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At 20^(@)C the radius of mercury drop i...

At `20^(@)C` the radius of mercury drop is 3 mm and its surface tension is `4.65 xx 10^(-1)` N/m .What is the excess of pressure inside the drop ?

A

`310 N//m^(2)`

B

`210 N//m^(2)`

C

` 110 N//m^(2)`

D

`10 N//m^(2)`

Text Solution

Verified by Experts

The correct Answer is:
A
` P= (2T)/r = (2xx4.65 xx 10^(-1))/(3xx10^(-3))= 210 " N/m"^(2)`
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