Determine the pressure difference in a tube of non-uniform cross-sectional area as shown in the figure.
ΔP = ?
d₁ = 5 cm, V₁ = 4, d₂ = 2 cm, V₂ = ?
(A) 304200 Pa
(B) 304500 Pa
(C) 302500 Pa
(D) 303500 Pa

To calculate the pressure difference in a tube of non-uniform cross-sectional area, we can use Bernoulli's equation and the principle of continuity.
Given Information:
• d1=5 cmd_1 = 5 \, \text{cm} (initial diameter of the tube),
• V1=4 m/sV_1 = 4 \, \text{m/s} (initial velocity of the fluid),
• d2=2 cmd_2 = 2 \, \text{cm} (final diameter of the tube),
• V2V_2 is the unknown velocity at the second cross-section,
• We need to find the pressure difference ΔP\Delta P between the two sections.
Step 1: Apply the Continuity Equation
The continuity equation states that for incompressible fluid flow, the product of the cross-sectional area and velocity is constant along the flow. Therefore,
A1V1=A2V2A_1 V_1 = A_2 V_2
Where A1A_1 and A2A_2 are the cross-sectional areas at the first and second points, respectively, and V1V_1 and V2V_2 are the velocities at these points. The cross-sectional area of the tube at each point is given by:
A=π(d2)2=πd24A = \pi \left( \frac{d}{2} \right)^2 = \frac{\pi d^2}{4}
So, we can write the continuity equation as:
πd124V1=πd224V2\frac{\pi d_1^2}{4} V_1 = \frac{\pi d_2^2}{4} V_2
This simplifies to:
d12V1=d22V2d_1^2 V_1 = d_2^2 V_2
Substitute the given values d1=5 cmd_1 = 5 \, \text{cm}, d2=2 cmd_2 = 2 \, \text{cm}, and V1=4 m/sV_1 = 4 \, \text{m/s}:
(5)2×4=(2)2×V2(5)^2 \times 4 = (2)^2 \times V_2 25×4=4×V225 \times 4 = 4 \times V_2 V2=1004=25 m/sV_2 = \frac{100}{4} = 25 \, \text{m/s}
Step 2: Apply Bernoulli's Equation
Now that we have the velocities at both points, we can use Bernoulli's equation to determine the pressure difference:
P1+12ρV12=P2+12ρV22P_1 + \frac{1}{2} \rho V_1^2 = P_2 + \frac{1}{2} \rho V_2^2
Rearranging this equation to solve for the pressure difference:
ΔP=P1−P2=12ρ(V22−V12)\Delta P = P_1 - P_2 = \frac{1}{2} \rho \left( V_2^2 - V_1^2 \right)
Here, ρ\rho is the density of the fluid, which we will assume to be that of water (ρ=1000 kg/m3\rho = 1000 \, \text{kg/m}^3).
Substitute the values V1=4 m/sV_1 = 4 \, \text{m/s}, V2=25 m/sV_2 = 25 \, \text{m/s}, and ρ=1000 kg/m3\rho = 1000 \, \text{kg/m}^3:
ΔP=12×1000×(252−42)\Delta P = \frac{1}{2} \times 1000 \times \left( 25^2 - 4^2 \right) ΔP=500×(625−16)\Delta P = 500 \times \left( 625 - 16 \right) ΔP=500×609\Delta P = 500 \times 609 ΔP=304500 Pa\Delta P = 304500 \, \text{Pa}
Final Answer:
The pressure difference is 304500 Pa304500 \, \text{Pa}.
Thus, the correct answer is (B) 304500 Pa.