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11th
Physics
Units and Measurement
Errors in Measurement
$Errors In Measurement Quest...$

Important questions on Errors In Measurement

Q: In an experiment, the following observations were recorded: L = 2.820 m, M = 3.00 kg, l = 0.087 cm , diameter D = 0.041 cm . Taking g = 9.81 m s^{-2} and using the formula, Y = \frac{4Mg}{\pi D^2 l} , find the maximum permissible error in Y.

y=\dfrac { 4Mgh }{ \Pi { D }^{ 2 }l } Maximum possible error in y is \dfrac { \Delta y }{ y } \times 100=\left( \dfrac { \Delta M }{ M } +\dfrac { \Delta g }{ g } +\dfrac { 2\Delta D }{ D } +\dfrac { \Delta L }{ L } +\dfrac { \Delta l }{ l } \right) \times 100 =\left( \dfrac { 1 }{ 3.00 } +\dfrac { 1 }{ 9.81 } +\dfrac { 1 }{ 41 } \times 2+\dfrac { 1 }{ 2820 } +\dfrac { 1 }{ 87 } \right) \times 100 =0.065\times 100 =6.5 %

Q: Which of the following is the most accurate instrument?

In choice (A), Least count is 0.1\ cm In choice (B), 10\ v.s.d. = 9\ m.s.d L.C. = 1\ m.s.d. - 1\ v.s.d =\dfrac{1}{10} \times 0.1 = 0.01\ cm =\dfrac{1}{10}\ m.s.d In choice (C), L.C. =\left \{ 1-\dfrac{49}{50} \right \} m.s.d =\dfrac{1}{50} m.s.d =\dfrac{1}{50} \times 0.1 = 0.002 cm In choice (D), L.C. =\left \{ 1-\dfrac{49}{50} \right \}\ m.s.d =\dfrac{1}{50}\ m.s.d =\dfrac{1}{50}\times 0.05 = 0.001 cm \Rightarrow Instrument (D) is most accurate.

Q: The radius of a sphere is measured to be 5.3 \pm 0.1 cm . Calculate the percentage error in the measurement of its volume.

The volume is given as V = (4/3)πr^3 so, V= (4/3) \times 3.14 \times 5.33 thus, V = 623.30 cm^3 Now, the error equation ΔV/V = 3(Δr/r) ΔV = V \times 3(Δr/r) thus, ΔV = 623.30 \times 3 \times (0.1/5.3) ΔV = 35.28 cm^3 thus, ΔV/V \times 100 = (35.28/623.30) \times 100 so, ΔV/V \times 100 = 5.66 %

Q: A beam balance of unequal arm length is used by an unscruplus trader. When an object is weighed on left pan. The weight i found to be {W}_{1} . When the object is weighed on the right plan, the weight is found to be {W}_{2} . If {W}_{1}\neq {W}_{2} , the correct weight of the object is

Let the beam balance be as shown.And let actual weight be W .(1) Wr_{L}=W_{1}r_{R} (2) W_{2}r_{L}=Wr_{R} Dividing Equation 1 by Equation 2, \cfrac{W}{W_{2}}=\cfrac{W_{1}}{W} \Rightarrow W=\sqrt{W_{1}W_{2}}

Q: An experiment measures quantities a, b, c and x is calculated from x = ab^2/c^3 . If the maximum percentage error in a, b and c are 1%, 3% and 2% respectively, the maximum percentage error in x will be

Error in a , \dfrac{\Delta a}{a}\times 100 = 1 %Error in b \dfrac{\Delta b}{b}\times 100 = 3 %Error in c \dfrac{\Delta c}{c}\times 100 = 2 %The quantity is given as x = \dfrac{ab^2}{c^3} Error in measurement of x \dfrac{\Delta x}{x}\times 100 = \dfrac{\Delta a}{a}\times 100+ 2\times \dfrac{\Delta b}{b}\times 100 +3\times \dfrac{\Delta c}{c}\times 100 \implies \ \dfrac{\Delta x}{x}\times 100 =1+2\times 3+3\times 2 = 13 %

Q: A physical quantity X is related to four measurable quantities a, b, c and d as follows: X=a^{2} b^{3} c^{5 / 2} d^{-2} The percentage error in the measurement of a, b, c and d are 1 \%, 2 \%, 3 \% and 4 \% respectively. What is the percentage error in quantity X ? If the value of X calculated on the basis of the above relation is 2.763 , to what value should you round off the result?

