An inverted (bell) lying at the bottom of a lake $47.6 m$ deep has $50 c m^{3}$ 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 $= 70 c m$ of $H g$ and density of $H g = 13.6 g / c m^{3}$ ).

350 c m^{3}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

300 c m^{3}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

250 c m^{3}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

22 c m^{3}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A $300 c m^{3}$
Volume of air trapped $\propto \frac{1}{P r e s s u r e}$
PV=constant
Then ,P $_{d e p t h} \times V_{d e p t h} = P_{s u r f a c e} \times V_{s u r f a c e}$
$(P_{a t m} + ρ g h) V_{d e p t h} = P_{a t m} \times V_{s u r f a c e}$

Thus putting values as given , V $_{s u r f a c e}$ =300 cm $^{3}$

Suggest Corrections

Similar questions

Q. 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)

Q. An inverted vessel (bell) lying at the bottom of a lake

50.6 m

deep has

50 c c

of air trapped in it. The bell is brought to the surface of lake. The volume of the trapped air will now be

Q. The volume of an air bubble becomes three times as it rises from the bottom of a lake to its surface. Assuming atmospheric pressure to be 75 cm of Hg and the density of water to be 1/10 of the density of mercury, the depth of the lake.

The volume of an air bubble becomes three times as it rises from the bottom of a lake to its surface. Assuming atmospheric pressure to be 75 cm of Hg and the density of water to be 1/10 of the density of mercury, the depth of the lake is

Join BYJU'S Learning Program

An inverted (bell) lying at the bottom of a lake 47.6 m deep has 50cm3 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).

The correct option is A 300 cm3Volume of air trapped ∝1Pressure PV=constantThen ,Pdepth×Vdepth=Psurface×Vsurface(Patm+ρgh)Vdepth=Patm×VsurfaceThus putting values as given , Vsurface=300 cm3

An inverted (bell) lying at the bottom of a lake $47.6 m$ deep has $50 c m^{3}$ 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 $= 70 c m$ of $H g$ and density of $H g = 13.6 g / c m^{3}$ ).