AP Physics Practice Questions: Fluids

AP Physics multiple choice questions on fluid statics and dynamics. Each question includes detailed solutions at the bottom of the page.

1. What is the pressure at a depth of 10 m in water? (ρ = 1000 kg/m³, g = 9.8 m/s², P_atm = 1.01×10⁵ Pa)

   A) 1.01×10⁵ Pa
   B) 1.99×10⁵ Pa
   C) 2.01×10⁵ Pa
   D) 2.99×10⁵ Pa
   E) 3.01×10⁵ Pa

2. A cube of side 0.1 m and density 800 kg/m³ floats in water (ρ = 1000 kg/m³). What fraction of its volume is submerged?

   A) 0.2
   B) 0.4
   C) 0.6
   D) 0.8
   E) 1.0

3. Water flows through a pipe of radius 2 cm at 1 m/s. What is the flow rate?

   A) 1.26×10⁻³ m³/s
   B) 3.14×10⁻³ m³/s
   C) 6.28×10⁻³ m³/s
   D) 12.6×10⁻³ m³/s
   E) 31.4×10⁻³ m³/s

4. A pipe narrows from 4 cm diameter to 2 cm diameter. If the velocity in the wide section is 2 m/s, what is the velocity in the narrow section?

   A) 2 m/s
   B) 4 m/s
   C) 6 m/s
   D) 8 m/s
   E) 10 m/s

5. According to Bernoulli's equation, for horizontal flow of an ideal fluid, which statement is true?

   A) Pressure increases where velocity increases
   B) Pressure decreases where velocity increases
   C) Pressure is constant everywhere
   D) Velocity is constant everywhere
   E) Both pressure and velocity are constant

Solutions

1. B) 1.99×10⁵ Pa
   \( P = P_{\text{atm}} + \rho g h = 1.01 \times 10^5 + 1000 \times 9.8 \times 10 \)
   \( = 1.01 \times 10^5 + 9.8 \times 10^4 = 1.01 \times 10^5 + 0.98 \times 10^5 = 1.99 \times 10^5 \, \text{Pa} \)

2. D) 0.8
   For floating: weight = buoyant force
   \( \rho_{\text{object}} V_{\text{object}} g = \rho_{\text{fluid}} V_{\text{submerged}} g \)
   \( \dfrac{V_{\text{submerged}}}{V_{\text{object}}} = \dfrac{\rho_{\text{object}}}{\rho_{\text{fluid}}} = \dfrac{800}{1000} = 0.8 \)

3. A) 1.26×10⁻³ m³/s
   Area: \( A = \pi r^2 = \pi (0.02)^2 = \pi \times 4 \times 10^{-4} = 1.257 \times 10^{-3} \, \text{m}^2 \)
   Flow rate: \( Q = A v = 1.257 \times 10^{-3} \times 1 = 1.257 \times 10^{-3} \, \text{m}^3/\text{s} \)

4. D) 8 m/s
   Continuity equation: \( A_1 v_1 = A_2 v_2 \)
   \( A \propto d^2 \), so \( v_2 = v_1 \dfrac{A_1}{A_2} = v_1 \dfrac{d_1^2}{d_2^2} = 2 \times \dfrac{4^2}{2^2} = 2 \times \dfrac{16}{4} = 2 \times 4 = 8 \, \text{m/s} \)

5. B) Pressure decreases where velocity increases
   Bernoulli's equation for horizontal flow: \( P + \dfrac{1}{2} \rho v^2 = \text{constant} \)
   So if \( v \) increases, \( P \) must decrease to keep the sum constant.