Civil Engineering - Surveying - Discussion

Discussion Forum : Surveying - Section 4 (Q.No. 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
- 0.000909 m
+ 0.0909 m
0.000909 m
none of these
Answer: Option
No answer description is available. Let's discuss.
27 comments Page 1 of 3.

Rana said:   8 months ago
I think option C will be the correct answer.

Shahab Mehsud said:   3 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.

Jinx said:   3 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.

Deeksha said:   3 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.

Vishal Vishwakarma said:   3 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.

Avik said:   3 years ago
The right Answer is +0.00909m.

Umesh said:   4 years ago
You are right, thanks, @Ajmal Khan.

A.Ahmed said:   4 years ago
All my friends are said to correct answer is C but they did not convert the unit of 30m chain into 30*100 cm.

A.Ahmed said:   4 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.

Bee said:   5 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.

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