Tag: physics

Questions Related to physics

The radius of sphere is measured to be $\displaystyle \left ( 2.1\pm 0.5 \right )$cm.
 Calculate its surface area with error limits.

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

  1. $\displaystyle S=\left ( 55.4\pm 26.4 \right )cm^{2}$

  2. $\displaystyle S=\left ( 55.4\pm 26.4 \right )mm^{2}$

  3. $\displaystyle S=\left ( 55.4\pm 13.2 \right )cm^{2}$

  4. None of these

Show answer
Correct Option: A
Explanation:

Surface area, $\displaystyle S=4\pi r^{2}=\left ( 4 \right )\left ( \frac{22}{7} \right )\left ( 2.1 \right )^{2}$$\displaystyle =55.44=55.4cm^{2}$
Further, $\displaystyle \frac{\Delta S}{S}=2.\frac{\Delta r}{r}$
              $\displaystyle \Delta S=2\left ( \frac{\Delta r}{r} \right )\left ( S \right )=\frac{2\times 0.5\times 55.4}{2.1}$$\displaystyle =26.38=26.4 cm^{2}$

$\displaystyle \therefore S=\left ( 55.4\pm 26.4 \right )cm^{2}$

Find density when a mass of 9.23 kg occupies a volume of $\displaystyle 1.1m^{3}$ with due regard to significant figures.

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

  1. $\displaystyle 8.4 kg/m^{3}$

  2. $\displaystyle 8.39 kg/m^{3}$

  3. $\displaystyle 8.3 kg/m^{3}$

  4. None of the above

Show answer
Correct Option: A
Explanation:

Mass $= 9.23\ kg$ .... 3 significant figures

Volume $= 1.1\  m^3$ .... 2 significant figures

Therefore, $density = mass/volume = 8.4\ kg/m^3$ 

The value due regard to significant figure  $ 4.0\times 10^{-4}-2.5\times 10^{-6}$ 

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

  1. $\displaystyle 4.0\times 10^{-4}$

  2. $\displaystyle 3.975\times 10^{-4}$

  3. $\displaystyle 3.9\times 10^{-4}$

  4. none of the above

Show answer
Correct Option: B
Explanation:
The value of this case is,
$4.0\times 10^{-4}-2.5\times 10^{-6}$
$=3.975\times 10^{-4}$.

Length, breadth and thickness of a rectangular slab are 4.234 m, 1.005 m and 2.01 m respectively. Find volume of the slab to correct significant figures.

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

  1. $\displaystyle 8.55 m^{3}$

  2. $\displaystyle 8.552 m^{3}$

  3. $\displaystyle 8.6 m^{3}$

  4. None of the above

Show answer
Correct Option: A
Explanation:

l = 4.234 m 4 significant figures

b = 1.005 m 4 significant figures
w = 2.01 m 3 significant figures 
Therefore, Volume = lbw= 8.55 .........3 significant figures

Write down the number of significant figures in the following value.

$0.0628 cm$

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

  1. $four$

  2. $three$

  3. $two$

  4. $one$

Show answer
Correct Option: B
Explanation:

Zeroes before, a non zero digit are not significant.

Therefore, (B) three significant figures in 0.0628

Round off following number upto four significant figures.

$2.00823$

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

  1. $2.008$

  2. $2.0082$

  3. $2.0083$

  4. $2$

Show answer
Correct Option: A
Explanation:

$2.00823 = 2.008$ because the digit 2 is less than 5.

Write down the number of significant figures in 12.00 N.

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

  1. four

  2. three

  3. two

  4. one

Show answer
Correct Option: A
Explanation:

According to general rule for determining of number of significant figures, zeroes placed after other digits but behind a decimal point are significant. Thus 12.00 has four significant figures. 

Write down the number of significant figures in 0.00628 cm.

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

  1. four

  2. three

  3. two

  4. one

Show answer
Correct Option: B
Explanation:

Zeroes before any non zero value are not significant.

Therefore, 0.00628 has three significant figures.

Write down the number of significant figures in the following.

$62.00m$

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

  1. $four$

  2. $three$

  3. $two$

  4. $one$

Show answer
Correct Option: A
Explanation:

Zeroes after decimal but after non zero value are significant.

Therefore, 62.00 has four significant figure.

Write down the number of significant figures in the following.
$6428$

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

  1. $four$

  2. $three$

  3. $two$

  4. $one$

Show answer
Correct Option: A
Explanation:

For a number with $NO \ \ ZERO$ to the right, the number of significant figures is simply the total number of digits.

Here the total significant figures is $4$.