Question

A physical quantity $Q$ is found to depend on observables $x,y$ and $z$ obeying relation $Q=\frac{x^3{y}^2}{z}$. The percentage error in the measurement of $x,y$ and $z$ are $1\%,2\%$ and $4\%$ respectively. What is the percentage error in the quantity $Q$ ?

• A. $4\%$
• B. $3\%$
• C. $11\%$
• D. $1\%$

Answer 1

The correct option is C

$11\%$

Step 1. Given data

$Q=\frac{x^3{y}^2}{z}$

$\frac{∆x}{x}=1\%$

$\frac{∆y}{y}=2\%$

$\frac{∆z}{z}=4\%$

Step 2. Finding the percentage error in $Q$

By using the percentage error formula,

$\frac{∆Q}{Q}=3\times \frac{∆x}{x}+2\times \frac{∆y}{y}+\frac{∆z}{z}$

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

$\frac{∆Q}{Q}=11\%$

Therefore, the percentage error in $Q$ is $11\%$.

Hence, the correct option is $\left(C\right)$.