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 ) /(sqrt2 ) `
D. ` (v_0)/(4sqrt2) `

Answer

D

Explanation

` v_ 0 = sqrt (2ga ) , v' = sqrt(2g (a)/(sqrt2)) = (v_0)/(4sqrt2 ) `