NEET-XII-Physics
P1 year:2019
- #7A soap bubble, having radius of 1 mm, is
blown from a detergent solution having a
surface tension of ``2.5 × 10^{-2}`` N/m.
The pressure inside the bubble equals at a
point ``Z_0`` below the
free surface of water in a container. Taking
``g = 10 m/s^2``, density of water = ``10^3 kg/m^3``,
the value of ``Z_0`` is :
(1) 100 cm
(2) 10 cm
(3) 1 cm
(4) 0.5 cmdigAnsr: 3Ans : ( 3 )
Sol. Excess pressure =
`` \frac{4T}{R} ``
Gauge pressure
=`` ρgZ_0``
``P_0 + \frac{4T}{R} = P_o + ρgZ_0``
Hence ``\rho_0 = \frac{4T}{R ρ g}``
``Z_0 = \frac{4 \times\ 2.5 \times 10^{-2} }{10^{-3} \times 1000 \times 10 } ``
= 1 cm.