Question

A large tank is filled with water to a height H. A small hole is made at the base of the tank. It takes $\eta$ is

• A. 2
• B. 3
• C. 4
• D. $2 \sqrt { 2 }$

Answer 1

Answer: (C)

4

Answer 2

$t = \frac { A } { a } \sqrt { \frac { 2 } { g } } \left[ \sqrt { H _ { 1 } } - \sqrt { H _ { 2 } } \right]$

Now, $T _ { 1 } = \frac { A } { a } \sqrt { \frac { 2 } { g } } \left[ \sqrt { H } - \sqrt { \frac { H } { \eta } } \right]$

and $T _ { 2 } = \frac { A } { a } \sqrt { \frac { 2 } { g } } \left[ \sqrt { \frac { H } { \eta } } - \sqrt { 0 } \right]$

According to problem $T _ { 1 } = T _ { 2 }$

∴$\sqrt { H } - \sqrt { \frac { H } { \eta } } = \sqrt { \frac { H } { \eta } } - 0$⇒ $\sqrt { H } = 2 \sqrt { \frac { H } { \eta } } \Rightarrow \eta = 4$