Question

A physical quantity ${}^{\prime }{y}^{\prime }$ is represented by the formula $y=m^2{r}^{-4}{g}^x{l}^{-\frac{3}{2}}$. If the percentage errors found in $y,m,r,l$ and $g$ are $18,1,0.5,4$ and $p$ respectively, then find the value of $x$ and $p$

• A. $5\text{and}\pm 2$
• B. $4\text{and}\pm 3$
• C. $\frac{16}{3}\text{and}\pm \frac{3}{2}$
• D. $8\text{and}\pm 2$

Answer 1

The correct option is C $\frac{16}{3}\text{and}\pm \frac{3}{2}$

Given, $y=m^2{r}^{-4}{g}^x{l}^{-\frac{3}{2}}$
As we know that, relative errors can be written as
$\frac{\Delta y}{y}=\frac{2△m}{m}+\frac{4△r}{r}+\frac{x△g}{g}+\frac{3}{2}\frac{△l}{l}$
On multiplying each term by $100$, each term will represent percentage error.
So, putting the given values in the above equation we get,
$18=2\left(1\right)+4\left(0.5\right)+xp+\frac{3}{2}\left(4\right)\Rightarrow 18=10+xp$
$\Rightarrow 8=xp$
By trial and error, only option (c) is valid for this scenario. Thus,$x=\frac{16}{3},p=\pm \frac{3}{2}$