A vertical triangular area with vertex downward and altitude 'h' has its base lying on the free surface of a liquid. The centre of pressure below the free surface is at a distance of -

- 1
  
  $\frac{h}{4}$
- 2
  
  $\frac{h}{3}$
- 3
  
  $\frac{h}{2}$
- 4
  
  $\frac{2 h}{3}$

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