Water flows into a large tank with flat bottom at the rate of  ${10}^{-4}{\text{\hspace{0.17em}m}}^{\text{3}}/\text{s}$. Water is also leaking out of a hole of area  $1{\text{\hspace{0.17em}cm}}^{\text{2}}$ at its bottom. If the height of the water  in the tank remains steady,  then this height (in cm) is :  $\left(\text{take\hspace{0.17em}}g=9.8\text{\hspace{0.17em}m}/{\text{s}}^{\text{2}}\right)$

Given
Rate of flow  $Q={10}^{-4}{\text{\hspace{0.17em}m}}^{\text{3}}/\text{s}$
Acceleration due to gravity  $g=9.8\text{\hspace{0.17em}m}/{\text{s}}^{\text{2}}$

Since height of water column is constant therefore, water inflow rate (Q_in) = water outflow rate 

$\begin{array}{l}{Q}_{\text{in }}={10}^{-4}{\mathrm{m}}^3/\mathrm{s}\\ Q_{\text{out }}=Au={10}^{-4}\times \sqrt{2gh}\\ {10}^{-4}={10}^{-4}\sqrt{19.6\times h}\\ h=\frac{1}{19.6}\mathrm{m}=5.1\mathrm{cm}\end{array}$

Therefore, the correct answer is 5.10