Question

Given $P_{atm}=76\text{cm of Hg}$.
When the tube is held at an angle of ${45}^{\circ }$ to the plane with the open end up, what should be the length of air column?

• A. 12 cm
• B. 11.17 cm
• C. 11.0 cm
• D. 10.8 cm

Answer 1

The correct option is B 11.17 cm

Applying Boyle's law,
$(12\text{cm}\times A)\times 76\text{cm}=(l\times A)\times (76+8\sin {45}^{\circ })$
$\therefore l=\frac{12\times 76}{81.65}\text{cm}=11.17\text{cm}$