Synthetic Division (Part A) | Mathematical Tool | By Naman Agarwal | JEE Mains and Advanced

The figure shown below is a screwgauge. When the circular scale division matches the main scale at 43, the 6.5 mm mark on the main scale is just visible. The main scale has (1)/(2)mm marks. In complete rotation, the screw advances by (1)/(2) mm and circular scale has 50 divisions. The reading of the screwgauge is

In an instrument, there are 25 divisions on the vernier scale which have length of 24 divisions of the main scale. 1 cm on main scale is divided in 20 equal parts. Find the least count.

Internal micrometer is a measuring intrument used to measure internal diameter (ID) of a large cylinder bore with high accuracy. Construction is shown in figure. There is one fixed rod B (to the right in figure) and one moved rod A (to the left in figure). It is based on the particle of advancement of a screw when it is rotated in a nut with internal threads. Main scale reading can be directly seen on the hub which is fixed with respect to rod B. When the cap is rotated rod A moves in or cut depending on direction of rotation. The circular scale reading is seen by checking which division of circular scale coincide with the references line This is to be multiplied by LC to get circular scale reading. Least count = value of 1 circular scale division = ("pitch")/("number of division on circular scale") Length of rod A is chosen to match the ID(PQ) to be measured. Zero error is checked by taking reading between standard blocks fixed at normal value of ID to be measured. Zero error is positive if cap end is one the right of the main scale and negative it is on the left side. During zero setting of the above instrument, the end of the cap is on left side of the zero of main scale (i.e. zero of main scale is not visible) and 41^(th) division of circular scale coincides with the reference line, the zero error is -

A spherometer has 100 equal divisions marked along the periphery of its disc, and one full rotation of the disc advances on the main scale by 0.01cm. Find the least count of the system.

In a vernier calipers, 1 cm of the main scale is divided into 20 equal parts. 19 divisions of the main scale coincide with 20 divisions on the vernier scale . Find the least count of the instrument.

In a vernier callipers, there are 10 divisions on the vernier scale and 1 cm on the main scale is divided in 10 parts. While measuring a length, the zero of the vernier lies just ahead of 1.8 cm mark and 4th division of vernier coincides with a main scale division FInd the length.

In a complete rotation, spindle of a screw gauge advances by (1)/(2)mm . There are 50 divisions on circular scale. The main scale has (1)/(2)mm marks to (is graduated to (1)/(2)mm ) If a wire is put between the jaws, 3 main scale divisions are clearly visible, and 20th division of circular scale coincides with reference line. Find diameter of wire in correct significant figures.

In a complete rotation, spindle of a screw gauge advances by (1)/(2)mm . There are 50 divisions on circular scale. The main scale has (1)/(2)mm marks to (is graduated to (1)/(2)mm ) If a wire is put between the jaws, 3 main scale divisions are clearly visible, and 20th division of circular scale coincides with reference line. Find diameter of wire in correct significant figures.

In a vernier callipers, there are 10 divisions on the vernier scale and 1 cm on the main scale is divided in 10 parts. While measuring a length, the zero of the vernier lies just ahead of 1.8 cm mark and 4th division of vernier coincides with a main scale division . If zero error of vernier callipers is -0.02 cm, what is the correct length ?