Linear Measurement MCQ Quiz in मल्याळम - Objective Question with Answer for Linear Measurement - സൗജന്യ PDF ഡൗൺലോഡ് ചെയ്യുക

Explanation:

The various instruments used in surveying and their purpose is given below in the tabulated form:

Concept:

True area = measured area × (R.F.)

For area,

$\mathrm{R}\mathrm{e}\mathrm{p}\mathrm{r}\mathrm{e}\mathrm{s}\mathrm{e}\mathrm{n}\mathrm{t}\mathrm{a}\mathrm{t}\mathrm{i}\mathrm{v}\mathrm{e}\mathrm{f}\mathrm{a}\mathrm{c}\mathrm{t}\mathrm{o}\mathrm{r}\left(\mathrm{R}.\mathrm{F}.\right)={\left(\frac{\mathrm{t}\mathrm{r}\mathrm{u}\mathrm{e}\mathrm{l}\mathrm{e}\mathrm{n}\mathrm{g}\mathrm{t}\mathrm{h}}{\mathrm{s}\mathrm{t}\mathrm{a}\mathrm{n}\mathrm{d}\mathrm{a}\mathrm{r}\mathrm{d}\mathrm{l}\mathrm{e}\mathrm{n}\mathrm{g}\mathrm{t}\mathrm{h}}\right)}^2$

Calculation:

Given:

Standard chain length = 100 m, (a 100m chain). Area measured = 100 acres.

Now,

Actual length = 100 + 0.8 = 100.8 m 

$\mathrm{R}\mathrm{e}\mathrm{p}\mathrm{r}\mathrm{e}\mathrm{s}\mathrm{e}\mathrm{n}\mathrm{t}\mathrm{a}\mathrm{t}\mathrm{i}\mathrm{v}\mathrm{e}\mathrm{f}\mathrm{a}\mathrm{c}\mathrm{t}\mathrm{o}\mathrm{r}\left(\mathrm{R}.\mathrm{F}.\right)={\left(\frac{\mathrm{t}\mathrm{r}\mathrm{u}\mathrm{e}\mathrm{l}\mathrm{e}\mathrm{n}\mathrm{g}\mathrm{t}\mathrm{h}}{\mathrm{s}\mathrm{t}\mathrm{a}\mathrm{n}\mathrm{d}\mathrm{a}\mathrm{r}\mathrm{d}\mathrm{l}\mathrm{e}\mathrm{n}\mathrm{g}\mathrm{t}\mathrm{h}}\right)}^2$

$\left(\mathrm{R}.\mathrm{F}.\right)={\left(\frac{100.8}{100}\right)}^2$=1.01606

True area = 100 × 1.01606

True area = 101.606 acres.

Tips and tricks:

For finding out the true length measured by a false chain-

True length × True chain length = Measured length × Actual length,

Concept:

$Truelengthline=\frac{l^{\prime }}{l}\times measuredlength$

Where, 

 l' is corrected length of chain

 l is actual length of chain

Calculation:

True length of line (L) = ?

Actual length of chain (l) = 20 m

Measured length of line (L') = 327 m

Corrected length of chain (l') = 20.03 m 

So, we know

$Truelengthline=\frac{l^{\prime }}{l}\times measuredlength$

$\Rightarrow L=\frac{20.03}{20}\times 327$

∴ L= 327.49 m

Concept:

Types of Error and Correction:

Pull correction may be Positive or negative. Hence, option 2 is incorrect.

Concept:

Representative Fraction (RF): The scale expressed in the same units. For example, 1 cm = 50 m is represented in RF as 1: 5000 or 1/5000. Here 50 m is converted as 5000 cm.

Shrinkage factor or Shrinkage ratio = Shrunk scale / Original scale or Shrunk length / Actual length

Note: The Shrinkage ratio is always less than one.

Calculation:

Given,

Original Scale, 1 cm = 19 m

Shrunk Scale, 1 cm = 20 m

Therefore,

RF of shrunk scale = 1: 2000 or 1 / 2000

RF of original scale = 1: 1900 or 1 / 1900

Shrinkage Factor = Shrunk scale / Original scale = (1/2000) / (1/1900) = 19/20

Shrinkage Factor = 19/20 and RF of shrunk scale = 1/2000

Explanation:

Original length of line = 10 cm

Length after shrinkage of line = 9.7 cm

$\mathbf{S}\mathbf{h}\mathbf{r}\mathbf{i}\mathbf{n}\mathbf{k}\mathbf{a}\mathbf{g}\mathbf{e}\mathbf{F}\mathbf{a}\mathbf{c}\mathbf{t}\mathbf{o}\mathbf{r}=\frac{\mathbf{S}\mathbf{h}\mathbf{r}\mathbf{u}\mathbf{n}\mathbf{k}\mathbf{l}\mathbf{e}\mathbf{n}\mathbf{g}\mathbf{t}\mathbf{h}}{\mathbf{O}\mathbf{r}\mathbf{i}\mathbf{g}\mathbf{i}\mathbf{n}\mathbf{a}\mathbf{l}\mathbf{l}\mathbf{e}\mathbf{n}\mathbf{g}\mathbf{t}\mathbf{h}}$

$\mathbf{S}\mathbf{h}\mathbf{r}\mathbf{i}\mathbf{n}\mathbf{k}\mathbf{a}\mathbf{g}\mathbf{e}\mathbf{F}\mathbf{a}\mathbf{c}\mathbf{t}\mathbf{o}\mathbf{r}=\frac{9.7}{10}$

Shrinkage Factor = 0.97

Explanation:

Offsets:

• The distance measured right or left of the chain line to locate details like boundaries, culverts, etc. are called offsets.
• Offset which can be judged by the naked eye or offset less than 15 m is called short offset and offset greater than 15 m is called long offset.
• Most commonly short offsets are preferred.

These are of two types:

(i) Perpendicular offset at right angles to the chain line

Explanation: 

The various instruments used in surveying and their purpose is given below in the tabulated form:

Concept:

Measurement by Pacing:

• Pacing is a rapid means of approximately checking more precise measurements of distance.
• Pacing over rough country may be done with a precision of one in one hundred(1 in 100).
• In average conditions, a person with some experience should have little difficulty in pacing with a precision of one in two hundred.
• Obviously, there is not much precision in this method and the procedure provides only an approximation of distance.
• The natural pace of each individual normally varies from 2 ½ to 3 ft.
• A convenient relation between the pace and the foot is 40 paces approximately equal 100 ft.
• Technicians involved in surveying standardize their pace by walking over known distances on level, sloping, and uneven ground. 

Correction for sag is given by:

$C_s=-\frac{w^2{l}^3}{24P^2}=-\frac{W^{2}l}{24P^2}$

W is the total weight of tape  = w × L

Where,

W = weight of tape per unit length (N/m)

l = length of tape suspended between supports (m)

p = pull applied (N)