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 6-10- 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

R.L of A = 880.88 m R.L of plane of collimation, =880.88+1.1=881.98 m B'B"CB"=tan 6^∘10' B'B"=2200×tan 6^∘10' =237.701 R.L of B = R.L of line of sight ...

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Byju's Answer

Standard X

Physics

Total Internal Reflection

A theodolite ...

Question

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)
642.105

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Solution

The correct option is A 642.105
R.L of A = 880.88 m

R.L of plane of collimation,

$= 880.88 + 1.1 = 881.98 m$

$\frac{B^{'} B "}{C B "} = tan 6^{\circ} 10^{'}$

$B^{'} B "= 2200 \times tan 6^{\circ} 10^{'}$

$= 237.701$

R.L of B = R.L of line of sight - B'B"

$- (2.5 - 0.0673 \times 2.22)$

$= 881.98 - 237.701 - (2.5 - 0.3257)$

$= 642.105 m$

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The correct option is A 642.105R.L of A = 880.88 m R.L of plane of collimation, =880.88+1.1=881.98 m B′B"CB"=tan6∘10′ B′B"=2200×tan6∘10′ =237.701 R.L of B = R.L of line of sight - B'B" −(2.5−0.0673×2.22) =881.98−237.701−(2.5−0.3257) =642.105m