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/s-2. the rate of flow of water through the open hole would be nearly:

Question

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/s-2. the rate of flow of water through the open hole would be nearly: (A) 2.23 × 10-6m3/s-1 (B) 6.4 × 10-6m3/s-1 (C) 12.6 × 10-6m3/s-1 (D) 8.9 × 10-6m3/s-1

Tardigrade Admin · Accepted Answer

Rate of flow liquid Q =au = a√2gh =2×10-6 m2 ×√2×10×2 m/s = 2×2 ×3.14 ×10-6 m3/s -1 = 12.56×10-6 m3/s -1 = 12.6 ×10-6 m3 /s -1

Q. A small hole of area of cross-section $2 m m^{2}$ 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:

$2.23 \times 1 0^{- 6} m^{3} / s^{- 1}$

$6.4 \times 1 0^{- 6} m^{3} / s^{- 1}$

$12.6 \times 1 0^{- 6} m^{3} / s^{- 1}$

$8.9 \times 1 0^{- 6} m^{3} / s^{- 1}$

Rate of flow liquid
$Q = a u = a 2 g h$
$= 2 \times 1 0^{- 6} m^{2} \times 2 \times 10 \times 2 m / s$
$= 2 \times 2 \times 3.14 \times 1 0^{- 6} m^{3} / s^{- 1}$
$= 12.56 \times 1 0^{- 6} m^{3} / s^{- 1}$
$= 12.6 \times 1 0^{- 6} m^{3} / s^{- 1}$

Q. A small hole of area of cross-section 2mm2 is present near the bottom of a fully filled open tank of height 2m. Taking g=10m/s−2. the rate of flow of water through the open hole would be nearly:

2.23×10−6m3/s−1

6.4×10−6m3/s−1

12.6×10−6m3/s−1

8.9×10−6m3/s−1