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)//.^(4)sqrt(2)`

Answer

D

Explanation

`v_(0)=sqrt(2ga)" "....(1)`
`v=sqrt(2gh)`
`v=sqrt(2gxx(a)/(sqrt(2)))" "....(2)`
From `(1) & (2) , " "v=(v_(0))/(2^(1//4))`