How much work will be done in increasing...

How much work will be done in increasing the diameter of a soap bubble from 2 cm to 5 cm. Surface tension of soap solution is `3.0xx10^(-1) Nm^(-1)`.

Text Solution

Verified by Experts

Soap bubble has two surfaces.
Hence, `W=T Delta A`
Here, `DeltaA=2[4pi{(2.5xx10^(-2))^(2)-(1.0xx10^(-2))^(2)}]`
`=1.32xx10^(-2)m^(2)`
`therefore " Work done" W=(3.0xx10^(-2))(1.32xx10^(-2))J=3.96xx10^(-4) J`

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How much work will be done in increasing the diameter of a soap bubble from 2cm to 5cm? Surface tension solution is 3.0xx10^(-2)N//m.

`3.96 xx 10^4 J`

`2.58 xx 10^4 J`

`1.46 xx 10^4 J`

`1.88 xx 10^4 J`

Work done in increasing the size of a soap bubble from a radius of 3 cm to 5 cm is nearly. (surface tension of soap solution = 0.3 Nm^(-1))

`4pimJ`

`0.2pimJ`

`2pimJ`

`0.4 pi mJ`

Work done in increasing the size of a soap bubble from a radius of 3cm to 5cm is nearly (Surface tension of soap solution =0.03Nm^-1 )

(a) `0.2pimJ`

(b) `2pimJ`

(d) `4pimJ`