Exams › NEET › Physics

The cylindrical tube of a spray pump has radius, R, one end of which has n fine holes, each of radius r. If the speed of the liquid in the tube is V, the speed of the ejection of the liquid through the holes is:

1. V²R² / nr²
2. V²R² / n³r²
3. V²R / nr
4. V²R² / n²r²

Correct answer: V²R² / nr²

Solution

Using the principle of conservation of volume flow rate, the flow rate in the tube (πR²V) must equal the total flow rate through the holes (n × πr²v). Solving for v, we get v = V(R² / nr²).

Related NEET Physics questions

• If the air density were uniform, then the height of the atmosphere above the sea level to produce a normal atmospheric pressure of \( 1.0 \times 10^{5} \) Pa is (density of air is \( 1.3 \mathrm{kg} / \mathrm{m}^{3}, \mathrm{g}=10 \mathrm{m} / \mathrm{s}^{2} \) ):
• 1atm \( = \)
• The atmospheric pressure at is 100,000 pascal.
• According to Newton, the viscous force acting between liquid layers of area A and velocity gradient ΔV/ΔZ is given by \( F = -ηA \frac{ΔV}{ΔZ} \) where η is constant called coefficient of viscosity. The dimensional formula of η is
• As we know that, surface tension(s) is:
• A small hole of area of cross-section 2 mm² is present near the bottom of a fully filled open tank of height 2 m. Taking g = 10 m/s², the rate of flow of water through the open hole would be nearly:

⚔️ Practice NEET Physics free + battle 1v1 →