The length \ell , breadth b and thickness t of a block of wood were measured with the help of a measuring scale. The results with permissible errors are \ell = 15.12 \pm 0.01 cm, t = 5.28 \pm 0.01 cm. b = 10.15 \pm 0.01 cm. The percentage error in volume upto proper significant figures is :

Volume, V=lbt \dfrac{dV}{V}=\dfrac{dl}{l}+\dfrac{db}{b}+\dfrac{dt}{t} The % error in volume, \dfrac{dV}{V}\times 100=(\dfrac{dl}{l}+\dfrac{db}{b}+\dfrac{dt}{t})\times 100=(\dfrac{0.01}{15.12}+\dfrac{0.01}{10.15}+\dfrac{0.01}{5.28})\times 100=0.36 %

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11th
Physics
Units and Measurement
Significant Figures
$Significant Figures Questio...$

Important questions on Significant Figures

Q: A student measures the time period of 100 oscillations of a simple pendulum four times. The data set is 90\ s, 91\ s, 95\ s and 92\ s . If the minimum division in the measuring clock is 1\ s , then the reported mean time should be:

Sum of all observation= 90 + 91 + 95 + 92 = 368 Average =368 /4 = 92 Sum of modulus errors= 2 + 1 + 3 = 6 Average error =6/4 = 1.5 ,rounded of 2 .Final answer = (92 ± 2)\ s

Q: The edge of a cube is measured using a vernier callipers. [9 divisions of the main scale is equal to 10 divisions of vernier scale and 1 main scale division is 1 mm]. The main scale division reading is 10 and 1 division of vernier scale was found to be coinciding with the main scale. The mass of the cube is 2.736 gm. Calculate the density in \mathrm{g}\mathrm{m}/\mathrm{c}\mathrm{m}^{3} up to correct significant figures.

Given, 9 MSD = 10 VSD1 MSD = 1 mm L.C. =1 MSD-I VSD =1 mm- \displaystyle \frac{9}{10} mm =\displaystyle \frac{1}{10} mmMeasured reading of edge = MSR + VSR \displaystyle \times(\mathrm{L}\mathrm{C}) =10+1\times\frac{1}{10}=10.1 mm Volume of cube =(10.1)^{3}=1030.301\mathrm{m}\mathrm{m}^{3} =1.030301\mathrm{c}\mathrm{m}^{3} Density of the cube =\displaystyle \frac{2.736}{1.030301}=2.66\mathrm{g}\mathrm{m}/\mathrm{c}\mathrm{m}^{3}

Q: The number of significant figures in 0.06900 is the

Significant figures in the measured value of a physical quantity tell the number of digits in which we have confidence. Larger the number of significant figures obtained in a measurement, greater is the accuracy of the measurement. The reverse is also true.The following rules are observed in counting the number of significant figures in a given measured quantity.1. All non-zero digits are significant.2. A zero becomes significant figure if it appears between two non¬zero digits.3. Leading zeros or the zeros placed to the left of the number are never significant.4. Trailing zeros or the zeros placed to the right of the number are significant.5. In exponential notation, the numerical portion gives the number of significant figures.Leading zeros or the zeros placed to the left of the number are never significant.Hence,number of significant figures are four.

Q: In an experiment the following observations were recorded L=2.890 metre \mathrm{M}=3.0 kg \ell=0.87 cm \mathrm{D}=0.041 cm \mathrm{g}=981\mathrm{c}\mathrm{m}/\sec^{2} and the formula used for calculatlon of Young's modulus ( \mathrm{Y} ) is \displaystyle \mathrm{Y}=\frac{\mathrm{M}\mathrm{g}\mathrm{L}}{\pi \mathrm{r}^{2}l} What is the maximum possible error expressed in percentage?

Y=\dfrac{4MgL}{\pi D^2 l} Hence, log(Y)=log4+log(M)+log(g)+log(L)-2log(D)-log(l) \dfrac{\Delta Y}{Y}=\dfrac{\Delta M}{M}+\dfrac{\Delta g}{g}+\dfrac{\Delta L}{L}+\dfrac{2\Delta D}{D}+\dfrac{\Delta l}{l} For maximum possible error, \Delta M=0.05kg \Delta g=0.5cm/sec^2 \Delta L=0.0005metre \Delta D=0.0005cm \Delta l=0.005cm Putting the values in \dfrac{\Delta Y}{Y}\times 100 Hence maximum possible error is 4.8%

Q: Calculate focal length of a spherical mirror from the following observations.Object distance, u=(50.1±0.5)u=(50.1±0.5) cmImage distance, v=(20.1±0.2) cm

Let focal length of the mirror be f cm. \dfrac{1}{v} + \dfrac{1}{u} = \dfrac{1}{f} \dfrac{1}{20.1} + \dfrac{1}{50.1} = \dfrac{1}{f} \implies f = 14.3 cmAs, f = \dfrac{uv}{u+v} \therefore Taking log and differentiating, we get \Delta f = f \bigg( \dfrac{\Delta u}{u} + \dfrac{\Delta v}{v} + \dfrac{\Delta u + \Delta v}{u+ v} \bigg) \Delta f = 14.3 \bigg( \dfrac{0.5}{50.1} + \dfrac{ 0.2}{20.1} + \dfrac{0.5 +0.2}{50.1+20.1} \bigg) = 0.4 cmThus focal length of the mirror = (14.3 \pm 0.4) cm

Q: The length and breadth of a metal sheet are 3.124 \ m and 3.002\ m respectively. The area of this sheet upto four correct significant figure is :

