A liquid of density 600 kg/m$$^3$$ flowing steadily in a tube of varying cross-section. The cross-section at a point $$A$$ is 1.0 cm$$^2$$ and that at $$B$$ is 20 mm$$^2$$. Both the points $$A$$ and $$B$$ are in same horizontal plane, the speed of the liquid at $$A$$ is 10 cm/s. The difference in pressures at $$A$$ and $$B$$ points is __________ Pa.

Solution :

Given :

Density of liquid :

$$\rho = 600\text{ kg m}^{-3}$$

Area at point $$A$$ :

$$A_A = 1.0\text{ cm}^2$$

$$= 10^{-4}\text{ m}^2$$

Area at point $$B$$ :

$$A_B = 20\text{ mm}^2$$

$$= 20 \times 10^{-6}\text{ m}^2$$

Speed at point $$A$$ :

$$v_A = 10\text{ cm s}^{-1}$$

$$= 0.1\text{ m s}^{-1}$$

Using equation of continuity :

$$A_Av_A = A_Bv_B$$

$$10^{-4}\times 0.1 = 20\times10^{-6}\times v_B$$

$$v_B = \frac{10^{-5}}{20\times10^{-6}}$$

$$= 0.5\text{ m s}^{-1}$$

Using Bernoulli’s equation :

$$P_A + \frac{1}{2}\rho v_A^2=P_B + \frac{1}{2}\rho v_B^2$$

Therefore,

$$P_A - P_B=\frac{1}{2}\rho(v_B^2-v_A^2)$$

$$=\frac{1}{2}(600)(0.5^2-0.1^2)$$

$$=300(0.25-0.01)$$

$$=300 \times 0.24$$

$$= 72\text{ Pa}$$

Final Answer :

$$72$$