Question

By what factor must the mass increase to reduce the sampling standard deviation by a factor of 2?

Short answer

The mass needs to be increased four times to reduce the sampling standard deviation by a factor of 2.

Step-by-step solution

Assumption

Let,

K_s= Sampling constant

m = mass of particle

R= relative standard deviation

σ = Standard deviation

m₁ = mass of particlefor 1^st case

R₁= relative standard deviationfor 1^st case

σ₁ = Standard deviationfor 1^st case

m₂ = mass of particle for 2^nd case

R₂= relative standard deviationfor 2^nd case

σ₂ = Standard deviationfor 2^nd case

As per the given information

${\sigma }_{n2}=\frac{{\sigma }_{n1}}{2}$

Determine relative standard deviation

$\begin{array}{rcl}{K}_s& =& m{R}^2\\ m& =& \frac{K_s}{R^2}\\ m_1& =& \frac{K_s}{{R_1}^2}\\ m_1& =& \frac{K_s}{{\left({\displaystyle \frac{{\sigma }_{n1}}{n}}\right)}^2}\\ m_2& =& \frac{K_s}{{R_2}^2}\\ m_2& =& \frac{K_s}{{\left({\displaystyle \frac{{\sigma }_{n2}}{n}}\right)}^2}\end{array}$

$$\begin{gathered} m_2=\frac{K_s}{{\left({\displaystyle \frac{{\sigma }_{n1}}{2n}}\right)}^2}As{\sigma }_{n2}=\frac{{\sigma }_{n1}}{2n} \\ {m}_2=\frac{K_s}{{\displaystyle \frac{1}{4}}{\left({\displaystyle \frac{{\sigma }_{n1}}{n}}\right)}^2} \\ {m}_2=\frac{4K_s}{{\left({\displaystyle \frac{{\sigma }_{n1}}{n}}\right)}^2} \\ {m}_2=4m_{1}As{m}_1=\frac{K_s}{{\left({\displaystyle \frac{{\sigma }_{n1}}{n}}\right)}^2} \end{gathered}$$

Therefore, the mass needs to be increased four times to reduce the sampling standard deviation by a factor of 2