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If the surface tension of a soap water is 0.04 N/m Then excess pressure inside a 10 mm diameter soap bubble in `N//m^(2)` will be

A

4

B

16

C

8

D

32

Text Solution

Verified by Experts

The correct Answer is:
D
` P_(i)- P_(0)= (4T)/r `
` = (4xx 0.04)/(5 xx 10^(-3))= 16/5 xx 10 = 160/5 = 32 `
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Knowledge Check

  • The surface tension of soap water is 0.04 N/m. The excess pressure inside a 10 mm diameter soap bubble in will be

    A
    4
    B
    8
    C
    16
    D
    32

  • The excess pressure inside a soap bubble is

    A
    inversely proportional to the surface tension
    B
    inversely proportional to its radius
    C
    directly proportional to square of its radius
    D
    inversely proportional to square of its radius

  • Surface tension of water is 0.072 N m^(-1) . The excess pressure inside a water drop of diameter 1.2 mm is

    A
    `240 Nm^(-2)`
    B
    `120 Nm^(-2)`
    C
    `60 Nm^(-2)`
    D
    `72 Nm^(-2)`

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