Question

A large tank filled with water to a height $H$. A small hole is made at the base of the tank. It takes $T_1$ time to decrease the height of water to $H/\eta ,(\eta >1)$ and it takes $T_2$ time to take out the rest of water. If $T_1=T_2$, then the value of $\eta$ will be

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

Answer 1

The correct option is C $4$

Velocity of water coming out of orifice is given by $\rho gh=\rho v^2/2\Rightarrow v=\sqrt{2gh}$

Volume flowing out per second is $A\times v=-d\left(Ah\right)/dt$

$\Rightarrow \sqrt{2}gdt=-dh/sqrth$

$\sqrt{2}g{\int }_0^{T_1}=-\left(2\sqrt{h}\right)|{}_H^{H/n}$

$Rightarrow\sqrt{2}g{T}_1=2\sqrt{H}-2\sqrt{H/n}$

as, $H/n\to 0\sqrt{2}g{T}_1=2\sqrt{H}$

$\sqrt{H}-\sqrt{H/n}=\sqrt{H/n}\Rightarrow n=4$