`W_(1)` is the work done in blowing a soap bubble of radius r from a soap solution at room temperature . The soap solution is then heated and second soap bubble of radius 2r is blown from the heated solution . If `W_(2)` is the work done in forming the second bubble , then

Options

A. `W_(2)=2W_(1)`
B. `W_()lt2W_(1)`
C. `W_(2)=4W_(1)`
D. `W_(2) lt 4W_(1)`

Answer

D

Explanation

The work done in forming a soap in forming a soap bubble of radius r is `W_(1) = 8pir^(2)T_(1)` where `T_(1)` is the surface tension at room temperature and the work done to form a soap bubble of radius 2r is `W_(2) =8pi (2r)^(2)xx T_(2) = 32 pir^(2) xx T_(2) ` for the heated solution .
` :. (W_(2))/(W_(1)) =(32pir^(2)T_(2))/(8pir^(2)T_(1)) = (4T_(2))/(T_(1))`
If `T_(1) = T_(2)` , then `W_(2) = 4W_(1)`
But it is given that the solution is heated . As the temperature is increased , the S.T of the soap solution is decreased . ` :. T_(2) lt T_(1) " " :. W_(2) lt 4W_(1)`