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#### Information

**Name: **Error in Measurement

**Subject: **Physics

**Topic: **Error in Measurement

**Questions: **15 Objective Type

**Time Allowed: **10 Minutes

**Important for: **IIT JEE, UPSEE, WBJEE Engineering Entrance exams, AIPMT and other PMT entrance test, +2 school students and B. Sc. / M. Sc. College Students.

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- Question 1 of 14
##### 1. Question

1 pointsWhat is the value of error of measurement =

CorrectWe know that the correct value of error of measurement is equal to the difference of true value and measured value.

IncorrectWe know that the correct value of error of measurement is equal to the difference of true value and measured value.

- Question 2 of 14
##### 2. Question

1 pointsThe degree of closeness of the measured value of a certain quantity with its true value is known as

CorrectWe know that the degree of closeness of the measured value of a certain quantity with its true value is known as accuracy.

IncorrectWe know that the degree of closeness of the measured value of a certain quantity with its true value is known as accuracy.

- Question 3 of 14
##### 3. Question

1 pointsThe mean diameter of a ball is to be measured with vernier calipers. A set of 100 measurements of diameter is expected to yield more accurate estimate than a set of 10 measurements, because

CorrectWe know that the minimize random errors, the measurement should be repeated many times. And arithmetic mean of all observations of the concerned measurement should be taken to get nearly true value of the measured quantity.

IncorrectWe know that the minimize random errors, the measurement should be repeated many times. And arithmetic mean of all observations of the concerned measurement should be taken to get nearly true value of the measured quantity.

- Question 4 of 14
##### 4. Question

1 pointsTe random error can be eliminated by

CorrectWe know that the random error can be eliminated by taking mean of a large no of readings. This method is generally used when more accurate observations are required.

IncorrectWe know that the random error can be eliminated by taking mean of a large no of readings. This method is generally used when more accurate observations are required.

- Question 5 of 14
##### 5. Question

1 pointsA student performs experiment with simple pendulum and measures time for 10 vibrations. The error in the measurement of time-period is reduced by a factor of

CorrectWe know that when the student measures time for 10 vibrations, then while taking mean of these 10 observation, random error is reduced by factor of 10.

IncorrectWe know that when the student measures time for 10 vibrations, then while taking mean of these 10 observation, random error is reduced by factor of 10.

- Question 6 of 14
##### 6. Question

1 pointsThe least count of an instrument is 0.01 cm. Taking all precautions, the most possible error in the measurement can be

CorrectGiven: Least count of instrument = 0.01 cm.

We know that taking all precautions, the most possible error in the measurement is equal to the least count of the instrument.*i.e.*, 0.01 cm.IncorrectGiven: Least count of instrument = 0.01 cm.

We know that taking all precautions, the most possible error in the measurement is equal to the least count of the instrument.*i.e.*, 0.01 cm. - Question 7 of 14
##### 7. Question

1 pointsA spherometer has a least count of 0.005 mm and its head scale is divided into 200 equal divisions. Distance between any two consecutive threads on the spherometer screw is

CorrectGiven: Least count = 0.005 mm and no. of equal divisions = 200.

We know that distance between any two consecutive threads on the spherometer screw = 200 × 0.005 = 1.0 mm.IncorrectGiven: Least count = 0.005 mm and no. of equal divisions = 200.

We know that distance between any two consecutive threads on the spherometer screw = 200 × 0.005 = 1.0 mm. - Question 8 of 14
##### 8. Question

1 pointsA spherometer has 250 equal divisions marked along the periphery of its disc. One full rotation of the disc advances on the main scale by 0.0625 cm. The least count of the system is

CorrectGiven: No of equal divisions = 250 and one full rotation of disc = 0.0625 cm.

We know that least count of the system = 0.0625 / 250 = 0.00025 cm = 2.5 × 10^{-4}cmIncorrectGiven: No of equal divisions = 250 and one full rotation of disc = 0.0625 cm.

We know that least count of the system = 0.0625 / 250 = 0.00025 cm = 2.5 × 10^{-4}cm - Question 9 of 14
##### 9. Question

1 pointsA simple pendulum completes 20 oscillations in 25s. The time is measured with a stop watch of least count 0.2s. The error in the measurement of time is

CorrectGiven: No of oscillations= 20;

Time-taken = 25s and least count of stop watch = 0.2s.

