If the distance between the instrument station and staff station is 1km, the correction for refraction in 'm' is

0.0121

No worries! We‘ve got your back. Try BYJU‘S free classes today!

0.0785

No worries! We‘ve got your back. Try BYJU‘S free classes today!

0.0112

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

0.0673

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is C 0.0112
Correction due to refraction is,

$C_{r} = + 0.0112 d^{2}$

Where, $C_{r}$ is in meters and d in kilometers.

Suggest Corrections

Similar questions

Q. A theodolite is set up at station A and 3 m long staff is is held vertically at station B. The depression angle reading at 2.5 m marking on the staff is

6^{\circ} 10^{'}

. The horizontal distance between A and B is 2200 m. Height of instrument at station A is 1.1 m and R.L of A is 880.88 m. Apply the curvature and refraction correction, and determine the R.L of B (in m)

Q. A train reaches Station D by overshooting Station C by

\frac{1}{4}

km. Station B lies exactly between stations A and D, and the total distance between Station A and Station D is

3 \frac{1}{5}

km. Calculate the distance between Station B to Station C.

Q. A theodolite is set up at station A. The RL of instrument axis is 212.250 m. The angle of elevation to the top of a 4 m long staff , held vertical at station B, is

7^{\circ}

. The horizontal distance between station A and B is 400 m. Neglecting the errors due to curvature of earth and refraction , the RL ( in m , round of to three decimal places ) of station B is

Q. An instrument was setup at P and angle of elevation to a vane 4 m above the foot of staff held at Q was

9^{\circ} 30^{'}

. The horizontal distance between P and Q was known to be 2000 m. What is the R.L. (in m) of staff station Q?
[Given that the R.L of instrument axis was 2650.38 m and consider refraction and curvature corrections]

All were so happy at the idea except the rail master. He told that before it was so easy to calculate the distance between the two stations, for e.g. distance between Station (1) and Station (5) is (5 - 1) = 4 km. But with this design, how to calculate the distance between the new stations i.e. Station (-1) and Station (-4)?

Then you explained: We move in the forward direction if we add two integers and we move towards the left of the number line if we subtract two integers. Using this we can find the distance between Station (-1) and Station (-4) equal to

Join BYJU'S Learning Program

If the distance between the instrument station and staff station is 1km, the correction for refraction in 'm' is

The correct option is C 0.0112Correction due to refraction is, Cr=+0.0112d2 Where, Cr is in meters and d in kilometers.

The correct option is C 0.0112
Correction due to refraction is,

$C_{r} = + 0.0112 d^{2}$

Where, $C_{r}$ is in meters and d in kilometers.