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With the bubble centered, a 100-m-length sight gives a reading of 1. 352 m. After

moving the bubble four divisions off center, the reading is 1. 410 m. For 2-mm vial

divisions, what is (a) the vial radius of curvature in meters, and (b) the angle in

seconds subtended by one division?

1 Answer

3 votes

Step-by-step explanation:

To find the vial radius of curvature and the angle subtended by one division, we can use the formula:

Radius of Curvature (R) = (L^2) / (8S)

where:

- L is the sight length (in meters)

- S is the distance the bubble is moved off center (in meters)

(a) Vial Radius of Curvature:

Using the given values, L = 100 m and S = 4 divisions × 2 mm/division = 8 mm = 0.008 m.

Substituting these values into the formula:

R = (100^2) / (8 × 0.008) = 125000 m

Therefore, the vial radius of curvature is 125,000 meters.

(b) Angle Subtended by One Division:

To find the angle subtended by one division, we can use the formula:

Angle (θ) = 2πR / (L × 3600)

Using the value of R calculated in part (a) and L = 100 m:

θ = (2π × 125000) / (100 × 3600) ≈ 0.001745 radians

To convert the angle to seconds, we can use the fact that 1 radian is equal to 206,265 seconds:

θ_seconds = 0.001745 × 206265 ≈ 359 seconds

Therefore, one division subtends an angle of approximately 359 seconds.

In summary:

(a) The vial radius of curvature is 125,000 meters.

(b) One division subtends an angle of approximately 359 seconds.

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