$A$ physical quantity $x$ is related as $x = \frac{\sqrt{a}b^3}{c^4d^2}$. Relative errors in the quantities $a, b, c$ and $d$ are $2\%, 1\%, 3\%$ and $4\%$ respectively. The relative error in $x$ will be (in $\%$):

The period of oscillation of a simple pendulum is given by $T = 2\pi\sqrt{\frac{l}{g}}$ where $l$ is about $80\text{ cm}$ and is known to have $0.1\text{ cm}$ accuracy. The period is about $1.5\text{ s}$. The time of $50$ oscillations is measured by a stop watch of least count $0.1\text{ s}$. The percentage error in $g$ is nearly

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:

For $z = a^{2} x^{3} y^{1/2}$,where $a$ is a constant. If the percentage error in the measurement of $x$ and $y$ are $4\%$ and $12\%$,respectively,then the percentage error for $z$ will be $........... \%$.

The error in the measurement of the length and the breadth of a rectangular table is $1 \%$. If the length and breadth of the table are $1 \ m$ and $50 \ cm$ respectively, then the area of the table including error is

$A$ physical quantity $P$ is related to four observables $a, b, c$ and $d$ as follows: $P = \frac{a^3 b^2}{\sqrt{c} d}$. The percentage errors of measurement in $a, b, c$ and $d$ are $1 \%, 3 \%, 4 \%$ and $2 \%$ respectively. What is the percentage error in the quantity $P$? If the value of $P$ calculated using the above relation turns out to be $3.763$,to what value should you round off the result?