In this post, I will share some important questions which are frequently asked in competitive examinations such as SSC, UPSC and state PSCs.
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The true length of a chain is known to be 500 m. The line was again measured with a 20 m tape and found to be 502 m. What is the correct length of the 20 m tape?
- 19.92 m
- 20.80 m
- 20.36 m
- 19.86 m
View AnswerAnswer: a
Since the true length (TL) is 500 m and Measured Length (ML) is 502 m. The correction required for the tape is negative.
Number of 20 m tapes required to measure 500 m = 500/20 = 25
That means the extra 2 m is accumulated in 25 settings of 20 m tape
Therefore, Extra Length measured during each setting = 20/25 = 0.08 m
The correct Length of 20 m tape = 20 - 0.08 = 19.92 m
Or one can use the following formula
$TL=ML\frac{\text{Correct Length of Tape}}{\text{Original length of Tape}}$ -
The length of a line measured with a 20 m chain was found to be 250 m. What is the true length of the line if the chain was 10 cm too long?
- 248.75 m
- 250.25 m
- 251.25 m
- 252.75 m
View AnswerAnswer: c
Since the chain is too long correction must be positive
Number of chain settings = 250/20 = 12.5
Therefore, corection = 12.5 * 10 cm = 125 cm
Correct Length of line = 250+1.25 = 251.25 m
Or,
$TL=ML\frac{\text{Correct Length of Chain}}{\text{Original length of Chain}}$
$TL=250\frac{\text{20.1}}{\text{20}}$
TL = 251.25 m -
Select the incorrect statement.
- Correction for temperature is either positive or correction.
- Correction for the pull is either positive or negative
- Correction for sag is either positive or negative
- none of the above
View AnswerAnswer: c
Correction for sag is always negative. -
Correction for sag is calculated from ...
- ${C}_{s}=\frac{L \times (w\times L)^2}{24\times P^2}$
- ${C}_{s}=\frac{L^2 \times W^2}{24\times P^2}$
- Both (a) and (b)
- none of the above
Where $w=$ Weight of the tape per unit length
$W=$ Total weight of the tape
$L=$ Length of the tape
$P=$ Pull appliedView AnswerAnswer: a
${C}_{s}=\frac{L \times (w\times L)^2}{24\times P^2}$
Or,
${C}_{s}=\frac{L \times W^2}{24\times P^2}$ -
The pull that equalises the correction due to pull and correction due to sag is called ____.
- Equalising Pull
- Normal Tension
- Normalizing Pull
- Normal Force
View AnswerAnswer: b -
Correction for slope is given by ...
- ${C}_{s}=L(1-\cos\theta)$
- ${C}_{s}=2L \times \sin^2 \frac{\theta}{2}$
- Both (a) and (b)
- None of the above
View AnswerAnswer: c -
Select the correct statement
- Correction for slope is always negative
- Correction for slope is always positive
- Correction for slope is either positive or negative
- None of the above
View AnswerAnswer: a
The distance measured along the slope is always greater than the horizontal distance and hence the correction is always NEGATIVE. -
Correction for bad ranging is given by ...
- ${C}_{h}=\frac{d^3}{3L}$
- ${C}_{h}=\frac{d^2}{4L}$
- ${C}_{h}=\frac{d^3}{2L}$
- ${C}_{h}=\frac{d^2}{2L}$
View AnswerAnswer: d -
What will be the total error after calculating various errors?
- It will be the summation of magnitudes all the errors
- It will be the average of all the errors
- It will be the algebraic sum of all the errors
- All are correct
View AnswerAnswer: c -
Select the correct statement.
- Check Lines are run in the field to check the accuracy of the work
- Tie Lines are run to take details of the nearby objects
- Tie Line can also be used as a Check Line
- All are correct
View AnswerAnswer: d -
What is the least angle of a well-conditioned triangle?
- $20^\circ$
- $30^\circ$
- $25^\circ$
- $40^\circ$
View AnswerAnswer: b -
Which of the following statement is true regarding offsets?
- They are Perpendicular to the chain line
- They are not perpendicular to the chain line
- They are always perpendicular to the chain line
- They are either perpendicular or oblique to the chain line.
View AnswerAnswer: d -
An offset is laid $3^\circ$ from its true direction on the field. If the scale of plotting is 20 m to 1 cm, find the maximum length of the offset so that the displacement of the point on the paper may not exceed 0.25 mm.
$\sin 3^\circ = 0.0523$; $\cos 3^\circ = 0.9986$
- 9.5 m
- 0.5 m
- 41.86 m
- 3.9 m
View AnswerAnswer: a
Displacemet of the point on the paper$=\frac{l\times \sin 3^\circ}{20}=0.025 \text{ cm}$
$l=9.5 m$
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