Question

An inverted bell lying at the bottom of a lake 47.6 m deep has 50 cm³ of air trapped in it. The bell is brought to the surface of the lake. The volume of the trapped air will be (atmospheric pressure = 70 cm of Hg and density of Hg = 13.6 g/cm³)

• A. 350 cm³
• B. 300 cm^3B
• C. 250 cm³
• D. 22 cm

Answer 1

Answer: (B)

300 cm^3B

Answer 2

According to Boyle's law, pressure and volume are inversely

proportional to each other i.e. $P \propto \frac { 1 } { V }$

⇒ $P _ { 1 } V _ { 1 } = P _ { 2 } V _ { 2 }$

⇒ $\left( P _ { 0 } + h \rho _ { w } g \right) V _ { 1 } = P _ { 0 } V _ { 2 }$

$\Rightarrow V _ { 2 } = \left( 1 + \frac { h \rho _ { w } g } { P _ { 0 } } \right) V _ { 1 }$

⇒ $V _ { 2 } = \left( 1 + \frac { 47.6 \times 10 ^ { 2 } \times 1 \times 1000 } { 70 \times 13.6 \times 1000 } \right) V _ { 1 }$

[As $= 70 \times 13.6 \times 1000$]