Question

A cylindrical tank having large diameter is filled with water to a height H. A hole of cross-sectional area 5 cm$^2$ in the tank allows water to drain out. If the water drains out at the rate of $2 \times 10^{-3$ m$^3$s$^{-1}$, then the value of H is} (acceleration due to gravity = $10$ ms$^{-2}$)

• A. 80 cm
• B. 120 cm
• C. 60 cm
• D. 90 cm

Hint:

Torricelli's theorem for speed of efflux: $v = \sqrt{2gH}$.
Volume flow rate: $Q = A \times v$.
Ensure all units are consistent (SI units are recommended).
Convert area from cm$^2$ to m$^2$: $1 \text{ cm}^2 = (10^{-2} \text{ m})^2 = 10^{-4} \text{ m}^2$.

Answer 1

The Correct Option is A

According to Torricelli's theorem, the speed of efflux ($v$) of a liquid through a hole at a depth $H$ below the surface is given by $v = \sqrt{2gH}$, where $g$ is the acceleration due to gravity. This assumes the diameter of the tank is large compared to the hole, so the velocity of the liquid surface falling is negligible. The volume flow rate ($Q$) of the liquid draining out is given by $Q = A_{hole} \times v$, where $A_{hole}$ is the cross-sectional area of the hole. So, $Q = A_{hole} \sqrt{2gH}$. Given values: Area of the hole $A_{hole} = 5 \text{ cm}^2 = 5 \times (10^{-2} \text{ m})^2 = 5 \times 10^{-4} \text{ m}^2$. Volume flow rate $Q = 2 \times 10^{-3} \text{ m}^3\text{s}^{-1}$. Acceleration due to gravity $g = 10 \text{ ms}^{-2}$. \[ \begin{itemize} \item[1.] Substitution: \begin{itemize} \item $2\!\times\!10^{-3} = 5\!\times\!10^{-4}\sqrt{20H}$ \end{itemize} \item[2.] Isolate radical: \begin{itemize} \item $\sqrt{20H} = \frac{2\!\times\!10^{-3}}{5\!\times\!10^{-4}} = 4$ \end{itemize} \item[3.] Solve for H: \begin{itemize} \item $20H = 16$ \item $H = 0.8$ m \end{itemize} \end{itemize} \] To convert to centimeters (cm), since $1 \text{ m} = 100 \text{ cm}$: $H = 0.8 \text{ m} \times 100 \frac{\text{cm}}{\text{m}} = 80 \text{ cm}$. \[ \boxed{80 \text{ cm}} \]