Question

A tank of height 5 m is full of water. There is a hole of cross sectional area 1cm21cm2 in its bottom. The value of water that will come out from this hole per second is (g=10m/s2g=10m/s2)

Answer 1

Answer:

10-3m3/s

Explanation:

v/t=A√2gh=10-4x√2x10x5=10-3m3/s

Answer 2

Question:

A tank of height $5 \mathrm{~m}$ is full of water. There is a hole of cross-sectional area $1 \mathrm{~cm}^{2}$ in its bottom. The initial volume of water that will come out from this hole per second is

Answer:

The volume flow rate is $10^{-3} \mathrm{~m} / \mathrm{s}$

Explanation:

Given:

A tank of height $5 \mathrm{~m}$ is full of water. There is a hole of cross-sectional area $1 \mathrm{~cm}^{2}$ in its bottom.

To find:

The initial volume of water that will come out from this hole per second.

Step 1

Since the velocity of water coming out $=\sqrt{2 \mathrm{gh}}$

$&=\sqrt{2 \times 10 \times 5}$

$&=10 \mathrm{~m} / \mathrm{s}$

Step 2

So the volume of water $=$ velocity $*$ area

$=10 \times 10^{-4} \mathrm{~m}^{3} / \mathrm{s}$

volume flow rate $=10^{-3} \mathrm{~m} / \mathrm{s}$

Therefore, the volume flow rate is $10^{-3} \mathrm{~m} / \mathrm{s}$.

#SPJ2