In a screw gauge the zero of main scale coincides with fifth division of circular scale, the circular scale contains 50divisions Now a ball placed between screws, it takes four 0.5mm on main scale in one rotation and it takes four rotatory. To measure the ball then find diameter of the ball is (Index line coincides with 25line on circular scale)

The distance moved by the screw of a screw gauge is $2 \mathrm{mm}$ in four rotations and there are 50 divisions on its cap. When nothing is put between its jaws, 20th division of circular scale coincides with reference line and zero of linear scale is hidden from circular scale when two jaws touch each other or zero of circular scale is lying above the reference line. When plate is placed between the jaws, main scale reads 2 divisions and circular scale reads 20 divisions. Thickness of plate is