A) \[11.0\times {{10}^{-2}}\text{N/m}\]
B) \[6.0\times {{10}^{-2}}\text{N/m}\]
C) \[3.0\times {{10}^{-2}}\text{N/m}\]
D) \[1.5\times {{10}^{-2}}\text{N/m}\]
Correct Answer: C
Solution :
Work done = Area increased\[\times \]surface tension so, surface tension, \[{{S}_{T}}=\frac{3\times {{10}^{-4}}}{2\times (10\times 11-10\times 6)\times {{10}^{-4}}}\] \[=3\times {{10}^{-2}}N\text{/}m\]You need to login to perform this action.
You will be redirected in
3 sec