Hello there,

Here is a solution to your question :

The cross-section area of the cylindrical tube A1=8cm2 =8×10−4m2

The speed of the liquid flow inside the tube V1=1.5 m/minute =601.5ms−1

Area of each hole =π(0.5×10−3)2 m2

Area of 40 holes A2=40π(0.5×10−3)2m2

A1V1=A2V2

V2=A2A1V1 =40π×(0.5×10−3)2×608×10−4×1.5=0.636ms−1.

Hence, the answer is 0.636ms−1.

Thank you.