*4 (ii)

                           (i)/(ii)

=1527/28=54 bottles (ANS.)

the volume of the liquid in the hemispherical bowl remains same when it is transfered to the cylindrical bottles. therefore,

volume of liquid in the bowl = total volume of liquid in the cylinders

$\frac {2}{3} * {\pi} * 9*9*9$ = n ${\pi}*\frac {3}{2}* \frac {3}{2}*4$ , where n is the number of cylinders .

hence, n = $\boxed {54}$

1/2.V ball = k.V bottle ---> 2/3 . phi . r ball^3 = k . phi . r bottle^2 . t_bottle ---> 2/3 . 9^3 = k . (3/2)^2 . 4 ---> 2 . 3 . 9^2 = k . 3^2 ---> k = 54. So, bottles will be needed to empty the bowl are 54. Answer : 54

To find the number of bottles needed to empty the bowl, we need to find the ratio of the volume of the bowl to the volume of each bottle. The volume of a sphere of radius $r$ is given by $\dfrac{4}{3}\pi r^{3}$ and the volume of a right cylinder is given by $\pi r^{2}h$ , where $r$ is the radius of the base and $h$ is the height of the cylinder.

$\dfrac{\dfrac{2}{3}\pi r^{3}}{\pi\left(\dfrac{3}{2}\right)^{2}4} = \boxed{54}$

Volume of bowl = (2/3pi x 9 x 9 x 9) = 486pi

Volume of one bottle = (pi x 3/2 x 3/2 x 4) = 9pi

No. of bottles = (486pi/9pi) = 54