KCET 2025 Physics Paper

To solve this problem, we apply the principles of fluid dynamics, specifically the continuity equation and Bernoulli's equation.
Using the Continuity Equation:
The continuity equation for fluid flow states that the product of the cross-sectional area of the pipe and the fluid velocity must remain constant:
A1V1=A2V2
Given:
‌A1=10cm2=10×10−4m2
‌V1=1m∕ s
‌A2=5cm2=5×10−4m2
Solving for V2 :
V2=‌

A1V1

A2

=‌

10×10−4×1

5×10−4

=2m∕ s
Applying Bernoulli's Equation:
Bernoulli's equation for steady, incompressible flow along a streamline is:
P1+‌

1

2

ρV12=P2+‌

1

2

ρV22
Where:
‌P1=2000‌Pa
‌ρ=1000‌kg∕m3‌ (density of water) ‌
Rearranging to find P2‌ : ‌
‌P2=P1+‌

1

2

ρ(V12−V22)
Substituting the known values:
‌P2=2000+‌

1

2

×1000×(12−22)
‌P2=2000+‌

1

2

×1000×(1−4)
‌P2=2000+500×(−3)=2000−1500=500‌Pa
Thus, the pressure of the water at the second point in the pipe is 500 Pa .