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NEET PHYSICSEasy

A small hole of an area of cross-section 2 mm22 \text{ mm}^2 is present near the bottom of a fully filled open tank of height 2 m2 \text{ m}. Taking g=10 m/s2g = 10 \text{ m/s}^2, the rate of flow of water through the open hole would be nearly:

A

6.4×10−6 m3/s6.4 \times 10^{-6} \text{ m}^3/\text{s}

B

12.6×10−6 m3/s12.6 \times 10^{-6} \text{ m}^3/\text{s}

C

8.9×10−6 m3/s8.9 \times 10^{-6} \text{ m}^3/\text{s}

D

2.23×10−6 m3/s2.23 \times 10^{-6} \text{ m}^3/\text{s}

Step-by-Step Solution

  1. Torricelli's Law: The speed of efflux (vv) of a fluid flowing out of an open tank from a hole at a depth hh below the free surface is given by the formula derived from Bernoulli's theorem: v=2ghv = \sqrt{2gh} .
  2. Rate of Flow: The rate of flow (or volume flux, QQ) is defined as the product of the cross-sectional area (AA) of the hole and the velocity (vv) of the fluid emerging from it: Q=A×vQ = A \times v .
  3. Given Data:
  • Area, A=2 mm2=2×10−6 m2A = 2 \text{ mm}^2 = 2 \times 10^{-6} \text{ m}^2.
  • Height, h=2 mh = 2 \text{ m}.
  • Acceleration due to gravity, g=10 m/s2g = 10 \text{ m/s}^2.
  1. Calculation:
  • Calculate velocity: v=2×10×2=40≈6.32 m/sv = \sqrt{2 \times 10 \times 2} = \sqrt{40} \approx 6.32 \text{ m/s}.
  • Calculate flow rate: Q=(2×10−6)×6.32=12.64×10−6 m3/sQ = (2 \times 10^{-6}) \times 6.32 = 12.64 \times 10^{-6} \text{ m}^3/\text{s}.
  • Rounding to one decimal place gives 12.6×10−6 m3/s12.6 \times 10^{-6} \text{ m}^3/\text{s}.
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