A mercury tube of length 100 cm , open at one end is kept vertically with open end upwards. It is filled with 10 cm of mercury to trap an air column of height 40 cm . What will be the length of trapped air if it is held vertically with open end downwards? [Assume atmospheric pressure to be 1 atm ]

Question

A mercury tube of length 100 cm , open at one end is kept vertically with open end upwards. It is filled with 10 cm of mercury to trap an air column of height 40 cm . What will be the length of trapped air if it is held vertically with open end downwards? [Assume atmospheric pressure to be 1 atm ] (A) 40.00 cm (B) 32.00 cm (C) 52.12 cm (D) 82.00 cm

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="Solution" src="https://cdn.tardigrade.in/q/nta/c-50popa0ht6mx6cq4.jpg" /> Now P1V1=P2V2 ⇒ (.76+10.)× 40=(.76-10.)× l ⇒ 86× 40=66× l ⇒ l=52.12 cm

$40.00 c m$

$32.00 c m$

$52.12 c m$

$82.00 c m$

Now $P_{1} V_{1} = P_{2} V_{2}$
$\Rightarrow (76 + 10) \times 40 = (76 - 10) \times l$
$\Rightarrow 86 \times 40 = 66 \times l$
$\Rightarrow l = 52.12 c m$

Q. A mercury tube of length 100cm , open at one end is kept vertically with open end upwards. It is filled with 10cm of mercury to trap an air column of height 40cm . What will be the length of trapped air if it is held vertically with open end downwards? [Assume atmospheric pressure to be 1atm ]

40.00cm

32.00cm

52.12cm

82.00cm

Now P1​V1​=P2​V2​ ⇒(76+10)×40=(76−10)×l ⇒86×40=66×l ⇒l=52.12cm

$40.00 c m$

$32.00 c m$

$52.12 c m$

$82.00 c m$

Now $P_{1} V_{1} = P_{2} V_{2}$
$\Rightarrow (76 + 10) \times 40 = (76 - 10) \times l$
$\Rightarrow 86 \times 40 = 66 \times l$
$\Rightarrow l = 52.12 c m$