An eye drop bottle is prepared consisting of a hemisphere, a cylinder and a conical cap, as shown in the given diagram. Height of the cylindrical and conical parts are each, equal to the diameter (7 cm). Find the :

(a) minimum height of the cylindrical box required to pack this bottle.

(b) volume of the liquid medicine (shaded part) in the bottle. Give your answer to the nearest whole number. (Use π = $\dfrac{22}{7}$)

(a) Given,

Height of the cylindrical and conical parts = 7 cm

Minimum height of the cylindrical box required to pack this bottle = Height of cylindrical part + Height of conical part = 7 + 7 = 14 cm.

Hence, minimum height of the cylindrical box required to pack this bottle = 14 cm.

(b) From figure,

Diameter of cylindrical part = 7 cm

Radius of cylindrical part (r) = $\dfrac{7}{2}$ = 3.5 cm

Volume of Hemisphere = $\dfrac{2}{3}πr^3$

= $\dfrac{2}{3} \times \dfrac{22}{7} \times (3.5)^3$

= $\dfrac{2}{3} \times \dfrac{22}{7} \times 42.875$

= 89.83 cm³

Volume of cylinder = πr²h

= $\dfrac{22}{7} \times (3.5)^2 \times 7$

= 22 × 12.25

= 269.5 cm³

Total volume of liquid medicine (shaded part) = Volume of cylinder - Volume of hemisphere

= 269.5 - 89.83

= 179.67 ≈ 180 cm³

Hence, volume of liquid medicine = 180 cm³.