Question

A physical quantity A is related to four observable a, b, c and d as follows, $A=\frac{a^2{b}^3}{c\sqrt{d}}$, the percentage errors of measurement in a, b, c and d are $1\%,3\%,2\%$ and $2\%$ respectively. The maximum error in the quantity A is

• A. $5\%$
• B. $7\%$
• C. $12\%$
• D. $14\%$

Answer 1

The correct option is C $12\%$

$\sqrt{(x^2+y^2)}=a.e^{ta{n}^{-1\left(\frac{y}{x}\right)}}\frac{1}{2\sqrt{x^2+y^2}}(2x+2y{y}^{\prime })=a\cdot e^{ta{n}^{-1\left(\frac{y}{x}\right)}}\times \frac{1}{(1+\frac{y^2}{x^2})}\times \frac{x{y}^{\prime }-y}{x^2}...\left(i\right)\Rightarrow \frac{x+y{y}^{\prime }}{2\sqrt{x^2+y^2}}=\sqrt{x^2+y^2}\times \frac{x^2}{(x^2+y^2)}\times \frac{x{y}^{\prime }-y}{x^2}$
[from Eq. (i)]
$\therefore x+y{y}^{\prime }=x{y}^{\prime }-y\Rightarrow y^{\prime }=\frac{x+y}{x-y}\therefore y"=\frac{2(x{y}^{\prime }-y)}{(x-y{)}^2}y"\left(0\right)=\frac{2(0-y(0\left)\right)}{{\{0-y(0\left)\right\}}^2}=\frac{-2}{y\left(0\right)}=\frac{-2}{a{e}^{\frac{\pi }{2}}}=\frac{-2}{a}{e}^{\frac{-\pi }{2}}$