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The surface tension of water at $20 ° C$ if 72.75 $dyne {cm}^{- 1}$ . Its value in SI system is

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0.07275 $N m^{- 1}$

None of the above

2.275 $N m^{- 1}$

0.7275 $N m^{- 1}$

answer is C.

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Detailed Solution

1 dyne = $10^{- 5}$

1Cm= $10^{- 2}$

Therefore

72.75 dyne= $\frac{10^{- 5}}{10^{- 2}}$

$\begin{array}{r} 72.75 \times 10^{- 3} {Nm}^{- 1} \end{array}$

Then 0.07275 ${Nm}^{- 1}$

Required option is C

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