Water rises in a vertical capillary tube up to a length of 10 cm. If the tube is inclined at $45^\circ$, the length of water risen in the tube will be

Question

KollegeApply · Accepted Answer

\text{ Given }: \text{ l = 10 cm } \alpha = {45}^0 \text{ Rise in water level after the tube is tilted = h} \Rightarrow \text{ l = h} \cos {45}^0 \Rightarrow h = \frac{l}{\cos {45}^0} = \frac{10}{\left( 1/$\sqrt$(2) \right)} = 10$\sqrt$(2)\text{ cm } [Solution diagram: 67365_solution_1.png]

AChoose Your Answer