1. 12.6×10⁻⁶ m ³ /s 2. 8.9×10⁻⁶ m ³ /s 3. 2.23×10⁻⁶ m ³ /s 4. 6.4×10⁻⁶ m ³ /s

Answer: 1. 12.6×10⁻⁶ m ³ /s By Torricelli's theorem the speed of efflux is v=√(2gh) . v=√(2×10×2)=√(40)≈6.32 m/s. Rate of flow =Av=2×10⁻⁶×6.32≈12.6×10⁻⁶ m ³ /s.

Topic: Bernoulli's principle and applications· Properties of Solids and Liquids

A small hole of area of cross-section 2 mm^2 is present near the bottom of a fully filled open tank of height 2 m. Taking g = 10 m/s^2, the rate of flow of water through the open hole would be nearly :

1. \(12.6\times10^{-6}\) m\(^3\)/s

2. \(8.9\times10^{-6}\) m\(^3\)/s

3. \(2.23\times10^{-6}\) m\(^3\)/s

4. \(6.4\times10^{-6}\) m\(^3\)/s

Doubtly Asked 30 Oct 2026 NEET UG 2019 Practice test neet Save WhatsApp Report

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Answer: 1. \(12.6\times10^{-6}\) m\(^3\)/s

By Torricelli's theorem the speed of efflux is \(v=\sqrt{2gh}\).

\(v=\sqrt{2\times10\times2}=\sqrt{40}\approx6.32\) m/s.

Rate of flow \(=Av=2\times10^{-6}\times6.32\approx12.6\times10^{-6}\) m\(^3\)/s.

Doubtly AI · 30 Oct 2026 AI-generated · not yet verified Report

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A small hole of area of cross-section 2 mm^2 is present near the bottom of a fully filled open tank of height 2 m. Taking g = 10 m/s^2, the rate of flow of water through the open hole would be nearly :

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