In an experiment, mass of an object is measured by applying a known force on it, and then measuring its acceleration. If in the experiment, the measured values of applied force and the measured acceleration are $F=10.0 \pm 0.2 \,N$ and $a=1.00 \pm 0.01 \,m / s ^2$, respectively. Then, the mass of the object is ............... $kg$
$10.0$
$10.0 \pm 0.1$
$10.0 \pm 0.3$
$10.0 \pm 0.4$
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 error in $a$, $b$ and $c$ are $3 \%, 4 \%$ and $5 \%$ respectively. Then, the percentage error in $\mathrm{Q}$ is :
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?
In an experiment to find acceleration due to gravity $(g)$ using simple pendulum, time period of $0.5\,s$ is measured from time of $100$ oscillation with a watch of $1\;s$ resolution. If measured value of length is $10\; cm$ known to $1\; mm$ accuracy. The accuracy in the determination of $g$ is found to be $x \%$. The value of $x$ is
If a copper wire is stretched to make its radius decrease by $0.1\%$ , then percentage increase in resistance is approximately .......... $\%$
While measuring the acceleration due to gravity by a simple pendulum, a student makes a positive error of $1\%$ in the length of the pendulum and a negative error of $3\%$ in the value of time period. His percentage error in the measurement of $g$ by the relation $g = 4{\pi ^2}\left( {l/{T^2}} \right)$ will be ........ $\%$