Civil Engineering - Surveying - Discussion
Discussion Forum : Surveying - Section 4 (Q.No. 46)
46.
A standard steel tape of length 30 m and cross-section 15 x 1.0 mm was standardised at 25°C and at 30 kg pull. While measuring a base line at the same temperature, the pull applied was 40 kg. If the modulus of elasticity of steel tape is 2.2 x 106 kg/cm2, the correction to be applied is
Discussion:
27 comments Page 1 of 3.
Deeksha said:
4 years ago
30 m steel tape was standardized under 60 N pull at 65° F.
It was suspended in 5 equal span during measurement. The mean temperature during measurement was 90° F and the pull exerted was 100 N. The area of the cross-section of the tape was 8 mm2. Find the true length of the tape, if, α = 6.3 * 10^6/F, E = 2 *105 N/mm2 and unit weight of steel = 78.6 kN/m3.
It was suspended in 5 equal span during measurement. The mean temperature during measurement was 90° F and the pull exerted was 100 N. The area of the cross-section of the tape was 8 mm2. Find the true length of the tape, if, α = 6.3 * 10^6/F, E = 2 *105 N/mm2 and unit weight of steel = 78.6 kN/m3.
(1)
SHANU said:
7 years ago
A steel tape20 m long standardized at 55oF with a pull of 10 kg was used for measuring a
baseline. Find the correction per tape length, if the temperature at the time of measurement
was 80oF and the pull exerted was 16 kg, weight of/ cubic cm of steel.
= 7.86 g, wt of the tape.
= 0.8 kg and F ; u kg/cmz- Coefficient of expansion of tape per l"F.
= 6.2x104.
baseline. Find the correction per tape length, if the temperature at the time of measurement
was 80oF and the pull exerted was 16 kg, weight of/ cubic cm of steel.
= 7.86 g, wt of the tape.
= 0.8 kg and F ; u kg/cmz- Coefficient of expansion of tape per l"F.
= 6.2x104.
Vishal Vishwakarma said:
5 years ago
Correction should be positive since we measured longer than actual it was, mean that suppose we have seen on tape length is 150m while measurement but actual it was more than that of 150m, let say 151 m actual therefore to obtain this from 150 to 151m.
We need correction positive.
So, the right option will be C.
We need correction positive.
So, the right option will be C.
Abhishek Anand said:
9 years ago
Ans must be C.
As the applied pull is more than std pull so the length of tape increases.Hence the distance measured becomes less than actual.
Correction = +'ve if app>std pull.
Correction = - ve if app<std pull.
Cp=(P-Ps)L/AE;
P=app pull.
Ps=std pull.
As the applied pull is more than std pull so the length of tape increases.Hence the distance measured becomes less than actual.
Correction = +'ve if app>std pull.
Correction = - ve if app<std pull.
Cp=(P-Ps)L/AE;
P=app pull.
Ps=std pull.
Ajmal Khan said:
9 years ago
No, @Satish the answer A is correct because the standard pull is 30 Kg and applied pull is 40 Kg, so the tape will measure long distance than actual, so the final result must be subtracted from measured distance. That's why it is negative one.
(1)
Jinx said:
4 years ago
The error is +ive but at the end of the question, "correction to be applied" has been asked. Which makes option A correct. The value of error is positive so we will deduct it from the final value, so the applied correction will be -0. 000909m.
A.Ahmed said:
6 years ago
(B) is the correction option.
Correction due to pull = (Pm-P0)*L/AE.
= (40-30)*30*100/1.5*0.1*2.2*10^6.
= +0.0909.
Correction due to pull = (Pm-P0)*L/AE.
= (40-30)*30*100/1.5*0.1*2.2*10^6.
= +0.0909.
Bee said:
6 years ago
@Ajmal khan nope option C is correct, as pull you applied is more so the length measured will be less than true length and correction will be positive and error will be negative.
Rajesh said:
8 years ago
When pull the tape more than standard then the measurement will be less than 30m. So sure the correction to be added to measured length.
Shahab Mehsud said:
4 years ago
I think option C is correct; as the tape elongated due to extra pull applied, hence the measured length will be less than the actual.
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