A physical quantity $z$ depends on four observables $a,$ $b,$ $c$ and $d ,$ as $z =\frac{ a ^{2} b ^{\frac{2}{3}}}{\sqrt{ c } d ^{3}} .$ The percentage of error in the measurement of $a, b, c$ and $d$ are $2 \%, 1.5 \%, 4 \%$ and $2.5 \%$ respectively. The percentage of error in $z$ is$......\%$
$12.5$
$14.5$
$16.5$
$13.5$
If the percentage errors in measuring the length and the diameter of a wire are $0.1 \%$ each. The percentage error in measuring its resistance will be:
In an experiment of simple pendulum time period measured was $50\,sec$ for $25$ vibrations when the length of the simple pendulum was taken $100\,cm$ . If the least count of stop watch is $0.1\,sec$ . and that of meter scale is $0.1\,cm$ then maximum possible error in value of $g$ is .......... $\%$
$Assertion$ : When percentage errors in the measurement of mass and velocity are $1\%$ and $2\%$ respectively, the percentage error in $K.E.$ is $5\%$.
$Reason$ : $\frac{{\Delta E}}{E} = \frac{{\Delta m}}{m} + \frac{{2\Delta v}}{v}$
A physical quantity $P$ is related to four observables $a, b, c$ and $d$ as follows: $P=\frac{a^{2} b^{2}}{(\sqrt{c} d)}$ 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?
The temperatures of two bodies measured by a thermometer are $t_{1}=20^{\circ} C \pm 0.5^{\circ} C$ and $t_{2}=50^{\circ} C \pm 0.5^{\circ} C$ Calculate the temperature difference and the error theirin.