When a point source of monochromatic light is at a distance of 0.2 m from a photoelectric cell, the cut-off voltage and the saturation current are 0.6 volt and 18 mA respectively. If the same source is placed 0.6 m away from the photoelectric cell, then

The stopping potential will be 0.6 V

Cut off voltage is independent of intensity and hence remains the same. Since distance becomes 3 times, so intensity (I) becomes $\frac{I}{9}$. Hence photo current also decreases by this factor i.e. becomes $\frac{18}{9}=2 mA$.