Two droplets each radius r of mercury merge and form a bigger drop . If surface tension of mercury is T.Then what will be the surface energy ?

Options

A. `2^((5)/(3))pir^(2)T`
B. `4pir^(2)T`
C. `2pir^(2)T`
D. `2^((8)/(3)pir^(2))T`

Answer

d

Explanation

Radius of small droplet is r and volume is `V_(1)` when two droplets are merged and form a bigger drop having radius R and volume is `V_(2)`
`thereforeV_(2)=V_(1)xx2`
`(4)/(3)piR^(3)=2xx(4)/(3)pir^(3)`
`R^(3)=2r^(3)`
`R=2^((1)/(3))xxr` ...(1)
Now , surface enerfy of bigger drop,
`E=4piR^(2)T=4pi(2^((2)/(3))*r^(2))T=pixx2^(2)xx2^((2)/(3))xxr^(2)T`
`=pixx2^(2+(2)/(3))xxr^(2)T`
`thereforeE=2^((8)/(3))pir^(2)T`