Bissoy.com
Doctors Medicines MCQs Q&A

When an instrument is at P the staff readings on P is 1.824 and on Q is 2.748. When the instrument at Q the staff readings on P is 0.928 and Q is 1.606. Distance between P and Q is 1010 mts. R.L. of P is 126.386. Find true R.L. of Q?

When an instrument is at P the staff readings on P is 1.824 and on Q is 2.748. When the instrument at Q the staff readings on P is 0.928 and Q is 1.606. Distance between P and Q is 1010 mts. R.L. of P is 126.386. Find true R.L. of Q? Correct Answer 125.585

When observations are taken from P the apparent difference in elevation between P and Q is 2.748 – 1.824 = 0.924. When observations are taken from Q the apparent difference in elevation between P and Q is 1.606 – 0.928 = 0.678. Hence true difference in elevation is (0.924 +0.678)/2 = 0.801 m. Therefore true elevation of Q is 126.386 – 0.801 = 125.585 m.

Related Questions

When an instrument is at P the staff readings on P is 2.748 and on Q is 1.824 when instrument at Q the staff readings on P is 1.606 and Q is 0.928. Distance between P and Q is 1010 mts. R.L. of P is 126.386. Find the angular error in collimation adjustment of the instrument?
When an instrument is at P the staff readings on P is 2.748 and on Q is 1.824 when instrument at Q the staff readings on P is 1.606and Q is 0.928. Distance between P and Q is 1010 mts. R.L. of P is 126.386. Find true R.L of Q?
When instrument is at P the staff readings on P is 1.824 and on Q is 2.748. When instrument at Q the staff readings on P is 0.928 and Q is 1.606. Distance between P and Q is 1010 mts. R.L. of P is 126.386. Find combined correction for curvature and refraction?
When an instrument is at P the staff readings on P is 1.824 and on Q is 2.748. When instrument at Q the staff readings on P is 0.928 and Q is 1.606. Distance between P and Q is 1010 mts. R.L. of P is 126.386. Find the error due to collimation?
When instrument is at P the staff readings on P is 2.748 and on Q is 1.824 when instrument at Q the staff readings on P is 1.606and Q is 0.928. Distance between P and Q is 1010 mts. R.L. of P is 126.386. Find combined effect correction for curvature and refraction?
When instrument is at P the staff readings on P is 2.748 and on Q is 1.824 when instrument at Q the staff readings on P is 1.606and Q is 0.928. Distance between P and Q is 1010 mts. R.L. of P is 126.386. Find error due to collimation?
Match the following, in the context of the leveling staff. a. Self-reading staff 1. The smallest division is 5 mm. its length is about 3 metres. b. Solid Staff 2. It has a moving target against which readings are taken by the staff man. The sliding target comes along with the vernier. c. Telescopic Staff 3. The readings are always taken from the telescope hence appear inverted. Therefore, readings are taken from downwards. d. Target staff 4. It has three telescopic lengths of 1.5, 1.5, and 2 meters when stretched out fully. Normally, its total length is 5 metres
The following are the staff readings given when the staff is held normal. The line of sight of instrument is placed at an angle +3024ꞌ. It being an anallactic lens, find the horizontal distance between staff and instrument station. Staff readings – 2.145 m, 1.925 m, 1.464 m.
Read the given statements and conclusions carefully. Assuming that the information given in the statement is true, even if it appears to be at variance with commonly known facts, decide which of the given conclusions logically follow (s) from the statements. Statement– 1. Some CGL are CHSL 2. Some CHSL are MTS 3. No MTS is JE Conclusion– 1. No MTS is CGL 2. Some CGL is not MTS 3. Some CHSL is not JE
Stadia readings were taken with a theodolite on a vertical staff with the telescope inclined at an angle of depression of 3o30′. The staff readings were 2.990, 2.055 and 1.120. The reduced level of the staff station is 100.000m, and the height of the instrument is 1.40 m. What is the reduced level of the ground at the instrument? Take constants as 100 and zero.