A 30 m long tape is standardized at $25^{\circ}\text{C}$. It was used to measure the length of a line which came out to be 200 m. The temperature during the measurement was $35^{\circ}\text{C}$. The coefficient of expansion of the tape was $11 \times 10^{-6}$ per $^{\circ}\text{C}$. The correction in the measured length (in mm) due to change in temperature is (in integer).

Question

What is the CORRECT match between the air pollutants and treatment techniques given in the table

Air Pollutants		Treatment techniques
P	NO₂	i	Flaring
Q	SO₂	ii	Cyclonic separator
R	CO	iii	Lime scrubbing
S	Particles	iv	NH₃ injection

Answer 1

A different way to reach the same correction is to first find how much the 30 m tape itself stretches at the field temperature, then scale that stretch up to however many tape lengths fit inside the measured 200 m line.

The tape's nominal length is $L_0 = 30$ m. At the field temperature of $35^\circ$C, which is $10^\circ$C above its standardization temperature of $25^\circ$C, its actual length becomes

\[ L_{actual} = L_0 (1 + \alpha \Delta T) = 30 \left(1 + 11\times10^{-6}\times 10\right) = 30 \times 1.00011 = 30.0033 \ \text{m} \]

So each time the tape is laid down, it actually measures out 30.0033 m of true ground distance, not the nominal 30 m marked on it, even though the surveyor reads off 30 m from its markings. The extra length per tape lay is

\[ \Delta L_0 = 30.0033 - 30 = 0.0033 \ \text{m} = 3.3 \ \text{mm per 30 m} \]

The line was recorded as 200 m, which corresponds to $200/30 = 6.667$ tape lengths. The total correction over the whole line is this extra length per tape lay multiplied by the number of lays:

\[ C_t = 3.3 \times 6.667 = 22.0 \ \text{mm} \]

This is the same figure reached by applying the correction formula directly to the full 200 m length in one step, confirming the tape ran long by 22 mm over the whole measurement. Since the tape was hotter and longer than standard, the ground distance it laid off was more than 200 m, so this correction is added to the recorded length.

\[ \boxed{22 \ \text{mm}} \]

Answer 2

A different way to reach the same correction is to first find how much the 30 m tape itself stretches at the field temperature, then scale that stretch up to however many tape lengths fit inside the measured 200 m line.

The tape's nominal length is $L_0 = 30$ m. At the field temperature of $35^\circ$C, which is $10^\circ$C above its standardization temperature of $25^\circ$C, its actual length becomes

\[ L_{actual} = L_0 (1 + \alpha \Delta T) = 30 \left(1 + 11\times10^{-6}\times 10\right) = 30 \times 1.00011 = 30.0033 \ \text{m} \]

So each time the tape is laid down, it actually measures out 30.0033 m of true ground distance, not the nominal 30 m marked on it, even though the surveyor reads off 30 m from its markings. The extra length per tape lay is

\[ \Delta L_0 = 30.0033 - 30 = 0.0033 \ \text{m} = 3.3 \ \text{mm per 30 m} \]

The line was recorded as 200 m, which corresponds to $200/30 = 6.667$ tape lengths. The total correction over the whole line is this extra length per tape lay multiplied by the number of lays:

\[ C_t = 3.3 \times 6.667 = 22.0 \ \text{mm} \]

This is the same figure reached by applying the correction formula directly to the full 200 m length in one step, confirming the tape ran long by 22 mm over the whole measurement. Since the tape was hotter and longer than standard, the ground distance it laid off was more than 200 m, so this correction is added to the recorded length.

\[ \boxed{22 \ \text{mm}} \]

Show Hint

Correct Answer: 22

Solution and Explanation

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