Measurement of two quantities along with the error is: $A = 2.5 \ ms^{-1} \pm 0.5 \ ms^{-1}$,$B = 0.10 \ s \pm 0.01 \ s$. The value of $AB$ will be:

The period of oscillation of a simple pendulum is $T = 2\pi(L/g)^{1/2}$. Measured value of $L$ is $10 \text{ cm}$ known to $1 \text{ mm}$ accuracy and time for $100$ oscillations is $50 \text{ s}$, using a watch of $1 \text{ s}$ resolution. The percentage error in the measurement of $g$ is (in $\%$)

The time period of a simple pendulum is given by $T = 2 \pi \sqrt{\frac{\ell}{g}}$. The measured value of the length of the pendulum is $10 \ cm$ known to a $1 \ mm$ accuracy. The time for $200$ oscillations of the pendulum is found to be $100 \ s$ using a clock of $1 \ s$ resolution. The percentage accuracy in the determination of $g$ using this pendulum is $x$. The value of $x$ to the nearest integer is ...........$\%$

The measurements of two quantities along with the precision of their respective measuring instruments are $A = 2.5 \, m/s \pm 0.5 \, m/s$ and $B = 0.10 \, s \pm 0.01 \, s$. The value of $AB$ will be:

If the random error in the arithmetic mean of $50$ observations is $\alpha$,then the random error in the arithmetic mean of $150$ observations would be

$A$ 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 $\%$)