The potential difference across the ends of a conductor is $(30 \pm 0.3) \ V$ and the current through the conductor is $(5 \pm 0.1) \ A$. The error in the determination of the resistance of the conductor is: (in $\%$)

$A$ physical quantity $S$ is related to four observables $a, b, c$, and $d$ as $S = \frac{\sqrt{a} b}{c^3 d^4}$. If the percentage errors of measurement in $a, b, c$, and $d$ are $2 \%$, $1 \%$, $1 \%$, and $1 \%$ respectively, then the percentage error in the quantity $S$ is: (in $\%$)

According to Joule's law of heating,heat produced $H = I^2Rt$,where $I$ is current,$R$ is resistance,and $t$ is time. If the errors in the measurement of $I, R$,and $t$ are $3\%, 4\%$,and $6\%$ respectively,then the error in the measurement of $H$ is:

$A$ certain body weighs $22.42 \; g$ and has a measured volume of $4.7 \; cc$. The possible errors in the measurement of mass and volume are $0.01 \; g$ and $0.1 \; cc$,respectively. Then the maximum percentage error in the density will be: (in $\%$)

Resistance of a rod is calculated by measuring its length with the help of a meter scale of least count $1 \ mm$. Its radius is measured with the help of a screw gauge having $50$ divisions on the circular scale and a pitch of $1 \ mm$. Resistivity of the material is exact. The length of the wire is found to be $20 \ cm$ and the diameter of the wire is $4 \ mm$. Find the percentage error in the calculation of resistance.

If two resistors have resistances $R_1 = (350 \pm 3) \text{ } \Omega$ and $R_2 = (140 \pm 4) \text{ } \Omega$, then the percentage error for the sum and difference of $R_1$ and $R_2$ are respectively: