By sucking through a straw, a student can reduce the pressure in his lungs to $750 m m$ of $H g$ (Density $= 13.6 g / {c m}^{3}$ ). Using the straw, he can drink water from a glass up to a maximum depth of

10 c m

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75 c m

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13.5 c m

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1.36 c m

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Solution

The correct option is A $13.5 c m$
Given,
$ρ_{H} = 13.6 g / c m^{3}$
$ρ_{w} = 1 g / c m^{3}$
$P_{0} = 760 m m o f H g$
$P_{l} = 750 m m o f H g$
The pressure difference between the lungs of student and the atmosphere is given by
$Δ P = P_{0} - P_{l}$
$Δ P = 760 - 750 = 10 m m o f H g$
$Δ P = 1 c m o f H g$
This pressure can be used for the drinking water.
$1 c m o f H g =$ Pressure difference due to water column
$ρ_{H} g h_{H} = ρ_{w} g H$
$1 \times 13.6 \times g = 1 \times g \times H$
$H = 13.6 c m$
The correct option is C.

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By sucking through a straw, a student can reduce the pressure in his lungs to 750mm of Hg (Density =13.6g/cm3). Using the straw, he can drink water from a glass up to a maximum depth of

By sucking through a straw, a student can reduce the pressure in his lungs to $750 m m$ of $H g$ (Density $= 13.6 g / {c m}^{3}$ ). Using the straw, he can drink water from a glass up to a maximum depth of