A square box of water has a small hole located the bottom corners. When the box is full and sitting on a level surface, complete opening of the hole results in a flow of water with a speed `v_(0)` as shown in Fig. (a). when the box is still half empty, it is tilted by `45^@` so that hole is at the lowest point. Now the water will flow out with a speed of

Options

A. `v_(0)`
B. `(v_(0))/(2)`
C. `(v_(0))/(sqrt(2))`
D. `(v_(0))/(4sqrt(2))`

Answer

D

Explanation

`V_(0)=sqrt(2gh)impliesV^(')=sqrt(2g.(h)/(sqrt(2)))=(V_(0))/(4sqrt(2))`