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A soap bubble, having radius of 1mm, is ...

A soap bubble, having radius of 1mm, is blown from a detergent solution having a surface tension of `2.5 xx 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

A

0.5cm

B

100cm

C

10cm

D

1cm

Text Solution

Verified by Experts

The correct Answer is:
D
Here, R=mm `=1 xx 10^(-1) cm= 1 xx 10^(-3)m`
Density of water `rho= 10^(3) kgm^(-3)`
`h= Z_(0)`
`T= 2.5 xx 10^(-2) N//m`

`g= 10 ms^(-2)`
Pressure inside the bubble, formed in air,

`P_(i) = P_(0) + (4T)/(R )= P + (4T)/(R )` ...(1) (Where P= atmospheric pressure)
If pressure at depth `Z_(0)` from the free surface of water is P. then,
`P.= P + rho gh= P + rho g Z_(0)` ...(2)
Comparing equations (1) and (2), `rho g Z_(0) = (4T)/(R )`
`:. Z_(0) = (4T)/(rho gR) = (4 xx 2.5 xx 10^(-2))/(10^(3) xx 10 xx 1 xx 10^(-3))`
`:. Z_(0) = 10^(-2)m= 1cm`
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