Given J-tube has $2.4 m L$ of air at a pressure of $1 a t m$ . On adding mercury, volume of air is reduced to $1.9 m L$ as shown. Difference in the level of mercury in two columns is ?

700 m m

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200 m m

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900 m m

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760 m m

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Solution

The correct option is B $200 m m$
$P_{1} = 760$ mmHg

$V_{1}$ =2.4mL, $V_{2}$ =1.9mL

$P_{1} V_{1} = P_{2} V_{2}$ or $P_{2} =$ $\frac{P_{1} \times V_{1}}{V_{2}}$ $= \frac{760 \times 2.4}{1.9} = 960$ mmHg

Now
$960 = 760 + h; h = 960 - 760 = 200$ mmHg

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Given J-tube has 2.4 mL of air at a pressure of 1 atm. On adding mercury, volume of air is reduced to 1.9 mL as shown. Difference in the level of mercury in two columns is ?

The correct option is B 200 mmP1=760 mmHg V1=2.4mL, V2 =1.9mL P1V1=P2V2 or P2= P1×V1V2=760×2.41.9=960 mmHg Now960=760+h;h=960−760=200 mmHg

Given J-tube has $2.4 m L$ of air at a pressure of $1 a t m$ . On adding mercury, volume of air is reduced to $1.9 m L$ as shown. Difference in the level of mercury in two columns is ?

The correct option is B $200 m m$
$P_{1} = 760$ mmHg

$V_{1}$ =2.4mL, $V_{2}$ =1.9mL

$P_{1} V_{1} = P_{2} V_{2}$ or $P_{2} =$ $\frac{P_{1} \times V_{1}}{V_{2}}$ $= \frac{760 \times 2.4}{1.9} = 960$ mmHg

Now
$960 = 760 + h; h = 960 - 760 = 200$ mmHg