\because \dfrac{\Delta X}{X} \times 100=\pm\left[2 \dfrac{\Delta a}{a}+3 \dfrac{\Delta b}{b}+\dfrac{5}{2} \dfrac{\Delta c}{c}+2 \dfrac{\Delta d}{d}\right] \times 100 \dfrac{\Delta \mathrm{X}}{\mathrm{X}} \times 100=\pm\left[\dfrac{2 \times 1}{100}+\dfrac{3 \times 2}{100}+\dfrac{5}{2} \times \dfrac{3}{100}+\dfrac{2 \times 4}{100}\right] 100 =\pm \dfrac{100}{100}\left[2+6+\dfrac{15}{2}+8\right] \dfrac{\Delta X}{X} \times 100=\pm\left[16+\dfrac{15}{2}\right]=\pm\left[\dfrac{32+15}{2}\right]=\pm \dfrac{47}{2}=\pm 23.5 \% Mean absolute error =\pm \dfrac{23.5}{100}=\pm 0.235 =0.24 (rounding off in significant figure)Again rounding off X=2.763 in two significant figure =2.8

Q: The time period of a simple pendulum is given by the relation T=2\pi \sqrt{\cfrac{L}{g}} . The measured values of length L and period of 50 oscillations are (20\pm 0.1) cm and (50 \pm 1) s respectively. The percentage error in the determination of g is

T=2\pi \sqrt{\cfrac{L}{g}} \Rightarrow T^2=4\pi^2\cfrac{L}{g} \Rightarrow g=\cfrac{4\pi ^2L}{T^2} \cfrac{\Delta g}{g}=\cfrac{\Delta L}{L}+\cfrac{2\Delta T}{T} \Rightarrow \cfrac{\Delta g}{g}=\cfrac{0.1}{20}+\cfrac{2}{50} \Rightarrow \cfrac{\Delta g}{g}=4.5\%

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easy81 Questionsmedium289 Questionshard42 Questions

HARD

If error in measuring the distance of a planet from sun is 0.4% and in measuring the mass of sun is 0.6% , then error in measuring the time period of planet around the sun is

1.8%

0.6%

1%

0.9%

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HARD

In an experiment, the following observations were recorded:
$L = 2.820 m, M = 3.00 k g, l = 0.087 c m$ , diameter $D = 0.041 c m$ . Taking $g = 9.81 m s^{- 2}$ and using the formula, Y = $\frac{4 M g}{π D ^{2} l}$ , find the maximum permissible error in Y.

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HARD

Which of the following is the most accurate instrument?

A scale whose least count is 0.1 cm.

A vernier caliper in which 10 vernier divisions coincide with 9 main scale divisions whose main scale is millimetre scale.

A microscope in which 50 vernier divisions coincide with 49 main scale divisions whose main scale is millimetre scale.

A microscope in which 50 vernier divisions coincide with 49 main scale divisions and in the main scale there are 20 divisions in 1 cm.

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HARD

Two resistors of resistances $R_{1} = 150 \pm 2 Ω$ and $R_{2} = 220 \pm 6 Ω$ areconnected in parallel combination.Calculate the equivalent resistance.

$180 Ω$

$90 Ω$

$18 Ω$

$10 Ω$

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HARD

A person measures the depth of a well by measuring the time interval between dropping a stone and receiving the sound of impact with the bottom of the well. The error in his measurement of time is $δ T = 0.01$ seconds and he measures the depth of the well to be $L = 20$ meters. Take the acceleration due to gravity $g = 10 m s^{- 2}$ and the velocity of sound $300 m s^{- 1}$ . Then the fractional error in he measurement $\frac{δ L}{L}$ , is closest to

This question has multiple correct options

$0.2 %$

$3 %$

$5 %$

$1 %$

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HARD

The radius of a sphere is measured to be $5.3 \pm 0.1 c m$ . Calculate the percentage error in the measurement of its volume.

$3.45$ %

$5.66$ %

$2.4$ %

$1.05$ %

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HARD

A beam balance of unequal arm length is used by an unscruplus trader. When an object is weighed on left pan. The weight $i$ found to be $W_{1}$ . When the object is weighed on the right plan, the weight is found to be $W_{2}$ . If $W_{1} \neq = W_{2}$ , the correct weight of the object is

$\frac{W _{1} + W _{2}}{2}$

$W_{1} W_{2}$

$W_{1}^{2} + W_{2}^{2}$

$W_{2}^{2} - W_{1}^{2}$

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HARD

An experiment measures quantities a, b, c and x is calculated from $x = a b^{2} / c^{3}$ . If the maximum percentage error in a, b and c are 1%, 3% and 2% respectively, the maximum percentage error in x will be

$13$ %

$17$ %

$14$ %

$11$ %

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HARD

A physical quantity $X$ is related to four measurable quantities $a, b, c$ and $d$ as follows: $X = a^{2} b^{3} c^{5 / 2} d^{- 2}$ The percentage error in the measurement of $a, b, c$ and $d$ are $1 %, 2 %, 3 %$ and $4 %$ respectively. What is the percentage error in quantity $X$ ? If the value of $X$ calculated on the basis of the above relation is $2.763$ , to what value should you round off the result?