We know that error in the measurement of time = [0.2 / 25] × 100 = 0.8%IncorrectGiven: No of oscillations= 20;

Time-taken = 25s and least count of stop watch = 0.2s.

We know that error in the measurement of time = [0.2 / 25] × 100 = 0.8% - Question 10 of 14
##### 10. Question

1 pointsIf the measurement of a length is made as (227 ± 1) mm, then percentage error in the measurement is

CorrectGiven: Measurement of length (

*l*) = 227 mm and absolute error (*∆l*) = 1 mm.

We know that percentage error in the measurement of length

(*l*) = [_{p}*∆l*/*l*] × 100 = [1 / 277] × 100 = 0.44 %IncorrectGiven: Measurement of length (

*l*) = 227 mm and absolute error (*∆l*) = 1 mm.

We know that percentage error in the measurement of length

(*l*) = [_{p}*∆l*/*l*] × 100 = [1 / 277] × 100 = 0.44 % - Question 11 of 14
##### 11. Question

1 pointsIf radius of the sphere is (5.3 ± 0.1) cm., then percentage error in its volume will be

CorrectGiven: Radium of sphere (

*r*) = 5.3 and error in radium (*∆**r*) = 0.1 cm.

We know that volume of the sphere = [4/3]×π×*r*^{3}.

Therefore percentage error in the volume = 3 × [*∆r / r*] × 100 = [3 × (0.1 / 5.3)] × 100 = 1.9%.IncorrectGiven: Radium of sphere (

*r*) = 5.3 and error in radium (*∆**r*) = 0.1 cm.

We know that volume of the sphere = [4/3]×π×*r*^{3}.

Therefore percentage error in the volume = 3 × [*∆r / r*] × 100 = [3 × (0.1 / 5.3)] × 100 = 1.9%. - Question 12 of 14
##### 12. Question

1 pointsIf surface tension of a liquid is measured to be 0.06 N-m

^{-1}and absolute error is 0.0015 N-m^{-1}, then percentage error in the measurement of surface tension will beCorrectGiven: Surface tension (

*T*) = 0.06 N-m^{-1}and absolute error (*∆T*) = 0.0015 N-m^{-1}

We know that percentage error in the measurement of surface tension (T_{p}) = [∆T / T] × 100

= [0.0015 / 0.06] × 100

= 2.5%IncorrectGiven: Surface tension (

*T*) = 0.06 N-m^{-1}and absolute error (*∆T*) = 0.0015 N-m^{-1}

We know that percentage error in the measurement of surface tension (T_{p}) = [∆T / T] × 100

= [0.0015 / 0.06] × 100

= 2.5% - Question 13 of 14
##### 13. Question

1 pointsTwo resistance are expressed as R1 = (4 ± 0.5) Ω and R2 = (12 ± 0.5) Ω. What is the absolute error in the net resistance, when they are connected in series?

CorrectGiven: Resistance of first resistor (R1) = 5 Ω; Tolerance of first resistor (∆R1) = 5%; Resistance of second resistor (R2) = 8 Ω; Tolerance of second resistor (∆R2) = 20%.

We know that resistance of the combination, when both the resistor are connected in series (R) =R1 + R2.

Therefore percentage error of the combination (∆R)= ∆R1 + ∆R2 = 5% + 20% = 25%.IncorrectGiven: Resistance of first resistor (R1) = 5 Ω; Tolerance of first resistor (∆R1) = 5%; Resistance of second resistor (R2) = 8 Ω; Tolerance of second resistor (∆R2) = 20%.

We know that resistance of the combination, when both the resistor are connected in series (R) =R1 + R2.

Therefore percentage error of the combination (∆R)= ∆R1 + ∆R2 = 5% + 20% = 25%. - Question 14 of 14
##### 14. Question

1 pointsThe errors in the measurement of mass and velocity of a moving body are 2% and 3% respectively. Error, in kinetic energy obtained by measuring mass and speed, will be

CorrectGiven: Error in mass (∆m) = 2% and error in velocity (∆v) = 3%.

We know that kindetic energy (K) = ½ mv2 ∝ mv2.

Therefore the error in the kinetic energy (∆K) =∆m + 2∆v = 2% + (2×3%) = 8%.IncorrectGiven: Error in mass (∆m) = 2% and error in velocity (∆v) = 3%.

We know that kindetic energy (K) = ½ mv2 ∝ mv2.

Therefore the error in the kinetic energy (∆K) =∆m + 2∆v = 2% + (2×3%) = 8%.