A $10 c m$ column of air is trapped by a column of $H g 4.00 c m$ long in capitary tube of uniform bore when the tube is held horizontally in a room at $1 a t m$ tube is held vertically with the open end up is:

9.50 c m

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1.52 c m

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3.53 c m

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4.61 c m

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Solution

The correct option is A $9.50 c m$
$P = P_{a t m} + 4 \times ρ q$
$= 76 + 4 = 80 m m$ of $H g$
No. of moles is same
$∴ P_{1} V_{1} = P_{2} V_{2}$
$\Rightarrow 76 \times 10 \times A = 80 \times x \times A$
$\Rightarrow x = 9.50 c m$

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A 10 cm column of air is trapped by a column of Hg 4.00cm long in capitary tube of uniform bore when the tube is held horizontally in a room at 1atm tube is held vertically with the open end up is:

The correct option is A 9.50cmP=Patm+4×ρq=76+4=80 mm of HgNo. of moles is same ∴P1V1=P2V2⇒76×10×A=80×x×A⇒x=9.50cm

A $10 c m$ column of air is trapped by a column of $H g 4.00 c m$ long in capitary tube of uniform bore when the tube is held horizontally in a room at $1 a t m$ tube is held vertically with the open end up is:

The correct option is A $9.50 c m$
$P = P_{a t m} + 4 \times ρ q$
$= 76 + 4 = 80 m m$ of $H g$
No. of moles is same
$∴ P_{1} V_{1} = P_{2} V_{2}$
$\Rightarrow 76 \times 10 \times A = 80 \times x \times A$
$\Rightarrow x = 9.50 c m$