View Answer

HARD

The time period of a simple pendulum is given by the relation $T = 2 π \frac{L}{g}$ . The measured values of length $L$ and period of $50$ oscillations are $(20 \pm 0.1)$ $c m$ and $(50 \pm 1)$ $s$ respectively. The percentage error in the determination of $g$ is

$2.5 %$

$4.5 %$

$6 %$

$7.5 %$

View Answer

Important questions on Errors In Measurement

If error in measuring the distance of a planet from sun is 0.4% and in measuring the mass of sun is 0.6% , then error in measuring the time period of planet around the sun is

1.8%

0.6%

1%

0.9%

In an experiment, the following observations were recorded: L=2.820m,M=3.00kg,l=0.087cm, diameter D=0.041cm. Taking g=9.81ms−2 and using the formula, Y =πD2l4Mg​, find the maximum permissible error in Y.

Which of the following is the most accurate instrument?

A scale whose least count is 0.1 cm.

A vernier caliper in which 10 vernier divisions coincide with 9 main scale divisions whose main scale is millimetre scale.

A microscope in which 50 vernier divisions coincide with 49 main scale divisions whose main scale is millimetre scale.

A microscope in which 50 vernier divisions coincide with 49 main scale divisions and in the main scale there are 20 divisions in 1 cm.

Two resistors of resistances R1​=150±2 Ω and R2​=220±6 Ω areconnected in parallel combination.Calculate the equivalent resistance.

180Ω

90Ω

18Ω

10Ω

0.2%

3%

5%

1%

The radius of a sphere is measured to be 5.3±0.1cm. Calculate the percentage error in the measurement of its volume.

3.45%

5.66%

2.4%

1.05%

A beam balance of unequal arm length is used by an unscruplus trader. When an object is weighed on left pan. The weight i found to be W1​. When the object is weighed on the right plan, the weight is found to be W2​. If W1​=W2​, the correct weight of the object is

2W1​+W2​​

W1​W2​​

W12​+W22​​

W22​−W12​​

An experiment measures quantities a, b, c and x is calculated from x=ab2/c3. If the maximum percentage error in a, b and c are 1%, 3% and 2% respectively, the maximum percentage error in x will be

13%

17%

14%

11%

The time period of a simple pendulum is given by the relation T=2πgL​​. The measured values of length L and period of 50 oscillations are (20±0.1) cm and (50±1) s respectively. The percentage error in the determination of g is

2.5%

4.5%

6%

7.5%

More important questions

In an experiment, the following observations were recorded:
$L = 2.820 m, M = 3.00 k g, l = 0.087 c m$ , diameter $D = 0.041 c m$ . Taking $g = 9.81 m s^{- 2}$ and using the formula, Y = $\frac{4 M g}{π D ^{2} l}$ , find the maximum permissible error in Y.

Two resistors of resistances $R_{1} = 150 \pm 2 Ω$ and $R_{2} = 220 \pm 6 Ω$ areconnected in parallel combination.Calculate the equivalent resistance.

$180 Ω$

$90 Ω$

$18 Ω$

$10 Ω$

$0.2 %$

$3 %$

$5 %$

$1 %$

The radius of a sphere is measured to be $5.3 \pm 0.1 c m$ . Calculate the percentage error in the measurement of its volume.

$3.45$ %

$5.66$ %

$2.4$ %

$1.05$ %

A beam balance of unequal arm length is used by an unscruplus trader. When an object is weighed on left pan. The weight $i$ found to be $W_{1}$ . When the object is weighed on the right plan, the weight is found to be $W_{2}$ . If $W_{1} \neq = W_{2}$ , the correct weight of the object is

$\frac{W _{1} + W _{2}}{2}$

$W_{1} W_{2}$

$W_{1}^{2} + W_{2}^{2}$

$W_{2}^{2} - W_{1}^{2}$

An experiment measures quantities a, b, c and x is calculated from $x = a b^{2} / c^{3}$ . If the maximum percentage error in a, b and c are 1%, 3% and 2% respectively, the maximum percentage error in x will be

$13$ %

$17$ %

$14$ %

$11$ %

The time period of a simple pendulum is given by the relation $T = 2 π \frac{L}{g}$ . The measured values of length $L$ and period of $50$ oscillations are $(20 \pm 0.1)$ $c m$ and $(50 \pm 1)$ $s$ respectively. The percentage error in the determination of $g$ is

$2.5 %$

$4.5 %$

$6 %$

$7.5 %$