Question

A spherical glass vessel has a cylindrical neck $ 7 $ cm long and $ 4cm $ in diameter. The diameter of the spherical part is $ 21 $ cm. Find the quantity of water it can hold. [ use $ \pi = \dfrac{{22}}{7} $ ]

Answer 1

Hint: Here we will find the volume of the cylinder and the volume of the spherical vessel, add both the volumes and then find the resultant value for the quantity of water held and convert it in litres.

Complete step-by-step answer:
Given that the spherical vessel has a cylindrical neck with $ 9cm $ long
Diameter, $ d = 4cm $
Radius, $ r = 2cm $
Diameter of the spherical part $ = 21cm $
To get the quantity of water that it can hold can be calculated by using the volume of the spherical vessel.
Volume of spherical vessel, $ = \dfrac{4}{3}\pi {r^3} $ cubic units
Also, the volume of spherical vessel can also be expressed as $ = \dfrac{1}{6}\pi {d^3} $ cubic units
Place the values in the above expression –
 $ V = \dfrac{1}{6} \times \dfrac{{22}}{7} \times {21^3} $
Simplify the above expression finding the multiples and division –
 $ V = 4851c{m^3} $
Now, the volume of the cylinder neck $ = \pi {r^2}h $ cubic units
Place the values in the above equation –
 $ V = \dfrac{{22}}{7} \times {2^2} \times 7 $
Simplify the above expression finding the multiples and division –
 $ V = 88c{m^3} $
Therefore, the total volume of the vessel is the sum of the volume of the spherical vessel and the volume of its cylinder.
 $
  V = 4851 + 88 \\
  V = 4939c{m^3} \;
  $
Total volume is $ = 4939c{m^3} $
 $ 1c{m^3} = 1000mL $
The amount of water the vessel can hold in litres $ = \dfrac{{4939}}{{1000}} $
Simplify the above expression –
The amount of water the vessel can hold in litres $ = 4.939 $ litres
So, the correct answer is “4.939 litres”.

Note: Always read the question twice, find the correlation between the given terms and unknown terms and use the correct formula to solve it. Remember the correct formulas for the volume of the solid figures. All the systems of units should be in the same format. Know the conversational ratios and apply it accordingly.