1.

A soap bubble is blown to a radius of 3 cm.If it is tob be further blown to a radius of 4 cm what is the work done ? (Surface tension of soap solution =3.06xx10^(-2)Nm^(-1))

Answer»

Solution :Initial RADIUS of the soap bubble `R_(1s)=3cm=3xx10^(-2)m`
Final radius of the soap bubble
`R_(2)=4cm=4xx10^(-2)m`
WORK done in blowing a soap bubble from radius
`R_(1)` to `R_(2)=W=8pi(R_(2)^(2)-R_(1)^(2))S`
`=8xx22/7xx3.06xx10^(-2)(16-9)xx10^(-4)`
`176xx3.06xx10^(-6)J=539.6xx10^(-6)J`


Discussion

No Comment Found