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1.Units, Dimensions and Measurement
easy
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
A
$\Delta P = \pm \left( { 3 \frac{{\Delta A}}{A} + \frac{5}{2}\frac{{\Delta B}}{B}} \right)P$
B
$\Delta P = \pm \left( { 3 \frac{{\Delta A}}{A} + \frac{5}{2}\frac{{\Delta B}}{B}} \right)$
C
$\Delta P = \pm \left( { 3 \frac{{\Delta A}}{A} - \frac{5}{2}\frac{{\Delta B}}{B}} \right)P$
D
$\Delta P = \pm \left( { 3 \frac{{\Delta A}}{B} - \frac{5}{2}\frac{{\Delta B}}{A}} \right)P$
Solution
Here $\frac{{\Delta P}}{P} = \pm \left( {3\frac{{\Delta A}}{A} + \frac{5}{2}\frac{{\Delta B}}{B}} \right)$
Standard 11
Physics
