$J m^{- 2}$ is the unit of:

Surface tension

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Viscosity

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Strain energy

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Power

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Solution

The correct option is A Surface tension
Surface tension $= \frac{F}{l} = \frac{N}{m} = \frac{N m}{m^{2}} = \frac{J}{m^{2}}$

Viscosity $= \frac{F o r c e}{A r e a \times \frac{d v}{d z}} = \frac{N}{m^{2} \frac{m}{s m}} = \frac{N m s}{m^{3}} = \frac{J s}{m^{3}}$

Strain energy $= J$
Power $= J / s$

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