A body of mass ‘m’ dropped from a height ‘h’ reaches the ground with a speed of $0.8 \sqrt{\mathrm{gh}}$ . The value of work done by air friction is

From the question,

Total Work done = change in kinetic energy

${\mathrm{W}}_{\mathrm{mg}}+{\mathrm{W}}_{\mathrm{air}-\mathrm{friction}}=\frac{1}{2}\mathrm{m}{(0.8\sqrt{\mathrm{gh}})}^2-\frac{1}{2}\mathrm{m}{\left(0\right)}^2$

${\mathrm{W}}_{\mathrm{air}-\mathrm{friction}}=\frac{0.64}{2}\mathrm{mgh}-\mathrm{mgh}=-0.68\mathrm{mgh}$

Hence the correct answer is -0.68mgh.