Hi Scientists/Engineers,

I just want to know if I'm tackling this problem in the right way or if I don't please help me. I was only given the density of mercury at 13600kgm^-3.

Saturated steam at 2 bars enters the top of a sealed cylindrical cavity and condenses, forming a pool of condensate. At the bottom of the cylinder is an exit pipe of diameter 5 cm. This exit pipe is open to the atmosphere, which consists of air at 1 bar. A cone of mass 1 kg blocks this exit, preventing any condensate from draining from the cylinder. The height of the cone that protrudes into the drain is 5 cm.

The volume of a cone of height H and base radius R is πR2H/3.

A manometer which is open to the atmosphere contains 50 cm of mercury. If the condensate level h is equal to 10cm, what if I now ask for the liquid level h where the cone will begin to float, allowing the condensate to exit the cylinder? For this problem, it's ok to ignore any friction that might occur between the cone and the orifice.

I tried equating the weight of the cone which is 9.8N to the weight of the condensate displaced (ρgV). After rearranging for V, I got V= 10^-3 m^-3. So my unknown is height and the radius of the truncated cone in water?

Working:

W = mg = 9.8N = ρgV

=> V = 10^-3 m^3

Thanks in advance.

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**Edited by Dudesons123, 14 August 2017 - 05:50 AM.**