An ideal fluid of density 800 kgm⁻³, flows smoothly through a bent pipe (as shown in figure) that tapers in cross-sectional area from a to a/2. The pressure difference between the wide and narrow sections of pipe is 4100 Pa. At wider section, the velocity of fluid is √x/6 ms⁻¹ for x = ___________. (Given g = 10 ms⁻²)

Solution

Physics sir · 5 steps

The wide end sits 1 m higher, so Bernoulli needs the ρ g h term along with the pressure difference.

1. Continuity first. The area halves, so the speed doubles: a v₁ = a/2 v₂ gives v₂ = 2v₁.
2. Bernoulli between ends. Take the narrow end as the reference level. The wide end is then at h = 1 m. Rearranging gives the equation below, with P₁ - P₂ = 4100 Pa.

   (P₁ - P₂) + ρ g h = 1/2ρ(v₂² - v₁²) = 3/2ρ v₁²

3. Put in numbers. The height term is ρ g h = 800 × 10 × 1 = 8000 Pa. So 4100 + 8000 = 12100 = 3/2 × 800 × v₁² = 1200 v₁². That gives v₁² = 121/12.
4. Match the form. v₁ = 11/√12 = 11/2√3 = 11√3/6 = √363/6 m s⁻¹, since 121 × 3 = 363.
5. So. Comparing with √x/6 gives x = 363.

The common slip: 123. Leaving out the 1 m height drop means only 4100 Pa drives the speed change. That gives v₁² = 41/12 and x = 123.

The idea to keep. In a pipe that changes height, Bernoulli has three terms on each side. Use continuity to write both speeds in terms of one, and check the figure for any height difference.

Answer from the NTA final answer key. Solution by PrepWiser.