All Exams Test series for 1 year @ ₹349 only
Question

An angle measured with theodolite is α with weight 2. The weight of \(\rm \frac{\alpha}{4}\) will be

The correct answer is 2 × 42

Calculating the Weight of a Scaled Angle from Theodolite Measurement

In surveying and measurements, the 'weight' assigned to an observation indicates its relative reliability or precision. A higher weight means a more reliable observation. Weight is inversely proportional to the variance (square of the standard deviation) of the observation.

The relationship between weight (\(w\)) and variance (\(\sigma^2\)) for an observation is given by:

\(w \propto \frac{1}{\sigma^2}\)

This can be written as \(w = \frac{k}{\sigma^2}\), where \(k\) is a constant related to the standard variance of unit weight.

Propagation of Weights for Scaled Observations

Consider an observation \(x\) with weight \(w_x\). If we define a new quantity \(y\) as a linear function of \(x\), say \(y = cx\), where \(c\) is a constant, the variance of \(y\) (\(\sigma_y^2\)) is related to the variance of \(x\) (\(\sigma_x^2\)) by:

\(\sigma_y^2 = c^2 \sigma_x^2\)

Now, let's find the weight of \(y\), denoted as \(w_y\). Using the inverse relationship between weight and variance:

\(w_y = \frac{k}{\sigma_y^2} = \frac{k}{c^2 \sigma_x^2}\)

Since \(w_x = \frac{k}{\sigma_x^2}\), we can substitute this into the equation for \(w_y\):

\(w_y = \frac{1}{c^2} \left(\frac{k}{\sigma_x^2}\right) = \frac{1}{c^2} w_x\)

So, if an observation is multiplied by a constant \(c\), its weight is multiplied by \(\frac{1}{c^2}\).

Applying the Concept to the Theodolite Angle

We are given an angle measured with a theodolite, denoted as \(\alpha\), and its weight is \(w_\alpha = 2\). We need to find the weight of \(\frac{\alpha}{4}\).

Let \(y = \frac{\alpha}{4}\). In this case, \(y\) is a scaled version of \(\alpha\) with the scaling constant \(c = \frac{1}{4}\).

Using the propagation of weights formula \(w_y = \frac{1}{c^2} w_x\), where \(y = \frac{\alpha}{4}\), \(x = \alpha\), and \(c = \frac{1}{4}\):

\(w_{\frac{\alpha}{4}} = \frac{1}{(1/4)^2} w_\alpha\)

Calculate the term \(\frac{1}{(1/4)^2}\):

\(\frac{1}{(1/4)^2} = \frac{1}{1/16} = 1 \times \frac{16}{1} = 16\)

Now substitute this back into the weight equation:

\(w_{\frac{\alpha}{4}} = 16 \times w_\alpha\)

We are given \(w_\alpha = 2\). Substitute this value:

\(w_{\frac{\alpha}{4}} = 16 \times 2 = 32\)

The weight of \(\frac{\alpha}{4}\) is 32.

Evaluating the Options

Let's compare our result (32) with the given options:

  1. \( \frac{2}{4} = 0.5 \)
  2. \( \frac{4}{2} = 2 \)
  3. \( 2 \times 4^2 = 2 \times 16 = 32 \)
  4. \( 2 \times 4 = 8 \)

Option 3 gives the value 32, which matches our calculated weight for \(\frac{\alpha}{4}\).

Therefore, the weight of \(\frac{\alpha}{4}\) is \(2 \times 4^2\).

The final answer is \(\mathbf{2 \times 4^2}\).

Original Quantity Scaled Quantity Scaling Constant (\(c\)) Original Weight (\(w_x\)) Formula for \(w_y\) Calculated \(w_y\) Matching Option
\(\alpha\) \(\frac{\alpha}{4}\) \(c = \frac{1}{4}\) \(w_\alpha = 2\) \(w_{\frac{\alpha}{4}} = \frac{1}{c^2} w_\alpha\) \(w_{\frac{\alpha}{4}} = \frac{1}{(1/4)^2} \times 2 = 16 \times 2 = 32\) \(2 \times 4^2 = 32\)

Revision Table: Surveying Measurement Weights

Concept Description Relationship
Weight (\(w\)) Measure of reliability/precision of an observation. Higher weight = higher reliability. Inversely proportional to variance.
Variance (\(\sigma^2\)) Measure of the spread or dispersion of observations around the mean. Lower variance = higher precision. Square of standard deviation (\(\sigma\)).
Scaling an Observation Multiplying an observation \(x\) by a constant \(c\) to get \(y = cx\). Variance scales by \(c^2\): \(\sigma_y^2 = c^2 \sigma_x^2\).
Weight Propagation for Scaling How the weight changes when an observation is scaled. Weight scales by \(\frac{1}{c^2}\): \(w_y = \frac{1}{c^2} w_x\).

Additional Information: Weights in Surveying and Theodolites

Weights are fundamental in combining multiple measurements, especially in techniques like Least Squares Adjustment, which is commonly used with theodolite data. Different observations might have different weights based on factors like:

  • The instrument used (e.g., precision of a theodolite).
  • Observation conditions (e.g., visibility, atmospheric refraction).
  • Number of repetitions of the measurement.
  • Geometric strength of the network (though this relates more to propagated error/variance).

When measurements are combined or transformed, understanding how their weights propagate is crucial for correctly assessing the precision of the final results. The principle \(w \propto \frac{1}{\sigma^2}\) is key to this propagation.

The question highlights a simple case of scaling an angle. In more complex scenarios, involving sums or other functions of observations, different error propagation rules apply, leading to corresponding rules for weight propagation.

Was this answer helpful?

Important Questions from Accuracy and Errors

  1. The clogging of chain rings with mud introduces (with ‘error’ defined in the standard way)

    1. Negative cumulative error

    2. Positive cumulative error

    3. Compensating error

  2. If the probable error in single observation is ± 0.04 m and that of the mean is ± 0.01 m, then the number of observations are

  3. Errors arising from carelessness of the observer are known as

  4. The errors such as sag in chain and chain not being horizontal during stepping are common in:

  5. The length of a line measured with a 30 m chain is 800.64 m. Afterwards it is found that the chain is 0.05 m too long. The true length of the line is:

Need Expert Advice?

Start Your Preparation with Prepp Mobile App

Download the app from Google Play & App Store
Download the app from Google Play & App Store
Prepp Mobile App