Limiting Error MCQ Quiz in मराठी - Objective Question with Answer for Limiting Error - मोफत PDF डाउनलोड करा

Last updated on Mar 17, 2025

पाईये Limiting Error उत्तरे आणि तपशीलवार उपायांसह एकाधिक निवड प्रश्न (MCQ क्विझ). हे मोफत डाउनलोड करा Limiting Error एमसीक्यू क्विझ पीडीएफ आणि बँकिंग, एसएससी, रेल्वे, यूपीएससी, स्टेट पीएससी यासारख्या तुमच्या आगामी परीक्षांची तयारी करा.

Latest Limiting Error MCQ Objective Questions

Top Limiting Error MCQ Objective Questions

Limiting Error Question 1:

A plot of land has measured dimensions of 50 by 150 m. The uncertainty in the 50 m dimension is ±0.01 m. Calculate the uncertainty with which the 150 m dimension must be measured to ensure that the total uncertainty in the area is not greater than 150 percent of that value it would have if 150 m dimension were exact.

  1. ±0.015 m
  2. ±0.033 m
  3. ±0.024 m
  4. ±0.045 m

Answer (Detailed Solution Below)

Option 2 : ±0.033 m

Limiting Error Question 1 Detailed Solution

Let the length of the plot is L and the breadth of the plot is B.

L = 150 m, B = 50 m

Area of the plot = A = LB

Uncertainty in area, 

When there is no uncertainty in the measurement of L.

w= 0

Uncertainty in area, 

When there is uncertainty in the measurement of L.

The uncertainty of area is not to exceed 1.5 × 1.5 = ±2.25 m2

⇒ w= ±0.0335 m

Limiting Error Question 2:

A non-ideal Si-based pn junction diode is tested by sweeping the bias applied across its terminals from -5 V to +5 V. The effective thermal voltage, VT, for the diode is measured to be (29 ± 2) mV. The resolution of the voltage source in the measurement range is 1 mV. The percentage uncertainty (rounded off to 2 decimal places) in the measured current at a bias voltage of 0.02 V is______

Answer (Detailed Solution Below) 5 - 6

Limiting Error Question 2 Detailed Solution

Concept:

The equation of a diode current is given by

By applying log on both sides

By differentiating with the above equation with respect to VT

For η = 1, 

Now, by differentiating the diode current equation with respect to VD:

For η = 1,  

% uncertainty 

Calculation:

VT = (29 ± 2) mV.

VT = (0.029 ± 0.002) V, VD = 0.02 V, WVD = 0.001 V, WVT = 0.002 V

Therefore, 

Limiting Error Question 3:

Resistance is determined by the voltmeter ammeter method. The voltmeter reads 100 V with a probable error of ±12 V and ammeter reads 10 A with a probable error of ±2 A. The probable error in the computed value of resistance will be nearly

  1. 2.33Ω
  2. 3.33Ω 

Answer (Detailed Solution Below)

Option 3 : 2.33Ω

Limiting Error Question 3 Detailed Solution

We have resistance:

Weighted probable error in the resistance due to voltage is:

Weighted probable error in the resistance due to current is:

 
Probable error in computed resistance is:

 

Limiting Error Question 4:

A resistor is measured by the voltmeter-ammeter method. The voltmeter reading is 123.4 V on the 250 V scale and the ammeter reading is 283.5 mA on the 500 mA scale. Both meters are guaranteed to be accurate within ±1 percent of full scale reading. Calculate the limits within which the result can be guaranteed. 

  1. 435.27 ± 16.5 Ω 
  2. 435.27 ± 6.5 Ω
  3. 435.27 ± 11.25 Ω
  4. 435.27 ± 3 Ω

Answer (Detailed Solution Below)

Option 1 : 435.27 ± 16.5 Ω 

Limiting Error Question 4 Detailed Solution

Given that voltmeter reading = 123.4 V

Full scale reading = 250 V.

Guaranteed accuracy error = ± 1% of FSC

Error = 250 × 0.01 = 2.5 V

V = 123.4 ± 2.5 V = [120.9, 125.9] V

Given that ammeter reading = 283.5 mA

Full scale reading = 500 mA

Guaranteed accuracy error = ±1% of FSD

Error = 500 × 0.01 = 5 mA

I = 283.5 ± 5 mA = [278.5, 288.5] mA

Indicated value

When V = 120. 9, I = 288.5 mA

 

When V = 125.9, I = 278.5 mA

 

Resistance range = [419.06, 452.06] Ω

R = [435.27 ± 16.5] Ω 

Limiting Error Question 5:

A variable ‘W’ is related to three other variables x,y,z as W = xy/z. The Variables are measured with meters of accuracy ± 0.5 % reading, ± 1% of full scale value and ± 1.5% reading the actual reading of the three meters are 80, 20 and 50 with 100 being the full scale value for all three. The maximum uncertainly in the measurement of ‘W’ will be.

