1.

A soap bubble (surface tension= 30 xx 10^(-3) N.m^(-1)) has radius 2 cm. The work done in doubling the radius is :

Answer»

0
`1.1305xx10^(-4)J`
`12.261xx10^(-4)J`
`4.403xx10^(-4)J`

Solution :WORK DONE = Surface tension `xx` Increase in area.
For soap bubble increase in area
`=2[4pi(2r)^(2)-4pir^(2)]=24pir^(2)`
`therefore` Work done = T`xx24pir^(2)`
`=30xx10^(-3)xx24xx22/7xx(2xx10^(-2))^(2)`
`=4.043xx10^(-4)J`
Hence correct CHOICE is (d).


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