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30 m tape actually measures 29.985 m, when it is supported at the two ends only, with the temperature at 30º C and a tension of 70N. Weight of the tape is 12N and has a cross-sectional area of 0.04 cm². if the field temperature is also 30° C, what tension should be applied so that the tape measures exactly 30 m when supported at the two ends.?
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30 m tape actually measures 29.985 m, when it is supported at the two ...
Understanding the Problem
The tape measures 29.985 m instead of the expected 30 m due to the combined effects of tension and temperature. To find the required tension that allows the tape to measure exactly 30 m, we need to consider the elongation caused by the existing tension.
Key Parameters
- **Original Length**: 30 m
- **Measured Length**: 29.985 m
- **Temperature**: 30º C
- **Current Tension**: 70 N
- **Weight of Tape**: 12 N
- **Cross-sectional Area**: 0.04 cm²
Calculating Elongation
The elongation (\( \Delta L \)) of the tape can be determined using the formula:
\[
\Delta L = \frac{F \cdot L_0}{A \cdot E}
\]
Where:
- \( F \) = force (tension minus weight)
- \( L_0 \) = original length
- \( A \) = cross-sectional area
- \( E \) = Young's modulus (assumed constant for the material)
To measure 30 m, the elongation needs to be zero. Therefore, the tension needs to be adjusted to balance the weight and maintain the desired length.
Required Tension Calculation
1. **Determine Effective Force**:
\[
F_{\text{effective}} = T - W
\]
Where \( T \) is the applied tension and \( W \) is the weight of the tape (12 N).
2. **Setting Up the Equation**:
To achieve 30 m, set \( \Delta L \) to zero:
\[
0 = \frac{T \cdot 30}{0.0004 \cdot E} - 12
\]
3. **Solve for T**:
Rearranging gives:
\[
T = \frac{12 \cdot 0.0004 \cdot E}{30}
\]
This equation allows for calculating the required tension \( T \) based on the specific Young's modulus \( E \) of the tape material.
Conclusion
To measure exactly 30 m, adjust the tension \( T \) using the derived formula, ensuring that the elongation due to the weight is counteracted, thus achieving the correct measurement at the given temperature.
The tape measures 29.985 m instead of the expected 30 m due to the combined effects of tension and temperature. To find the required tension that allows the tape to measure exactly 30 m, we need to consider the elongation caused by the existing tension.
Key Parameters
- **Original Length**: 30 m
- **Measured Length**: 29.985 m
- **Temperature**: 30º C
- **Current Tension**: 70 N
- **Weight of Tape**: 12 N
- **Cross-sectional Area**: 0.04 cm²
Calculating Elongation
The elongation (\( \Delta L \)) of the tape can be determined using the formula:
\[
\Delta L = \frac{F \cdot L_0}{A \cdot E}
\]
Where:
- \( F \) = force (tension minus weight)
- \( L_0 \) = original length
- \( A \) = cross-sectional area
- \( E \) = Young's modulus (assumed constant for the material)
To measure 30 m, the elongation needs to be zero. Therefore, the tension needs to be adjusted to balance the weight and maintain the desired length.
Required Tension Calculation
1. **Determine Effective Force**:
\[
F_{\text{effective}} = T - W
\]
Where \( T \) is the applied tension and \( W \) is the weight of the tape (12 N).
2. **Setting Up the Equation**:
To achieve 30 m, set \( \Delta L \) to zero:
\[
0 = \frac{T \cdot 30}{0.0004 \cdot E} - 12
\]
3. **Solve for T**:
Rearranging gives:
\[
T = \frac{12 \cdot 0.0004 \cdot E}{30}
\]
This equation allows for calculating the required tension \( T \) based on the specific Young's modulus \( E \) of the tape material.
Conclusion
To measure exactly 30 m, adjust the tension \( T \) using the derived formula, ensuring that the elongation due to the weight is counteracted, thus achieving the correct measurement at the given temperature.
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30 m tape actually measures 29.985 m, when it is supported at the two ends only, with the temperature at 30º C and a tension of 70N. Weight of the tape is 12N and has a cross-sectional area of 0.04 cm². if the field temperature is also 30° C, what tension should be applied so that the tape measures exactly 30 m when supported at the two ends.?
Question Description
30 m tape actually measures 29.985 m, when it is supported at the two ends only, with the temperature at 30º C and a tension of 70N. Weight of the tape is 12N and has a cross-sectional area of 0.04 cm². if the field temperature is also 30° C, what tension should be applied so that the tape measures exactly 30 m when supported at the two ends.? for UPSC 2024 is part of UPSC preparation. The Question and answers have been prepared according to the UPSC exam syllabus. Information about 30 m tape actually measures 29.985 m, when it is supported at the two ends only, with the temperature at 30º C and a tension of 70N. Weight of the tape is 12N and has a cross-sectional area of 0.04 cm². if the field temperature is also 30° C, what tension should be applied so that the tape measures exactly 30 m when supported at the two ends.? covers all topics & solutions for UPSC 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for 30 m tape actually measures 29.985 m, when it is supported at the two ends only, with the temperature at 30º C and a tension of 70N. Weight of the tape is 12N and has a cross-sectional area of 0.04 cm². if the field temperature is also 30° C, what tension should be applied so that the tape measures exactly 30 m when supported at the two ends.?.
30 m tape actually measures 29.985 m, when it is supported at the two ends only, with the temperature at 30º C and a tension of 70N. Weight of the tape is 12N and has a cross-sectional area of 0.04 cm². if the field temperature is also 30° C, what tension should be applied so that the tape measures exactly 30 m when supported at the two ends.? for UPSC 2024 is part of UPSC preparation. The Question and answers have been prepared according to the UPSC exam syllabus. Information about 30 m tape actually measures 29.985 m, when it is supported at the two ends only, with the temperature at 30º C and a tension of 70N. Weight of the tape is 12N and has a cross-sectional area of 0.04 cm². if the field temperature is also 30° C, what tension should be applied so that the tape measures exactly 30 m when supported at the two ends.? covers all topics & solutions for UPSC 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for 30 m tape actually measures 29.985 m, when it is supported at the two ends only, with the temperature at 30º C and a tension of 70N. Weight of the tape is 12N and has a cross-sectional area of 0.04 cm². if the field temperature is also 30° C, what tension should be applied so that the tape measures exactly 30 m when supported at the two ends.?.
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30 m tape actually measures 29.985 m, when it is supported at the two ends only, with the temperature at 30º C and a tension of 70N. Weight of the tape is 12N and has a cross-sectional area of 0.04 cm². if the field temperature is also 30° C, what tension should be applied so that the tape measures exactly 30 m when supported at the two ends.?, a detailed solution for 30 m tape actually measures 29.985 m, when it is supported at the two ends only, with the temperature at 30º C and a tension of 70N. Weight of the tape is 12N and has a cross-sectional area of 0.04 cm². if the field temperature is also 30° C, what tension should be applied so that the tape measures exactly 30 m when supported at the two ends.? has been provided alongside types of 30 m tape actually measures 29.985 m, when it is supported at the two ends only, with the temperature at 30º C and a tension of 70N. Weight of the tape is 12N and has a cross-sectional area of 0.04 cm². if the field temperature is also 30° C, what tension should be applied so that the tape measures exactly 30 m when supported at the two ends.? theory, EduRev gives you an
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