  1. ± 0.5 % rdg
  2. ± 5.5 % rdg
  3. ± 6.7 % rdg
  4. ± 7.0 % rdg

Answer (Detailed Solution Below)

Option 4 : ± 7.0 % rdg

Limiting Error Question 5 Detailed Solution

X = ± 0.5 % of reading 80 (Limiting error)

Y = ± 1% of FSV (G.A.E)

Z = ± 1.5 % of reading 50 (L.E.)

L.E of Y is

Limiting Error Question 6:

A 0-150 V voltmeter has a guaranteed accuracy of 1% of full scale reading. The voltage measured by this instrument is 75 V. What is the percentage of limiting error ?

  1. 1%
  2. 2%
  3. 3%
  4. 4%

Answer (Detailed Solution Below)

Option 2 : 2%

Limiting Error Question 6 Detailed Solution

Concept: 

Calculation:

Given-

Full scale reading = 150 V

Guaranteed accuracy = 1% of full-scale reading = 0.01 × 150 = 1.5 V

Measured value = 75 A

Limiting error 

Limiting Error Question 7:

In a DC motor, the input is measured as 2 kW ± 4% and the losses are 200 ± 5 W.

The percentage limiting error of the output power is 

  1. 9%
  2. 6.5%
  3. 1.5%
  4. 4.7%

Answer (Detailed Solution Below)

Option 4 : 4.7%

Limiting Error Question 7 Detailed Solution

Input = 2 kW ± 4%

Error in the input 

Losses = 200 ± 5 W

Output = Input – Loses

= (2000 ± 80) – (200 ± 5)

= 1800 ± 85 W

Limiting error in the measurement of output

Losses  

Limiting Error Question 8:

The resistance of the circuit is found by measuring current flowing and power fed into the circuit. If limiting errors in the measurement of power and current are ± 2.5% and ± 1.5%. Respectively, the limiting error in the measurement of resistance will be _______ (in percentage)

  1. ± .5%
  2. ± 5.5%
  3. ± 4%
  4. ± 6.5%

Answer (Detailed Solution Below)

Option 2 : ± 5.5%

Limiting Error Question 8 Detailed Solution

%ER = 2.5 + 1.5 × 2 = ± 5.5%

Limiting Error Question 9:

A (0 - 100)V voltmeter has a guaranteed accuracy of 1.2% on full scale reading. The voltage measured by this instrument is equal to 3/4th of its full scale value. Then the percentage limiting error on the measured value is equal to

  1. 1.2%
  2. 1.6%
  3. 0.12%
  4. 0.16%

Answer (Detailed Solution Below)

Option 2 : 1.6%

Limiting Error Question 9 Detailed Solution

Concept:

Limiting Error: The maximum allowable error in the measurement is specified in terms of true value, is known as limiting error. It will give a range of errors. It is always with respect to the true value, so it is a variable error.

1) If two quantities are getting added, then their limiting errors also gets added.

2) If two quantities are getting multiplied or divided, then their percentage limiting errors gets added.

Calculation:

The voltmeter reads 0V to 100V .

GAE of voltmeter = 1.2%

Error in voltmeter measurement on FSD = 0.012 × 100 = 1.2 V

Percentage error at a meter indication of 3/4th of its full scale, i.e. 75 V

Limiting error 

Note:

Absolute Error: The deviation of the measured value from the true value (or) actual value is called error.

Absolute error (E) = Am – At

Am = Measured value

At = True value

Relative Static Error: The ratio of absolute error to the true value is called relative static error.

Guaranteed Accuracy Error: The allowable error in measurement is specified in terms of full-scale value is known as a guaranteed accuracy error. It is a constant error seen by the instrument since it is with respect to full-scale value.

Limiting Error Question 10:

The current I flowing through a resistance of value 100 Ω ± 0.2% is I = 4A ± 0.5% the uncertainty in measurement of power is

  1. 1600 W ± 0.01%
  2. 1600 W ± 0.02%
  3. 1600 W ± 0.05%
  4. 1600 W ± 1.002%

Answer (Detailed Solution Below)

Option 4 : 1600 W ± 1.002%

Limiting Error Question 10 Detailed Solution

Concepts

When many variables are in valued is done on the same basis as is done for error analysis when the result

are expressed as standard deviation or probable errors.

Let x = F(x1, x2, x3 …..)

Wx1, Wx2, …. Wxn be the uncertainties of x1, x2, … xn then uncertainty of x is given by

Calculation:

R = 100 ± 0.2% = 100 ± 0.2

I = 4 ± 0.5% = 4 ± 0.02

P = 100 × 42 = 1600 W

P = I2R

In % term

Hot Links: teen patti all teen patti rules all teen patti master teen patti casino apk teen patti gold