The viscous liquid flows in a narrow capillary with the radius r_(0) How many times the volume passing through the capillary in 1 second will change if the radius of capillary increases two times (that is r=2r_(0) ) and the pressure difference remains the same? Please select one answer option V=2V_(0) V=16V_(0) V=8V_(0) V=4V_(0)

To solve this problem, we can use the Hagen-Poiseuille equation, which describes the flow rate of a viscous liquid through a narrow capillary. The equation is given by:<br /><br />\[ Q = \frac{\pi r^4 \Delta P}{8 \eta L} \]<br /><br />where:<br />- \( Q \) is the flow rate,<br />- \( r \) is the radius of the capillary,<br />- \( \Delta P \) is the pressure difference across the capillary,<br />- \( \eta \) is the dynamic viscosity of the liquid,<br />- \( L \) is the length of the capillary.<br /><br />Given that the pressure difference remains the same and the length of the capillary does not change, we can compare the flow rates for the original capillary with radius \( r_0 \) and the new capillary with radius \( 2r_0 \).<br /><br />Let \( Q_0 \) be the flow rate for the original capillary and \( Q \) be the flow rate for the new capillary. Using the Hagen-Poiseuille equation for both cases, we have:<br /><br />\[ Q_0 = \frac{\pi (r_0)^4 \Delta P}{8 \eta L} \]<br />\[ Q = \frac{\pi (2r_0)^4 \Delta P}{8 \eta L} \]<br /><br />We can simplify the equation for \( Q \):<br /><br />\[ Q = \frac{\pi (16r_0^4) \Delta P}{8 \eta L} \]<br /><br />\[ Q = 16 \cdot \frac{\pi (r_0)^4 \Delta P}{8 \eta L} \]<br /><br />\[ Q = 16 \cdot Q_0 \]<br /><br />Therefore, the flow rate for the new capillary is 16 times the flow rate for the original capillary. Since the flow rate is directly proportional to the volume passing through the capillary per second, the volume passing through the capillary in 1 second will also increase by a factor of 16.<br /><br />Thus, the correct answer is:<br /><br />\[ V = 16V_0 \]