Chapter 20, Problem 66A

Try this in the bathtub or while washing dishes. Lower a glass, mouth downward, over a small floating object as shown. What happens? How deep would the glass have to be pushed to compress the enclosed air to, half its volume? (Hint: You can’t do this in your bathtub unless it’s 10.3 m deep!)

To determine

To explain: The effect if a glass is lowered down over a floating object.

To find: The depth at which the glass should be pushed in order to compress the enclosed air.

Answer to Problem 66A

If glass is lowered down then air pocket shrinks.

Glass should be push double of its height to compress the air to half of its volume that time

Explanation of Solution

Introduction:

Buoyant force:The force exerted by water or any liquid .

When the glass is pushed downward over a floating object in water that at time air pocket shrinks. Air present in glass will put force on water and pressure on glass will increases due to which it is difficult to push it downward .

Consider

Height of glass = H

When air is compressed to half of its volume that time

Density of water= 2ρ

Density of air = ρ

By pascals law,

  $\begin{array}{l}{\rho }_{1}g{h}_1={\rho }_{2}g{h}_2\\ \rho gH=\frac{1}{2}\rho g{h}_2\\ 2H=h_2\end{array}$

So glass should be push double of its height to compress the air to half of its volume that time.