A large tank filled with water to a height ' h ' is to be emptied through a small hole at the bottom. The ratio of time taken for the level of water to fall from h to ( h /2) and from ( h /2) to bottom is

Question

A large tank filled with water to a height ' h ' is to be emptied through a small hole at the bottom. The ratio of time taken for the level of water to fall from h to ( h /2) and from ( h /2) to bottom is (A) √2 (B) (1/2) (C) √2-1 (D) (1/√2-1)

Tardigrade Admin · Accepted Answer

According to the relation t=(A/A0) √(2/g) H-(A/A0) √(2 H'/g) t =( A / A o ) √(2/ g ) ×(√ H -√ H') t propto(√ H -√ H') ( t 1/ t 2)=(√ H -√( H /2)/√( H )2-√0)=(1-(1/√2)/(1)√2)=√2-1

Q. A large tank filled with water to a height ' $h$ ' is to be emptied through a small hole at the bottom. The ratio of time taken for the level of water to fall from $h$ to $\frac{h}{2}$ and from $\frac{h}{2}$ to bottom is

$2$

$\frac{1}{2}$

$2 - 1$

$\frac{1}{2 - 1}$

According to the relation
$t = \frac{A}{A _{0}} \frac{2}{g} H - \frac{A}{A _{0}} \frac{2 H ^{'}}{g}$
$t = \frac{A}{A _{o}} \frac{2}{g} \times (H - H^{'})$
$t \propto (H - H^{'})$
$\frac{t _{1}}{t _{2}} = \frac{H - \frac{H}{2}}{\frac{H}{2} - 0} = \frac{1 - \frac{1}{2}}{\frac{1}{2}} = 2 - 1$

Q. A large tank filled with water to a height ' h ' is to be emptied through a small hole at the bottom. The ratio of time taken for the level of water to fall from h to 2h​ and from 2h​ to bottom is

2​

21​

2​−1

2​−11​

According to the relation t=A0​A​g2​H​−A0​A​g2H′​​ t=Ao​A​g2​​×(H​−H′​) t∝(H​−H′​) t2​t1​​=2H​​−0​H​−2H​​​=2​1​1−2​1​​=2​−1

Q. A large tank filled with water to a height ' $h$ ' is to be emptied through a small hole at the bottom. The ratio of time taken for the level of water to fall from $h$ to $\frac{h}{2}$ and from $\frac{h}{2}$ to bottom is

$2$

$\frac{1}{2}$

$2 - 1$

$\frac{1}{2 - 1}$

According to the relation
$t = \frac{A}{A _{0}} \frac{2}{g} H - \frac{A}{A _{0}} \frac{2 H ^{'}}{g}$
$t = \frac{A}{A _{o}} \frac{2}{g} \times (H - H^{'})$
$t \propto (H - H^{'})$
$\frac{t _{1}}{t _{2}} = \frac{H - \frac{H}{2}}{\frac{H}{2} - 0} = \frac{1 - \frac{1}{2}}{\frac{1}{2}} = 2 - 1$