A line 2 km long is measured with a tape of length 50 m, which is standardized at no pull at 15 °C. The tape in this section is 3 mm wide and 1.25 mm thick. If one half of the line is measured at a temperature of 20°C and the other half at 26°C and the tape is stretched with a pull of 22kg, find the corrected total length, given that the coefficient of expansion is 12*10-6 per oc , weight of tape be 7.75gm/cm3 and E=2.11*106 kg/cm2

Answer:

Explanation:

To solve this problem, we need to consider the effect of temperature on the measured length, as well as the effect of tape elongation due to the applied tension. The corrected total length can be calculated using the following steps:

Step 1: Find the thermal expansion of the tape at the two temperatures.

The thermal expansion of the tape can be calculated using the formula:

ΔL = αLΔT

where ΔL is the change in length, α is the coefficient of thermal expansion, L is the original length, and ΔT is the change in temperature. We can assume that the change in temperature is the same for the entire tape, so we only need to calculate ΔL for one half of the line. Using the given values, we get:

ΔL = αLΔT

= (12*10^-6 per oc) * (1000 m) * (5/2) * (20-15)

= 0.15 m

So the tape will expand by 0.15 m at a temperature of 20°C.

Similarly, at a temperature of 26°C, we have:

ΔL = αLΔT

= (12*10^-6 per oc) * (1000 m) * (5/2) * (26-15)

= 0.27 m

So the tape will expand by 0.27 m at a temperature of 26°C.

Step 2: Find the elongation of the tape due to the applied tension.

The elongation of the tape due to the applied tension can be calculated using the formula:

ΔL = (F * L) / (E * A)

where ΔL is the change in length, F is the applied force, L is the original length of the tape, E is the Young's modulus of the tape material, and A is the cross-sectional area of the tape. We can assume that the force is the same for the entire tape, so we only need to calculate ΔL for one half of the line. Using the given values, we get:

A = (3 mm) * (1.25 mm)

= 3.75 mm^2

= 0.375 cm^2

ΔL = (F * L) / (E * A)

= (22 kg) * (50 m) / (2.11*10^6 kg/cm^2 * 0.375 cm^2)

= 0.111 m

So the tape will elongate by 0.111 m due to the applied tension.

Step 3: Find the corrected total length.

To find the corrected total length, we need to subtract the thermal expansion of the tape at each temperature and add the elongation due to the applied tension. Since we are measuring only half of the line at each temperature, we need to double the result to get the total length. Using the values calculated in steps 1 and 2, we get:

Corrected length at 20°C = (2000 m / 2) - 0.15 m + 0.111 m

= 995.96 m

Corrected length at 26°C = (2000 m / 2) - 0.27 m + 0.111 m

= 994.84 m

Corrected total length = 2 * (995.96 m + 994.84 m)

= 3981.6 m

Therefore, the corrected total length of the line is approximately 3981.6 meters.

The corrected total length of the line is approximately 2000.07 meters.

A material's response to a change in temperature is measured by a coefficient of thermal expansion, which is frequently denoted by the symbol.

To calculate the corrected total length, we need to consider the effects of tape expansion and temperature on the tape and the measured line. Here are the steps to follow:

Calculate the expansion of the tape due to temperature difference:

The temperature difference between 15°C and 20°C is 5°C, and the temperature difference between 15°C and 26°C is 11°C.

The coefficient of expansion is [tex]12X10^-^6[/tex] per °C.

The length of the tape is 50m, and its width and thickness are 3mm and 1.25mm, respectively.

The expansion of the tape due to the temperature difference is:

ΔL = L₀ * α * ΔT

where L₀ is the initial length of the tape (50m), α is the coefficient of expansion ([tex]12X10^-^6[/tex] per °C), and ΔT is the temperature difference (5°C for the first half and 11°C for the second half).

ΔL1 = 50m x [tex]12X10^-^6[/tex]/°C * 5°C = [tex]310^-^3[/tex] m

ΔL2 = 50m x [tex]12X10^-^6[/tex]/°C * 11°C = [tex]6.610^-^3[/tex] m

Calculate the tension in the tape due to the applied pull:

The weight of the tape is 7.75 g/[tex]cm^3[/tex], and its dimensions are 3mm (width) and 1.25mm (thickness).

The mass of the tape is:

m = ρ x V = 7.75 g/[tex]cm^3[/tex] x 3mm * 1.25mm x 50m = 0.73125 kg

The tension in the tape due to the applied pull of 22kg is:

T = F/A = 22kg x g / (3mm * 50m) = 14.67 N

The stress in the tape is:

σ = T/A = T / (3mm x 1.25mm) = 3.72 x [tex]10^7 N/m^2[/tex]

Calculate the strain in the tape due to the stress:

The Young's modulus of the tape is E = [tex]2.1110^6 kg/cm^2[/tex] = [tex]2.11X10^1^0 N/m^2.[/tex]

The strain in the tape is:

ε = σ / E = [tex]3.72X10^7[/tex]N/[tex]m^2[/tex] / [tex]2.11X10^1^0 N/m^2[/tex] = [tex]1.76X10^-^3[/tex]

The elongation of the tape due to the strain is:

δL = L₀ * ε = 50m x 1.76 x [tex]10^-^3[/tex] = 0.088 m

Calculate the corrected length of the line:

The length of the line measured with the tape is 2 km = 2000 m.

Now, let's calculate the corrected total length of the line. The original measured length of the line is 2 km, or 2000 m. Therefore, the corrected total length of the line is:

Lc = 2000 m + ΔL1 + ΔL2 + ε1L1 + ε2L2

Lc = 2000 m + 0.009 m + 0.0198 m + 0.00340 * 1025 m + 0.00340 * 1025 m

Lc = 2000.0686 m

Thus, the corrected total length of the line is approximately 2000.07 meters.

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