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The radii of two mercury drops are R(1) ...

The radii of two mercury drops are `R_(1) and R_(2)` .Under isothermal conditions , a drop is formed from them. The radius (R ) of the resultant rop is given by

A

`R=R_(1)+R_(2)`

B

`R=(R_(1)+R_(2))/2`

C

`R^(3)=R_(1)^(3)+R_(2)^(3)`

D

`R^(2)=R_(1)^(2)+R_(2)^(2)`

Text Solution

Verified by Experts

The correct Answer is:
C
` 4/3 pi R^(3) = 4/3 pi[R_(1)^(3)+R_(2)^(3)] " " :. R^(3)=R_(1)^(3)+R_(2)^(3)`
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