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Question

Question: A large tank filled with water to a height ‘h’ is to be emptied through a small hole at the bottom. ...

A large tank filled with water to a height ‘h’ is to be emptied through a small hole at the bottom. The ratio of times taken for the level of water to fall from h to h2\frac { h } { 2 } and from h2\frac { h } { 2 } to zero is

A

2\sqrt { 2 }

B

12\frac { 1 } { \sqrt { 2 } }

C

21\sqrt { 2 } - 1

D

121\frac { 1 } { \sqrt { 2 } - 1 }

Answer

21\sqrt { 2 } - 1

Explanation

Solution

Time taken for the level to fall from H to HH ^ { \prime }

t=AA02g[HH]t = \frac { A } { A _ { 0 } } \sqrt { \frac { 2 } { g } } \left[ \sqrt { H } - \sqrt { H ^ { \prime } } \right]

According to problem- the time taken for the level to fall from

h to t1=AA02g[hh2]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 h2\frac { h } { 2 } to zero

t2=AA02g[h20]t _ { 2 } = \frac { A } { A _ { 0 } } \sqrt { \frac { 2 } { g } } \left[ \sqrt { \frac { h } { 2 } } - 0 \right]