Question

Air is trapped in a horizontal glass tube he tby $40cm$ mercury column as shown below:
If the tube is held vertical keeping the open end up. length of air column shrink to 19cm. Length in cm by which the mercury column shifts down is?

• A. 8
• B. 10
• C. 15
• D. 20

Answer 1

Answer: (B)

Final pressure=$1+\frac{40}{76}=\frac{116}{76}$ atm

Final height=$19cm$

Initial pressure= $1$ atm

So, $P_i{V}_i=P_t{V}_t=K$

$1hi.A=\frac{116}{76}\times 19A$

$hi=29$

So, length by which mercury column shifts down

$=h_i-h_t$

$=29-19$

$=10cm$