A line measured on a uniform slope with a 30 m tape pulled with a force of 5 kg. The length of line was found to be 525.80 m. The temperature at the time of measurement was $30^{\circ} C$ and the tape is correct at $24^{\circ} C$ when there is no pull. Determine the horizontal distance between the end points of the line. The difference of levels of the two ends of the line is 9.14 m. The coefficient of thermal expansion of the material of the tape is 0.00001 per $^{\circ} C$ . Cross-sectional area of the tape is 2 $m m^{2}$ and its modulus of elasticity is $21 \times 10^{5} k g / c m^{2}$ .
525.8146

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Solution

The correct option is A 525.8146
$T e m p e r a t u r e c o r r e c t i o n = L \times (T - T_{0}) α$

$= 525.8 \times (30 - 24) \times 10^{- 5}$

$= 0.0315 m$

$P u l l c o r r e c t i o n = \frac{(P - P_{0}) \times L}{A E}$

$= \frac{(5 - 0) \times 525.8}{2 \times 10^{- 2} \times 21 \times 10^{5}}$

$= 0.0626 m$

$Slope correction = - L (1 - cos θ)$

$N o w, s i n θ = \frac{9.14}{525.80}$

$∴ θ = 0^{\circ} 59^{'} 45.69 "$

$∴ slope correction = - 525.8 \times (1 - cos (59^{'} 45.69 "))$

$= - 0.0795 m$

$Total correction = 0.0315 + 0.0626 - 0.0795 = 0.0146 m$

$∴ Correct horizontal distance = + 525.8 + 0.0146$

$= 525.8146 m$

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