Question

A physical quantity Q is given by $Q=\frac{A^2.B^{3/2}}{C^4{D}^{1/2}}$. The percentage error in A, B, C, D are $1\%,2\%,4\%,2\%$ respectively. Find the percentage error in Q.

Answer 1

The percentage errors for $A,B,C,D$ are respectively $1$%, $2$%, $4$% and $2$%

The formula for Q is, $Q=\frac{A^2{B}^{\frac{3}{2}}}{C^4{D}^{\frac{1}{2}}}$

So from the formula of error calculation:

The percentage error in $Q=$$\frac{\Delta Q}{Q}\times 100$

$\frac{\Delta Q}{Q}\times 100=2\frac{\Delta A}{A}\times 100+3/2\frac{\Delta B}{B}\times 100+4\frac{\Delta C}{C}\times 100+1/2\frac{\Delta D}{D}\times 100$

$=2\times 1+\frac{3}{2}\times 2+4\times 4+\frac{1}{2}\times 2$

$=22$% = $(20+2)$