Tag: units and measurement: error analysis

Quizzes (17) Questions (163)

Questions Related to units and measurement: error analysis

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

The length,breadth and thickness of a rectangular copper sheet are $4 cm, 2.5 cm $ and $1.56 cm $ respectively. Find the area of the sheet to corrected significant figures.

  1. $40.3 cm^2$

  2. $40.28 cm^2$

  3. $40 \times 10^2 cm^2$

  4. $0.4\times 10^2 cm^2$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

If $l$ be the length, $b$ be the breadth and $t$ be the thickness of the sheet,
then the area of sheet is $A=2(lb+bt+tl)=2[(4\times 2.5)+(2.5\times 1.56)+(1.56\times 4)]=40.28 cm^2$
As length has least significant figures i.e one so the answer must be expressed as one significant figures. 
Thus according rule of determining significant figures, the area will be $0.4\times 10^2 cm^2$

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

Express 0.006006 into scientific notation in three significant digits:

  1. $6.01 \times 10^{-3}$

  2. $6.0006 \times 10^{-3}$

  3. $6.00 \times 10^{-3}$

  4. $6.0 \times 10^{-3}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
The number  $0.006006$ has 6 significant figures.
We need to round off the given number to convert into 3 significant figure. Thus,  the correct answer is  $6.01\times 10^{-3}$ as it has 3 significant figure including rounding off.
Multiple choice physics units and measurement: error analysis significant figures significant figures and rounding of digits units and measurements

Which of the following quantifies has the least number of significant digits? 

  1. $0.80760 $

  2. $0.08765$

  3. $5.7423 \times 10^2$

  4. $80.760$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
The number $0.80760$ has $5$ significant digits, $0.08765$ has $4$ significant digits, $5.7423\times 10^{2}$ has $5$ significant digits and $80.760$ has $5$ significant digits.
Hence option B is correct.
Multiple choice physics units and measurement: error analysis significant figures significant figures and rounding of digits units and measurements

A person was weighing $102.1\  kg$ last week and gained $0.28 \ kg$ this week. His weight as of now is correctly expressed as :

  1. $102.38\  kg$

  2. $102.3\  kg$

  3. $102.4\  kg$

  4. $102\  kg$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

We know that number of significant figures after addition will be equal to least number of significant figures present in any of the operand.
Therefore total weight after weight gain is $102.1+0.28=102.38 kg$
Now least number of significant figure present after the decimal in the operand is 1, present in $102.1\ kg$ .
Therefore, the number of significant figures present after the calculation should also be $1$ after the decimal. Hence, when we round the answer $102.38$ becomes $102.4\  kg$
Therefore correct answer is option (C)

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

Subtract $3.2 \times 10^{-6}$ from $4.7 \times 10^{-4}$ with due regard to significant figures.

  1. $~4.7 \times 10^{-4}$

  2. $~7 \times 10^{-4}$

  3. $~5.7 \times 10^{-4}$

  4. $~4.7 \times 10^{-5}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$4.7\times { 10 }^{ -4 }-3.2\times { 10 }^{ -6 }$

$=4.7\times { 10 }^{ -4 }-0.032\times { 10 }^{ -4 }$
$=4.668\times { 10 }^{ -4 }\approx 4.7\times { 10 }^{ -4 }$.

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

Subtract $1.5 \times 10^3$ from $4.8 \times 10^4$ with due regard to significant figures.

  1. $~4.6 \times 10^{4}$

  2. $~4.6 \times 10^{5}$

  3. $~4.6 \times 10^{6}$

  4. $~5 \times 10^{4}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$4.8\times { 10 }^{ 4 }-1.5\times { 10 }^{ 3 }$

$=4.8\times { 10 }^{ 4 }-0.15\times { 10 }^{ 4 }$
$\Rightarrow 4.65\times { 10 }^{ 4 }\approx 4.7\times { 10 }^{ 4 }$

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

Add $3.8 \times 10^{-6} to 4.2 \times 10^{-5}$ with due regard to significant figures.

  1. $~4.6 \times 10^{-5}$

  2. $~4.6 \times 10^{-6}$

  3. $~5.6 \times 10^{-6}$

  4. $~6.6 \times 10^{-5}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$3.8\times { 10 }^{ -6 }+42\times { 10 }^{ -6 }=45.8\times { 10 }^{ -6 }=4.58\times { 10 }^{ -5 }$

                                                                      $=4.6\times { 10 }^{ -5 }$.

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

The radius of the earth is $6.37 \times 10^6 m$ and its mass is $5.975 \times 10^{24} kg$. Find the earth's average density to appropriate significant figures.

  1. $5 \times 10^3 kg m^{-3}$

  2. $5.52 \times 10^3 kg m^{-3}$

  3. $2 \times 10^3 kg m^{-3}$

  4. $5.52 \times 10^3 kg m^{-4}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Density = Mass / Volume. Volume = (4/3) * pi * r^3. Density = 5.975e24 / ((4/3) * pi * (6.37e6)^3) = 5518 kg/m^3, which is 5.52e3 kg/m^3.

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

The number of significant figures in $5.69 \times 10^{15}$kg is

  1. $1$

  2. $2$

  3. $3$

  4. $18$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

No. of significant figures in $\underbrace { 5.69 } \times { 10 }^{ 15 }$ Kg is $3$.

                                           Digits that cannot meaning to measurement and measurement resolution.