The radius of a sphere is $(5.3 \pm 0.1) \,cm$. The percentage error in its volume is
$\frac{{0.1}}{{5.3}} \times 100$
$3 \times \frac{{0.1}}{{5.3}} \times 100$
$\frac{{0.1 \times 100}}{{3.53}}$
$3 + \frac{{0.1}}{{5.3}} \times 100$
The current voltage relation of diode is given by $I=(e^{1000V/T} -1)\;mA$, where the applied voltage $V$ is in volts and the temperature $T$ is in degree Kelvin. If a student makes an error measuring $ \mp 0.01\;V$ while measuring the current of $5\; mA$ at $300\; K$, what will be the error in the value of current in $mA$ ?
The amount of heat produced in an electric circuit depends upon the current $(I),$ resistance $(R)$ and time $(t).$ If the error made in the measurements of the above quantities are $2\%, 1\%$ and $1\%$ respectively then the maximum possible error in the total heat produced will be ........... $\%$
$Assertion$ : The error in the measurement of radius of the sphere is $0.3\%$. The permissible error in its surface area is $0.6\%$
$Reason$ : The permissible error is calculated by the formula $\frac{{\Delta A}}{A} = \frac{{4\Delta r}}{r}$
A cylindrical wire of mass $(0.4 \pm 0.01)\,g$ has length $(8 \pm 0.04)\,cm$ and radius $(6 \pm 0.03)\,mm$.The maximum error in its density will be $......\,\%$
A sliver wire has mass $(0.6 \pm 0.006) \; g$, radius $(0.5 \pm 0.005) \; mm$ and length $(4 \pm 0.04) \; cm$. The maximum percentage error in the measurement of its density will be $......\,\%$