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A line was measured with a steel tape which was exactly 30m at 25$^\circ$C at a pull of 10kg. The measured length being 1700m. The temperature during measurement was 34$^\circ$C and pull applied ...
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Given - $P_m=\text{pull applied} =18kg , P_o=\text{pull} =10kg , \text{ cross section area}= 0.025 cm^2 . E= 2.1× 106 kg/cm^2\text{ Total length} =1700m, α=\text{ coefficient of expansion}^\circ \text{C}=3.5×10^{-6}/ ^\circ C$

$T_m=340$

$T_0=250C$

$L=30m$

1. $\text{Temperature correction per tape length}$ ( $C_t$ ) = $\alpha (T_m - T_0 )L$ $\phantom{Temperature correction per tape length ($C_t$) =}= 3.5×10-6 ( 34 – 25 ) × 30$ $\phantom{Temperature correction per tape lengthsdgfg}$( $C_t$)$=9.45 × 10^{-4} (+VE)$

2. $\text{To find pull correction per tape length}\\ \text{Pull correction per tape length} = \frac{p_{m-P_0 } L}{AE}=\frac{(18-10)30}{(0.025 × 2.1× 10^6 ) } = 4.57× 10^{-3}$

3. $\text{To find combine correction}$ $\text{Combine correction} = 9.45 × 10^{-4} + 4.57× 10^{-3} = 5.515× 10^{-3} m$

4. $\text{To find true length of the tape}$

$\text{True length of the tape} =30 + \text{combine correction}$ $\phantom{\text{True length of the tape}}= 30 + 5.515× 10^{-3}$

$\phantom{\text{True length of the tape}}= 30.0055m$

5. $\text{To find true length of the line}$

$\text{True length of line} = \frac{30.0055}{30}×1700 = 1700.312m$