Tag: significant figures and rounding of digits

Questions Related to significant figures and rounding of digits

Write the number of significant digits in 0.6464 :

physics units and measurement: error analysis significant figures significant figures and rounding of digits units and measurements

  1. 4

  2. 3

  3. 5

  4. None of the above

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Correct Option: A
Explanation:

All non-zero numbers are always significant.All zeroes before a non zero number are insignificant. All zeroes which are simultaneously to the right of the decimal point and at the end of the number are significant. The only non significant digit here is the zero before the decimal point.
Hence, number of significant digits is 4.

Write the number of significant digits in 6.032 ?

physics units and measurement: error analysis significant figures significant figures and rounding of digits units and measurements

  1. 4

  2. 3

  3. 2

  4. none of the above

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Correct Option: A
Explanation:

The significant figures of a number are those digits that carry meaning contributing to its measurement resolution. This includes all digits except:
1. All leading zeros;
2. Trailing zeros when they are merely placeholders to indicate the scale of the number (exact rules are explained at identifying significant figures); 
3. Spurious digits introduced, for example, by calculations carried out to greater precision than that of the original data, or measurements reported to a greater precision than the equipment supports.

According to rules,
$6.032$ 
All the digits are significant. (no leading and trailing zero)
No. of significant digits $= 4$

The respective number of significant figures for the numbers 6.320, 6.032, 0.0006032 are then

physics units and measurement: error analysis significant figures significant figures and rounding of digits units and measurements

  1. 3, 4, 8

  2. 4, 4, 8

  3. 4, 4, 4

  4. 4, 3, 4

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Correct Option: C
Explanation:

According to the rules of significant figures $6.320$ has four significant figures.
$6.032$ has four significant figures.
$0.0006032$ has four significant figures.

The radius of a sphere is 1.41 cm. Its volume to an appropriate number of significant figures is then

physics units and measurement: error analysis significant figures significant figures and rounding of digits units and measurements

  1. 11.73 $cm^3$

  2. 11.736 $cm^3$

  3. 11.7 $cm^3$

  4. 117 $cm^3$

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Correct Option: C
Explanation:
Radius of the sphere, $r = 1.41 cm$ ($3$ significant figures)
Volume of the sphere,
$\displaystyle V = \dfrac {4}{3} \pi r^3 = \dfrac{4}{3} \times 3.14 \times (1.41)^3\,cm^3 = 11.736\, cm^3$
Rounded off upto $3$ significant figures $= 11.7 cm^3$.

The number of significant figure In the numbers $4.8000 \times 10^4$ and $48000.50 $ are respectively then

physics units and measurement: error analysis significant figures significant figures and rounding of digits units and measurements

  1. $5$ and $6.$

  2. $5$ and $ 7$.

  3. $2$ and $7.$

  4. $2$ and $6$.

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Correct Option: B
Explanation:
If any number has more than one digit after decimal place are significant. Power of 10 does not affect number of significant figures.
$4.8000 \times 10^{4} = 5$ significant figure $(4,8,0,0,0)$
$48000.50 = 7$ significant figure $(4,8,0,0,0,5, 0)$

The length and breadth of a rectangular sheet are 16.2 cm and 10.1 cm, respectively. The area of the sheet in appropriate significant figures and error is:

physics units and measurement: error analysis significant figures significant figures and rounding of digits units and measurements

  1. $164 \pm 3 cm^2$

  2. $163.62 \pm 2.6 cm^2$

  3. $163.6 \pm 2.6 cm^2$

  4. $163.62 \pm 3 cm^2$

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Correct Option: A
Explanation:

Let the length and breadth of a rectangular sheet are measured by using a meter scale as 16.2 cm and 10.1 cm respectively. Each measurement has three significant figures.
$\therefore$ Length l can be written as
$l = 16.2 \pm 0.1 cm = 16.2 cm \pm 0.6$%

Similarly, the breadth b can be written as
$b = 10.1 \pm 0.1 cm = 10.1 cm \pm 1$%

Area of the sheet,
$A = l \times b = 163.62 cm^2 \pm 1.6% =163.62 \pm 2.6 cm^2$
Therefore, as per rule, the area will have only three significant figures and errors will have only one significant figure.
Rounding off, we get $A = 164 \pm 3 cm^2$

The number of significant figures in 0.06900 is the

physics units and measurement: error analysis significant figures significant figures and rounding of digits units and measurements

  1. 5

  2. 4

  3. 2

  4. 3

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Correct Option: B
Explanation:
The significant figures of a number are those which adds the valuable importance to the number. The prefixing and suffixing of zeros in decimal number and normal number value have some significant figure value.
Therefore, given is the decimal number which has zero in front and at the last. The zero in front makes the decimal attain a value but the last end zeros are mere bulky figure without importance.
Therefore, the number of significant figures in $0.06900$ is $4$ because of actual valuable number to be $0.069.$

The value of 3.00 km in yards upto correct significant figures and the scientific notation is, (1m=1.094 yards)

physics units and measurement: error analysis significant figures significant figures and rounding of digits units and measurements

  1. $3.282$

  2. $3.28\times { 10 }^{ 2 }$

  3. $3.28\times { 10 }^{ 3 }$

  4. none of these

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Correct Option: C
Explanation:
The right option is B
Explanation:
As we know
$1$km$=1093.61$yards
Therefore,
$3$km$=3\times1093.61$
$=3280.84$yards
$=32.8\times10^2$yards
Hence, In the scientific notation value of $3$km in yards is $32.8\times10^2$

The number of significant figures in $0.00210$ is

physics units and measurement: error analysis significant figures significant figures and rounding of digits units and measurements

  1. five

  2. six

  3. four

  4. three

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Correct Option: D
Explanation:

In the given number  $0.00210$,  only last digits $210$ are considered as significant figures. Thus the given number has three significant figures.

the length and breath of a metal sheet are $2.325\ m$ and $3.142\ m$ respectively. What is the area of this sheet using proper number of significant figures

physics units and measurement: error analysis significant figures significant figures and rounding of digits units and measurements

  1. $7.30515\ m^{2}$

  2. $7.3051\ m^{2}$

  3. $7.305\ m^{2}$

  4. $7.31\ m^{2}$

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Correct Option: A