Question

A vertical narrow glass tube of uniform cross-section, closed at one end has air filled in it upto $40cm$ of height, with a column of mercury of height $10cm$ above $i$. If the tube is gradually inclined in the vertical plane and brought to a position making an angle of ${60}^o$ with the vertical, then new length of air column entrapped within the tube is ( atmosphereic pressure=76cm of $Hg$)

• A. $45.47cm$
• B. $43.47cm$
• C. $42.47cm$
• D. $44.47cm$

Answer 1

The correct option is D $44.47cm$

A vertical narrow glass tube of uniform cross-section

The height $=40cm$ of height

Column of mercury of height $=10cm$

Angle$=60°$

$P=h\rho gP\times 40\times A={\rho }_1\times \left(50\right)\times A\Rightarrow {\rho }_1=\frac{45}{45}+1\times (45-x)\times {\rho }_2{h}_1=44.47cm$