The correct relation between variance and standard deviation (S.D) of a variable X is _________
A - S.D = Var
B - ${\text{S}}{\text{.D = }}\left[ {{\text{Var(X}}{{\text{)}}^{\dfrac{1}{2}}}} \right]$
C - ${\text{S}}{\text{.D = }}{\left[ {{\text{Var(X)}}} \right]^2}$
D - None of the above

Hint: As we know that the formula of standard deviation is $\sigma = \sqrt {\dfrac{{\sum\limits_{i = 1}^n {{{\left( {{x_i} - \mathop x\limits_{}^\_ } \right)}^2}} }}{n}} $ it will told how the data will deviated from its mean and the formula of the variance is the $\dfrac{{\sum\limits_{i = 1}^n {{{\left( {{x_i} - \mathop x\limits_{}^\_ } \right)}^2}} }}{n}$ hence from both these we can find the solution .

Complete step-by-step answer:
In this we have to make relation between the Standard deviation and the Variance ,
Firstly the standard deviation is the measure of how spread out numbers are; it is equal to the square root of the arithmetic mean of the squares of the deviations measured from the arithmetic mean of the data .
Standard deviation formula is $\sigma = \sqrt {\dfrac{{\sum\limits_{i = 1}^n {{{\left( {{x_i} - \mathop x\limits_{}^\_ } \right)}^2}} }}{n}} $ .......(i)
${x_i}$ the ${i_{th}}$ data point
$\mathop x\limits^\_ $ the mean of all data points
n = the number of data points​
Variance in statistics is a measurement of the spread between numbers in a data set. That is, it measures how far each number in the set is from the mean and therefore from every other number in the set. The average of the squared differences from the Mean.
Variance is equal to the = $\dfrac{{\sum\limits_{i = 1}^n {{{\left( {{x_i} - \mathop x\limits_{}^\_ } \right)}^2}} }}{n}$ ......(ii)
where
${x_i}$ the ${i_{th}}$ data point
$\mathop x\limits^\_ $ the mean of all data points
n = the number of data points​
Hence from (i) and (ii) we can find that the ${\text{S}}{\text{.D = }}\left[ {{\text{Var(X}}{{\text{)}}^{\dfrac{1}{2}}}} \right]$ .
So option B is the correct answer .

Note: Standard deviation along median is being calculated as the formula is $\sigma = \sqrt {\dfrac{{\sum\limits_{i = 1}^n {{{\left( {{x_i} - M} \right)}^2}} }}{n}} $
${x_i}$ the ${i_{th}}$ data point
 M the median of all data points
n = the number of data points
It will show how much the data is deviated from Median .