Air is trapped in a horizontal glass tube he tby $40 c m$ 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?

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Solution

The correct option is B 10
Final pressure= $1 + \frac{40}{76} = \frac{116}{76}$ atm
Final height= $19 c m$
Initial pressure= $1$ atm
So, $P_{i} V_{i} = P_{t} V_{t} = K$
$1 h i . A = \frac{116}{76} \times 19 A$
$h i = 29$
So, length by which mercury column shifts down
$= h_{i} - h_{t}$
$= 29 - 19$
$= 10 c m$

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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?

The correct option is B 10Final pressure=1+4076=11676 atmFinal height=19cmInitial pressure= 1 atmSo, PiVi=PtVt=K1hi.A=11676×19Ahi=29So, length by which mercury column shifts down=hi−ht=29−19=10cm

Air is trapped in a horizontal glass tube he tby $40 c m$ 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?