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 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

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 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 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 :

What is cut-off potential ?

When a point source of light is at a distnace of 1 m from a photo cell, the cut off voltage is is found to be V. if the same source is placed at 3m distance from photo cell, the cut of voltage will be

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

When a point source of light is at a distance of 50 cm from a photoelectric cell , the cut-off voltage is found to be V_(0) . If the same source is placed at a distance of 1 m from the cell , then the cut-off voltage will be