The figure shows a water-filled syringe with a 4.0-cm-long needle. What is the gauge pressure of the water at the point P, where the needle meets the wider chamber of the syringe? The viscosity of water is 1.0×10-3Pa ∙ s.

The gauge pressure of the water at point P is;

P_p = 104525 Pa

We are given;

Length of needle; L = 4 cm = 0.04 m

Viscosity of water; η = 1 × 10^(-3) Pa.s

Radius; r = 1 mm = 0.001 m

Speed; v = 10 m/s

From poiseuille's equation of viscous flow of liquid the flow rate through a cylindrical pipe is;

Q = (ΔP•πr⁴)/(8ηL)

Now, flow rate is also expressed as;

Q = πr²v

Thus;

πr²v = (ΔP•πr⁴)/(8ηL)

This reduces to;

v = (ΔP•r²)/(8ηL)

Plugging in the relevant values;

10 = (ΔP × 0.001²)/(8 × 10^(-3) × 0.04)

ΔP = (10 × 8 × 10^(-3) × 0.04)/(0.001²)

ΔP = 3200 Pa

Now, this means that;

P_p - P_atm = ΔP

Where;

P_p is gauge pressure of the water at point P

P_atm is atmospheric pressure = 101325 Pa

Thus;

P_p - 101325 = 3200

P_p = 101325 + 3200

P_p = 104525 Pa

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Pressure at P point is given as

Explanation:

As we know by poiseuille's equation of viscous flow of liquid through cylindrical pipe

here we know that

L = 4 cm

r = 1 mm

Q = flow rate

now we have

v 154.9= 0.9 length length s little over 154.9cm

f=ma newton

f=100x2.5=250n