In an experiment,the values of refractive indices of glass were found to be $1.54, 1.53, 1.44, 1.54, 1.56$ and $1.45$ in successive measurements. The mean absolute error is:

$A$ physical quantity $P$ is given by $P = \frac{A^3 B^{1/2}}{C^{-4} D^{3/2}}$. The quantity which brings in the maximum percentage error in $P$ is

If the radius of a sphere is $(5.3 \pm 0.1) \; cm$,then the percentage error in its volume will be:

Two resistances $60 \pm 0.36 \ \Omega$ and $30 \pm 0.09 \ \Omega$ are connected in parallel. The equivalent resistance is

Zero error of an instrument introduces

Let $x = \frac{a^2 b^2}{c}$ be a physical quantity. If the percentage errors in the measurement of physical quantities $a, b,$ and $c$ are $2\%, 3\%,$ and $4\%$ respectively,then the percentage error in the measurement of $x$ is: (in $\%$)