Water is filled up to a height h in a cylindrical vessel. It takes time t to completely drain the vessel by means of a small hole at the bottom. If water is filled up to a height 4h then the time it takes to completely drain the vessel is

Question

Water is filled up to a height h in a cylindrical vessel. It takes time t to completely drain the vessel by means of a small hole at the bottom. If water is filled up to a height 4h then the time it takes to completely drain the vessel is (A) t (B) 4t (C) 2t (D) (t/4)

Tardigrade Admin · Accepted Answer

Time required to empty the tank, t=(A/A0)√(2 H/g)⇒ (t2/t1)=√(H2/H1)=√(8 h/2 h)=2 ∴ t2=2t

Q. Water is filled up to a height $h$ in a cylindrical vessel. It takes time $t$ to completely drain the vessel by means of a small hole at the bottom. If water is filled up to a height $4 h$ then the time it takes to completely drain the vessel is

$t$

$4 t$

$2 t$

$\frac{t}{4}$

Time required to empty the tank,
$t = \frac{A}{A _{0}} \frac{2 H}{g} \Rightarrow \frac{t _{2}}{t _{1}} = \frac{H _{2}}{H _{1}} = \frac{8 h}{2 h} = 2$
$∴ t_{2} = 2 t$

Q. Water is filled up to a height h in a cylindrical vessel. It takes time t to completely drain the vessel by means of a small hole at the bottom. If water is filled up to a height 4h then the time it takes to completely drain the vessel is

t

4t

2t

4t​

Time required to empty the tank, t=A0​A​g2H​​⇒t1​t2​​=H1​H2​​​=2h8h​​=2 ∴t2​=2t

Q. Water is filled up to a height $h$ in a cylindrical vessel. It takes time $t$ to completely drain the vessel by means of a small hole at the bottom. If water is filled up to a height $4 h$ then the time it takes to completely drain the vessel is

$t$

$4 t$

$2 t$

$\frac{t}{4}$

Time required to empty the tank,
$t = \frac{A}{A _{0}} \frac{2 H}{g} \Rightarrow \frac{t _{2}}{t _{1}} = \frac{H _{2}}{H _{1}} = \frac{8 h}{2 h} = 2$
$∴ t_{2} = 2 t$