Question

A large tank is filled with water upto height H. A small hole is made at the base of the tank. It takes $T_1$ times to decreases the height of water to $vandittakes{T}_2$ time to take out rest of the water.If $T_1=T_2$ then value of X is.

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

Answer 1

The correct option is C 4

$\text{We know that}$

$t=\frac{A}{a}\sqrt{\frac{2}{g}}(\sqrt{H_1}-\sqrt{H_2})$

$t_1=\frac{A}{a}\sqrt{\frac{2}{g}}[\sqrt{H}-\sqrt{\frac{H}{x}}]$

$t_2=\frac{A}{a}\sqrt{\frac{2}{g}}[\sqrt{\frac{H}{x}}-0]$

$t_1=t_2$

$\sqrt{H}-\sqrt{\frac{H}{x}}=\sqrt{\frac{4}{x}}$

$\frac{1}{2}=\frac{1}{\sqrt{x}}\Rightarrow x=4$

Answer:- c