Question

If an air bubble rises from bottom of a mercury tank to te top of its volume becomes 1 and half times . when normal pressure is 76 cm of hg then the depth of the tank is ?

Answer 1

Answer: 38 cm

Explanation:

Answer 2

Given that,

Initial volume = V

Final volume $V_{2}=1\dfrac{1}{2}V$

Pressure $P_{1} =76\ cm$

Pressure $P_{2}=76+h$

Tank is at constant pressure.

We need to calculate the depth of tank

Using formula of isothermal

$P_{1}V_{1}=P_{2}V_{2}$

Where, $P_{1}$ = Initial pressure

$P_{2}$ = Finally pressure

$V_{1}$ = Initial volume

$V_{2}$ = Final volume

Put the value into the formula

$76\times \dfrac{3}{2}V=(76+H)\times V$

$76\times\dfrac{3}{2}V=76\times V+V\times H$

$76\times\dfrac{3}{2}=(76+H)$

$H=\dfrac{76\times3}{2}-76$

$H=38\ cm$

Hence, The depth of tank is 38 cm