Question

A $100cm$ long thin tube (sealed at both ends) lies horizontally, in the middle $0.1m$ containing mercury and the two ends containing air at standard atmospheric pressure. If the tube is turned to a vertical position, by what amount will the mercury be displaced?

Answer 1

When the tube is horizontally placed:(shown in fig.)

When the tube is vertically placed:(shown in fig.)

Let the area of the tube be 'a'

Boyle's law:$P_1{V}_1=P_2{V}_2$

For the chamber A: $(76\times 45a)=P_A\times (45+d)a$

$P_A=\frac{76\times 45}{45+d}$

For the chamber B: $(76\times 45a)=P_B\times (45-d)a$

$P_B=\frac{76\times 45}{45-d}$

Also $P_A+10=P_B$

$=\frac{76\times 45+(450+10d)}{45+d}=\frac{76\times 45}{45-d}$

Cross multiplying we get, $10d^2+6840d-20250=0$

$d=3cm$