a source of light in placed at a distance `4m` from a photocell and the stopping potential is then `7.7` volt. If the distence is helved the stopping potential now will be

A source of light is placed at a distance of 50 cm from a photo cell and the stopping Potential is found to be V_0 . If the distance between the light surface and photo cell is made 25 cm, the new stopping potential will be

A source of light is placed at a distance of 1m. From a photocell and the cut off potential is found to be V_0. If the distance is doubled what will be the cut off potential?

A source of light is placed at a distance x from a photocell and the cut off potential is V . If the distance of the same source from the photocell is doubled, then cut off potential will be :

A source of light is placed at a distance of 50cm from a photocell and the cutt-off potential is found to be V_(0) . If the distance between the light source and the photocell is made 25cm, what will be the new cutt-off potential? Justify your answer.

A source of light is placed at a distance of 1 m from a photo cell and the cut off potential is V_0 . If the source of light is placed at a distance of 50 cm, the cut off potential will be :

A source of light is placed at a distance of 1 m from a photo cell and the cut off potential is found to be 5 V. If the source is kept at 1.5 m from the cell, then the cut off potential will be

When light source is placed at 1 m distant from photo electric cell , then value of stopping potential is obtained 4 volt. If it is placed at 4 m distant, then value of stopping potential becomes -

When a monochromatic source of light is at a distance of 0.2 m from a photocell, the stopping potential (cut off voltage) and the saturation current are found to be respectively 1 V and 27 mA. If the same source is placed at a distance 0.6 m from the cell, then

When point source is kept at 20 cm distance from photcell ,stopping voltage obtained is V_(0) When source distance is changed to 1m, stopping potential will be ……..