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 zero is
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(c)Time taken for the level to fall from H to $H'$ $t = \frac{A}{{{A_0}}}\sqrt {\frac{2}{g}} \,\,\left[ {\sqrt H - \sqrt {H'} } \right]$
According to problem- the time taken for the level to fall from h to $\frac{h}{2}$ ${t_1} = \frac{A}{{{A_0}}}\sqrt {\frac{2}{g}} \;\;\left[ {\sqrt h - \sqrt {\frac{h}{2}} } \right]$
and similarly time taken for the level to fall from $\frac{h}{2}$ to zero ${t_2} = \frac{A}{{{A_0}}}\sqrt {\frac{2}{g}} \;\left[ {\sqrt {\frac{h}{2}} - 0} \right]$ $\therefore \;\frac{{{t_1}}}{{{t_2}}} = \frac{{1 - \frac{1}{{\sqrt 2 }}}}{{\frac{1}{{\sqrt 2 }} - 0}} = \sqrt 2 - 1.$
According to problem- the time taken for the level to fall from h to $\frac{h}{2}$ ${t_1} = \frac{A}{{{A_0}}}\sqrt {\frac{2}{g}} \;\;\left[ {\sqrt h - \sqrt {\frac{h}{2}} } \right]$
and similarly time taken for the level to fall from $\frac{h}{2}$ to zero ${t_2} = \frac{A}{{{A_0}}}\sqrt {\frac{2}{g}} \;\left[ {\sqrt {\frac{h}{2}} - 0} \right]$ $\therefore \;\frac{{{t_1}}}{{{t_2}}} = \frac{{1 - \frac{1}{{\sqrt 2 }}}}{{\frac{1}{{\sqrt 2 }} - 0}} = \sqrt 2 - 1.$
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