Area of metal sheet = 3.124 \times 3.002 = 9.378248 Now, the result must have significant figures equal to the least of figure being multiplied. So, area of metal sheet = 9.378\ m^{2}

Q: The mass of a body is 10.000 gm and its volume is 10.00 \mathrm{cm}^{3} . If the measured values are expressed upto correct significant figures, then the maximum error in the measurement of density is :

Given : Mass, M = 10.000 gm Volume, V = 10.00 cm^3 \therefore Density, d = \dfrac{M}{V} = \dfrac{10.000}{10.00} = 1 gm/cm^3 Error in mass, \Delta M = 0.001 gmError in volume, \Delta V = 0.01 cm^3 Error in density, \Delta d = d \bigg( \dfrac{\Delta M}{M} + \dfrac{\Delta V}{V} \bigg) = 1 \bigg( \dfrac{0.001}{10.000} + \dfrac{0.01}{10.00} \bigg) = 0.0011 gm/cm^3

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HARD

A student measures the time period of $100$ oscillations of a simple pendulum four times. The data set is $90 s, 91 s, 95 s$ and $92 s$ . If the minimum division in the measuring clock is $1 s$ , then the reported mean time should be:

$92 \pm 2 s$

$92 \pm 5.0 s$

$92 \pm 1.8 s$

$92 \pm 3 s$

View Answer

HARD

The diameter of a sphere is 3.34 m. Calculate is Volume with due regard to significant figures.

View Answer

$9.378248 m^{2}$

$9.3782 m^{2}$

View Answer

HARD

Solve with due regard to significant figures:
$\frac{5 . 4 2 \times 0 . 6 7 5 3}{0 . 0 8 5}$

View Answer

HARD

The mass of a body is 10.000 gm and its volume is $10.00$ $c m^{3}$ . If the measured values are expressed upto correct significant figures, then the maximum error in the measurement of density is :

$0.0012$ $g m / c m^{3}$

$0.001$ $g m / c m^{3}$

$0.0001$ $g m / c m^{3}$

$0.021$ $g m / c m^{3}$

View Answer

Important questions on Significant Figures

A student measures the time period of 100 oscillations of a simple pendulum four times. The data set is 90 s,91 s,95 s and 92 s. If the minimum division in the measuring clock is 1 s, then the reported mean time should be:

92±2 s

92±5.0 s

92±1.8 s

92±3 s

The diameter of a sphere is 3.34 m. Calculate is Volume with due regard to significant figures.

The number of significant figures in 0.06900 is the

5

4

2

3

In an experiment the following observations were recorded L=2.890 metre M=3.0 kg ℓ=0.87 cm D=0.041 cm g=981cm/sec2 and the formula used for calculatlon of Young's modulus (Y) is Y=πr2lMgL​ What is the maximum possible error expressed in percentage?

6.36%

4.8%

3%

1%

Calculate focal length of a spherical mirror from the following observations.Object distance, u=(50.1±0.5)u=(50.1±0.5) cm Image distance, v=(20.1±0.2) cm

(14.3±0.4) cm

(14.3±0.2) cm

(12.3±0.4) cm

(12.3±0.2) cm

The length ℓ, breadth b and thickness t of a block of wood were measured with the help of a measuring scale. The results with permissible errors are ℓ = 15.12 ± 0.01 cm, t = 5.28 ± 0.01 cm. b = 10.15 ± 0.01 cm. The percentage error in volume upto proper significant figures is :

0.28%

0.36%

0.48%

0.64%

The length and breadth of a metal sheet are 3.124 m and 3.002 m respectively. The area of this sheet upto four correct significant figure is :

9.378 m2

9.37 m2

9.378248 m2

9.3782 m2

Solve with due regard to significant figures: 0.0855.42×0.6753​

The mass of a body is 10.000 gm and its volume is 10.00 cm3. If the measured values are expressed upto correct significant figures, then the maximum error in the measurement of density is :

0.0012 gm/cm3

0.001 gm/cm3

0.0001 gm/cm3

0.021 gm/cm3

More important questions

A student measures the time period of $100$ oscillations of a simple pendulum four times. The data set is $90 s, 91 s, 95 s$ and $92 s$ . If the minimum division in the measuring clock is $1 s$ , then the reported mean time should be:

$92 \pm 2 s$

$92 \pm 5.0 s$

$92 \pm 1.8 s$

$92 \pm 3 s$

The number of significant figures in $0.06900$ is the

$5$

$4$

$2$

$3$

Calculate focal length of a spherical mirror from the following observations.
Object distance, u=(50.1±0.5) cm

Image distance, v=(20.1±0.2) cm

$(14.3 \pm 0.4)$ cm

$(14.3 \pm 0.2)$ cm

$(12.3 \pm 0.4)$ cm

$(12.3 \pm 0.2)$ cm

The length and breadth of a metal sheet are $3.124 m$ and $3.002 m$ respectively. The area of this sheet upto four correct significant figure is :

$9.378 m^{2}$

$9.37 m^{2}$

$9.378248 m^{2}$

$9.3782 m^{2}$

Solve with due regard to significant figures:
$\frac{5 . 4 2 \times 0 . 6 7 5 3}{0 . 0 8 5}$

The mass of a body is 10.000 gm and its volume is $10.00$ $c m^{3}$ . If the measured values are expressed upto correct significant figures, then the maximum error in the measurement of density is :

$0.0012$ $g m / c m^{3}$

$0.001$ $g m / c m^{3}$

$0.0001$ $g m / c m^{3}$

$0.021$ $g m / c m^{3}$