$A$ physical quantity $Q$ is found to depend on quantities $a, b, c$ by the relation $Q = \frac{a^4 b^3}{c^2}$. The percentage errors in $a, b,$ and $c$ are $3 \%, 4 \%,$ and $5 \%$ respectively. Then,the percentage error in $Q$ is: (in $\%$)

The energy of a system as a function of time $t$ is given as $E(t)=A^2 \exp(-\alpha t)$,where $\alpha=0.2 \ s^{-1}$. The measurement of $A$ has an error of $1.25 \%$. If the error in the measurement of time is $1.50 \%$,the percentage error in the value of $E(t)$ at $t=5 \ s$ is

In an experiment of a simple pendulum,the errors in the measurement of the length of the pendulum $(L)$ and time period $(T)$ are $3 \%$ and $2 \%$,respectively. The maximum percentage error in the value of $\frac{L}{T^2}$ is (in $\%$)

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 $\%$)

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

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: