A steady flow of a liquid of density ρ is shown in figure. At point 1 , the area of crosssection is 2 A and the speed of flow of liquid is 2   m   s - 1 . At point 2 , the area of crosssection is A . Between the points 1 and 2 , the pressure difference is 100   N   m - 2…

Question

A steady flow of a liquid of density &#961; is shown in figure. At point 1 , the area of crosssection is 2 A and the speed of flow of liquid is 2 &#160; m &#160; s - 1 . At point 2 , the area of crosssection is A . Between the points 1 and 2 , the pressure difference is 100 &#160; N &#160; m - 2 and the height difference is 10 &#160; cm . The value of &#961; is (Acceleration due to gravity = 10 &#160; m &#160; s - 2 )

Aicharya · Accepted Answer

According to the equation of continuity,
$A_{1} v_{1} = A_{2} v_{2} \Rightarrow v_{2} = 2 \sqrt{2} m s^{- 1}$
For a steady flow, applying Bernoulli's equation at point $1$ and $2$ ,
$P_{1} + ρ g h_{1} + \frac{1}{2} ρ {v_{1}}^{2} = P_{2} + ρ g h_{2} + \frac{1}{2} ρ {v_{2}}^{2} \Rightarrow ρ [g (h_{1} - h_{2}) + \frac{({v_{1}}^{2} - {v_{2}}^{2})}{2}] = P_{2} - P_{1} \Rightarrow ρ [10 \times 0.1 + \frac{2 - 8}{2}] = - 100 \Rightarrow ρ = 50 kg m^{- 3}$

A steady flow of a liquid of density ρ is shown in figure. At point 1 , the area of crosssection is 2 A…

Solution

A steady flow of a liquid of density &#961; is shown in figure. At point 1 , the area of crosssection is 2 A…

Solution

A steady flow of a liquid of density ρ is shown in figure. At point 1 , the area of crosssection is 2 A…