A cylindrical can open at the bottom end lying at the bottom of a lake $47.6 m$ deep has $50 {cm}^{3}$ of air trapped in it. The can is brought to the surface of the lake. The volume of the trapped air will become $($ atmospheric pressure $= 70 cm$ of Hg and density of Hg $= 13.6 g/cc)$ :

350 {cm}^{3}

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300 {cm}^{3}

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250 {cm}^{3}

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22 {cm}^{3}

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Solution

The correct option is C $300 {cm}^{3}$
$P_{o} = 70 cm$ of Hg $= 70 \times 10^{- 2} \times 13600 \times 9.8 = 93296 Pa$
Using Boyle's law: $P_{1} V_{1} = P_{2} V_{2}$
$\Rightarrow (P_{o} + H ρ g) \times 50 \times 10^{- 6} = P_{o} \times V_{2}$
$\Rightarrow (93296 + 47.6 \times 1000 \times 9.8) \times 50 \times 10^{- 6} = 93296 \times V_{2}$
$\Rightarrow (93296 + 466480) \times 50 \times 10^{- 6} = 93296 \times V_{2}$
$\Rightarrow V_{2} = 300 \times 10^{- 6} m^{3} = 300 {cm}^{3}$

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A cylindrical can open at the bottom end lying at the bottom of a lake 47.6 m deep has 50 cm3 of air trapped in it. The can is brought to the surface of the lake. The volume of the trapped air will become (atmospheric pressure =70 cm of Hg and density of Hg =13.6 g/cc):

The correct option is C 300 cm3Po=70 cm of Hg =70×10−2×13600×9.8=93296 PaUsing Boyle's law: P1V1=P2V2⇒(Po+Hρg)×50×10−6=Po×V2⇒(93296+47.6×1000×9.8)×50×10−6=93296×V2⇒(93296+466480)×50×10−6=93296×V2⇒V2=300×10−6 m3=300 cm3

A cylindrical can open at the bottom end lying at the bottom of a lake $47.6 m$ deep has $50 {cm}^{3}$ of air trapped in it. The can is brought to the surface of the lake. The volume of the trapped air will become $($ atmospheric pressure $= 70 cm$ of Hg and density of Hg $= 13.6 g/cc)$ :