If $P = \frac{{{A^3}}}{{{B^{5/2}}}}$ and $\Delta A$ is absolute error in $A$ and $\Delta B$ is absolute error in $B$ then absolute error $\Delta P$ in $P$ is
$\Delta P = \pm \left( { 3 \frac{{\Delta A}}{A} + \frac{5}{2}\frac{{\Delta B}}{B}} \right)P$
$\Delta P = \pm \left( { 3 \frac{{\Delta A}}{A} + \frac{5}{2}\frac{{\Delta B}}{B}} \right)$
$\Delta P = \pm \left( { 3 \frac{{\Delta A}}{A} - \frac{5}{2}\frac{{\Delta B}}{B}} \right)P$
$\Delta P = \pm \left( { 3 \frac{{\Delta A}}{B} - \frac{5}{2}\frac{{\Delta B}}{A}} \right)P$
The period of oscillation of a simple pendulum is $T=2 \pi \sqrt{L / g}$ Measured value of $L$ is $20.0 \;cm$ known to $1\; mm$ accuracy and time for $100$ oscillations of the pendulum is found to be $90 \;s$ using a wrist watch of $1\; s$ resolution. What is the accuracy in the determination of $g in \% ?$
The dimensions of a cone are measured using a scale with a least count of $2 mm$. The diameter of the base and the height are both measured to be $20.0 cm$. The maximum percentage error in the determination of the volume is. . . . .
Can error be completely eliminated ?
The maximum percentage errors in the measurement of mass (M), radius (R) and angular velocity $(\omega)$ of a ring are $2 \%, 1 \%$ and $1 \%$ respectively, then find the maximum percenta? error in the measurement of its rotational kinetic energy $\left(K=\frac{1}{2} I \omega^{2}\right)$
The energy of a system as a function of time $t$ is given as $E(t)=A^2 \exp (-\alpha t)$, where $\alpha=0.2 s ^{-1}$. The measurement of $A$ has an error of $1.25 \%$. If the error in the measurement of time is $1.50 \%$, the percentage error in the value of $E(t)$ at $t=5 s$ is