Two capacitors with capacitance values $C_1 = 2000 \pm 10 \text{ pF}$ and $C_2 = 3000 \pm 15 \text{ pF}$ are connected in series. The voltage applied across this combination is $V = 5.00 \pm 0.02 \text{ V}$. The percentage error in the calculation of the energy stored in this combination of capacitors is . . . . . .

Two resistors of resistances $R_{1} = 100 \pm 3 \ \Omega$ and $R_{2} = 200 \pm 4 \ \Omega$ 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 $\frac{\Delta R^{\prime}}{R^{\prime 2}} = \frac{\Delta R_{1}}{R_{1}^{2}} + \frac{\Delta R_{2}}{R_{2}^{2}}$.

Four persons measure the length of a rod as $20.00 \ cm$,$19.75 \ cm$,$17.01 \ cm$,and $18.25 \ cm$. The relative error in the measurement of the average length of the rod is:

$A$ 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 $\%$)

$Assertion$ : When percentage errors in the measurement of mass and velocity are $1\%$ and $2\%$ respectively,the percentage error in $K.E.$ is $5\%$.
$Reason$ : $\frac{{\Delta E}}{E} = \frac{{\Delta m}}{m} + \frac{{2\Delta v}}{v}$

The percentage error in the measurement of mass and velocity are $3 \%$ and $4 \%$ respectively. The percentage error in the measurement of kinetic energy is (in $\%$)