A physical quantity is $A = P^2/Q^3.$ The percentage error in measurement of $P$ and $Q$ is $x$ and $y$ respectively. Maximum error in measurement of $A$ is

Two resistors of resistances $R_{1}=100 \pm 3$ $ohm$ and $R_{2}=200 \pm 4$ $ohm$ are connected $(a)$ in series, $(b)$ in parallel. Find the equivalent resistance of the $(a)$ series combination, $(b)$ parallel combination. Use for $(a)$ the relation $R=R_{1}+R_{2}$ and for $(b)$ $\frac{1}{R^{\prime}}=\frac{1}{R_{1}}+\frac{1}{R_{2}}$ and $\frac{\Delta R^{\prime}}{R^{\prime 2}}=\frac{\Delta R_{1}}{R_{1}^{2}}+\frac{\Delta R_{2}}{R_{2}^{2}}$

A physical quantity $'x'$ is calculated from the relation $x = \frac{{{a^2}{b^3}}}{{c\sqrt d }}$ in $a$,$b$,$c$ and $d$ are $2\%$, $1 \%$, $3\%$ and $4\%$ respectively, what is the percentage error in $x$.

Error in the measurement of radius of a sphere is $1\%$. The error in the calculated value of its volume is  ……… $\%$

The current voltage relation of diode is given by $I=(e^{1000V/T} -1)\;mA$, where the applied voltage $V$ is in volts and the temperature $T$ is in degree Kelvin. If a student makes an error measuring $ \mp 0.01\;V$ while measuring the current of $5\; mA$ at $300\; K$, what will be the error in the value of current in $mA$ ?