Question

3. The volume of the air bubble increases by
10%, when it rises from bottom to the
surface of the water pond. If the
temperature of the water is constant, find
the depth of the pond. ( 1 atm = 10m of
water column)​

Answer 1

AnSwer :

Volume of the air bubble at the bottom of the pond (V₁) = V

Volume of the air bubble at the surface of the pond V₂ = V × 110/100

Pressure at the bottom of the pond (P₁) = (h + H)cm of water column

where 'h' is the depth of the pond and 'H' is the atmospheric pressure

Pressure on the surface of the pond (P₂) = H cm of water column.

From Boyle's law,

${\sf {P_1 V_1 = P_2V_2}}$

${ \sf{(h + H)V =H \bigg(V \times \dfrac{110}{100} \bigg) }}$

${ \sf{h + H =H \times \dfrac{110}{100} }}$

${ \sf{h = H \times \dfrac{110}{100} - H}}$

${\sf{h = H \times \dfrac{10}{100} }}$

${ \sf{h = \dfrac{H}{10}}}$

Therefore, Depth of the pond = H/10 cm.

= 1000/10 = 100cm = 1m.

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

#sanvi….