The zero line of the Vernier scale lies between divisions 20 and 21 of the main scale. The 4th Vernier scale division exactly coincides with a main scale division. The 5 divisions of the Vernier scale are equal to 4 divisions of the main scale. If one main scale division is 1 mm, the measured value (in mm) is ........... (Rounded off to one decimal place)}
The question involves reading a measurement from a Vernier scale. Given the details: 5 divisions of the Vernier scale equal 4 divisions of the main scale, and one main scale division equals 1 mm.
The relationship between the Vernier scale and main scale is defined as: if 5 Vernier scale divisions (VSD) equal 4 main scale divisions (MSD), one division of the Vernier scale is slightly smaller than one division of the main scale:
VSD = (4/5) MSD = 0.8 mm per VSD
The 4th Vernier scale division coincides with a main scale division, indicating:
4 × VSD = n × MSD
where n is the coinciding main scale division value. Thus,
4 × 0.8 mm = n mm
n = 3.2
This represents the difference between the main scale zero point and the Vernier scale zero point. This discrepancy is called the Vernier constant or least count (LC), calculated as:
LC = 1 MSD - 1 VSD = 1 mm - 0.8 mm = 0.2 mm
The zero of the Vernier lies between divisions 20 and 21 on the main scale, signifying the base measurement as:
Base measurement = 20 mm
The coinciding division (4th Vernier) alters the reading by adding the Vernier coincidence value:
Measured value = Base measurement + (Number of Vernier divisions × LC)
Measured value = 20 mm + (4 × 0.2 mm) = 20.8 mm
Thus, the final measured value is 20.8 mm, which falls within the specified range of 20.8 to 20.8 mm.
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Three plants P1, P2, and P3 produce 6, 1, and 9 thousand liters of fruit juice, respectively. The produced fruit juice is transported to three distribution centers D1, D2, and D3 with a requirement of 7, 5, and 4 thousand liters of juice, respectively. The transportation cost (in hundreds of Rupees) from each plant to each distribution center is given in the table. The total transportation cost (in hundreds of Rupees) in the initial basic feasible solution using Vogel’s approximation method is ............. (Answer in integer) 
A steel plate is fastened to a channel using three identical bolts as shown in the figure. The bolts are made of carbon steel of permissible yield strength in shear as 400 N/mm². The plate is subjected to a force of 12 kN. Neglect the weight of the plate. The magnitude of the resultant shear force (in N) on bolt 2 is ............. (Answer in integer) 